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PROCESS INSTRUMENTATION I 

MODULE CODE: EIPIN1B 

STUDY PROGRAM: UNIT 2 

VUT 

Vaal University of Technology 

2/10

EIPINI Chapter 4: Level Measurement Page 4-1 

4. LEVEL MEASUREMENT 

This chapter aims to introduce students to some of the level measurement technologies 

used to measure the level of liquids and granulars in a container. 

4.1 INTRODUCTION 

On the 28th of March, 1979, thousands of people fled from Three Mile Island when 

the cooling system of a nuclear reactor failed. This dangerous situation developed 

because the level controls turned off the coolant flow to the reactor when they detected 

the presence of cooling water near the top of the tank. Unfortunately, the water 

reached the top of the reactor vessel not because there was too much water in the tank, 

but because there was so little that it boiled and swelled to the top. From this example, 

we can see that level measurement is more complex than simply the determination of 

the presence or absence of a fluid at a particular elevation. 

Most level measurement techniques can be categorized into one of two groups – 

direct and indirect (inferred) methods. Direct methods involve measuring the height of 

fluid directly as for example with a dipstick, overflow pipe, float or sight glass. With 

indirect methods, another variable is measured that correlate to the liquid level. 

Measuring the weight of a substance in a container or the pressure exerted on the 

bottom of a tank or transmitting an ultrasonic beam to the level surface and measuring 

the time of flight of the transmitted and received signal, is indirectly related to the 

level height. 

Level measurement may also be continuous or “point-level”. A continuous method 

refers to a technique whereby the device measures level on a constant basis, displaying 

or transmitting the actual level of the liquid as it changes. Point-level devices measure 

liquid at specific points within the tank. As the liquid level rises and falls, it passes 

through definite points during it's transit. Continuous methods may however be 

programmed to output alarms at specific points also, in addition to transmitting a 

continuous level measurement. 

Level measuring devices are also described as either a contact or non-contact type of 

instrument. A contact type device, such as a float, makes physical contact with the 

liquid in the container, in order to determine the level. A non-contact device, such as 

ultrasonic or radar, does not require contact with the material in the container to 

measure the level. 

4.2 THE SIGHT GLASS 

The sight glass consists of a strengthened glass tube, attached to the container as 

shown in Figure 4-1, through which the fluid level in the container can easily be 

observed by the operator. Monitoring the level from a distance, is facilitated when a 

magnetic float inside the tube, is allowed to rise and fall with the liquid level, causing 

metallic flags (or louvers) to flip and expose a different colour, indicating the level. 

These devices are available with continuous monitoring equipment, allowing both a 

local visible indication as well as an external signal for remote monitoring.


Valve 

Sight glass 

Magnetic 

float 

Metallic 

flaps 

Valve 

Container with sight glass 

Sight glass with magnetic indicator 

Figure 4-1 

4.3 FLOAT TYPE LEVEL INDICATORS 

4.3.1 Chain Float 

This type of float is linked to a 

rotating drum, by means of a 

chain, as shown in Figure 4-2. 

The chain engages a sprocket, 

which turns the drum, and with 

it the level indicator. A tape, 

that wraps around the drum, is 

also used, instead of a chain. A 

weight is attached to the other 

end of the chain or tape, to 

keep the chain pulled straight 

while the float moves up or 

down with the changing level. 

Level indicator 

Drum 

Seal 

Chain 

Weight 

Float 

Figure 4-2 

4.3.2 Ball Float 

The float is attached to a rod 

and rotary shaft, operating 

through a packing and bearing 

in the container wall, to the 

level indicator (Figure 4-3). 

Practical considerations, limit 

the shaft rotation to ± 30º from 

the horizontal, and therefore the 

range of the instrument as well. 

Figure 4-3 

Level 

indicator 

Rotary shaft 

Packing and 

bearing 

Float


4.3.3 Magnetic Float (Magnetic Coupled Float and Follower) Level Meter 

For this type of indicator, a dip 

pipe, made of non-magnetic 

material, is permanently installed in 

the container and sealed off, as 

shown in Figure 4-4. A doughnut 

shaped float with high strength 

magnet, is fitted around the dip 

pipe, to rise and fall vertically with 

the liquid level. A similar magnet 

that magnetically bonds to the 

outer magnet, is suspended within 

the dip pipe and attached to a 

rod, chain and pulley or similar 

arrangement, to the level indicator. 

This system is advantageous where 

leakages cannot be tolerated – such 

as with toxic, explosive or 

flammable liquids, as no packing 

or stuffing box is required and 

only the dip tube is required to 

withstand vessel pressure and 

temperature conditions. 

Figure 4-4 


Indicator rod 

Non-magnetic 

dip tube 

Doughnut float 

with 

outer magnet 

Follower 

with 

inner magnet 

4.3.4 Magnetic Float Switch 

The magnetic float switch 

(Figure 4-5) is a point level device. 

When the level reaches a certain 

point, the float magnet activates 

the magnetic reed switch. The 

electric contacts are safely isolated 

from the inside (wet side) of 

the container by non-magnetic 

material that allows magnetic 

interaction between the float 

magnet and magnetic switch. The 

contacts may be used to switch a 

pump on or off, to sound an alarm 

or for other control purposes. 

Float 

Figure 4-5 

Float magnet 

Non-magnetic 

housing 

Magnetic 

reed switch 

Swivel pin


4.3.5 Flexure Tube Displacer (Torque tube) Level Meter 

The displacer type liquid level measuring instrument is not a float as such, for the 

displacer is heavier than the process fluid and the displacer moves very little during 

changes in tank level (a definite advantage over other float types). According to 

Archimedes’s law (illustrated in the figure below), the apparent weight of the 

displacer when immersed in a liquid, is its nominal weight in air minus the weight 

of the displaced liquid. The weight of the displacer will thus vary linearly from its 

weight in air (when the tank is empty) to its apparent weight when fully immersed 

in the liquid (when the tank is full). The weight of the displacer acting on the 

torque arm, will cause an angular displacement of the free end of the flexible torque 

tube and this movement will be transmitted to the outside world, by the torque rod. 

Torque tube 

Torque tube 

flange 


Torque rod 

Archimedes’s Principle 

Torque arm 

Chain 

Displacer 

Figure 4-6 

Note: This instrument may also 

be fitted in a separate 

measuring chamber, fitted 

to the main vessel. 

Main vessel 

Measuring 

chamber with 

displacer, torque 

tube and measuring 

equipment 

Example 4-1 

During calibration of a displacer type level meter, it is found that the torque registered 

by the meter when the tank is empty, is 10 N-m. If the torque arm is 0.1 m, calculate 

the torque that will be generated when the displacer, with volume 0.002 m 3 , is fully 

immersed in the process fluid with density 1000 kg/m 3 . 

Remembering that torque is given by T = Fr, the weight of the displacer in air is 

10/0.1 = 100 N. Using Archimedes’ law, the apparent weight of the displacer when it 

is completely immersed in the liquid, is 100 – 0.002×1000×9.81 = 80.38 N. The torque 

generated then, will be 80.38×0.1 = 8.038 N-m.


4.4 DIFFERENTIAL PRESSURE MEASUREMENT OF LEVEL 

Perhaps the most frequently used technique of measuring level, is the method of 

measuring differential pressure. The primary benefit of this method is that the 

equipment measuring the differential pressure, can be externally installed or 

retrofitted to an existing vessel. It can also be isolated safely from the process 

using block valves for maintenance and testing. 

There are, however, some disadvantages to this method. One vessel penetrations 

near the bottom of the vessel is needed, where leak paths could be the cause of 

many problems. Measurement errors also occur due to changes in liquid density. 

Density variations are caused by temperature changes or change of product. These 

variations must always be compensated for, to maintain accurate measurements. 

4.4.1 Open Containers 

To determine the fluid level H (Figure 4-7) in an open container, it is only necessary to 

measure the difference between the pressure at the bottom of the container and 

atmospheric pressure. In Figure 4-7 (a), the pressure difference is measured with a 

differential pressure transmitter while in Figure 4-7 (b), a u tube manometer is used. 

The pressure instruments may be installed with their zero lines level with the bottom 

of the container or at a distance z, below the bottom of the container. 

P atm 

P atm 

P atm 

H 

P atm 

H 

z 

HP 

LP 

DP 

transmitter 

Zero line 

z 

HP 

LP 

h 

U tube 

manometer 

Figure 4-7 (a) 

Figure 4-7 (b)


Example 4-2 

A mercury (density 13600 

P 

kg/m 3 atm P atm 

) u-tube manometer is 

used to measure the level of a 

liquid (density 1000 kg/m 3 ) in 

5 m 

a 5 meter high container, with 3 m 

the zero level of the 

manometer, exactly in line 

h Zero line 

with the bottom of the tank. 

Calculate the manometer 

X 

Y 

reading h, if the tank level is 3 

meter. 

Equating pressures on the XY line: U-tube 

P X = P Y 

∴P atm + 1000×(3 + ½h)×9.81 = P atm + 13600×h×9.81 

∴3 + ½h = 13.6h 

∴13.1h = 3 

∴h = 0.229 m 

= 229 mm 

Example 4-3 

A mercury (density 13600 

kg/m 3 ) well type manometer 

with A 2 /A 1 = 0.01 is used to 

measure the level of a liquid 

(density 1000 kg/m 3 ) in a 5 

meter high container, with the 

zero level of the manometer, 

exactly in line with the 

bottom of the tank. Calculate 

the manometer reading h if 

the tank level is 3 meter. 

5 m 

3 m 

We can approach this problem in three ways: 

a) Compare the pressures on the XY line, the 

same way as with the u tube manometer. 

b) Calculate P 1 and P 2 exactly on the well type 

manometer liquid meniscuses and utilise 

Equation 2-5 (Chapter 2). 

P ATM 

c) Ignore the level change (d) in the well (because it is so small), assume P 1 acts on 

the zero line and P 2 on the manometer liquid in the tube, and use Equation 2-5 to 

arrive at an approximate answer for h. 

Methods a) and b) will produce the exact value of h while method c) will give us an 

approximate value of h, although very nearly correct. 

X 

d 

P 1 

P 2 = P ATM 

h 

Zero line 

Y 

Well type manometer 

(Students must please take 

note of the orientation of 

the manometer. The well is 

always connected to the 

higher pressure.)


We will now use method a), b) and c) to solve this problem, but we will prefer the 

approximate but easier method c) for the remaining well type manometer problems 

and also for closed tanks involving well type manometers. 

a) Comparing pressures on the XY line: 

P ATM + 1000×3×9.81 + 1000×d×9.81 = P ATM + 13600×h×9.81 + 13600×d×9.81 

∴1000×3×9.81 + 1000×d×9.81 = 13600×h×9.81 + 13600×d×9.81 

∴29430 + 9810d = 133416h + 133416d 

∴29430 = 133416h + 123606d………………………………………………..(a) 

But from Equation (2), Chapter 2, Page 2-7: d = (A 2 /A 1 )h 

∴d = 0.01h……………………………………………………………………..(b) 

(b) in (a): 

29430 = 133416h + 123606×(0.01h) 

∴29430 = 133416h + 1236.06h 

∴134652.06h = 29430 

∴h = 0.21856 m 

b) P 1 = P ATM + 1000×3×9.81 + 1000×d×9.81 

= P ATM + 29430 + 9810×d = P ATM + 29430 + 9810×(0.01h) 

∴P 1 = P ATM + 29430 + 98.1h …………………………………….…………… (a) 

And P 2 = P ATM ………………………………………………………………… (b) 

From (a) and (b): 

P 1 – P 2 = 29430 + 98.1h ………………………………………………………. (c) 

But from Equation 2-5 (Chapter 2): 

P 1 – P 2 = ρhg(1 + A 2 /A 1 ) …………………………………...…………………. (d) 

(c) in (d): 

29430 + 98.1h = 13600×h×9.81(1 + 0.01) 

∴29430 + 98.1h = 133416×h×1.01 

∴29430 + 98.1h = 134750.16h 

∴134652.06h = 29430 

∴h = 0.21856 m 

c) P 1 = P ATM + 1000×3×9.81 (neglecting d) 

∴P 1 = P ATM + 29430 ………………………………………….…………….… (a) 

And P 2 = P ATM …………………………………………………….……...…… (b) 

From (a) and (b): 

P 1 – P 2 = 29430 ………………………………………………….……………. (c) 

And from Equation 2-5 (Chapter 2): 

P 1 – P 2 = ρhg(1 + A 2 /A 1 ) …………………………………...…………………. (d) 

(c) in (d): 

29430 = 13600×h×9.81(1 + 0.01) 

∴29430 = 134750.16h 

∴h = 0.2184 m


Example 4-4 

A DP transmitter must be calibrated to measure the level of a liquid in an open tank. 

The density of the liquid is 1000 kg/m 3 . The DP transmitter will be mounted one meter 

below the bottom of the tank. The tank is full when the height of the liquid in the tank 

is 5 meter and it is empty when there is only liquid in the high pressure line connected 

to the DP transmitter. Determine the necessary calibration specifications for an output 

signal of 4 to 20 mA. 

Empty: 

P 1 = P atm + ρhg 

= P atm + 1000×1×9.81 

= P atm + 9810 

∴P 1 – P 2 = 9810 Pa 

Full: 

P 1 = P atm + ρhg 

= P atm + 1000×6×9.81 

= P atm + 58860 

∴P 1 – P 2 = 58860 Pa 

Empty 

1 m 

P atm 

P atm 

P 1 P 2 

Full 

5 m 

1 m 

DP cell 

4 mA 20 mA 

P atm 

P 1 P 2 

DP cell 

Calibration specification: Output = 4 mA when input = 9810 Pa (empty condition) 

Output = 20 mA when input = 58860 Pa (full condition) 

Example 4-5 

If, in example 4-4, the DP cell is replaced by a 

u-tube manometer using mercury (density 13600 

kg/m 3 ), with its zero level 1 meter below the 

bottom of the container, calculate the full and 

empty readings, h empty and h full . 

Empty: 

Equating pressures on the XY line: 

P atm +1000×(1+½h)×9.81 = P atm +13600×h×9.81 

∴1 + ½h = 13.6h 

∴h empty = 0.07634 m = 76.34 mm. 

Full: 

Equating pressures on the XY line: 

P atm +1000×(6+½h)×9.81 = P atm +13600×h×9.81 

∴6 + ½h = 13.6h 

∴h full = 0.458 m. = 458 mm. 

Empty 

ZL 

Full 

5 m 

ZL 

1 m 

1 m 

P atm 

X 

P atm 

X 

P atm 

P atm 

h empty 

Y 

U tube 

P atm 

h full 

Y 

U tube


Example 4-6 

A DP transmitter, correctly calibrated to measure the level of 

a liquid in a 5 meter high open container, delivers an output 

pressure of 50 kPa. Calculate the liquid level in the tank. 

Liquid level H = [(50-20)/(100-20)]×5 = 1.875 m. 

Or alternatively: 50 = 16×H + 20 ⇒ H = 1.875 m 

Example 4-7 

A mercury (density 13600 kg/m 3 ) u tube manometer 

is used to measure the level of a liquid (density 

1000 kg/m 3 ) in a 5 meter high container, with the 

zero level of the manometer, 1 meter below the 

bottom of the tank. Calculate the level in the tank if 

the manometer reading is 300 mm (0.3 m). 

Comparing pressure on the XY line: 

P atm +1000×(H+1+0.15)×9.81=P atm +13600×0.3×9.81 

∴H + 1.15 = 13.6×0.3 

∴H = 4.08 – 1.15 

= 2.93 meter 

Example 4-8 

A mercury (density 13600 kg/m 3 ) well type 

manometer with A 2 /A 1 = 0.01, is used to measure 

the level of a liquid (density 1000 kg/m 3 ) in a 

5 meter high container, with the zero level of the 

manometer, 1 meter below the bottom of the tank. 

Calculate the level in the tank if the manometer 

reading is 200 mm (0.2 m). 

The input pressure, P 1 , on the well side of the 

manometer, with respect to the zero line, is: 

P 1 = P atm + 1000×(H + 1)×9.81 

and P 2 = P atm 

∴P 1 – P 2 = P atm + 1000×(H + 1)×9.81 – P atm 

= 1000×(H+1)×9.81 = 9810(H+1) 

Furthermore, from Equation 2-5 (Chapter 2): 

P 1 – P 2 = ρhg(1 + A 2 /A 1 ) = 13600×0.2×9.81×(1 + 0.01) = 2747×9.81 = 26948 

Therefore 9810(H+1) = 26948 ⇒ H+1 = 2.747 

∴H = 1.747 m. 

Note: If we did wish to include d, then d = (A 2 /A 1 )h = 0.01×0.2 = 0.002 m 

P 1 =P atm +1000×(H+1+d)×9.81=P atm +1000×(H+1+0.002)×9.81=9810(H+1.002) 

And P 1 – P 2 = 9810(H+1.002) = 26948 ⇒ H + 1.002 = 2.747 

∴H = 1.745 m 

H 

H 

ZL 

ZL 

100 

50 

20 

1 m 

d 

P o 

P o =16×Level+20 

0 H 5 Level 

1 m 

P atm 

X 

P atm 

P 1 

5 m 

5 m 

Well type 

manometer 

P atm 

0.3 m 

Y 

U tube 

P 2 =P atm 

0.2 m

4.4.2 Closed Containers 


For a closed container (Figure 4-8), the outer leg of the differential pressure 

measuring equipment, can not be open to atmosphere, as the tank may be 

pressurised and the tank pressure P t , could differ from atmospheric pressure. The 

outer leg is therefore brought back into the top of the vessel, to ensure that the 

pressure inside the tank and the outer leg pressure, are both equal to P t . Liquid 

vapour will typically condense in the outer leg and if an effort is made to keep this 

leg devoid of process fluid, it will result in what is called a dry leg system. If, on 

the other hand, the outer leg is allowed to be completely filled with the same 

process fluid as in the tank, a wet leg system will result. We will focus on wet leg 

closed tank systems. 

Differential pressure level measurement in a closed vessel (wet leg), is 

fundamentally different from level measurement in an open vessel. The high 

pressure input will always be connected to the outer leg while the low pressure 

input is obtained from the bottom of the tank. Another difference is that the 

differential pressure is zero when the tank is full, while the maximum differential 

pressure is obtained when the tank is empty. Students must ensure that they 

understand this critical difference between differential pressure measurement in an 

open vessel and a closed vessel, very clearly. 

Figure 4-8(a) illustrates a DP transmitter measuring the pressure difference between 

the outer leg and the tank. It may be connected in line with the bottom of the tank or 

below. In Figure 4-8 (b), a u tube manometer is used to measure the pressure 

difference. In the case where a manometers is used, it is clear from Figure 4-8 (b), that 

the zero line must be lower than the bottom of the tank. 

P t 

P t 

P t 

P t 

H 

L 

H 

L 

z 

LP 

HP 

DP transmitter 

z 

LP 

HP 

ZL 

h 

U tube 

manometer 


Figure 4-8 (b)


Example 4-9 

A DP transmitter is used to measure the level of a liquid, with density 1000 kg/m 3 in a 

closed tank, that can store liquid to a maximum level of 5 meter. The DP transmitter is 

installed 1 meter below the bottom of the tank. Calculate the input differential pressure 

to the DP transmitter when the tank is empty and when it is full. 

