fluid_mechanics
196 Chapter 5 ■ Finite Control Volume Analysis For steady flow involving more than one stream of a specific fluid or more than one specific fluid flowing through the control volume, a m# in a m # out The variety of example problems solved above should give the correct impression that the fixed, nondeforming control volume is versatile and useful. F l u i d s i n t h e N e w s New 1.6 GPF standards Toilets account for approximately 40% of all indoor household water use. To conserve water, the new standard is 1.6 gallons of water per flush (gpf). Old toilets use up to 7 gpf; those manufactured after 1980 use 3.5 gpf. Neither are considered low-flush toilets. A typical 3.2 person household in which each person flushes a 7-gpf toilet 4 times a day uses 32,700 gallons of water each year; with a 3.5-gpf toilet the amount is reduced to 16,400 gallons. Clearly the new 1.6-gpf toilets will save even more water. However, designing a toilet that flushes properly with such a small amount of water is not simple. Today there are two basic types involved: those that are gravity powered and those that are pressure powered. Gravity toilets (typical of most currently in use) have rather long cycle times. The water starts flowing under the action of gravity and the swirling vortex motion initiates the siphon action which builds to a point of discharge. In the newer pressure-assisted models, the flowrate is large but the cycle time is short and the amount of water used is relatively small. (See Problem 5.32.) Some problems are most easily solved by using a moving control volume. V W V CV 5.1.3 Moving, Nondeforming Control Volume It is sometimes necessary to use a nondeforming control volume attached to a moving reference frame. Examples include control volumes containing a gas turbine engine on an aircraft in flight, the exhaust stack of a ship at sea, and the gasoline tank of an automobile passing by. As discussed in Section 4.4.6, when a moving control volume is used, the fluid velocity relative to the moving control volume 1relative velocity2 is an important flow field variable. The relative velocity, W, is the fluid velocity seen by an observer moving with the control volume. The control volume velocity, V cv , is the velocity of the control volume as seen from a fixed coordinate system. The absolute velocity, V, is the fluid velocity seen by a stationary observer in a fixed coordinate system. These velocities are related to each other by the vector equation V W V cv (5.14) as illustrated by the figure in the margin. This is the same as Eq. 4.22, introduced earlier. For a system and a moving, nondeforming control volume that are coincident at an instant of time, the Reynolds transport theorem 1Eq. 4.232 for a moving control volume leads to DM sys Dt 0 0t cv r dV cs rW nˆ dA (5.15) From Eqs. 5.1 and 5.15, we can get the control volume expression for conservation of mass 1the continuity equation2 for a moving, nondeforming control volume, namely, 0 0t cv r dV cs rW nˆ dA 0 (5.16) Some examples of the application of Eq. 5.16 follow. E XAMPLE 5.6 Conservation of Mass—Compressible Flow with a Moving Control Volume GIVEN An airplane moves forward at a speed of 971 kmhr as shown in Fig. E5.6a. The frontal intake area of the jet engine is 0.80 m 2 and the entering air density is 0.736 kgm 3 . A stationary observer determines that relative to the earth, the jet engine exhaust gases move away from the engine with a speed of 1050 kmhr. The engine exhaust area is 0.558 m 2 , and the exhaust gas density is 0.515 kgm 3 . FIND Estimate the mass flowrate of fuel into the engine in kghr.
5.1 Conservation of Mass—The Continuity Equation 197 V plane = 971 km/hr Control volume m • fuel in (a) V plane = 971 km/hr W 1 = 971 km/hr Section (1) Section (2) V 2 = 1050 km/hr W 2 = 1050 + 971 = 2021 km/hr (b) F I G U R E E5.6 SOLUTION The control volume, which moves with the airplane 1see Fig. E5.6b2, surrounds the engine and its contents and includes all fluids involved at an instant. The application of Eq. 5.16 to these contents of the control volume yields 0 1flow relative to moving control volume is considered steady on a time-average basis2 Assuming one-dimensional flow, we evaluate the surface integral in Eq. 1 and get or 0 0t cv r dV cs rW nˆ dA 0 m # fuel in m # fuel in r 1 A 1 W 1 r 2 A 2 W 2 0 r 2 A 2 W 2 r 1 A 1 W 1 We consider the intake velocity, , relative to the moving control volume, as being equal in magnitude to the speed of the airplane, 971 kmhr. The exhaust velocity, W 2 , also needs to be measured relative to the moving control volume. Since a fixed W 1 (1) (2) observer noted that the exhaust gases were moving away from the engine at a speed of 1050 kmhr, the speed of the exhaust gases relative to the moving control volume, is determined as follows by using Eq. 5.14 or and is shown in Fig. E5.6b. From Eq. 2, m # fuel in m # fuel in V 2 W 2 V plane W 2 , W 2 V 2 V plane 1050 kmhr 1971 kmhr2 2021 kmhr 10.515 kgm 3 210.558 m 2 212021 kmhr2 11000 mkm2 10.736 