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234 Chapter 5 ■ Finite Control Volume Analysis<br />

H out<br />

h L<br />

h s<br />

h s + h L<br />

h s<br />

h s – h L<br />

H in<br />

h L = 0, h s = 0<br />

h L > 0, h s = 0<br />

h L = 0, h s > 0<br />

h L > 0, h s > 0<br />

h L = 0,<br />

h s < 0<br />

h L > 0,<br />

h s < 0<br />

F I G U R E 5.8<br />

<strong>fluid</strong> flows.<br />

Total-head change in<br />

Some important possible values of H out in comparison to H in are shown in Fig. 5.8. Note that h L<br />

(head loss) always reduces the value of H out , except in the ideal case when it is zero. Note also that<br />

h L lessens the effect of shaft work that can be extracted from a <strong>fluid</strong>. When h L 0 (ideal condition)<br />

the shaft work head, h s , and the change in total head are the same. This head change is sometimes<br />

called “ideal head change.” The corresponding ideal shaft work head is the minimum required<br />

to achieve a desired effect. For work out, it is the maximum possible. Designers usually strive to<br />

minimize loss. In Chapter 12 we learn of one instance when minimum loss is sacrificed for survivability<br />

of fish coursing through a turbine rotor.<br />

E XAMPLE 5.25<br />

Energy—Head Loss and Power Loss<br />

GIVEN The pump shown in Fig. E5.25a adds 10 horsepower<br />

to the water as it pumps water from the lower lake to the upper<br />

lake. The elevation difference between the lake surfaces is 30 ft<br />

and the head loss is 15 ft.<br />

FIND Determine<br />

(a) the flowrate and<br />

(b) the power loss associated with this flow.<br />

SOLUTION<br />

Section (2)<br />

Control volume<br />

Flow<br />

Pump<br />

Section (1)<br />

30 ft<br />

(a)<br />

The energy equation (Eq. 5.84) for this flow is<br />

where points 2 and 1 (corresponding to “out” and “in” in Eq.<br />

5.84) are located on the lake surfaces. Thus, p 2 p 1 0 and<br />

V 2 V 1 0 so that Eq. 1 becomes<br />

h s h L z 2 z 1<br />

(2)<br />

where z 2 30 ft, z 1 0, and h L 15 ft. The pump head is obtained<br />

from Eq. 5.85 as<br />

where<br />

h s<br />

p 2<br />

g V 2 2<br />

2g z 2 p 1<br />

g V 2 1<br />

2g z 1 h s h L<br />

h s W # shaft net in Q<br />

110 hp21550 ft#lbshp2162.4 lbft 3 2 Q<br />

88.1Q<br />

is in ft when Q is in ft 3 /s.<br />

(1)<br />

or<br />

Flow<br />

F I G U R E E5.25a<br />

Hence, from Eq. 2,<br />

88.1Q 15 ft 30 ft<br />

Q 1.96 ft 3 /s<br />

(Ans)<br />

COMMENT Note that in this example the purpose of the<br />

pump is to lift the water (a 30-ft head) and overcome the head loss<br />

(a 15-ft head); it does not, overall, alter the water’s pressure or<br />

velocity.

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