P t 

P t 

P 1 

P t 

P t 

P 1 

5 m 

5 m 

P 2 

1 m 1 m 

Empty: 

DP transmitter Full: DP transmitter 

P 1 = P t + 1000×(5 + 1)×9.81 P 1 = P t + 1000×(5 + 1)×9.81 

P 2 = P t + 1000×1×9.81 P 2 = P t + 1000×(5 + 1)×9.81 

P 1 – P 2 = 49050 Pa = 49.05 kPa. 

P 1 – P 2 = 0 Pa. 

Example 4-10 

A u tube mercury (δ = 13.6) manometer is used to measure the level of a liquid (δ=1), 

in a closed tank, that can store liquid to a maximum level of 5 meter. The zero level of 

the manometer is 1 meter below the bottom of the tank. Calculate the manometer 

readings when the tank is empty and full. 

P 2 

P t 

P t 

5 m 

P t 

P t 

5 m 

u-tube 

1 m 1 m 

h empty ZL u-tube 

X 

Y 

X 

h full 

ZL 

Y 

Empty: 

Full: 

Equating pressures on the XY line: Equating pressures on the XY line: 

P t +1000×(1 - ½h)×9.81 + 13600×h×9.81 P t +1000×(5+1-½h)×9.81+13600×h×9.81 

= P t + 1000×(5 + 1 + ½h)×9.81 = P t + 1000×(5 + 1 + ½h)×9.81 

∴(1 - ½h) + 13.6h = 6 + ½h ∴(6 - ½h) + 13.6h = 6 + ½h 

∴13.6h – ½h - ½h = 6 – 1 ⇒ 12.6h = 5 ∴13.6h – ½h - ½h = 0 ⇒ 12.6h = 0 

∴h empty = 0.3968 m = 396.8 mm. ∴h full = 0 (as we knew all along)

Example 4-11 

The level of a liquid (density 1000 

kg/m 3 ) in a closed tank, 5 meter 

high, is measured with a mercury 

(density 13600 kg/m 3 ) well type 

manometer (A 2 /A 1 =0.01). The 

manometer is installed with its 

zero line 1 meter below the bottom 

of the tank. If the reading on the 

manometer is 0.2 m, calculate the 

level H of the fluid in the tank. 

Example 4-12 

The level of a liquid, with density ρ l , stored in a 

closed tank, L meters high, is measured with a u 

tube manometer, using a manometer liquid, with 

density ρ m (ρ m > ρ l ). The manometer is installed 

with its zero line a distance z meter below the 

bottom of the tank. When the fluid level in the 

container is H meter, the reading obtained from 

the manometer is h meter. Derive an equation 

that can be used to calculate the liquid level H, in 

terms of the manometer reading h and in terms of 

the constants ρ l , ρ m , L, and z. 


1 m P 2 

0.2 m 

P 1 

d=.002 

Zero line 

P 1 = P t + 1000×(5+1)×9.81 = P t + 58860 

X 

Y 

P 2 = P t + 1000×[H+(1-0.2)]×9.81 = P t + 9810×(H+0.8) 

∴P 1 – P 2 = 58860 – 9810(H+0.8) = 58860 – 9810H – 7848 = 51012 – 9810H…(1) 

And P 1 – P 2 = ρhg(1 + A 2 /A 1 ) = 13600×0.2×9.81×(1 + 0.01) = 26950 ………… (2) 

From (1) and (2): 51012 – 9810H = 26950 ⇒ 9810H = 51012 – 26950 = 24062 

∴H = 2.453 m. 

(The exact value of H may be obtained if we equate pressures on the XY line: 

P t +1000×[H+(1–0.2)]×9.81+13600×(0.2+0.002)×9.81=P t +1000×(5+1+0.002)×9.81 

∴[H+0.8] + 13.6×0.202 = 6.002 ⇒ H + 0.8 + 2.747 = 6.002 ⇒ H + 3.547 = 6.002 

∴H = 2.455 m.) 

Equating pressures at points X and Y: 

P t + ρ l ×(H + z - ½h)×g + ρ m ×h×g = P t + ρ l ×(L + z + ½h)×g 

∴ρ l ×(H + z - ½h)×g + ρ m ×h×g = ρ l ×(L + z + ½h)×g 

∴ρ l ×(H + z - ½h) + ρ m ×h = ρ l ×(L + z + ½h) 

∴ρ l H + ρ l z - ½ρ l h + ρ m h = ρ l L + ρ l z + ½ρ l h 

∴ρ l H = ρ l L + ρ l z + ½ρ l h - ρ l z + ½ρ l h - ρ m h = ρ l L + ρ l h - ρ m h 

ρ 

⎛ ρ 

∴H = L + h - m h = L + 1 h 

ρ 

ρ m ⎞ 

⎜ ⎟ 

⎜ − ⎟ (interestingly, z doesn’t play a role) 

l 

l 

⎝ 

H 

⎠ 

P t 

H 

ρ l 

5 m 

P t 

z 

X 

ρ m 

P t 

L 

Please note that 

the well (high 

pressure input) 

is connected to 

the outside tube 

h 

P t 

ZL 

Y


4.5 BUBBLE METERS 

A bubbler level meter (air purge system or gas flushing system), provides a simple 

and inexpensive but less accurate (±1-2%) level measurement system for virtually 

any liquid, and typically for corrosive or slurry-type applications. A constant flow 

of compressed air or an inert gas (usually nitrogen), is introduced through a dip 

tube. The pressure required to force bubbles through the tube at a constant rate 

(normally 60 bubbles per minute), is proportional to the hydrostatic pressure at the 

bottom end of the dip tube. A bubbler sight glass, facilitate with adjusting the 

bubble rate. The level is determined from the difference between the bubbler 

pressure (back pressure) and the pressure at the surface of the liquid, by means of a 

DP transmitter or manometer. Disadvantages of the bubbler system are 1) that the 

liquid’s density, influences measurement accuracy and 2) the need for compressed 

air or gas. The end of the dip tube may become plugged or clogged, and the dip 

tube is periodically purged (release of large amounts of air to clear the tube), to 

combat this situation. Figure 4-9 (a) and (b), depicts the measurement arrangement 

for an open tank and a closed tank respectively, using a u-tube manometer. 

Bubbler 

sight glass 

Pressure 

regulator 

Filter 

Air/gas 

supply 

Dip tube 


Bubbler 

sight glass 

Pressure 

regulator 

Filter 

Air/gas 

supply 

Dip tube 

Figure 4-9 (b)

EIPINI Chapter 5: Temperature Measurement Page 5-1 

5. TEMPERATURE MEASUREMENT 

The purpose of this chapter is to introduce students to fundamental concepts related to 

temperature and to discuss important industrial methods used to measure temperature. 


Galileo Galilei (picture to the left), was born in the year 

1564 in Pisa, Italy. This brilliant scientist and astronomer is 

most famous for his early development of the telescope, 

being the first to see the moons of Jupiter and other celestial 

objects. Through his work in astronomy, Galileo supported 

the theory originated by Copernicus, that the earth moved 

around the Sun. Eventually, and under immense pressure 

from the Church, he publicly retracted his support for the 

Copernicus theory. Even so, Galileo was placed under house 

arrest and spent the last 10 years of his life in almost 

complete seclusion, having dared to offend the Church. 

Galileo was indeed an incredible man. He was the first to realise how the swinging 

of a pendulum, could be used to measure time and he is also credited to have 

developed the first device to indicate changes in temperature. 

This instrument did not measure temperature as such, and is 

therefore called a thermoscope. (Vincenzo Viviani states in 

his Vita di Galileo, that the thermoscope was conceived by 

Galileo in 1597.) Galileo’s thermoscope was based on the 

fascinating idea that air will expand when heated and contract 

when cooled. The instrument consisted of a glass bottle about 

the size of an egg, with a long glass neck. Air in the bottle 

was heated with the hands and the open end side of the tube 

immersed partially in a vessel containing water or wine. As 

the air in the tube cools, it contracts and the liquid is drawn 

into the tube. Once an equilibrium point is established, a rise 

Cold 

Hot 

in temperature increases the volume of the air, forcing fluid back down; a fall in 

temperature reduces the volume of the air, drawing more fluid into the tube. 

Early in the 1700’s, Gabriel Daniel Fahrenheit developed a 

more reliable temperature measuring device. A thin glass 

tube holds a volume of mercury in a bulb at the bottom. 

When heated, the mercury expands upward in the tube. The 

tube could be sealed and allowed for accurate, reproducible 

temperature readings. Because this device could be marked 

with a numerical scale, it was called a thermometer. 

Cold 

Hot


As the inventor of a reliable thermometer, Fahrenheit was afforded the luxury of 

developing a temperature scale to his liking. He defined a temperature scale on which 

0 was the coldest temperature he could reproduce in the lab (the temperature of an icesalt 

mixture) and 12 as his body temperature (as 12 inches in a foot). Fahrenheit soon 

discovered that 12 divisions between the two set points were not fine enough for good 

measurement so he doubled the divisions first to 24, then to 48 and finally to 96 (the 

average person’s body temperature is actually 98.6ºF). On the Fahrenheit temperature 

scale water freezes at 32º and boils at 212º. Indeed the unit, degrees, may arise from 

the fact that there are 180 increments between freezing and boiling of water. 

The Swedish astronomer, Anders Celsius, arbitrarily selected 100 to be the freezing 

point of water and 0 to be the boiling point of water. Later the end points were 

reversed and the centigrade scale was born (centigrade since there were 100 

increments between the freezing point and boiling point). In 1948 the name was 

officially changed from the centigrade scale to the Celsius scale. 

The Fahrenheit and Celsius scales are termed relative scales. An object can be colder 

than zero on the Fahrenheit scale or the Celsius scale and negative temperatures on 

both scales are common. Since temperature is related to the motion of atoms or 

molecules, zero should represent the absolute zero of temperature. Such a scale would 

be an absolute scale. Two absolute temperature scales do exist. The Kelvin scale 

(named after William Thomson, later Lord Kelvin) uses the same size increments as 

the Celsius scale, but has its zero at absolute zero temperature. The Rankine scale 

(named after William John Macquorn Rankine) is the absolute scale whose increments 

are the same size as those in the Fahrenheit scale. 

5.2 UNITS OF TEMPERATURE 

Definition of Temperature 

Temperature is defined as the degree of heat of a body. The SI unit for 

temperature is Kelvin (K). 

To set a temperature scale, two fixed temperature points must be defined. The distance 

between the two fixed points, is called the fundamental interval. The two fixed points 

used, are the following: 

Bottom fixed point: The temperature of ice (prepared from distilled water) 

mixed with distilled water, at standard atmospheric pressure of 760 

mm. mercury. 

Top fixed point: The temperature of distilled water that boils at standard 

atmospheric pressure of 760 mm. mercury. 

The Celsius scale 

The bottom fixed point is 0 degrees Celsius (0 °C). 

The top fixed point is 100 degrees Celsius (100 °C) 

The fundamental interval of the Celsius scale is thus divided into 100 even parts 

to express 1 °C.

The Fahrenheit scale 


The bottom fixed point is 32 degrees Fahrenheit (32 °F). 

The top fixed point is 212 degrees Fahrenheit (212 °F). 

The fundamental interval of the Fahrenheit scale is thus divided into 180 even 

parts to express 1 °F. 

The Kelvin scale 

Zero Kelvin (-273.15 °C) corresponds to the absolute zero of temperature. The 

melting point of ice is 273.15 Kelvin (K) and the boiling point of water is 373.15 K. 

A temperature change of 1 °C, corresponds to a temperature change of 1 K. 

The Rankine scale 

Zero Rankin (-459.67 °F) corresponds to the absolute zero of temperature. The 

melting point of ice is 491.67 Rankine (R) and the boiling point of water is 671.67 R. 

A temperature change of 1 °F, corresponds to a temperature change of 1 R. 

212 °F 672 R Water boils 100 °C 373 K 

68 °F 528 R Room temperature 20 °C 293 K 

32 °F 492 R Water freezes 0 °C 273 K 

-460 °F 0 R Absolute zero -273 °C 0 K 

Fahrenheit Rankine Celsius Kelvin 

Relationship between Fahrenheit and Celsius 

To determine the relationship between Fahrenheit and Celsius, we could draw a graph 

of F (temperature in °F) versus C (temperature in °C), for C in the range 0 ≤ C ≤ 100. 

Using the general equation for a 

straight line: 

y = m x + c, 

180 

F = C + 32 100 

∴F = 5 

9 C + 32 Equation 5-1 

We can now also express Celsius in 

terms of Fahrenheit. 

From Equation 5-1: 

9 C = F – 32 

5 

212 

F (°Fahrenheit) 

5 

∴C = (F – 32) Equation 5-2 

C 

0 100 

9 (°Celsius) 

32 

100 

180


Example 5-1 

Convert the absolute zero of temperature (–273.15 °C), to degrees Fahrenheit. 

F = (9/5)C + 32 = (9/5)×(-273.15) + 32 = -459.67 °F ≈ -460 °F 

Example 5-2 

Convert 86 °F into degrees Celsius, Rankine and Kelvin. 

C = (5/9)(F – 32) = (5/9)×(86 – 32) = 30 °C 

R = F + 460 = 86 + 460 = 546 R 

K = C + 273 = 30 + 273 = 303 K 

Example 5-3 

Determine the temperature in degrees Fahrenheit that would translate into the same 

value in degrees Celsius. 

In Equation 5-2 (or Equation 5-1), set C = F: 

∴F = (5/9)(F – 32) ⇒ 9F = 5F – 160 ⇒ 4F = -160 ⇒ F = -40 °F 

Example 5-4 

Convert 0 °F to degrees Celsius. 

C = (5/9)(F – 32) = (5/9)(0 – 32) = -17.78 °C. 

5.3 THE INTERNATIONAL TEMPERATURE SCALE 

This scale is based on a number of fixed temperature points and serves as calibration 

reference. All temperatures are measured at standard atmospheric pressure of 760 mm. 

mercury. 

• The oxygen point: 

• The ice point: 

• The boiling point: 

• The sulphur point: 

• The silver point: 

• The gold point: 

The boiling point of liquid oxygen: –182.97 °C. 

The melting point of pure ice: 0 °C. 

The boiling point of pure water: 100 °C. 

The boiling point of pure sulphur: 444.6 °C. 

The melting point of silver: 961.78 °C. 

The melting point of gold: 1064.18 °C. 

5.4 LIQUID IN GLASS THERMOMETERS 

The liquid in glass thermometer, is the most commonly used device to measure 

temperature and it is inexpensive to make and easy to use. The liquid in glass 

thermometer has a glass bulb attached to a sealed glass tube (also called the stem 

or capillary tube). A very thin opening, called a bore, exists from the bulb and 

extends down the centre of the tube. The bulb is typically filled with either 

mercury or red-coloured alcohol and is free to expand and rise up into the tube 

when the temperature increases, and to contract and move down the tube when 

the temperature decreases.


The background of the glass tube is covered with white enamel and the front of 

the glass tube forms a magnifying glass that enlarges the liquid column and 

facilitates with reading the temperature. 

In Figure 5-1 (a), an all glass thermometer is depicted, with its scale etched 

into the stem. Liquid in glass thermometers are fragile and for industrial use, the 

thermometer is mounted in a protective housing and the scale is engraved on a 

separate plate that is part of the protective case. An industrial thermometer is 

shown in Figure 5-1 (b). 

Protective case 

Scale 

(etched) 

Scale 

(on plate) 

Bore 

Lens front capillary 

Tube (stem) 

Bore 

Lens front capillary 

tube (stem) 

Liquid column 

Liquid column 

Socket 

Bulb 

Bulb chamber 


Figure 5-1 (b) 

Bulb 

Heat conducting 

medium 

Liquids used in glass thermometers 

Liquid 

Temperature range (Celsius) 

Mercury -35 to +510 

Alcohol -80 to +70 

Pentane -200 to +30 

Toluene -80 to +100 

Creosote -5 to +200


5.5 MERCURY IN STEEL THERMOMETERS 

The mercury in steel thermometer 

system, allows for rugged construction 

and is used extensively in industrial 

applications. The thermometer consists 

of a steel bulb, a steel capillary tube 

and Bourdon tube, as shown in Figure 

5-2. An advantage of the mercury in 

steel thermometer is that measurements 

can be taken a distance away from the 

application, as the steel tube can be 

made fairly long and flexible . The 

whole system is completely filled with 

mercury under pressure and sealed off. 

When the temperature around the bulb 

increases, the mercury inside the bulb 

will expand. The effect of the mercury 

trying to increase its volume within a 

confined space, will be an increase in pressure, transmitted via the capillary tube, 

to the coiled Bourdon tube. The increase in mercury volume and pressure inside 

the coiled Bourdon tube, will result in the Bourdon tube starting to uncoil, 

proportional to the temperature. A pointer, linked to the free end of the Bourdon 

tube, will subsequently move over the scale to indicate the temperature. 

5.6 GAS FILLED THERMOMETERS 

Figure 5-2 

Pointer 

and scale 

Bourdon 

tube 

Steel tube 

(capillary) 

Mercury 

Steel bulb 

The thermometer consists of a steel 

bulb, a steel tube and Bourdon tube, as 

shown in Figure 5-3. The whole system 

is filled with a gas at a high pressure 

and sealed off. The gas commonly used 

is nitrogen which is an inert gas, with a 

high cubical expansion coefficient and 

which is readily available. It may be 

assumed that the volume of the gas 

remains nearly constant. For an ideal 

gas, the gas law applies: 

PV = nRT. 

It is clear from this equation that if V is 

constant, the pressure is proportional to 

the temperature T. Temperature is thus 

converted to pressure, which is 

detected by the Bourdon tube. 

Figure 5-3 

Pointer 

and scale 

Bourdon 

tube 

Steel tube 

(capillary) 

Gas 

(nitrogen) 

Steel bulb


5.7 VAPOUR PRESSURE THERMOMETERS 

The thermometer consists of a steel 

bulb, a steel tube and Bourdon tube, as 

shown in Figure 5-4. The bulb is 

partly filled with a volatile liquid, 

evacuated and sealed off. With rising 

temperature, the average velocity of 

the molecules in the liquid will 

increase. As a result, more liquid 

molecules will acquire enough energy 

to escape from the surface of the liquid 

and saturate the evacuated area. The 

result will be an increase in vapour 

pressure in the capillary tube and 

Bourdon tube. The Bourdon tube, 

sensitive to the changes in pressure, 

will record the temperature via the 

pointer that is linked to the moving 

end of the Bourdon tube. 

5.8 BI-METAL THERMOMETERS 

Bimetal strips consists of two different 

metals welded together to form a cantilever 

as shown in Figure 5-5 (a). When heated, 

both metals expand, but the bottom strip 

(assuming that the lower strip has the higher 

expansion coefficient), expands more than 

the top metal. The result is that the bottom 

of the strip becomes longer than the top, and 

the cantilever curls upwards as shown in 

Figure 5-5 (b). 

Figure 5-4 

Pointer and 

scale 

Bourdon 

tube 

Steel tube 

(capillary) 

Vapour 

Steel bulb 

Volatile 

liquid 

Pointer 

and scale 

Socket 

Bearing 



Shaft 

Guide 

A bi-metal thermometer uses a bi-metal strip, shaped in 

a helix or spiral form, as shown in Figure 5-6. The one 

end is fixed and the other end is free to rotate as the helix 

curls in or out with changing temperature. A shaft and 

pointer is linked to the rotating helix, to indicate the 

temperature. The stem is filled with silicone fluid, to 

provide damping and thermal conductivity between the 

stem and bi-metal strip. 