kgm 3 210.80 m 2 21971 kmhr211000 mkm2 1580,800 571,7002 kghr 9100 kghr (Ans) COMMENT Note that the fuel flowrate was obtained as the difference of two large, nearly equal numbers. Precise values of and are needed to obtain a modestly accurate value of m # W 2 W 1 fuel. in E XAMPLE 5.7 Conservation of Mass—Relative Velocity GIVEN Water enters a rotating lawn sprinkler through its base at the steady rate of 1000 ml/s as sketched in Fig. E5.7. The exit area of each of the two nozzles is 30 mm 2 . Control volume A 2 = 30 mm 2 Section (2) FIND Determine the average speed of the water leaving the nozzle, relative to the nozzle, if (a) the rotary sprinkler head is stationary, (b) the sprinkler head rotates at 600 rpm, and (c) the sprinkler head accelerates from 0 to 600 rpm. Section (3) Section (1) Q W 2 Sprinkler head F I G U R E E5.7 Q = 1000 ml/s
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196 Chapter 5 ■ Finite Control Volume Analysis<br />
For steady flow involving more than one stream of a specific <strong>fluid</strong> or more than one specific<br />
<strong>fluid</strong> flowing through the control volume,<br />
a m# in a m # out<br />
The variety of example problems solved above should give the correct impression that the<br />
fixed, nondeforming control volume is versatile and useful.<br />
F l u i d s i n t h e N e w s<br />
New 1.6 GPF standards Toilets account for approximately 40%<br />
of all indoor household water use. To conserve water, the new<br />
standard is 1.6 gallons of water per flush (gpf). Old toilets use<br />
up to 7 gpf; those manufactured after 1980 use 3.5 gpf. Neither<br />
are considered low-flush toilets. A typical 3.2 person household<br />
in which each person flushes a 7-gpf toilet 4 times a day uses<br />
32,700 gallons of water each year; with a 3.5-gpf toilet the<br />
amount is reduced to 16,400 gallons. Clearly the new 1.6-gpf<br />
toilets will save even more water. However, designing a toilet<br />
that flushes properly with such a small amount of water is not<br />
simple. Today there are two basic types involved: those that are<br />
gravity powered and those that are pressure powered. Gravity<br />
toilets (typical of most currently in use) have rather long cycle<br />
times. The water starts flowing under the action of gravity and the<br />
swirling vortex motion initiates the siphon action which builds to<br />
a point of discharge. In the newer pressure-assisted models, the<br />
flowrate is large but the cycle time is short and the amount of<br />
water used is relatively small. (See Problem 5.32.)<br />
Some problems are<br />
most easily solved<br />
by using a moving<br />
control volume.<br />
V<br />
W<br />
V CV<br />
5.1.3 Moving, Nondeforming Control Volume<br />
It is sometimes necessary to use a nondeforming control volume attached to a moving reference<br />
frame. Examples include control volumes containing a gas turbine engine on an aircraft in flight,<br />
the exhaust stack of a ship at sea, and the gasoline tank of an automobile passing by.<br />
As discussed in Section 4.4.6, when a moving control volume is used, the <strong>fluid</strong> velocity relative<br />
to the moving control volume 1relative velocity2 is an important flow field variable. The relative<br />
velocity, W, is the <strong>fluid</strong> velocity seen by an observer moving with the control volume. The control<br />
volume velocity, V cv , is the velocity of the control volume as seen from a fixed coordinate system.<br />
The absolute velocity, V, is the <strong>fluid</strong> velocity seen by a stationary observer in a fixed coordinate system.<br />
These velocities are related to each other by the vector equation<br />
V W V cv<br />
(5.14)<br />
as illustrated by the figure in the margin. This is the same as Eq. 4.22, introduced earlier.<br />
For a system and a moving, nondeforming control volume that are coincident at an instant<br />
of time, the Reynolds transport theorem 1Eq. 4.232 for a moving control volume leads to<br />
DM sys<br />
Dt<br />
0 0t cv<br />
r dV cs<br />
rW nˆ dA<br />
(5.15)<br />
From Eqs. 5.1 and 5.15, we can get the control volume expression for conservation of mass<br />
1the continuity equation2 for a moving, nondeforming control volume, namely,<br />
0<br />
0t cv<br />
r dV cs<br />
rW nˆ dA 0<br />
(5.16)<br />
Some examples of the application of Eq. 5.16 follow.<br />
E XAMPLE 5.6<br />
Conservation of Mass—Compressible Flow with<br />
a Moving Control Volume<br />
GIVEN An airplane moves forward at a speed of 971 kmhr as<br />
shown in Fig. E5.6a. The frontal intake area of the jet engine is<br />
0.80 m 2 and the entering air density is 0.736 kgm 3 . A stationary<br />
observer determines that relative to the earth, the jet engine<br />
exhaust gases move away from the engine with a speed of<br />
1050 kmhr. The engine exhaust area is 0.558 m 2 , and the exhaust<br />
gas density is 0.515 kgm 3 .<br />
FIND Estimate the mass flowrate of fuel into the engine in<br />
kghr.