Stem 

Helical 

bi-metal 

element 

Figure 5-6


5.9 RESISTANCE TEMPERATURE DETECTORS (RTD’s) 

5.9.1 Temperature coefficient of resistance (TCR) 

Resistance thermometers are based on the principle that the resistance of a metal 

increases with temperature. The temperature coefficient of resistance (TCR) for 

resistance thermometers (denoted by α o ), is normally defined as the average resistance 

change per °C over the range 0 °C to 100 °C, divided by the resistance of the 

thermometer, R o , at 0 °C. 

α o = 

R 100 − R 0 

R 0 × 100 

Equation 5-3 

where, 

R 0 = resistance of wire at 0 °C (ohm), and 

R 100 = resistance of wire at 100 °C (ohm), 

As a first approximation, the relationship 

between resistance and temperature, may 

then be expressed as (see Figure 5-7): 

R t (Ω) 

R o 

Note: Starting with a straight line 

in standard form, y = mx + c: 

R t = mt + c where c = R o 

and m = (R 100 - R o )/100 = R o α o . 

∴R t = R o α o t + R o = R o (1 + α o t) 

R t = R o (1+α o t) 

t (°C) 

R 100 

Figure 5-7 

R t = R o (1 + α o t), Equation 5-4 

where: 

R t = resistance of wire at 

temperature t (ohm), 

R o = resistance of wire at 0 °C (ohm), and 

α o = temperature coefficient of resistance (TCR) at 0 °C (per °C) 

Example 5-5 

A platinum thermometer measures 100 Ω at 0 °C and 139.1 Ω at 100 °C. 

a) Calculate the TCR for platinum. 

b) Use Equation 5-4 and calculate the resistance of the thermometer at 50 °C. 

c) Use Equation 5-4 and calculate the temperature when the resistance is 110 Ω. 

R 

a) From Equation 5-3: α o = 100 − R 0 139.1-100 

= = 0.00391 /°C. 

R 0 × 100 100× 

100 

b) From Equation 5-4: R 50 = R o (1 + α o t) = 100(1 + 0.00391×50) = 119.55 Ω 

c) From Equation 5-4: R t = R o (1 + α o t) ⇒ 110 = 100(1 + 0.00391t) 

∴1 + 0.00391t = 1.1 ⇒ 0.00391t = 0.1 ⇒ t = 25.58 °C. 

5.9.2 Resistance curves 

In general, the relationship between resistance and temperature, is not linear and 

Equation 5-4 does not describe the resistance / temperature behaviour of metals, 

adequately. In Figure 5-8, the resistance curves for nickel, copper and platinum (with 

R o = 100 Ω) are shown and it is immediately clear that higher order equations must be 

used to describe the resistance / temperature relationship. The resistance / temperature 

characteristic for a platinum thermometer, for example, can be described by the 

equation: 

0 

100


⎪ 

⎧ 

⎨ 

⎪⎩ 

Ro[1+ 

At + Bt + Ct − 

< < ° 

= 

3 (t 100)] - 200 t 0 C 

R 2 

t 

Ro[1+ 

At + Bt 2 , 

] 

0 ≤ t < 850 ° C 

where: 

R t is the resistance at temperature t 

R o is the ice point resistance and 

A, B and C are coefficients describing the particular thermometer. 

800 

600 

400 

200 

-200 

R (ohm) 

Nickel 

Copper 

Platinum 

0 200 400 600 800 

Figure 5-8 

t (°C) 

An excerpt from the resistance table for a Minco 

platinum thermometer (code PB), is given in Table 5-1. 

Example 5-6 

In the temperature range 0 °C ≤ t < 850 °C, the 

resistance / temperature relation of a platinum 

thermometer with ice point resistance of 100 Ω, is 

described by the second order expression: 

Partial temperature/ 

resistance table for 

platinum (R o =100Ω) 

Temperature 

(°C) 

Resistance 

(Ω) 

-200 17.26 

-100 59.64 

-80 67.83 

-60 75.96 

-40 84.03 

-20 92.04 

0 100.00 

20 107.92 

40 115.78 

60 123.60 

80 131.38 

100 139.11 

200 177.04 

300 213.81 

400 249.41 

500 283.84 

600 317.09 

700 349.18 

R t = 100×[1 + 0.0039692×t – (5.8495×10 -7 )×t 2 ]. 

Table 5-1 

a) Calculate the resistance of the thermometer at 50 °C. 

b) Calculate the temperature when a resistance of 110 Ω is measured. 

a) R 50 = 100×(1 + 0.0039692×50 – 5.8495×10 -7 ×50 2 ) 

= 100×(1 + 0.19846 - 0.001462375) = 119.7 Ω. 

b) R t = 100[1 + (3.9692×10 -3 )×t – (5.8495×10 -7 )t 2 ]. 

∴110 = 100[1 + (3.9692×10 -3 )×t – (5.8495×10 -7 )t 2 ] 

∴110 = 100 + 0.39692t – 5.8495×10 -5 t 2 

∴(5.8495×10 -5 )t 2 + (– 0.39692)t + 10 = 0 

∴t = 

= 

0.39692 ± 

( 0.39692 ) 

2× 

( 5.8495× 

10 

0.39692 ± 0.39396 

−6 

116.99× 

10 

2 

− 4× 

( 5.8495× 

10 

−5 

) 

−5 

) × 10 

= 25.301 °C (ignoring 6760 °C.) 

[ax 2 +bx+c=0] 

[x= 

− b ± 

b 

2 

− 4ac 

2a 

]


5.9.3 Thermometer construction 

Depending upon their intended 

application, resistance thermometers are 

available in different shapes and forms. 

A typical structure is shown in Figure 

5-9. The resistance winding is located in 

the lower end of the protecting tube or 

stem, which is sealed. The upper part of 

the stem terminates in the terminal 

housing (head) with the resistor winding 

leads. The resistance winding must be in 

good thermal contact with the stem, for 

fast heat transfer from the medium to the 

winding while electrical isolation must 

be ensured. The fundamental design 

problem with resistance thermometers is 

to achieve high electrical insulation and 

minimum thermal insulation. 

Terminal 

cap 

Connector 

conduit 

Socket 

Resistor bulb 

(resistance winding 

or thin film) 

Figure 5-9 

Stem 

Leads 

5.9.4 Measuring temperature with resistance thermometers 

The simplest way to measure the 

resistance of a RTD, is to inject a 

constant current into the thermometer 

and to measure the voltage that develops 

across the thermometer. A Wheatstone 

bridge circuit, shown in Figure 5-10, is 

however generally used to detect the 

changes in resistance of a resistance 

thermometer. The values of the fixed 

resistors, R 1 ,R 2 and R 3 , are very 

accurately known, while RT represents 

the resistance thermometer with leads a 

and b. The bridge is said to be in null 

balance, when the voltage across points 

A and B is zero. This occurs when RT = R 3 × 

R 

1 

R 

2 

, causing VAC = V BC , resulting in 

the reading on M, to become zero. The zero condition would correspond to the zero 

point or set point of the resistance thermometer output. As the temperature 

increases, the resistance RT, of the resistance thermometer will increase, causing 

the bridge to become unbalanced, and meter M to show a reading. The meter M 

may be calibrated in temperature units or V AB may be converted into a standard 4 to 

20 mA or 1 to 5 V signal. The current flowing through the thermometer must be 

kept as low as possible (< 1 mA) to minimise errors caused by I 2 R losses and 

associated temperature rise in the thermometer itself. 

E 

B 

R 2 

R 3 

M 

C 

Figure 5-10 

a 

R 1 

b 

RT 

A


Example 5-7 

The Wheatstone bridge in Figure 5-10, is supplied from a 10 V battery, and used to 

measure temperature with a platinum resistance RT. The value of each of the fixed 

resistors R 1 , R 2 and R 3 , is 100 Ω (in a practical circuit, R 1 and R 2 would be chosen 

much higher to improve bridge linearity and minimize errors caused by I 2 R losses in 

the thermometer). The meter M measures the thermometer output voltage V, which 

is given by: V = V AB = V AC - V BC . At 0 °C, the resistance RT, of the platinum RTD, 

is 100 Ω, while at 100 °C, the resistance RT increases to 139.1 Ω. Calculate the 

output voltage V, of the thermometer, when the temperature is 0 °C and when the 

temperature is 100 °C. Assume that the measuring device draws negligible current 

from the circuit. 

0 °C: 

I 2 

I 1 

R 

I 1 = 10/(100 + 100) = 50.00 mA. 

2 R 1 

∴V AC = RT×I 1 = 100×(50×10 -3 ) 

10V 100Ω V 100Ω 

= 5 volt 

B 

A 

I 2 = 10/(100 + 100) = 50.00 mA. 

R 3 

V 

∴V BC = R 3 ×I 2 = 100×(50×10 -3 AB 

RT 

V BC V AC 

) 

100Ω 100Ω 

= 5 volt 

∴V = V AB = V AC – V BC (using Kirchoff’s 

C 

0 °C 

C 

= 5 – 5 = 0 volt voltage law: V AB -V AC +V BC =0) 

We of course expected this result, as the bridge is indeed balanced at 0 °C. 

100 °C: 

I 1 =10/(100+139.1)=41.82 mA. 

∴V AC = RT×I 1 = 139.1×(41.82×10 -3 ) 

= 5.817 volt 

I 2 = 10/(100 + 100) = 50.00 mA. 

∴V BC = R 3 ×I 2 = 100×(50×10 -3 ) 

= 5 volt 

∴V = V AB = V AC – V BC 

= 5.817 – 5.000 = 0.817 volt 

Example 5-8 

The Wheatstone bridge in Figure 5-10, is supplied from a 10 V battery, and used to 

measure temperature with a platinum resistance RT. The value of each of the fixed 

resistors R 1 , R 2 and R 3 , is 100 Ω. The meter M measures the thermometer output 

voltage V, which is given by: V = V AB = V AC - V BC . 

a) Derive an expression for the resistance RT of the platinum element, in terms of 

the thermometer output voltage V. 

b) Derive an expression for the temperature t in terms of RT, if in the temperature 

range 0 °C ≤ t < 850 °C, the relationship between the resistance RT of the 

thermometer and the temperature t, is given by: 

RT = 100[1 + (3.9692×10 -3 )×t – (5.8495×10 -7 )t 2 ]. 

c) Calculate the measured temperature when the bridge output voltage is 0, 0.2, 0.4, 

0.6, 0.8, and 1.0 volt. 

10V 

I 2 

V BC 

B 

C 

R 2 

100Ω 

R 3 

100Ω 

V 

100 °C 

R 1 

100Ω 

RT 

139.1Ω 

I 1 

A 

C 

V AC


10 

a) I 1 = ampere 

RT + 100 

I 2 

I 1 

10× 

RT 

R 2 R 1 

∴V AC = I 1 ×RT = volt 

RT + 100 

10V 100Ω V 100Ω 

10 

and I 2 = = 50 milliamp. 

B 

A 

100 + 100 

R 3 

∴V BC = I 2 ×R 3 = (50×10 -3 )×100 

V BC 

100Ω RT 

= 5 volt 

∴V = V AB = V AC - V BC 

C 

C 

10× 

RT 10× 

RT 

∴V = - 5 ⇒ = V + 5 ⇒ 10×RT = (RT + 100)×(V + 5) 

RT + 100 RT + 100 

∴10×RT = V×RT + 100×V + 5×RT + 100×5 

∴10×RT – 5×RT - V×RT = 100×V + 100×5 

∴(10 – 5 – V)×RT = 100×(V + 5) ⇒ (5 – V)×RT = 100×(V + 5) 

100(V + 5) 

∴RT = 

…………………………….………………. Equation (a) 

5 - V 

b) RT = 100[1 + (3.9692×10 -3 )×t – (5.8495×10 -7 )t 2 ] 

∴RT = 100 + 0.39692×t – (5.8495×10 -5 )×t 2 

∴(5.8495×10 -5 )×t 2 - 0.39692×t + RT – 100 = 0 

5.8495× 

10 

∴ 

-5 

5.8495× 

10 -5 ×t 2 0.39692 

1 

- 

5.8495× 

10 -5 ×t + 

5.8495× 

10 -5 ×(RT-100) = 0 

∴ t 2 - 6786t + [17095×(RT-100)] = 0 

[x 2 +bx+c=0] 

2 

6786 − (6786) − 4× 

17095× 

(RT −100) 

− b ± b 

2 

− 4c 

∴t = 

[x= ] 

2 

2 

6786 − 46.05× 

10 6 − 68380(RT −100) 

∴t = 

…….………….. Equation (b) 

2 

(Note: From Example 5-6 we already know that we can ignore the second 

solution associated with the positive square root.) 

c) V = 0: From Equation (a): RT = 100(0 + 5)/(5 - 0) = 100 Ω 

From Equation (b): t = {6786 - √[46.05×10 6 - 68380(100 - 100)]}/2 

= [6786 - 6786]/2 = 0 °C 

V = 0.2: From Equation (a): RT = 100(0.2 + 5)/(5 - 0.2) = 108.33 Ω 

From Equation (b): t = {6786 - √[46.05×10 6 -68380(108.33-100)]}/2 

= [6786 - √45.48×10 6 ]/2 = 21.05 °C 

V = 0.4: RT = 100(0.4 + 5)/(5 - 0.4) = 117.39 Ω 

t = {6786 - √[46.05×10 6 - 68380(117.39 - 100)]}/2 = 44.09 °C 

V = 0.6: RT = 100(0.6 + 5)/(5 – 0.6) = 127.27 Ω 

t = {6786 - √[46.05×10 6 - 68380(127.27 - 100)]}/2 = 69.40 °C 

V = 0.8: RT = 100(0.8 + 5)/(5 - 0.8) = 138.1 Ω 

t = {6786 - √[46.05×10 6 - 68380(138.1 - 100)]}/2 = 97.37 °C 

V AC


V = 1.0: 

RT = 100(1 + 5)/(5 - 1) = 150 Ω 

t = {6786 - √[46.05×10 6 - 68380(150 - 100)]}/2 = 128.4 °C 

The results from Example 

5-8, highlight again that the 

t (°C) 

relationship between the 140 

thermometer output and the 

measured temperature, is not 

120 

linear (a graph of our 100 

calculated temperatures versus 

thermometer output, is shown 

to the right). The non-linear 

behaviour arises from the nonlinear 

relationship between RT 

and V in Equation (a) and 

from the non-linear 

80 

60 

40 

20 

V (volt) 

relationship between the 

temperature t and RT in 

Equation (b). 

0 

0.2 0.4 0.6 0.8 1.0 

In Example 5-8, temperature and resistance were related by a simple quadratic 

equation. When converting resistance to temperature, thermometer manufacturers may 

in certain cases, be required to find the root of a third or fourth order polynomial. In 

the case of nickel, individual manufacturers have developed different step-wise 

approximations to calculate temperature from resistance. Fortunately today, many 

instruments use microprocessors and manufacturers are using resistance / temperature 

lookup tables stored in read only memory, with linear interpolation for measurements 

falling between table entries. 

5.9.5 Ambient temperature compensation 

If it is necessary to perform the temperature measurement, some distance away 

from where the resistance thermometer bulb is installed, ambient temperature can 

have a detrimental effect on the integrity of the measurement. The reason is that the 

leads that connect the thermometer to the instrument, will have resistance of their 

own, and what is more, this resistance will change with changing ambient 

temperature. The error introduced by lead resistance, can of course be minimised, 

if leads with very low resistance is used. 

An ingenious approach to the lead resistance problem, is the three wire method, 

illustrated in Figure 5-11. The three wire method attempts to cancel the effect of 

the lead resistance, by introducing the same resistance in both branches (wires c 

and a) of the Wheatstone bridge. The three wire method is used extensively for 

industrial RTD’s. Three leads a, b and c, connect to the resistance thermometer and 

the resistance of wires a and c must be matched. This method assumes that the 

meter is a high impedance device, and that essentially no current is flowing in the b 

wire, which only acts as a voltage sense lead.


R 1 

E 

A 

R 2 

V 

B 

c 

b 

R lead 

R lead 

a 

R lead 

RT 

R 3 

C 

Figure 5-11 

The four wire method, depicted in Figure 5-12, alleviates many problems 

associated with the Wheatstone bridge. Instead of using a Wheatstone bridge 

configuration, a current source is employed to supply a constant current I, to the 

thermometer, through wires a and d. A high impedance voltmeter M, measures the 

voltage developed across the thermometer, via wires b and c. The measured 

voltage is directly proportional to the resistance of the thermometer, so only the 

conversion from resistance to temperature is necessary. Wires b and c only act as 

voltage sense leads, and with M a high impedance meter, virtually no current flows 

in wires b and c, and therefore no voltage drop in these leads and thus no lead 

resistance error in the measurement. The disadvantage of the four wire method is 

that we need one more wire than with the three wire method, but it is a small price 

to pay if we are at all concerned with the accuracy of the temperature 

measurement. The four wire method is however gaining in popularity in industry. 

a 

R lead 

b 

R lead 

I 

M 

c 

R lead 

RT 

Figure 5-12 

d 

R lead 

Example 5-9 

A Wheatstone bridge, with fixed resistors R 1 , R 2 and R 3 , of 100 Ω and supplied from a 

10 V battery, is used to measure a temperature of 100 °C with a platinum resistance 

RT, that has a resistance of 139.1 Ω at 100 °C . The thermometer is connected to the 

bridge with leadwires that have a resistance R lead of 10 Ω at 20 °C and the thermometer 

is correctly calibrated for an ambient temperature of 20 °C. Assume that the lead 

resistance changes to 11 Ω when the ambient temperature rises to 30 °C.


Calculate the temperature error introduced at an ambient temperature of 30 °C, when 

using a) the two wire method and b) the three wire method 

a) Measuring with two leads 

Calibrated bridge voltage at 

Bridge output voltage at 

20 °C ambient temperature 30 °C ambient temperature 

I 2 

100Ω I 1 

100Ω 

I 2 

100Ω I 1 

100Ω 

10V 

Process 

temberature 

100°C 

I 1 = 10/(100+10+139.1+10) = 0.0386 A I 1 = 10/(100+11+139.1+11) = 0.0383 A 

I 2 = 10/(100+100) = 0.05 A 

I 2 = 10/(100+100) = 0.05 A 

V = (10+139.1+10)×0.0386 - 100×0.05 V = (11+139.1+11)×0.0383 - 100×0.05 

= 6.141 – 5 = 1.141 V (calibrated 100°C) = 6.17 – 5 = 1.17 V (105°C) 

Therefore, if the bridge was correctly calibrated for an ambient temperature of 20 °C, 

an error of 0.029 V will occur when the ambient temperature rises to 30 °C. So instead 

of reading 100 °C, it would measure approximately 105 °C, an error of 5 °C or 5%. 

b) Measuring with three leads 

Calibrated bridge voltage at 

Bridge output voltage at 

20 °C ambient temperature 30 °C ambient temperature 

10V 

100Ω 

I 2 

I 1 

100Ω 

100Ω 

V 

V 

b 

a 

100Ω 

c 

b 

a 

10Ω 

139.1Ω 

RT 

10Ω 

10Ω 

10Ω 

139.1Ω 

RT 

10Ω 

Process 

temberature 

100°C 

I 2 

I 1 

100Ω 

100Ω 

100Ω 

Process 

temberature 

100°C 

Process 

temberature 

100°C 

I 1 = 10/(100+10+139.1+10) = 0.0386 A I 1 = 10/(100+11+139.1+11) = 0.0383 A 

I 2 = 10/(100+100) = 0.05 A 

I 2 = 10/(100+100) = 0.05 A 

V = (139.1 + 10)×0.0386 - 100×0.05 V = (139.1 + 11)×0.0383 - 100×0.05 

= 5.755 – 5 = 0.755 V (calibrated 100°C) = 5.749 – 5 = 0.749 V (99.13°C) 

Therefore, measuring with the three wire method, an error of only 0.006 V will 

occur. If the thermometer was correctly calibrated to read 100 °C at an ambient 

temperature of 20 °C, then if the ambient temperature rises to 30 °C, the reading of 

0.749 V would correspond to a value of 138.76 Ω for RT with no rise in ambient 

temperature, or in terms of temperature, 99.13 °C, an error of 0.87 °C or 0.87%. This 

is clearly a remarkable improvement over the two wire method. 

10V 

10V 

100Ω 

V 

V 

b 

a 

c 

b 

a 

11Ω 

139.1Ω 

RT 

11Ω 

11Ω 

11Ω 

139.1Ω 

RT 

11Ω


5.10 THERMOCOUPLES 

5.10.1 The Seebeck effect 

If two dissimilar metals are joined together to form a closed loop, and if one 

junction is kept at a different temperature from the other, an electromotive force is 

generated and electric current will flow in the closed loop. 

Metal A 

Metal B 

Seebeck emf 

Metal B 

This very important discovery in the field of thermometry, was made by T.J. Seebeck 

in the year 1821. Experiments by Seebeck and others, have shown that the generated 

emf (called the Seebeck voltage in his honor), is relative, in a predictable manner, to 

the difference in temperature between the two junctions. So, if the temperature of one 

junction is kept at a known value, the temperature of the other junction can be 

determined by the amount of voltage produced. This discovery resulted in the 

temperature sensor that we know today as the thermocouple. 

The Seebeck voltage is made up of two components: the Peltier voltage generated at 

the junctions, plus the Thomson voltage generated in the wires by the temperature 

gradient. The Peltier voltage is proportional to the temperature of each junction while 

the Thomson voltage is proportional to the square of the temperature difference 

between the two junctions. It is the Thomson voltage that accounts for most of the 

observed voltage and non-linearity in thermocouple response. 

5.10.2 Thermocouple laws 

The following empirically derived thermocouple laws, are useful to understand, 

diagnose and utilise thermocouples. 

a) Law of homogeneous circuits 

If two thermocouple junctions are at T 1 and T 2 , then the thermal emf generated is 

independent and unaffected by any temperature distribution along the wires 

T 3 

T 1 

T 4 

T 2 

Figure 5-13 

In Figure 5-13, a thermocouple is shown with junction temperatures at T 1 and T 2 . 

Along the thermocouple wires, the temperature is T 3 and T 4 . The thermocouple emf is, 

however, still a function of only the temperature gradient T 2 – T 1 .


b) Law of intermediate metals 

The law of intermediate metals states that a third metal may be inserted into a 

thermocouple system without affecting the emf generated, if, and only if, the 

junctions with the third metal are kept at the same temperature. 

Metal A 

T 1 

T 2 

Metal B 

Metal C 

Metal B 

T 3 

Figure 5-14 

When thermocouples are used, it is usually necessary to introduce additional metals 

into the circuit This happens when an instrument is used to measure the emf, and when 

the junction is soldered or welded. It would seem that the introduction of other metals 

would modify the emf developed by the thermocouple and destroy its calibration. 

However, the law of intermediate metals states that the introduction of a third metal 

into the circuit will have no effect upon the emf generated so long as the junctions of 

the third metal are at the same temperature, as shown in Figure 5-14. 

c) Law of intermediate temperatures 

The law of intermediate temperatures states that the sum of the emf developed by a 

thermocouple with its junctions at temperatures T 1 and T 2 , and with its junctions at 

temperatures T 2 and T 3 , will be the same as the emf developed if the thermocouple 

junctions are at temperatures T 1 and T 3 . 

T 1 

T 2 

T 2 T 3 T 1 

T 3 

emf (T 2 -T 1 ) emf (T 3 -T 2 ) emf (T 3 -T 1 ) 

Figure 5-15 

This law, illustrated in Figure 5-15, is useful in practice because it helps in giving a 

suitable correction in case a reference junction temperature other than 0 °C is 

employed. For example, if a thermocouple is calibrated for a reference junction 

temperature of 0 °C and used with a junction temperature of 20 °C, then the correction 

required for the observation would be the emf produced by the thermocouple between 

0 °C and 20 °C. 

5.10.3 Thermocouple types 

Any two dissimilar metals can in theory be made into a thermocouple. However, 

certain metals have been selected over time that make ideal thermocouples for various 

applications. These metals have been chosen for their emf output and their ability to 

operate under various conditions. There are several types of these “standard” 

thermocouples in use today. Some are listed Table 5-2.


Type Positive Isolation Negative Isolation Outer Temperature 

element colour element colour isolation range (°C) 

K Chromel Yellow Alumel Red Yellow -200 to 1200 

E Chromel Purple Constantan Red Purple -200 to 800 

J Iron White Constantan Red Black -200 to 750 

T Copper Blue Constantan Red Blue -200 to 350 

S 

90% Platinum 

10% Rhodium 

Black Platinum Red Green 0 to 1500 

Table 5-2 

Types K, E, J and T, are the 'general purpose' thermocouples. Type K is the most 

popular thermocouple in use today while type T is used for lower temperature 

applications. The conventional type J thermocouple, even with its unfavourable iron 

lead, is still popular, mainly because of its widespread use in older instruments. Type 

E is the most sensitive of the standard thermocouples (68μV/°C). Noble metal type S, 

has low sensitivity(10μV/°C), is expensive and used in the higher temperature range. 

Thermocouple alloys referred to in Table 5-2, are chromel (chrome and nickel), alumel 

(aluminium and nickel) and constantan (copper and nickel). 

Different colour codes have been adopted for thermocouple wire and product 

identification and the colour codes listed in Table 5-2, are used in the United States. 

5.10.4 Thermocouple construction 

Thermocouple thermometers are available in different shapes and in different 

coverings. Three basic types of construction can be recognised: 

Wire construction 

The most basic construction of the thermocouple 

is the two dissimilar metals joined (welded) 

together to form the measuring junction. In this 

form, the exposed junction offers good response 

times, but may suffer from environmental 

damage. A bare thermocouple element is shown 

in Figure 5-16. 

Sheathed construction 

To improve mechanical strength, mineral 

insulated thermocouples were developed. The 

thermocouple wires are embedded in compressed 

mineral oxide powder and enclosed in a metal 

sheath, usually stainless steel or inconel (nickelchromium-iron 

alloy). A completely insulated 

thermocouple is shown in Figure 5-17 (a). 

Bare thermocouple junction 

Magnesium 

oxide 

Figure 5-16 

Insulated junction 

Thermocouple 

wires 


Inconel 

sheath


To improve response time, the junction may be grounded as shown in Figure 

5-17 (b), or the thermocouple tip could be exposed as shown in Figure 5-17 (c). 

Grounded junction 


Protecting tube construction 

Protecting tubes (or thermowells) 

are used to shield thermocouple 

sensing elements against 

mechanical damage, thermal shock 

and corrosive or contaminating 

atmospheres. Various types of 

constructions are available. A 

typical construction method, 

similar to that of RTD’s, is shown 

in Figure 5-18. Cast iron protection 

tubes are used for example, in 

molten aluminum, magnesium and 

zinc applications while ceramic 

tubes are used in industries such as 

iron and steel, glass, cement and 

lime processing. 

Exposed junction 

Figure 5-17 (c) 

Cover 

Conduit 

entrance 

Socket 

Metal or ceramic 

protecting tube 

Sheathed 

thermocouple 

Figure 5-18 

5.10.5 Measuring temperature with thermocouples 

The voltage generated by a thermocouple is a function of the temperature difference 

between the measurement and reference junctions. Traditionally the reference junction 

was held at 0 °C by an ice bath, as shown in Figure 5-19. The thermocouple emf is 

measured with a high impedance voltmeter. 

Hot junction 

v 

Cold junction 

Ice bath 

Figure 5-19


The relationship between thermocouple voltage and temperature is unfortunately not 

linear, and it is necessary to use thermocouple temperature conversion tables to find 

temperature from the measured voltage. An extract from the voltage / temperature 

table for a type K thermocouple (0 °C reference), is given in Table 5-3. 

Temperature (°C) versus emf (μV) for type K thermocouple with 0 °C reference. 

deg C 0 1 2 3 4 5 6 7 8 9 

0 0 39 79 119 158 198 238 277 317 357 

10 397 437 477 517 557 597 637 677 718 758 

20 798 838 879 919 960 1000 1041 1081 1122 1163 

30 1203 1244 1285 1326 1366 1407 1448 1489 1530 1571 

40 1612 1653 1694 1735 1776 1817 1858 1899 1941 1982 

50 2023 2064 2106 2147 2188 2230 2271 2312 2354 2395 

60 2436 2478 2519 2561 2602 2644 2685 2727 2768 2810 

70 2851 2893 2934 2976 3017 3059 3100 3142 3184 3225 

80 3267 3308 3350 3391 3433 3474 3516 3557 3599 3640 

90 3682 3723 3765 3806 3848 3889 3931 3972 4013 4055 

100 4096 4138 4179 4220 4262 4303 4344 4385 4427 4468 

Table 5-3 

We could store these look-up table values in a computer and use the table to convert 

between emf and temperature. A more viable approach used by manufacturers 

however, is to approximate the table values using a power series polynomial (see 

Example 5-14) and allow the instrument’s microprocessor or process computer, to 

calculate temperature from emf or the emf from temperature (inverse polynomial). 

The ice bath cold junction is not considered practical anymore. Instead the terminals 

connecting the thermocouple to the measuring device, are now assumed to play the 

role of the reference junction or 'cold junction', as it is still called today. The reference 

junction temperature may now be kept, for example, at room temperature, where the 

junction temperature is measured with an auxiliary temperature sensor, such as a RTD. 

According to the law of intermediate temperatures, the thermocouple voltage that 

corresponds to the cold junction temperature, may be added to the measured 

thermocouple voltage. The true temperature of the hot junction, with respect to 0 °C, 

can then be determined from this augmented voltage. 

Example 5-10 

Calculate the average sensitivity (μV/°C) of a type K thermocouple in the temperature 

range 0 °C to 100 °C. 

From Table 5-3 the change in emf developed by a type K thermocouple from 0 °C to 

100 °C, is 4096 μV. The average sensitivity is therefore 4096/100 = 40.96 μV/°C.


Example 5-11 

The cold junction of a type K thermocouple is kept at 0 °C. Use Table 5-3 to 

determine the temperature if the measured voltage is a) 798 μV and b) 2602 μV. 

a) 20 °C b) 64 °C 

Example 5-12 

The relationship between emf and temperature for a certain (imaginary) thermocouple, 

is described by the relation: v = t 2 , where v is the generated thermocouple emf in 

microvolt (μV), and t the temperature difference in °C, between the hot junction and 

0 °C. If the thermocouple emf reading is 3000 microvolt and the temperature of the 

cold junction is 25 °C, calculate the temperature of the hot junction. 

Emf corresponding to (25 – 0) °C = 25 2 = 625 μV 

Total emf (T - 0) = 3000 + 625 = 3625 μV 

According to the law of intermediate temperatures: 

Hot junction temperature T = √v = √3625 = 60.21 °C 

(T is NOT = √3000 + 25 = 54.77 + 25 = 79.77 °C) 

Example 5-13 

An unknown temperature is measured with a type K thermocouple. A thermocouple 

voltage of 2602 μV is measured. If the cold junction temperature is 20 °C, calculate 

the process temperature, measured by the hot junction side of the thermocouple. 

From Table 5-3, the cold junction (20°C) emf is 

798 μV. According to the law of intermediate 

temperatures, the correction voltage of 798 μV 

should be added to the measured voltage of 

2602 μV, to obtain 3400 μV. The corrected voltage 

represents the thermocouple emf that would be 

obtained, if the reference junction was kept at 0 °C. 

Again from Table 5-3, the temperature that 

corresponds to 3400 μV, is somewhere between 

83 °C and 84 °C. To find the correct temperature, 

we must use linear extrapolation between these 

two values. The difference between 3433 μV 

(84 °C) and 3391 μV (83 °C) is 42 μV, while 

3400 μV is 9 μV more than 3391 μV (83 °C). 

Therefore the temperature we are looking for is 

83 °C plus (9/42) °C, which is 83.2143 °C. 

The way NOT to calculate the hot junction 

temperature, is to look up the measured voltage 

(2602 μV) as 64 °C and then to add the cold 

junction temperature of 20 °C, to obtain 84 °C. 

This is NOT CORRECT. 

T 

0°C 25°C 25°C 

X 1 9 = ∴X = × 1 

9 42 42 

625μV 3000μV 

0°C T 

3625μV 

0°C 20°C 20°C T 

798μV 

2602μV 

0°C T 

3400μV 

84°C 

1°C 

 

X 

83°C 

9μV 

3391μV 

42μV 

3400μV 3433μV


Example 5-14 

The following equations are provided for a type K thermocouple, to calculate 

temperature (t in °C) from emf (v in μV) in the temperature range 0 °C to 500 °C: 

t = (2.508355×10 -2 )v + (7.860106×10 -8 )v 2 - (2.503131×10 -10 )v 3 

+ (8.315270×10 -14 )v 4 – (1.228034×10 -17 )v 5 + (9.804036×10 -22 )v 6 

- (4.413030×10 -26 )v 7 + (1.057734×10 -30 )v 8 – (1.052755×10 -35 )v 9 , 

Equation (a) 

and to calculate emf (v in μV) from temperature (t in °C) in the temperature range 

0 °C to 1372 °C: 

v = -17.600413686 + 38.921204975t + (1.8558770032×10 -2 )t 2 

- (9.9457592874×10 -5 )t 3 + (3.1840945719×10 -7 )t 4 

- (5.6072844889×10 -10 )t 5 + (5.6075059059×10 -13 )t 6 

- (3.2020720003×10 -16 )t 7 + (9.7151147152×10 -20 )t 8 

− 4 

2 

– (1.2104721275×10 -23 )t 9 + 118.5976e 

−1.183432× 

10 (t −126.9686) 

Equation (b) 

a) Use Equation (a) to calculate the temperature when the emf is 2602 μV. 

b) Use Equation (b) to calculate the emf when the temperature is 20 °C. 

a) t = (2.508355×10 -2 )×2602 + (7.860106×10 -8 )×2602 2 - (2.503131×10 -10 )×2602 3 

+ (8.315270×10 -14 )×2602 4 – (1.228034×10 -17 )×2602 5 + (9.804036×10 -22 )×2602 6 

- (4.413030×10 -26 )×2602 7 + (1.057734×10 -30 )×2602 8 – (1.052755×10 -35 )×2602 9 

= 65.26740 + 0.5321609 – 4.409664 + 3.811584 – 1.464694 + 0.3042627 

- 0.03563592 + 0.002222464 – 0.05755631 

= 63.95 °C 

b) v = -17.600413686 + 38.921204975×20 + (1.8558770032×10 -2 )×20 2 

- (9.9457592874×10 -5 )×20 3 + (3.1840945719×10 -7 )×20 4 

- (5.6072844889×10 -10 )×20 5 + (5.6075059059×10 -13 )×20 6 

- (3.2020720003×10 -16 )×20 7 + (9.7151147152×10 -20 )×20 8 

− 4 

2 

– (1.2104721275×10 -23 )×20 9 + 118.5976e 

−1.183432× 

10 (20 −126.9686) 

= -17.600414 + 778.4241 + 7.423508 – 0.7956607 + 0.05094551 

- 0.001794331 + 35.88804×10 -6 – 409.8652×10 -9 + 2.487069×10 -9 

- 6.197617×10 -12 + 30.61899 

= 798.1 microvolt 

(Compare these answers to those of Example 5-11) 

5.10.6 Compensating leads 

Thermocouple thermometers are normally installed some distance away from the 

voltmeter or computer that measures the emf generated by the thermocouple. For 

this purpose, cheaper and lower grade thermocouple wires, called extension wire or 

compensating leads, are used to connect the thermocouple to the measuring device 

at the reference junction. Compensating leads have the same thermoelectric


properties as the thermocouple and do not introduce a significant error into the 

temperature measurement. Compensating leads must be matched to the 

thermocouple and for each type of thermocouple, corresponding extension leads 

are available. 

5.11 THERMISTORS 

A thermistor is similar to a resistance thermometer, but a semiconductor material is 

used instead of a metal. A distinct advantage of thermistors over resistance 

thermometers, is that their temperature coefficient of resistance is approximately 

ten times higher than that of a resistance thermometer, causing thermistors to be 

much more sensitive temperature detectors. The resistance change with 

temperature is however very nonlinear, and unlike RTD’s, most thermistors have a 

negative temperature coefficient of resistance, that is, the resistance of a thermistor 

decreases with increasing temperature. 

In Figure 5-20, the resistance curve of a typical thermistor is shown. The following 

equation, called the Steinhart and Hart equation, is used to describe the resistance / 

temperature relation for thermistors: 

1 

= A + Bln(R) + ClnR 3 , 

T 

where T is absolute temperature in Kelvin, R is the thermistor resistance at 

temperature T and A, B and C are coefficients that describe a given thermistor. 

Resistance R 

Figure 5-20 

Temperature T 

The resistivity of thermistors is also much larger than that of RTD’s, and 

thermistors can therefore be made very small which means they will respond 

quickly to temperature changes. 

Thermistors cannot be used to measure high temperatures, compared to RTDs. In 

fact, the maximum temperature of operation is sometimes only 100 or 200 o C. The 

high resistivity as well as high temperature coefficient of resistance of the 

thermistor, makes it feasible to use the simpler two wire technique when measuring 

thermistor resistance.


For example, a common thermistor value is 5000 ohms at 25°C. With a typical 

temperature coefficient of resistance of 0.04/°C, a measurement lead resistance of 

10 Ω, produces only 0.05 °C error. This error is a factor of 500 times less than the 

equivalent RTD error. Thermistors have not gained nearly the popularity of RTDs or 

even thermocouples in industry due to their limited span as well as other 

disadvantages. Since thermistors are semiconductor devices, they are quite susceptible 

to permanent decalibration when exposed to high temperatures. In addition, 

thermistors are quite fragile and great care must be taken to mount them so that they 

are not exposed to shock or vibration. 

5.12 RTD, Thermocouple or Thermistor 

Resistance temperature detectors (RTDs) 

An RTD sensing element consists of a wire coil or deposited film of pure metal. The 

element’s resistance increases with temperature in a known and repeatable manner. 

RTDs exhibit excellent accuracy over a wide temperature range and represent the 

fastest growing segment among industrial temperature sensors. Their advantages 

include: 

• Temperature range: Models cover temperatures from -260 to 850°C. 

• Repeatability and stability: Industrial RTDs typically drift less than 0.1°C/year. 

• Sensitivity: The voltage drop across an RTD provides a much larger output than a 

thermocouple. 

• Linearity: Platinum and copper RTDs produce a more linear response than 

thermocouples or thermistors. RTD non-linearities can be corrected through 

proper design of resistive bridge networks. 

• Low system cost: RTDs use ordinary copper extension leads and require no cold 

junction compensation. 

Thermocouples 

A thermocouple consists of two wires of dissimilar metals welded together into a 

junction. At the other end of the signal wires, usually as part of the input instrument, is 

another junction called the reference junction. Heating the sensing junction generates a 

thermoelectric potential (emf) proportional to the temperature difference between the 

two junctions. This millivolt-level emf, when compensated for the known temperature 

of the reference junction, indicates the temperature at the sensing tip. Published 

millivolt tables assume the reference junction is at 0°C. Thermocouples are simple and 

familiar. Designing them into systems, however, is complicated by the need for special 

extension wires and reference junction compensation. Thermocouple advantages 

include: 

• Extremely high temperature capability: Thermocouples with precious metal 

junctions may be rated as high as 1800°C. 

• Ruggedness: The inherent simplicity of thermocouples makes them resistant to 

shock and vibration. 

• Small size/fast response: A fine-wire thermocouple junction takes up little space 

and has low mass, making it suitable for point sensing and fast response.


Thermistors 

A thermistor is a resistive device composed of metal oxides formed into a bead and 

encapsulated in epoxy or glass. A typical thermistor shows a large negative 

temperature coefficient. Resistance drops dramatically and non-linearly with 

temperature. Sensitivity is many times that of RTDs but useful temperature range is 

limited. There are wide variations of performance and price between thermistors from 

different sources. Typical benefits are: 

• Low sensor cost: Basic thermistors are quite inexpensive. However, models with 

tighter interchangeability or extended temperature ranges often cost more than 

RTDs. 

• High sensitivity: A thermistor may change resistance by tens of ohms per degree 

temperature change, versus a fraction of an ohm for RTDs. 

• Point sensing: A thermistor bead can be made the size of a pin head for small area 

sensing. 

RTD 

Sensor 

type 

Thermocouple 

Thermistor 

Temperature 

range 

-260 to 

850°C 

-270 to 

1800°C 

-80 to 

150°C 

(typical) 

Sensor 

cost 

System 

cost 

Stability Sensitivity Linearity Specify for: 

Moderate Moderate Best Moderate Best 

Low High Low Low Moderate 

Low Moderate Moderate Best Poor 

General purpose 

sensing 

Highest accuracy 

Highest 

temperatures 

Best sensitivity 

Narrow ranges 

(e.g. medical) 

Point sensing

EIPINI Chapter 6: Process Control Page 6-1 

6. PROCESS CONTROL 

The purpose of this chapter is to introduce students to industrial process control 

methods used in industry to ensure that manufactured products meet predetermined 

quality requirements. This is accomplished by continuously monitoring the production 

process and automatically correcting or minimizing any deviations from the required 

specifications, that may be detected during the manufacturing process. 


The history of automatic control goes back many centuries. Water level control may 

already be identified in water clocks used in the middle ages while mechanical clocks, 

making their appearance in the 1200’s, used an escapement mechanism that may be 

described in terms of feedback control. With the advent of the steam engine, it was 

evident that some sort of speed control was needed. In what is generally considered a 

major event in the history of feedback control, James Watt, completed 

the design of the centrifugal flyball governor for regulating the 

speed of the rotary steam engine, in 1788. This device 

employed two pivoted rotating flyballs which were flung 

outward by centrifugal force. As the speed of rotation increased, 

the flyweights swung further out, operating a steam flow throttling 

valve which slowed the engine down. Thus, a constant speed was 

achieved automatically. From this point onwards, many new developments followed 

to improve control systems as well as mathematical research to understand physical 

processes. A major invention, however, was the introduction of proportional, integral 

and derivative (PID) control, which was formulated in 1922 by Nicholas Minorsky 

(1885-1970). Observing the way in which a helmsman steered a ship and compensate 

for the disturbances from the ocean, motivated this threefold control strategy. 

Although advanced tools, such as model predictive controllers, are available today, 

the Proportional, Integral, Derivative (PID) control strategy, is still the most widely 

used in modern industry, controlling more than 95% of closed loop industrial 

processes. 

6.1.1 A SIMPLE WATER LEVEL CONTROL 

In Figure 6-1, an operator has the simple task of 

keeping the water level in a tank at a prescribed 

value, for instance half full. If the water level 

should drop below the desired level due to 

increased water outflow, the operator must 

increase the water inflow by opening the inflow 

valve more until the desired level is restored. 

Similarly, if the water level should increase due 

to a decrease in outflow, the operator must 

decrease the inflow by closing the inflow valve 

more, until the required level is restored. 

Inflow 

valve 

Water inlet 

Inflow 

Figure 6-1 

Outflow 

Outflow 

valve


If the water level reaches the desired value (also called the set point), the operator 

must try to keep the inflow equal to the outflow. This manual control example 

illustrates essentially how we would expect an automatic control system to behave. 

In Figure 6-2, an 

arrangement for 

automatic control 

of the water level 

in the container, 

is shown. A 

level measuring 

instrument senses 

the water level in 

the container, and 

Inflow 

valve 

feeds this value back to the controller. The basic function of the controller is to 

calculate the difference between the desired water level and the measured water level, 

and to determine a sensible control action, that will minimize this difference. Once the 

controller has calculated an appropriate control action, this value is transmitted as a 

pneumatic signal to the inflow valve, which is adjusted accordingly. A well-designed 

controller should respond quickly to changes in the water level and always steer the 

level in the water container back to the required level, in a disciplined way. 

6.1.2 CONTROL TERMINOLOGY 

The water container system in Figure 

6-2 could be represented as an 

input/output system or process, as 

shown in Figure 6-3. The quantity that 

we want to regulate is the water level 

in the container, which is called the 

controlled variable. The controlled 

Level 

controller 

Water inlet 

Water inflow 

(Input) 

Level 

detector 

Water outflow 

Figure 6-2 

Water level 

control 

Process 

(plant) 

Figure 6-3 

Inflow 

Outflow 

Outflow 

valve 

Water level 

(Output) 

variable is also called the output variable, the process variable or the measured 

variable. The rate of water inflow into the container (a system input) with which we 

may change or manipulate the water level in the container, is called the manipulated 

variable. The manipulated variable is also called the controller output or control 

variable. A change in the rate at which the water flows out of the container (a system 

input), will cause the water level in the container to change or to be disturbed, and is 

called a disturbance variable. (Note: there could be other disturbance variables that 

may influence the controlled variable, for instance a leak in the container.) 

Three values are associated with the controlled variable. Firstly, the measured value, 

which is the current water level in the container, determined by the level sensor. 

Secondly, the desired value (normally locally programmed into the controller), which 

is the water level that we need or require, for example 50% full. The desired value is 

also called the set point or reference value. Thirdly the error value which is the 

difference between the desired value and the measured value of the water level.


1. CONTROLLED VARIABLE: Process output variable that is maintained 

between specified limits. 

2. MEASURED VALUE: Actual value of the controlled variable, as determined 

by the instrumentation. 

3. DESIRED VALUE: Required value of the controlled variable (set point). 

4. ERROR VALUE: The difference between the desired value and the measured 

value. 

5. MANIPULATED VARIABLE: Process input variable that is adjusted, to steer 

the controlled variable towards the desired value. 

6. DISTURBANCE VARIABLE: Process input variable that can cause the 

controlled variable to deviate from the desired value. 

Example 6-1 

A control system for a heat exchanger is shown in Figure 6-4. Cold water enters the 

container at point 1 and is heated by steam, entering the steam line 3, at point 2. The 

heated water leaves the system at point 7. The temperature of the hot water is 

measured by a thermometer at point 6 and presented to the temperature controller, 

block 5. The controller operates a steam valve 4, such that when the hot water is below 

the required temperature, the steam flow rate through the steam line is increased, 

allowing more heat transfer to the cold water, and when the hot water is above the 

required temperature, the steam flow is reduced. Identify: a) the controlled variable, b) 

the measured value, c) the manipulated variable and d) some disturbance variables. 

2 

Steam 

supply 

4 

Steam 

control 

valve 

Cold water 

1 

3 

TIC 

Steam outlet 

6 

5 

Figure 6-4 

Heat 

exchanger 

7 

Hot water 

a) Controlled variable: The temperature of the hot water delivered at 7. 

b) Measured value: Temperature indicated by the temperature detector 6. 

c) Manipulated variable: The steam flow rate through the steam line 3, adjusted by 

valve 4. 

d) Disturbance variables: 

i) Change in hot water demand (this is an energy change at the output 7 and is 

known as an output or demand load disturbance). 

ii) Change in steam supply pressure (this is an energy change at the input 2, and 

is known as an input or supply load disturbance). 

iii) More heat loss through the heat exchanger walls, because of colder ambient 

conditions.


6.1.3 OPEN LOOP AND CLOSED LOOP SYSTEMS 

Automatic control systems may operate under open or closed loop control. The input 

of an open loop system is not influenced by its output. This is in contrast with a closed 

loop system that monitors its output and then manipulates the input, to ensure that the 

output meets set conditions. An example of an open loop system is a bullet fired from 

a rifle and an example of a closed loop systems is a cruiser missile, fired from a ship, 

continuously monitoring and adjusting its trajectory. 

Open loop control strategies may perform well in situations where the behaviour of 

the system is very well defined and modelled. For example a room temperature 

controller set to a certain on/off heater schedule, will produce the desired room 

temperature. However, it is difficult to suggest an open loop control strategy for the 

water level control in Figure 6-2, as the factors that influence the water level are 

difficult to predict, and a closed loop control strategy, as indicated in Figure 6-2, is 

clearly the most sensible solution. 

Open loop system: The input to the system is not determined by the output. 

Closed loop system: The input to the system is determined by the output. 

6.1.4 FEEDBACK AND FEEDFORWARD CONTROL 

Feedback closed loop control systems measure changes in the controlled variable 

directly and feed them back as input variables to the controller. Feedforward control 

systems on the other hand, do not measure changes in the controlled variable but 

rather other variables, termed intermediate variables, which are indicative of the 

controlled variable. 

The interrelationship between the elements of a general closed loop feedback control 

system, is illustrated with the block diagram in Figure 6-5. The water level control 

system in Figure 6-2 is an example of a closed loop feedback control system. The 

water level provided by the level sensor, is fed back to the error detector which 

calculates the error. The controller uses the error value to compute a suitable 

command for the inflow valve, thereby regulating the water level close to the desired 

value. 

Disturbance 

variables 

Desired 

value 

Comparator 

Measured 

value 

Error 

value 

Control 

unit 

Manipulated 

variable 

Sensor 

Process 

Controlled 

variable 

Output 

Figure 6-5 Block diagram of a feedback control system


Feedforward control systems, on the other hand, will measure disturbance variables 

that directly influence the controlled variable. It is conceivable, for example, to 

measure the outflow in the water level control system in Figure 6-2, and to base the 

inflow control on the value of this disturbance variable. Because the source of the 

level disturbance is monitored and acted upon, a change in water level is anticipated 

and corrected before it even occurs. 

Whereas feedback systems must wait for a deviation to occur before corrective 

action is taken, the principle advantage of feedforward control systems is that they act 

before deviations occur. It is clear that feedforward control alone would not be 

adequate for the water level control system, as the water level would eventually drift 

away from set point because of measuring and modelling errors. In general, 

feedforward control is used to complement feedback control and to enhance system 

performance. 

Feedback control: Measure the controlled variable to determine the control strategy. 

Feedforward control: Measure disturbance variables to determine the control 

strategy. 

6.1.5 DIRECT ACTING AND REVERSE ACTING CONTROL 

A control valve that opens more when the pressure input is increased (air to open) is 

called a reverse acting valve. For the water level control system depicted in Figure 6-2, 

we have assumed that a reverse acting control valve is used. This means that when the 

water level drops below set point, the controller must increase its output to the control 

valve to increase the water inflow. This type of control action is called reverse acting, 

because when the water level decreases, the controller output must increase. If a direct 

acting (air to close) control valve was used, the control action would be direct acting, 

as the controller output would decrease when the water level decreases. 

A heating system would typically be reverse acting, as a decrease in temperature 

would demand an increased controller output to the heating element. A cooling 

system, on the other hand, would typically be direct acting because an increase in 

temperature would require an increased output to the cooling element. 

Direct acting control: A control arrangement in which the controller output 

increases if the measured value rises above the set point. 

Reverse acting control: A control arrangement in which the controller output 

increases if the measured value drops below the set point. 

6.1.6 PROCESS TIME LAGS 

Most processes exhibit a time delay between its input and output. For example, if the 

heating element of a temperature control system is switched on, the temperature will 

not immediately rise to the maximum value. The inlet flow valve in the level control 

system in Figure 6-2 is another example. When it receives a signal to open 100%, the 

flow rate will not immediately increase to 100% due to delays inherent in the 

operation of the valve.


Figure 6-6 shows the typical response to a step input of a process such as a control 

valve receiving a command to open from 50 % to 60 %, resulting in the increase in 

inflow from 50 % to 60 %. 

Input 

60 % 

50 % 

Output 

60 % 

50 % 

t d 

t f 

Time 

Dead time and first order lag 

Time 

Figure 6-6 

The response curve in Figure 6-6 illustrates two kinds of time delays that may be 

identified. Firstly a dead time t d , when the system does not respond at all, and 

secondly a first order delay t f , when the system output changes to its new final value 

in an exponential fashion. 

The dead time is associated with a delay in the transportation of material or 

information. For example, it may take 20 minutes to send a signal to the Rover on 

Mars. During this time, the vehicle can not respond to this control command, because 

the control information is still being transported to the vehicle. The dead time is also 

known as a distance-velocity lag, and the reason for this is illuminated in 

Example 6-2. Another term used for the dead time, is transportation lag. 

The first order lag is associated with a delay in the transfer of energy. For example, 

when you step on the accelerator of a car, the car will take time to convert the 

chemical energy in the fuel, into mechanical kinetic energy, and the speed increase 

will be exponential in nature. The first order lag is also known as a transfer lag, 

exponential lag or resistance-capacitance lag. 

Dead time: Delay due to the time it takes information or material to be transported 

from one point to another. 

First order lag: Delay due to the time it takes energy to be transferred from one 

point or form to another. 

Example 6-2 

In the system depicted Figure 6-7, a conveyor belt is used to transport material that 

was deposited by a feeder, to the delivery station. A request for more material is given 

to the feeder. Calculate the time it will take the sensor at the delivery site to detect a 

reaction to the increased demand for material. The velocity of the conveyor belt is v 

meter per second and the distance between the feeder and delivery point is d meter.


Feeder 

v 

Delivery 

station 

Figure 6-7 

Sensor 

d 

Any increase in material delivered, must travel a distance d with a velocity v, before it 

reaches the delivery point, where the sensor will detect the increase. The time it takes 

the material to be transported to the sensor is t = distance/velocity = d/v seconds. The 

dead time lag in this case is called a distance-velocity lag and it is a problem not easily 

handled by a PID controller. 

6.2 CONTROL SCHEMAS 

We will now discuss some of the common 

techniques (on-off and PID control modes) 

used to control closed loop feedback 

systems. For this purpose we will use the 

water level control system in Figure 6-2 as 

basis. Central to our system is the water 

container shown in Figure 6-8. We will use 

a container with height 5 m and base area 

of 1 m 2 . The level of the water in the 

container is H, with a maximum value of 

5 m. The rate of water outflow is Q OUT , 

with a maximum value of 0.01 m 3 /sec. The 

rate of water inflow is Q IN . In order to have 

any control over the water level in the tank whatsoever, the maximum inflow should 

be more or equal to the maximum outflow. We will assume that they are equal, that is, 

when the demand is a maximum of 0.01 m 3 /sec. and the inflow is a maximum of 

0.01 m 3 /sec, the water level will remain constant. 

To simplify matters, we will use percentage values for all variables. The variable M 

will be used to represent the water level, expressed as a percentage value. The water 

level will be measured by an electronic DP transmitter, delivering 4 mA when the tank 

is empty (H = 0 m or M = 0 %) and 20 mA when the tank is full (H = 5 m or 

M = 100 %). We will also assume that the level detector output M will immediately 

(with no time delay) reflect the actual water level H. 

The variable C will be used to represent the controller output, expressed as a 

percentage value. The controller will transmit 20 kPa to the control valve to close the 

valve completely (C = 0 %) and 100 kPa to open the valve fully (C = 100 %). We will 

also assume that a linear relationship (with no time delay) exists between the 

controller output and the water inflow. This simplification means that we can use the 

same variable C for the controller output and the inflow (manipulated variable). 

H 

Q IN 

Figure 6-8 

Maximum inflow: 

0.01 m 3 /sec. 

1 m 2 

5m 

Q OUT 

Maximum 

level: 

5 meter 

Maximum outflow: 

0.01 m 3 /sec.


Therefore, if the controller output C is 0 %, the water inflow Q IN will immediately 

cease (Q IN = 0 m 3 /sec.) and when C is 100 %, the water inflow will immediately rise 

to its maximum value of Q IN = 0.01 m 3 /sec. 

The outflow demand or load Q OUT will be represented by the percentage variable L. 

L will be 0 % when the outflow Q OUT = 0 m 3 /sec and L will equal 100 % when the 

outflow is the maximum value of Q OUT = 0.01 m 3 /sec. 

The complete water level control system is shown in Figure 6-9. 

Inflow C (0% – 100%) 

100% 

Water inflow C 

Controller 

Output C 

100% 

Inflow 

valve 

C 

Level 

controller 

M 

Level 

detector 

Water level M 

(0% – 100%) 

Water supply 

Figure 6-9 

Water level 

control system 

Outflow L (0 – 100%) 

Before we can use our water tank, we must find out how it works. This means that 

we need to understand how the inflow and outflow influence the water level in the 

container. Clearly when C = L, the water level M will stay constant. When C > L, the 

water level will rise and when C < L, the water level will drop. It is therefore easy to 

Δ M 

argue that the rate at which the water level changes, , is proportional to C – L. In 

Δt 

the example given at the end of this chapter in Appendix 6-1, the exact relationship 

between M, C and L, was found to be: 

dM = 0.002×(C – L) percent per second. 

dt 

Finally, the error value will play a fundamental part in the operation of on-off and 

PID control. The percentage error value E is defined as: 

E = S – M ………….…………..………..…………………… Equation 6-1 

The set point is denoted by S and is expressed as a percentage of the maximum 

measured value. For example, a set point value S = 50 %, will imply a required water 

level of 2.5 m or half filled tank for the water level control system. 

6.2.1 ON-OFF OR TWO-POSITION CONTROL 

With on-off control mode (also called bang-bang control), the controller output C, can 

take on only two values, on or off. Referring to the level control in Figure 6-9, either


the inflow is switched on fully (100%) when the water level drops below a certain 

level, say 40%, or it is closed completely (0%), when the water level rises above a 

certain level, say 60%. With the aid of Equation 6-1, we could express the control law 

for the water level on-off control as: 

0% if E ≤ -10% (or M ≥ 60%) 

C = unchanged if -10% < E < 10% (or 60% > M > 40%) 

100% if E ≥ 10% (or M ≤ 40%) 

A possible response of the water level to on-off control as well as the controller 

action, is shown in Figure 6-10. On average, the water level remains 50%. 

Water level (M) 

60% 

40% 

Water inflow (C) 

100% 

Time 

0% 

Time 

Figure 6-10 Typical on/off control performance 

On-off control will also work well for a room temperature control – the heater is 

fully on, until the room temperature increases to a prescribed value, after which the 

heater is switched off, and so on. Not all systems are suited to on-off control. Imagine 

a speed cruise control in a car, set to 120 km/h, using on-off control. 118 km/h: full 

acceleration, 122 km/h: no throttle. Surely not very comfortable driving. 

A disadvantage of on-off control is the wear on the final control element, such as the 

control valve in the present example that continually moves from one extreme position 

to the other. 

Example 6-3 

A water container is 5 meter high and has a base area of 1 m 2 . The maximum water 

inflow rate, as well as the maximum water outflow rate, is 0.01 m 3 /sec. On-off control 

is used to regulate the water level in the tank, between 40% and 60% of its maximum 

level (5 meter). Assume that the rate of water outflow, is regulated and constant at 

60% of the maximum rate of outflow. Calculate: 

a) The time period that the inflow valve is closed 

b) The time period that the inflow valve is open. 

Solution: 

a) Water level when the inflow valve must close: 60% of 5 m = 0.6×5 = 3 meter. 

Water level when the inflow valve must open: 40% of 5 m = 0.4×5 = 2 meter. 

Constant demand from tank: 60% of 0.01 = 0.6×0.01 = 0.006 meter 3 /second.


The inflow valve closes when the 

water level reaches the 3 m mark, and 

stays closed until the water level 

drops to the 2 m mark. Total volume 

of water that must leave the container 

before the 2 m mark is reached: 

V = Ah = 1×1 = 1 m 3 . 

Rate of outflow Q, is 0.006 m 3 /s. 

But volume = flow rate×time 

∴V = Qt [(m 3 /sec)×(sec)] 

∴t closed = V/Q = 1/0.006 

= 166.7 sec = 2.778 min 

b) The inflow valve opens when the 

water level reaches the 2 m mark and 

stays open until the water level rises 

to the 3 m mark. Total volume of 

water that must enter the container, is 

1 m 3 . The resultant rate at which 

water flows into the tank is the rate of 

inflow, minus the outflow rate. 

Q = 0.01–0.006 = 0.004 m 3 /sec. 

∴t open = V/Q = 1/0.004 

= 250 sec = 4.167 min. 

5m 

Max inflow (100%) = 0.01 m 3 /s 

Min level=40%=2 m 

Controlled and 

1 m 2 constant outflow 

Outflow=60% of max (0.01) = 0.006 m 3 /s 

M 

60% 

40% 

C 

100% 

0% 

2.778 

min 

Max level=60%=3 m 

4.167 

min 

t 

t 

On-off control: A control strategy in which the controller output switches the final 

control element fully on or off to keep the controlled variable near set point. 

6.2.2 PROPORTIONAL CONTROL 

The proportional control strategy provides more smooth control than on-off control. 

The control effort is proportional to the magnitude of the error, which makes sense if 

one thinks about a person driving a car and trying to obey a 120 km/h speed limit. The 

more the speed falls below 120 km/h during an uphill climb, the more he will step on 

the accelerator, and when the speed tends to go above 120 km/h as he goes downhill, 

he will start to release the accelerator. When reaching 120 km/h on a straight road, he 

will keep the accelerator in just the right position to maintain his target speed of 

120 km/h. 

This is very important to understand, the driver does not release the accelerator 

completely when the target is reached. For proportional control, this control effort that 

is maintained when the error is zero, is called the bias. We will denote the bias by the 

percentage variable R. When designing a proportional controller, a fixed bias R must 

be chosen and this choice depends on the typical load on the system. For our water 

level control example, we will assume that the typical outflow demand is 50%, and a 

suitable choice for the bias, is R = 50%.


A simple and logical proportional control law for the water level system can now be 

formulated. The inflow equals 100% when the tank is empty, the inflow equals 50% 

(bias) when the tank is half full (equal to the desired value) and the inflow equals 0% 

when the tank is full (overflowing). This control strategy is depicted in Figure 6-11. 

Water inflow C (%) 

100 % 

50 % 

R (Bias) 

Figure 6-11 

Basic 

proportional 

control law 

This is obviously a reverse acting control strategy, because the controller output 

increases when the measured value decreases. A negative slope of the C-E line would 

indicate a direct acting control strategy. It is now a simple matter to write down an 

expression for this proportional control law, as the graph of C versus E, is a straight 

line of the form: y = mx + c. From Figure 6-11: C = E + 50. The slope of this line is 

clearly 1, because the controller output C, changes by 100% when the error E, changes 

by 100%. The slope of the line is called the proportional gain K P , of the controller. 

For the controller characteristic in Figure 6-11, the gain is therefore K P = 1. 

If the controller gain, K P , is changed, the slope of the straight line, describing the 

relationship between controller output C, and error value E, will also change. Graphs 

of C versus E, for three different controller gains, K P = 2, K P = 1 and K P = 0.5, are 

shown in Figure 6-12. (Note: Although a gain of 0.5, prescribes a theoretical error of 

100% before the output reaches 100 %, such an error could of course not occur in our 

water level control system.) 

100% 

-50 % 

Tank full, 

M = 100% 

C (%) 

100% 

0 % 50 % 

Tank half full, 

M = 50% 

C (%) 

Tank empty, 

M = 0% 

100% 

75% 

C (%) 

Error (%) 

(E=50-M) 

R=50% 

R=50% 

R=50% 

E(%) 

E(%) 

25% 

E(%) 

-25% 0% 25% -50% 0% 50% -100% -50% 0% 50% 100% 

K P = 2 K P = 1 K P = 0.5 

Figure 6-12 Proportional control with different proportional gains 

An alternative representation of the relationship between C and E, for K P = 2, K P = 1 

and K P = 0.5, is shown in Figure 6-13. The movement of the left hand side of a


Figure 6-13 Proportional band 

C E 

E 

0% 50% 

25% 

0 

-25% 

50% 

100% 

0 

-50% 

C 

0% 

50% 

100% 

0 

E 

100% 

C 

0% 

50% 

100% 

K P = 2 

K 

-100% 

P = 1 K P = 0.5 

pivoted beam, represents changes in the error E, while the right hand side, sweeps out 

corresponding values of the controller output C. For K P = 2, an error range of 50% 

(from -25% to 25%), sweeps out the complete output range of 100%. For K P = 1, an 

error range of 100% results in an output change of 100%. For K P = 0.5, the error value 

must change 200% to cover the whole output range. The error range that results in the 

total change in output range, is called the proportional band, of the controller, and is 

denoted by the percentage variable PB. 

For a high gain (narrowband) system (K P =2 in Figure 6-13 for example), a small 

error would cause a large reaction from the controller in his effort to correct the error. 

A low gain (wideband) system (K P = 0.5 in Figure 6-13 for example), would act more 

gently, because large errors, will cause a milder controller reaction. The relationship 

between proportional band PB (in percent) and proportional gain K P , is given by: 

PB = 

100 percent. …………...……………………….. Equation 6-2 

K 

P 

We can now obtain a general expression for the output C, of a proportional 

controller, in terms of the proportional band PB, the set point S, the measured value M 

and the bias R. In general, for any gain K P and bias R, the relationship between the 

controller output C, and the process error E, is a straight line of the form, y = mx + c, 

with the slope of the line equal to the controller gain K P and the y intersect, equal to 

the controller bias R. 

C (%) 

∴C = K P E + R ………………..………...….… (1) 

From Equation 6-1: E = S – M …...............….. (2) 

R (%) 

100 Slope=K P 

And from Equation 6-2: K P = ………..…. (3) 

PB 

(2) and (3) in (1): 

E (%) 

100 

C = K P E + R = (S – M) + R ……………………..…..…. Equation 6-3 

PB 

100 

In a typical application, S = 50 and R = 50, therefore C = (50 – M) + 50. PB


Example 6-4 

A process is controlled by a proportional controller. The controller is programmed for 

a positive gain (reverse acting controller) and proportional band of 80 %, a set point of 

50 % and a bias of 50 %. The system error is calculated from the difference between 

set point and the measured value i.e. E = 50 – M. 

a) Calculate the proportional gain of the controller. 

b) If the measured value of the process is indicated as 36 %, calculate the output of 

the controller at that instant. 

c) If the measured value of the process is indicated as 65 %, calculate the output of 

the controller at that instant. 

d) For the given settings, give the proportional control law (C as a function of E). 

e) Draw a graph of the controller control law (C versus E). 

a) From Equation 6-2: d) C = K P E + Bias 

K P = 100/PB = 100/80 = 1.25. ∴ C = 1.25E + 50 

b) From Equation 6-3: e) 

100 

C 

C = ×(S – M) + R 

100% 

ΔC 

PB 

Slope = 

ΔE 

100 

= ×(50 – 36) + 50 = 67.5%. 

100 

80 

= 

c) From Equation 6-3: 

50% 

80 

= 1.25 

100 

C = ×(S – M) + R 

PB 

E 

100 

= ×(50 – 65) + 50 = 31.25%. 

80 

-40% 0% 40% 

Example 6-5 (students must study this important example very carefully) 

The water level in a container is controlled by a proportional controller with K P = 1 

and bias R = 50%. At a given moment, the outflow demand L, is 50% and the inflow 

C, is 50%. The water level in the tank is also stable at its set point of 50% (the error is 

zero, therefore the inflow is 50% which is equal to the outflow). The outflow demand 

suddenly increases to L = 60%. Determine the new stable water level in the tank. 

Solution: The moment 

the outflow increases, the 

water level will start to 

drop, and it will continue 

to drop until the inflow C, 

also increases to 60%. 

The inflow is driven by 

the error value according 

to C = K P E + R = E + 50. 

C=50% 

L=50% 

S=50% 

M=50% 

C=60% 

S=50% 

L=60% 

M=40% 

Offset 

(E=10%) 

Therefore the controller output will reach 60% when E = 10%. This means, from 

E = 50–M, that the water level M, must drop to 40%. Although trying to keep the 

water level close to 50%, this example clearly illustrates that a proportional controller 

is not able to force the measured value back to set point, when the demand is


different from the bias value. Decreasing the proportional band (increasing K P ), will 

however keep the level closer to set point. For example, if K P = 2, the controller will 

need an error of 5 % to increase its output to 60 %. The water level will thus drop to 

only 45%. 

The difference between the set point and the measured value that may occur in 

proportional control, is called the offset, and it is a serious disadvantage of 

proportional only control. It is also important to note that we could have restored the 

level to 50 %, by changing the bias value of the controller to 60 %, if the outflow 

remained constant at 60 %. 

Example 6-6 

The water level in a container is controlled by a proportional controller and control 

valve. The controller gain is 1 and the bias is 50 %. Assume that the water inflow 

delivered by the valve is equal (no time delay) to the control signal C, and that the 

level detector instantly reflects the level M, of the water in the container. The 

relationship between the water level M, and the inflow C and outflow L, is given by: 

dM = 0.002×(C – L). 

dt 

a) Sketch the installation. 

b) Draw a block diagram of the system. 

c) The system is initially in the state: C = 50%, L = 50%, M = 50% and S = 50%. 

When time equals zero, the outflow demand L is suddenly increased to 60%. 

Determine a time expression for the water level M. (This will demonstrate the 

system’s response to a disturbance and is also called regulatory control action). 

a) 

S=50 

Inflow = C 

Inflow 

valve 

C 

Level 

controller 

M 

Level 

detector 

M 

b) 

S=50 

Water supply 

E=50-M 

K P =1 & R=50 

Control unit 

C=E+50 

C 

Control 

valve 

C 

L 

Outflow demand = L 

Water container 

dM =0.002(C–L) 

dt 

M 

M 

Controller 

Sensor 

c) The output of the proportional controller is given by: 

C = K P E + R = K P ×(S – M) + R = 1×(50 – M) + 50 = 100 – M. 

dM 

and = 0.002(C – L) = 0.002[(100-M) – 60] = 0.08 – 0.002M 

dt


dM 

∴ +0.002M = 0.08 and with M(0) = 50 gives the solution M = 40 + 10e 

-t/500 

dt 

A graph of the response of the water level M to the disturbance, is shown below. 

Demand L 

60 % 

50 % 

Time (sec.) 

Water level M 

50 % 

40 % 

t=0 

M = 40 + 10e -t/500 

43.7% 

t=500 sec. 

Offset 

Time (sec.) 

The behaviour of the water level may be compared with our findings in Example 6-5. 

Proportional control: A control strategy in which the controller output is 

proportional to the magnitude of the error. {C = K P E + R = (100/PB)×(S-M) + R} 

Proportional gain: Ratio of controller output change to error value change ΔC/ΔE. 

Proportional band: The error range that causes 100 % change in controller output. 

Offset: The steady state difference between the set point and the measured value. 

6.2.3 PROPORTIONAL AND INTEGRAL CONTROL 

Examples 6-5 and 6-6 illustrated very clearly the 

L 

basic weakness of proportional only control. When 

60% 

a disturbance causes an error, proportional control 

uses that same error to combat the disturbance. By 

50% 

its very nature, proportional control cannot M 

eliminate the offset caused by a disturbance. In the 

50% 

past, control engineers discovered that the offset 

40% 

could be eliminated by slowly changing the bias 

value up or down until the measured value is equal E 

to the desired value. This adjustment was called 

10% 

resetting the controller. In Example 6-5, for 

instance, shifting the bias to 60 % would reset the 0% 

level back to 50 %. This manual reset is in essence 

the same as automatic reset or integral reset. The C 

offset problem is highlighted again in Figure 6-14. 60% 

At instant A, the demand increases from 50% to 50% 

60%. With proportional only control, the level M 

drops to 40 %, the error E rises to 10 % and the 

inflow C, driven by the error value, rises to 60 %. 

Time 

Time 

Time 

A B Time 

Figure 6-14


The graphs from point B onwards, suggest a possible procedure with which we 

could eliminate the offset (perhaps someone gave us a hosepipe with running water 

and the task to restore the water level to 50 %). Firstly, we must keep the total inflow 

more than 60 % so that the level starts returning to 50 %. While we are doing this, we 

must remember that the error is getting smaller and we will get less and less assistance 

from the proportional controller. Secondly, after the level is restored to 50 %, the 

inflow from the proportional controller will only be 50 % and we need to supply the 

extra 10 % to equal the demand of 60 % so that the level remains at 50 %. 

Integral action gives an output proportional to the time integral of the error. This 

action can accomplish the two tasks mentioned above, namely, restoring the measured 

value to set point and providing the extra bias to match the demand. Integral control is 

rarely used alone but proportional control together with integral control, or PI control, 

is widely used in industry because it provides the important practical advantage of 

eliminating the offset. The controller output C, of a PI controller, can be expressed as: 

C = K P E + K I Edt + R, ……………….……………………… Equation 6-4 

∫ 

where K P is the proportional gain, K I is defined as the integral gain and R is the bias. 

(Note: For PI control mode the bias term is optional and may be disabled.) 

Comparison of Integral Control with Proportional Control 

In Figure 6-15 the load change behaviour of the water level control system with 

proportional plus integral control, is compared with the behaviour of the same system 

with only proportional control (refer to Example 6-6). For this illustration, S = 50 %, 

K P = 1 and R = 50 % for both controllers. The integral gain K I of the PI controller, 

was chosen as K I = 0.0005, which is small and produces a very sluggish but easilly 

understandable response. The integral component is denoted by I = 0.0005 

∫ Edt . 

The fundamental difference between P control and PI control is that with P control, 

the error persists while with PI control, the error vanishes. With P control it is indeed 

the error that sustains the final increase in inflow of 10 % to satisfy the increased 

demand of 60 %. With PI control it is the area under the error curve, accumulated by 

the integral action I = K I∫ Edt , that sustains the final increase in inflow of 10 % to 

satisfy the increased demand of 60 %. In a sense, the final scenario resembles a P only 

controller that shifted its bias automatically from 50 % to 60 %. In contrast with P 

control, the magnitude of the error has no effect on the inflow in the end, because the 

error is erased by the integral action. It is however clear from Figure 6-15, that in the 

early stage after the disturbance, the error and ascociated proportional action is 

playing a critical part in assisting the integral action to provide the extra inflow needed 

to replace the water lost during the initial drop in water level and to restore the water 

level back to 50 %. For the integral action to carry this burden alone, would typically 

result in the water level oscillating around the set point and may very well be one of 

the reasons why integral action is rarely used without proportional action.


Proportional Control 

Demand L 

60% 

50% 

Water level M 

50% 

40% 

Error E=50-M 

10% 

0% 

Time 

Time 

Time 

Proportional plus Integral Control 

Demand L 

60% 

50% 

Water level M 

50% 

40% 

Error E=50-M 

10% 

0% 

I=0.0005 

∫ Edt 

Time 

Time 

Time 

10% 

0% 

Time 

Bias R 

50% 

Controller output 

C = E + R 

60% 

50% 

Time 

Time 

Bias R 

50% 

Controller output 

C = E + I + R 

60% 

50% 

Time 

Time 

Figure 6-15 P and PI control comparison 

Repeat time and reset time 

Many commercial PI controllers do not directly allow for the adjustment of the 

integral gain K I , but rather the adjustment of the repeat time or reset time. Imagine an 

error signal that suddenly changes from 0 to 10 % and remains constant after that, as 

shown in Figure 6-16. 

Error E 

Area = 10×T R 

10% 

0% 

Time 

PI controller output 

C=K P E+K I 

∫ Edt 

Integral effect = K I 

∫ Edt = K I∫ 10dt = 10K IT R 

Proportional effect = K P ×E = K P ×10 

50% 

Time 

T 

Figure 6-16 

R 

(Repeat or reset time) 

Reset time 

If this step in error signal is presented to the input of a PI controller, that initially


had an output of 50 %, there would be a sudden increase in controller output because 

the proportional term K P E tracks the error instantaneously. Furthermore, as the 

integral term, K I ∫ Edt accumulates the area under the error curve, it wil gradually add 

to the output signal, in a linear fashion. There will come a time when the contribution 

of the integral term, equals the proportional contribution. The time it takes for this to 

happen is called the repeat time, as it indicates the time it takes the integral action to 

‘repeat’ the action of the proportional action. 

It is immediately clear from Figure 6-16 that the integral and proportional effects are 

equal when 10K I T R = 10K P or K I T R = K P . The repeat/reset time is therefore given by: 

T R = 

51 

K 

P 

K 

I 

or K I = 

K 

P ………..……….………..………….. Equation 6-5 

T 

R 

Controller tuning 

In a practical situation, different values for K P and K I would be chosen for the water 

level control system so that the water level would be corrected much faster after a 

disturbance. Choosing suitable values for K P and K I is called tuning of the controller. 

Guidelines that help with controller tuning do exist and even self tuning systems are 

available. One such rule of thumb suggests the following settings for a liquid level 

control: PB < 100 % and reset time of 10 minutes. If we should choose a PB of 80 % 

(K P = 1.25), then a reset time of 10 minutes (600 sec.) would translate into an integral 

gain of K I = 0.002083 (from Equation 6-5: K I = K P /T R = 1.25/600). Using these values 

for K P and K I , the plots of the water level M and controller output C, shown in 

Figure 6-17, were obtained with Matlab. The water level M drops to 45% and returns 

to setpoint within 25 minutes. 

M(%) 

C(%) 

70 

50 

49 

48 

65 

60 

47 

46 

55 

45 

0 500 1000 1500 2000 2500 3000 3500 4000 

50 

0 500 1000 1500 2000 2500 3000 3500 4000 

t (sec) 

t (sec) 

a) Water level Figure 6-17 b) Controller output 

System response to demand change from 50% to 60% with PI control (K P =1.25, K I =0.002083) 

Integral control: A control strategy in which the controller output is proportional 

to the integral of the error. 

{For PI control: C = K P E + K I 

∫ Edt + R = K ⎛ 

⎞ 

⎜ 1 ⎟ 

P E + Edt + R} 

⎜ T ∫ ⎟ 

⎝ R ⎠ 

Reset or repeat time: Time taken for the integral control action to equal the 

proportional control action under the influence of a constant error. {T R = K P /K I }


6.2.4 PROPORTIONAL AND DERIVATIVE CONTROL 

Derivative or differential control action (also called rate control or pre-act control) 

gives an output which is proportional to the derivative of the error. Because derivative 

action depends on the slope of the error curve, it has no effect if the error is constant. 

Derivative control action therefore only adds to the controller output when the error 

changes, that is when the set point or the measured value changes. Derivative control 

can thus not be used alone because the controller would not combat the error if it is not 

changing, even when the error is large. Derivative control action is usually found in 

combination with proportional control to form a PD controller. The output of a PD 

controller can be expressed as: 

dE 

C = K P E + K D + R, …………….…………….…………… Equation 6-6 

dt 

where K D is defined as the derivative gain. The basic effect of derivative control is to 

oppose any change in the error value (if the water level M, in our example, would 

drop, dM/dt would be negative and dE/dt = d/dt(S-M) would be positive and the 

inflow C would immediately increase). As such, derivative action will always have a 

stabilizing effect on the system by combating aggressive and unstable behaviour. 

Derivative control reacts to changes in the error and starts with corrective action much 

earlier than the proportional (and integral) action. Derivative action was in fact named 

‘pre-act’ because of its ability to ‘anticipate’ the future movement of the measured 

variable and the direction in which the process is heading. A PD controller, however, 

does not have the capability to eliminate the offset and will also add to the wear on the 

final control element such as a control valve. 

Rate time 

Commercial PD controllers normally do not directly allow for the adjustment of the 

derivative gain K D , but rather the adjustment of the rate time. Imagine a linear error 

signal that gradually changes from 0 to 10 % in T D seconds, as shown in Figure 6-18. 

Error E 

Figure 6-18 

10% 

Rate time 

10 

0% 

Time 

T D 

PD controller output 

Proportional effect = K 

dE 

P ×E = K P ×10 

C=K P E+K D 

dt 

dE 10 

Derivative effect = K D × = KD × 

50% 

dt T 

D 

Time 

T D 

If this ramp error signal is presented to the input of a PD controller, that initially had 

an output of 50 %, there would be a sudden increase in controller output because the 

dE 

derivative term, K D , immediately takes on a value proportional to the slope of the 

dt 

error curve. Furthermore, as the proportional term, K P E, follows the error value, it will 

increasingly add to the output signal, in a linear fashion. The time, T D , that it takes the


contribution of the proportional term to equal the derivative contribution is called the 

rate time, and it gives an indication of how much faster the derivative action is 

compared to the proportional action. 

It is immediately clear from Figure 6-18 that 10×K I T R = 10×K P = 10×K D /T D or 

K P T D = K D . The rate time is therefore given by: 

K 

T D = 

D 

or K D = K P T D ……..……….…..………………….. Equation 6-7 

K 

P 

Derivative control: A control strategy in which the controller output is 

proportional to the derivative of the error. 

dE ⎛ dE ⎞ 

{For PD control: C = K P E + K D + R = KP ⎜E 

+ T ⎟ + R } 

dt 

D dt ⎠ 

Rate time: Time taken for proportional action to equal derivative action under the 

influence of a linearly changing error. {T D = K D /K P } 

6.2.5 PROPORTIONAL, INTEGRAL AND DERIVATIVE CONTROL 

A proportional, integral and derivative controller (PID controller), gives an output that 

is determined by proportional action, integral action and derivative action. 

Mathematically the output of a PID controller can be expressed as: 

C = K P E + K I 

∫ Edt + dE KD + R. …………….……………… Equation 6-8 

dt 

Controller tuning 

Adding integral and derivative action to a controller increases the complexity of the 

system. Choosing suitable P, I and D gains for a PID controller (tuning the controller), 

becomes complicated. Even for the very simple water level control system in Figure 

6-9, analysis of the complete PID system is difficult. We will however conclude with 

one example of PID control of the water level control system. The controller gains 

chosen were K P = 5, K I = 0.08 (reset time = 1.04 min.) and K D = 500 (rate time = 1.7 

min.). The plots in Figure 6-19, were obtained with Matlab. The response curves in 

Figure 6-19, show the water level (although oscillating), not even dropping below 

49% and returning back to setpoint in less than seven minutes. 

M(%) 

50.6 

50.4 

50.2 

C(%) 

65 

⎝ 

50 

49.8 

60 

49.6 

49.4 

49.2 

0 500 1000 1500 2000 2500 3000 3500 4000 

a) Water level 

Figure 6-19 

t (sec) 

55 

0 500 1000 1500 2000 2500 3000 3500 4000 

b) Controller output 

System response to demand change from 50 % to 60 % 

with PID control (K P = 5, K I = 0.08 and K D = 500) 

t (sec)

6.3 PID CONTROLLERS 

6.3.1 Pneumatic PID controller 

One may justifiably ask 

whether pneumatic controllers 

still have a place in a digital 

age. The answer is yes and 

some of the reasons that can be 

put forward are familiarity, 

simplicity, a large installed 

base, no fear of electricity and 

above all, they are inherently 

explosion-proof. 

A typical front panel of a 

pneumatic controller is shown 

in Figure 6-20. Depending on 

the manufacturer, controllers 

will have slightly different 

configurations of pushbuttons, 

meters and switches. 


Manual 

pushbutton 

(decrease 

output) 

Manual 

button with 

indicating 

light 

MV 

0 

CLOSE 

MANUAL 

100 

90 

80 

70 

60 

50 

40 

30 

20 

10 

0 SP 

OUTPUT 

50 

100 

OPEN 

AUTOMATIC 

Set point 

thumbwheel 

Manual 

pushbutton 

(increase 

output) 

Automatic 

button with 

indicating 

light 

Figure 6-20 

Pneumatic implementation of the PID function is achieved in a variety of ways using 

diaphragms and valves in different mechanical designs. One possible arrangement 

revolves around the force balance principle utilizing a flapper-nozzle and bellows 

assembly. The pressure variable that represents the measured value is normally 

connected to the force bar (beam) by means of a bourdon tube or a bellows element. 

The set point value, the bias value and the proportional band settings are adjusted in 

several ways by different manufacturers using various mechanical or pneumatic 

techniques. 

i) Proportional control 

The controller arrangement for proportional control is shown in Figure 6-21. For the 

purpose of this illustration, we will assume that all variables are applied to the force 

balance system by means of bellows elements and that the pivot is positioned at a 

distance ‘a’ units from the S/M bellows and at a distance ‘b’ units from the C/R 

bellows. The position of the proportional bellows in Figure 6-21 will result in reverse 

acting control because a decrease in measured value will cause the flapper to move 

towards the nozzle and balance will be restored at a higher controller output pressure. 

Given that the flapper and nozzle will always maintain equilibrium, the controller 

operation can very easily be described by equating moments around the pivot point, 

assuming that all the bellows elements have a cross-sectional area of one unit. 

a 

a×S + b×R = a×M + b×C ⇒ bC = a(S – M) + bR ⇒ C = (S – M) + R 

b 

Using E = S-M and K P = (a/b), it follows that C = K P E + R. 

The device therefore provides a proportional control output with the proportional 

gain determined by the position of the pivot point.


Set point S 

(20-100 kPa) 

50% = 60 kPa 

Pilot 

relay 

Restriction 

Controller 

output C 

Air 

supply 

S 

Beam 

Proportional 

bellows 

C 

Measured 

value M 

(20-100 kPa) 

M 

a 

Pivot 

b 

Reset 

bellows 

R 

Bias value R 

(20-100 kPa) 

Flapper and 

nozzle 

Figure 6-21 

Pneumatic 

proportional 

controller 

ii) Proportional plus integral control 

With proportional control, the fixed bias value R that is applied to the reset bellows, is 

called ‘manual reset’ because one way to reset the controlled variable back to set point 

in a proportional only control system, is to manually change the bias value. In order to 

change the bias value automatically with the aim to reset the controlled variable 

automatically to set point, the reset bellows must now obtain its pressure from the 

controller output pressure C. To achieve automatic reset the output pressure must be 

applied via an adjustable needle valve restriction, as shown in Figure 6-22. 

Set point S 

(20-100 kPa) 

50% = 60 kPa 

C 

Pilot 

relay 

Restriction 

Controller 

output C 

Air 

supply 

Measured 

value M 

(20-100 kPa) 

S 

M 

Beam 

a 

C 

Pivot 

Reset time 

adjustment 

(r) 

Proportional 

bellows 

b 

Reset 

bellows 

Needle 

valve 

R 

C 

R 

Flapper and 

nozzle 

Automatic reset R 

Figure 6-22 

Pneumatic 

proportional 

plus 

integral 

controller


The restriction will prevent the pressure R in the reset bellows to immediately 

follow the output pressure C and causes a delay in the pressure build up (or decay) 

in the reset bellows which will depend on the needle valve setting. This delay in 

reset pressure will cause integral action to be added to the controller and it can be 

a ra 

shown that the controller output is C = (S – M) + (S- 

M)dt . Using E = S–M, 

b b 

K P = b 

a and KI = b 

ra , this equation then becomes C = KP E + K I 

∫ Edt . 

The controller thus provides a proportional plus integral output. The dynamic 

integral term effectively replaces the fixed bias with an ‘automatic bias’. Commercial 

controllers do however allow a fixed bias to be enabled for PI control, if so required. 

iii) Proportional plus derivative control 

Derivative action may be added to the proportional controller in Figure 6-21, by 

inserting a variable restriction in the pressure line that feeds the proportional bellows, 

as shown in Figure 6-23. The proportional bellows now receives a pressure P d that is a 

delayed version of the output pressure C. 

∫ 

P d 

Needle valve 

C 

Set point S 

(20-100 kPa) 

50% = 60 kPa 

Rate time 

adjustment 

(d) 

Pilot 

relay 

Restriction 

Controller 

output C 

Air 

supply 

S 

Beam 

Proportional 

bellows 

P D 

Measured 

value M 

(20-100 kPa) 

M 

a 

Pivot 

b 

Reset 

bellows 

R 

Bias value R 

(20-100 kPa) 

Flapper and 

nozzle 

Figure 6-23 

Pneumatic 

proportional 

plus 

derivative 

controller 

Again it can be shown that for the configuration in Figure 6-23, the controller output 

a a d(S- M) 

a a 

is given by C = (S – M) + + R. Using E = S - M, K P = and KD = , 

b bd dt 

b bd 

dE 

this equation then becomes C = K P E + K D + R. dt 

It is now clear that inserting a delay in the proportional pressure line, causes 

derivative action to be added to the proportional action. 

Adding delay actions simultaneously in both the reset and the feedback line, will 

result in a complete PID controller and such a structure is shown in Appendix 6-2.


6.3.2 Electronic PID controller 

Figure 6-24 shows an electronic implementation of the basic PID function. The first 

operational amplifier generates the error signal from the reference S and the measured 

value M. The error signal is fed to the P, I and D blocks to create the three PID 

functions which are added together to form the final PID control function C. 

R PF 

Proportional 

action 

E 

R PI 

Op-amp 

V P 

100k 

Gnd 

R 

V P = - PF E 

R 

PI 

= -K P E 

Error 

M 

S 

100k 

100k 

100k 

E 

Op-amp 

E=S-M 

100k 

Gnd 

E 

R I 

C I 

Op-amp 

Gnd 

Integral 

action 

V I 

100k 

100k 

Op-amp 

1 

V I = - 

R C ∫ Edt 

I I 

= -K I 

∫ Edt Gnd 

Adder 

C 

C = - V P – V I – V D 

E 

C D 

R D 

Op-amp 

Derivative 

action 

V D 

100k 

= K P E 

P 

+ K I 

∫ Edt 

I 

dE 

+ K D 

dt 

D 

Gnd 

dE 

V D = -R D C D 

dt 

dE 

= -K D 

dt 

Figure 6-24 

Electronic PID 

controller


6.3.3 Digital PID controller 

Digital or software implementation of the PID control algorithm uses a digital 

computer to calculate the control action as shown in Figure 6-25. Computers are 

discrete machines and not capable of calculating the control action on a continuous 

basis as the analog structure in Figure 6-24. It will instead try to approximate Equation 

6-8, using numerical methods. Therefore the measured value M must be sampled with 

sufficiently small sampling period T s and digitised with an analog to digital converter 

to deliver the sampled values M 0 , M 1 , M 2 , M 3 , M 4 , etc. to the computer at time instants 

0, 1, 2, 3, 4. At each time instant, the proportional, integral and derivative component 

must be calculated and added together to form the sequence of corresponding 

controller outputs C 0 , C 1 , C 2 , C 3 , C 4 , ….. 

M 

T S 

M 0 

Analog to digital 

converter 

M 1 M 2 

Computer 

Figure 6-25 

Digital to analog 

converter 

C C 3 

2 

M 3 M C 1 4 

C 0 

For instance, referring to Figure 6-26, at time instant 4, just after M 4 was sampled, 

the error E 4 may be obtained from 50–M 4 , given that the set point is 50%. From E 4 the 

proportional action can be calculated as P 4 = K P ×E 4 . If we assume that the controller 

started its operation at time equals zero, then a very simple method to calculate the 

integral component I 4 is to approximate the area A, under the error curve with 

4 

rectangles to obtain A= 

∫ × T 

S Edt ≈ T s ×E 0 + T s ×E 1 + T s ×E 2 + T s ×E 3 . The integral 

0 

action I 4 may then be expressed as I 4 = K I ×(T s ×E 0 + T s ×E 1 + T s ×E 2 + T s ×E 3 ) or 

I 4 =K I ×T s ×(E 0 + E 1 +E 2 +E 3 ). The derivative component D 4 , depends on the slope of the 

error curve at time instant 4, and may be approximated with D 4 = K D 

M 

S=50% 

E 

E 2 

E 3 

E 

T S 

E − E 

4 3 

T 

s 

C 

C 4 

E 0 

E 1 E 1 

E 

M 2 

0 

E 3 E 4 E 

M 0 

1 

M 2 M 3 M 4 

0 1 2 3 4 Time 0 1 2 3 4 

T s T s T s T s T s 

T s 

Figure 6-26 

E 3 ×T s E 4 

E 2 ×T s 

E 3 

E 1 ×T s 

E 0 ×T s 

T s T s 

Time 

C 4 = K P E 4 + K I ×T S ×(E 0 + E 1 + E 2 + E 3 ) + 

P 

I 

0 1 2 3 4 Time 

T s T s 

T s 

E − E 

K D × 4 3 

T 

s 

D 

T s


6.4 CONTROL VALVES 

Control valves are used to regulate the flow rate of a medium and 

serve as the correcting element in many control systems. 

A pneumatic control valve consists of a diaphragm actuator and a 

valve body with plug and seat through which the fluid is 

regulated. The principle of operation of a control valve is very 

simple. A diaphragm is bolted to a dished metal head to form a 

pressure tight compartment. A pneumatic signal is applied to this 

compartment to move the diaphragm which is opposed by a 

spring. Attached to the diaphragm is the valve stem and plug so 

that any movement of the diaphragm, results in a corresponding 

movement of the plug, thereby controlling the flow. A typical 

reverse acting control valve assisted by a valve positioner (section 

6.4.4) to enhance valve operation, is illustrated in Figure 6-28. 

6.4.1 Actuators 

Actuators may be categorized 

as ‘direct acting’ or ‘reverse 

acting’ and some configurations 

are indicated in Figure 6-27. In 

a reverse acting actuator an 

increase in the pneumatic 

pressure applied to the 

diaphragm lifts the valve stem 

(in a normally seated valve this 

will open the valve and is 

called ‘air to open’). In a directacting 

actuator, an increase in 

the pneumatic pressure applied 

to the diaphragm extends the 

valve stem (for a normally 

seated valve this will close the 

valve and is called ‘air to 

close’). The choice of valve 

action is dictated by safety 

considerations. In one case it 

may be desirable to have the 

valve fail fully open when the 

pneumatic supply fails. In 

another application it may be 

considered better if the valve 

fails fully shut. 

Reverse acting 

Direct acting 

Figure 6-27


Spring tension adjust 

(Valve closed at 20 kPa) 

Spring 

Diaphragm 

plate 

Diaphragm 

Actuator 

(Motor) 

Air 

supply 

Output 

Open 

½ 

Close 

Stem connector 

(stroke adjust 

20 – 100 kPa) 

Travel 

indicator 

Stem 

Valve 

positioner 

Connector 

arm 

Yoke 

Instrument 

signal 

(20-100 kPa) 

Gland and 

packing 

Bonnet (yoke 

hold down) nut 

Bonnet 

Gasket 

Plug 

Seat 

Valve 

body 

Figure 6-28 

Reverse acting (air to open) pneumatic control valve


6.4.2 Valve types 

The essential function of a valve is to start, throttle and stop a fluid flow. Throttling, 

as shown in Figure 6-29, means regulating or controlling the rate of fluid flow. Globe 

valves, gate valves, needle valves, pinch valves and diaphragm 

valves use a linear stem movement to operate. A gate valve is 

illustrated in Figure 6-30 (a) as an example of linear stem 

movement. Ball valves, plug valves and butterfly valves function 

with a 90° rotational movement and the operation of a ball valve is 

indicated in Figure 6-30 (b). The valve components inside the 

valve body are collectively referred to as the trimming. 

Closed Throttling Open 

Figure 6-29 

Ball valve 

Gate valve 


Linear valve movement 


Rotational valve movement


1. Globe valves 

Globe valves are the most widely used in industry for flow control in both on/off and 

throttling service. They typically have rounded bodies from 

which their name is derived. They are linear motion valves 

with a tapered plug or disk attached to the stem that closes onto 

a seating surface to act as flow control element, as shown in 

Figure 6-31. The liquid must make two 90° turns when passing 

through the valve and because of that, the pressure drop in the 

globe valve is significant, even when fully open. The high Figure 6-31 

pressure drop is the main disadvantage of the globe valve. 

Globe valve 

2. Gate valves 

Gate valves (also known as knife valves or slide valves) are the most common valve 

used for on/off service but could be used for throttling. They are linear motion valves 

and a flat disk or wedge slides into the flow stream 

to act as flow control element, as shown in Figure 

6-32. The direction of fluid flow is not changed by 

the valve. When fully open, the wedge completely 

clears the flow path creating minimum pressure drop. 

Gate valves are advantageous in applications 

involving slurries, as their ‘gates’ can cut right 

through the slurry. They are also used in applications 

that involve viscous liquids such as heavy oils, light 

grease, varnish, molasses, honey and cream. 

3. Needle valves 

A needle valve, shown in Figure 6-33, is used to make relatively 

fine adjustments to the fluid flow and can be used for on/off and 

throttling service. They are linear motion valves with a tapered 

‘needle like’ cone shaped plug that acts as the flow control 

element. The fluid enters from below into the seat which forms 

part of the flow path. The needle plug permits the flow opening 

in this channel to be increased or decreased very gradually. 

Needle valves are widely used for steam, air, gas, water or other 

non-viscous liquids. Disadvantages of the needle valve are a 

large pressure loss and possible clogging of the flow orifice. 

4. Pinch valves 

The relatively inexpensive pinch valve, illustrated 

In Figure 6-34, is the simplest of all the valve 

designs and may be used for on/off and throttling 

service. Pinch valves are linear motion valves 

that use a flexible tube or sleeve to effect valve 

closure. Pinch valves are ideally suited for the 

handling of slurries, liquids with large amounts 

of suspended solids and corrosive chemicals. 

Parallel 

disk 

Wedge 

Knife 

Figure 6-32 

Gate valve 

Figure 6-33 

Needle valve 

Figure 6-34 

Pinch valve


5. Diaphragm valves 

A diaphragm valve, illustrated in Figure 6-35, is related to pinch 

valves and one of the oldest types of valve known. Leather 

diaphragm valves were used by Greeks and Romans to control 

the temperature of their hot baths. Diaphragm valves use linear 

stem movement for on/off and throttling and are excellent for 

controlling fluid flow containing suspended solids. A stud moves 

a flexible and resilient diaphragm into the flow thereby acting as 

flow control element. The diaphragm valve is used primarily for 

handling viscous and corrosive fluids as well as slurries. 

6. Plug valves 

Plug valves, illustrated 

in Figure 6-36, are also 

called plug cock or stop 

cock valves and they 

also date back to ancient 

Close Throttle Open 

times, when used by Romans in plumbing systems. Today they 

remain one of the most widely used valves for both on/off and 

throttling services. A plug valve is a rotary moving valve that uses a cylindrical or 

tapered plug as the flow control element. The opening through the plug may be 

rectangular or round. Flow is regulated from closed to open during a 90° turn. If the 

opening is the same size or larger than the pipe’s inside diameter, it is referred to as a 

full port otherwise as standard round port. Full port plug valves offer little resistance 

to flow when fully open, resulting in small pressure loss. A disadvantage off plug 

valves is its poor throttling characteristics. 

7. Ball valves 

Ball valves, illustrated in Figure 6-37, are related to plug valves 

and are used in situations where tight shut-off is required. Ball 

valves are rotational motion valves and are used for on/off and 

throttling service. The flow control element is a sphere with a 

round opening rotating in a spherical seat. Flow is regulated 

from closed to open during a 90° turn. Full port ball valves 

create a minimum pressure loss when fully open. A disadvantage 

of the ball valve is its poor throttling characteristics. 

8. Butterfly valves 

A butterfly valve, illustrated in Figure 6-38, is a rotary 

movement valve used for on/off flow control and especially 

in throttling applications. The flow control element in 

butterfly valves is a circular disk with its pivot axis at right 

angles to the direction of flow. A 90° turn of the axis moves 

the valve from closed to open. Butterfly valves are well 

suited to handle large flows of liquids or gasses as well as 

slurries and liquids with large amounts of suspended solids. 

Figure 6-35 

Diaphragm valve 

Figure 6-36 

Plug valve 

Figure 6-37 

Ball valve 

Figure 6-38 

Butterfly valve


6.4.3 Valve characteristics 

In essence a valve simply functions as a restriction in a flow line, 

with flow areas A 1 and A 2 and pressure differential p 1 - p 2 , as shown 

in Figure 6-39. As such, the flow equation (Equation 3-8(e), 

Chapter 3) will in principle describe the flow rate of a fluid through 

a valve. The form of this equation used by valve suppliers is: 

q 

p 1 

p2 

A 1 

A 2 

Figure 6-39 

q = C 

Δp 

G 

Equation 6-9 

where C is called the valve flow coefficient, q the flow rate of the liquid through the 

valve, Δp = p 1 - p 2 the pressure difference across the valve and G the specific gravity 

(relative density) of the fluid. For historical reasons, the flow rate q, in Equation 6-9, is 

measured in gallons per minute (gpm) and the pressure difference Δp, in pounds per 

square inch (psi). When we compare Equations 3-8(e) and 6-9, it becomes clear that C 

is fundamentally determined by the effective flow areas A 1 and A 2 of the valve. The 

value of C will therefore change from zero (when the valve is fully closed) to a 

maximum value (when the valve is fully opened). 

Assume that a valve is opened a certain amount x (with x denoting the fractional 

valve opening or valve travel between 0 and 1). The flow rate q, of the liquid (we will 

assume water with G = 1 in Equation 6-9), flowing through the valve, is now allowed 

to increase until the pressure Δp across the valve, reaches a maximum of say 4 psi. We 

may conceivably expect the results depicted in Figure 6-40, shown for x = 0 (valve 

fully closed), x = 0.25 (valve one quarter open), x = 0.5 (valve half open), x = 0.75 

(valve three quarters open) and x = 1 (valve fully open). 

q 

0% open 25% open 50% open 75% open 100% open 

x = 0 x = ¼ x = ½ x = ¾ x = 1 

q q q q 

Δp 

Δp 

Δp 

Δp 

Δp 

q=0× 

Δ p 

q=10× 

Δ p 

q=20× 

Δp 

q=30× 

Δ p 

q=40× 

Δp 

(C=0) 

(C=10) 

(C=20) 

(C=30) 

(C=40) 

q (gpm) 

q (gpm) 

q (gpm) 

q (gpm) 

q (gpm) 

80 

60 

40 

20 

0 

4 

Δp 

psi 

0 

4 

Δp 

psi 

0 

4 

Δp 

psi 

0 

4 

Δp 

psi 

0 

4 

Δp 

psi 

Figure 6-40


The special value of C that corresponds to the valve fully open, is called the 

characteristic flow coefficient of the valve and denoted by C V . The characteristic flow 

coefficient C V , is an extremely important parameter of a specific valve, and is used 

extensively when choosing the correct valve for a certain application. For the valve 

characteristics in Figure 6-40, for example, C V = 40. 

The way q varies when the pressure is kept constant (normally the rated pressure 

drop across the valve for maximum flow) and the way C changes as the valve opening 

x changes, may be available for a particular valve, provided by the manufacturer in 

tabulated format. For example, the graph of q (with Δp = 4 psi) as a function of x and 

the graph of C as a function of x, are shown in Figure 6-41 (a) and (b) respectively, for 

the flow characteristics of our hypothetical valve, depicted in Figure 6-40. 

Figure 6-41(a) 

Figure 6-41(b) 

q (gpm) 

C 

80 

40 

60 

40 

30 

20 

Valve 

Valve 

20 

10 

opening 

opening 

x 

x 

0 

0 

¼ ½ ¾ 1 

¼ ½ ¾ 1 

(0%) (25%) (50%) (75%) (100%) 

(0%) (25%) (50%) (75%) (100%) 

The different values of C obtained as the valve travels through its full range (or 

stroke as it is also called) from closed to open, is generally expressed in the following 

format, by valve manufacturers: 

C = C V ×f(x) Equation 6-10 

where f(x) is called the inherent valve characteristic of a particular valve. For 

example, from the graph of C, in Figure 6-41(b), we could express C as: 

C = 40×x, 

and we conclude therefore from Equation 6-10, that C V = 40 and f(x) = x. 

The function f(x) varies from 0 (valve closed with x equal to 0) to 1 (valve fully 

open with x equal to 1). The inherent valve characteristic f(x), is intimately linked to 

the flow rate q through a valve, as we can see if we replace C with C V f(x) (as per 

Equation 6-10), to rewrite the valve Equation 6-9 in the form: 

q = C V f(x) 

ΔP 

G 

Equation 6-11 

It is indeed clear from Equation 6-11 that if we keep Δp constant (for a given liquid, G 

is constant as well), q and f(x) will have exactly the same shape.


A valve for which f(x) = x, is called a linear valve because the flow rate will change 

linearly with valve opening x. Depending however on the valve design, f(x) may also 

reflect quick opening or equal percentage (slow opening) characteristics, as shown in 

Figure 6-42. 

100% 

Graphs also 

represent 

flow rate q 

(% of max. 

flow) if Δp 

is constant. 

0% 

q 

f(x) 

1 

Quick 

opening 

Linear 

Figure 6-42 

Inherent valve 

characteristics 

(Ratio of 

Equal 

valve travel 

percentage x to maximum 

0 

valve travel) 

0 1 

(0% open) (100% open) 

Quick 

opening 

Linear 

Equal 

percentage 

The flow behaviour of a valve is dictated by the manner in which the flow areas A 1 

and A 2 changes with valve position and therefore by the style and design of the valve 

trimming and in particular the design of the valve seat and closure member (plug). The 

quick opening flow characteristic provides for maximum change in flow rate at low 

valve travels with a nearly linear relationship. Additional increases in valve travel 

gives sharply reduced changes in flow rate. The linear flow characteristic curve allows 

the flow rate to be directly proportional to the valve travel (Δq/Δx equals a constant) or 

in terms of the inherent valve characteristic, f(x) = x. An equal percentage valve starts 

initially with a slow increase in flow rate with valve position which dramatically 

increases as the valve opens more. 

The term equal percentage for a slow opening characteristic curve may at first be 

confused with the description of a linear characteristic curve. However, for an equal 

percentage valve, Δq/Δx at any stage, is proportional to the flow rate q at that moment. 

This is in contrast with a linear characteristic for which Δq/Δx is constant. That Δq/Δx 

is proportional to q, may be rephrased as Δq/q is proportional to Δx. This means the 

percentage change Δq with respect to the current flow rate q (that is (Δq/q)×100), is 

equal at every valve travel position x for the same change in valve travel Δx, hence the 

term ‘equal percentage’. The inherent valve characteristic for an equal percentage 

valve is exponential in nature and is normally given by valve manufacturers in the 

form f(x) = R x-1 , where R is a constant for the valve. The exponential behaviour of an 

equal percentage valve is explored further in Example 6-7. 

Example 6-7 

An equal percentage valve delivers 8 gpm of water when the valve is 50% open 

(x=0.5). When the valve is 60% open (x=0.6), the flow rate increases to 16 gpm.


Estimate the flow rate through the valve when it is 70% open (x=0.7). Assume that the 

pressure drop across the valve remains constant. 

Answer: When the valve opens from 

50% to 60%, the flow rate changes 

q (gpm.) 

from 8 gpm to 16 gpm. This means 

that Δq is 8 gpm. when Δx is 0.1 and 

=% 

q = 8. Therefore Δq/q = 8/8 = 1 and 

the percentage change is 100% 

 

(Δq/q×100) at x = 0,5. For an equal 

Δq= 

percentage valve, the percentage 16 

change in flow rate when the valve 8 

Δq=8 

x 

opens from 50% to 60%, (Δx = 0.1) 

must be equal to the percentage 

change in flow rate when the valve 

0 

0 .5 .6 .7 1 

opens from 60% to 70% (the same 

Δx=.1 Δx=.1 

Δx of 0.1). Therefore Δq/q at x = 0.6 must also be 1 (or 100%) for Δx = 0,1. But q is 

16 gpm. when the valve is 60% open and we conclude that Δq should be 16 gpm. 

when the valve opening changes from 60% to 70%. The flow rate through the valve is 

therefore approximately 32 gpm. at 70% valve opening. 

Typical application of quick opening, linear and equal percentage valves 

i) Quick opening valve: 

a) Frequent on-off service. 

b) Used for systems where ‘instant’ large flow is needed (safety or cooling 

water systems). 

ii) Linear valve: 

a) Liquid level and flow control loops. 

b) Used in systems where the pressure drop across the valve is expected to 

remain fairly constant. 

iii) Equal percentage valve (most commonly used valve): 

a) Temperature and pressure control loops. 

b) Used in systems where large changes in pressure drop across the valve are 

expected. 

Installed flow characteristic 

When valves are installed with pumps, piping and fittings, and other process 

equipment, the pressure drop across the valve will vary as the valve travel changes. 

When the actual flow in a system is plotted against valve opening, the curve is called 

the installed flow characteristic and it will differ from the inherent valve characteristic 

which assumed constant pressure drop across the valve. When in service, a linear 

valve will in general resemble a quick opening valve while an equal percentage valve 

will in general resemble a linear valve.


Valve sizing 

The characteristic flow coefficient C V of a valve and the inherent valve characteristic 

f(x), completely describe the flow-pressure profile of the valve and play an important 

role during the valve selection or valve sizing process. Although the characteristic 

flow coefficient (C V ) is the most fundamental and important parameter influencing the 

user when selecting a valve, it is not the only consideration. Conditions such as piping 

particulars, fluid type, laminar or turbulent flow, may be included as additional factors 

in Equation 6-9. Nevertheless, to select an appropriate valve for a certain 

implementation, the user must specify the maximum flow rate required and the 

pressure drop expected with the valve fully open. With this information, the necessary 

characteristic valve flow coefficient C V , can be calculated from Equation 6-9 by 

setting C = C V , q = Q Rated (rated maximum flow rate with valve fully opened) and 

Δp = ΔP Full (pressure across fully opened valve at maximum rated flow rate): 

C V = 

ΔP 

Q Full ………………...…………………. Equation 6-12 

Rated G 

Example 6-8 

A system is pumping water from one tank to another through a piping system with 

total pressure drop of 150 psi. The maximum design flow rate is 150 gpm. Calculate 

the characteristic flow coefficient required for a control valve to be used in this system. 

Solution: The usual rule of thumb is that a valve should be designed to use 10-15% of 

the total pressure drop or 10 psi, whichever is greater, when fully open. For this 

system 10% of the total pressure drop is 15 psi, which is what we must use. From 

Equation 6-12 (remember that specific gravity G is the same as relative density or in 

other words, G = δ): 

C V = Q Rated / ΔP Full 

/G = 150/√(15/1) = 38.72 ≈ 39 

(Note: C V is normally given as a dimensionless number but it really is a dimensional 

quantity with units [gallon-inch/minute-poundforce ½ ] from Equation 6-12.) 

Example 6-9 

The user in Example 6-8 has a choice between a valve with C V = 35 which would be 

too small and the next one with C V = 45 which could be too large. So he decides to 

check what the valve travel positions will be for the larger valve if he wants to control 

the water flow between a maximum flow rate of 150 gpm and a minimum flow rate of 

40 gpm. He expects a pressure drop of 15 psi across the valve at full flow and a rise in 

pressure drop to 25 psi, when the flow rate drops to 40 gpm. The valve he intends to 

purchase is a linear valve with inherent valve characteristic, f(x) = x. 

Solution: From Equation 6-11 (q = C V f(x) ΔP/G ), 

with q max = 150 gpm. and Δp qmax = 15 psi: 150 = 45×x × 15/1 

∴x = 150/(45×√15) = 0.8607 (86%) 

with q min = 40 gpm. and Δp qmin = 25 psi: 40 = 45×x × 25/1 

∴x = 40/(45×√25) = 0.1778 (18%) 

This seems to be good enough, as the aim should be to keep valve movement between 

20% and 80% of maximum travel.


6.4.4 Valve positioners 

A 20-100 kPa controller signal travelling perhaps a long distance to a control valve to 

exercise the required control action, may possibly not be powerful enough to quickly 

move a large valve to its new position. It may therefore be advantageous to allow the 

controller signal to control the air supply locally at the valve rather than to operate the 

valve by itself. The device that may be added to enhance the controller signal to a 

valve, is called a valve positioner. A valve positioner is essentially a pneumatic 

controller that senses the valve stem position, compares it to the incoming controller 

signal and adjusts the pressure to the actuator until the stem position corresponds to 

the controller signal. Positioners can be used to overcome stem friction, high fluid 

pressure, viscous or dirty fluids, or to improve slow system dynamic response. 

Valve actuator 

Elastic 

Forcebalance 

beam 

Pivot 

Cam 

Instrument 

bellows 

Flapper and 

nozzle 

Valve stem 

Controller signal 

(Instrument signal) 

20-100 kPa 

Pilot 

relay 

Restriction 

Figure 6-43 Valve positioner 

Actuator 

output 

Air supply 

Figure 6-43 shows a typical pneumatic valve positioner, assisting a reverse acting 

valve. The operation revolves around a force balance flapper and nozzle 

arrangement. The valve stem position is communicated to the force balance beam 

by means of a lever and cam. If the controller signal increases, with the intention to 

open the valve more, the instrument bellows will push the flapper (a flexible plate) 

towards the nozzle. This will increase the valve actuator pressure and the valve 

stem will start to move upwards. This will turn the cam which will subsequently 

push the flapper away from the nozzle, until balance is reached so that the valve 

remains in the new more opened position. If the controller signal decreases, the 

flapper will move away from the nozzle and the actuator pressure will be reduced. 

The valve will begin to close and the valve stem will begin to move downwards, 

relieving cam pressure on the beam that kept the flapper away from the nozzle. The 

flapper will tend to move towards the nozzle thereby restoring balance with the 

valve more closed than before.


APPENDIX 6-1 

A water container is 5 meter high and has a base area of 1 meter 2 . The rate, measured 

in cubic meter per second, at which water flows into the container at any instant, is 

Q IN . The rate, measured in cubic meter per second, at which water flows out of the 

container at any instant, is Q OUT . The level, measured in meter, of the water in the 

container at any instant, is H. The maximum value of Q IN and Q OUT , is 0.01 cubic 

meter per second, and the maximum value of H is 5 meter. 

a) Express H in terms of a percentage M, of the maximum level. 

b) Express Q IN in terms of a percentage C, of the maximum inflow. 

c) Express Q OUT in terms of a percentage L, of the maximum outflow. 

d) Calculate the volume V, of the water in the container, in terms of M. 

e) Calculate the resultant rate Q, at which water accumulates into the tank or drains 

out of the container. Express your answer in terms of C and L. 

Δ V 

f) Calculate the rate , at which the volume changes, in terms of C and L. 

Δt 

Δ M 

g) Calculate the rate , at which the water level changes, in terms of C and L. 

Δt 

Solution: 

0.01m 3 C (inflow in %) 

/s 

Q 

M Max in (inflow in m 3 /s) 

a) H = ×5 = 0.05×M meter …...……… (1) 

100 

C 

5m max 

b) Q in = ×0.01 = 0.0001×C m 3 /s ….….. (2) M (%) 

100 

H (m) 

L 

c) Q out = ×0.01 = 0.0001×L m 3 /s ….…. (3) 

100 

L (outflow in %) 0.01m 3 /s 

Q 

d) Volume V, of water in tank: 

out (outflow in m 3 /s) max 

V = Height×Area = H×1 

Q=0.0001×(C-L) 

From (1): V = 0.05×M×1 

C 

∴V = 0.05×M cubic meter ……..…..….. (4) 

e) Resultant rate Q = Q in - Q out 

M V=0.05M 

From (2) and (3): 

∴Q = 0.0001×C–0.0001×L 

∴Q = 0.0001(C – L) m 3 /s ….....……....... (5) 

Area = 1 m 2 L 

f) The rate at which the water volume in the container increases or decreases, is the 

same as the rate at which water is accumulated in or drained from the tank. 

ΔV 

ΔV 

∴ = Q. Therefore from (5): = 0.0001(C - L) cubic meter per second. 

Δ t Δt 

∴ΔV = 0.0001(C - L)×Δt meter 3 ….….... (6) 

g) From (4): V = 0.05×M 

∴ΔV = 0.05×ΔM meter 3 ……….…......... (7) 

Substituting ΔV from (7) into (6): 0.05×ΔM = 0.0001(C – L)×Δt 

ΔM 

0.0001 

dM Δ M 

∴ = ×(C – L), or, using the proper notation for , 

Δ t 0.05 

dt Δt 

dM = 0.002×(C – L) percent per second. 

dt


APPENDIX 6-2 

C 

Set point S 

(20-100 kPa) 

50% = 60 kPa 

P d 

P d 

Needle valve 

Rate time 

adjustment 

(d) 

Pilot 

relay 

Restriction 

Controller 

output C 

Air 

supply 

S 

Beam 

Proportional 

bellows 

P d 

M 

a 

Pivot 

b 

Reset 

bellows 

R 

Flapper and 

nozzle 

Measured 

value M 

(20-100 kPa) 

Reset time 

adjustment 

(r) 

Needle 

valve 

Automatic reset R 

Pneumatic PID controller 

P d 

R 

a(r + d) ar 

C = (S-M) + 

bd 

b 

Using E = S-M, K P = 

a(r + d) 

bd 

C = K P E + K I 

∫ Edt + dE 

KD 

dt 

∫ 

a 

(S- 

M)dt (S-M) + bd 

d(S- M) 

dt 

ar a 

, K I = and KD = : 

b bd

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