Microsoft Word - Cengel and Boles TOC _2-03-05_.doc

give a negative result when work is done on the system. To avoid the nega-
tive sign, Eq. 7–51 can be written for work input to steady-flow devices
such as compressors and pumps as

(7–53)

The resemblance between the vdPin these relations and P dvis striking.
They should not be confused with each other, however, since P dvis associ-
ated with reversible boundary work in closed systems (Fig. 7–41).
Obviously, one needs to know vas a function of Pfor the given process
to perform the integration. When the working fluid is incompressible, the
specific volume vremains constant during the process and can be taken out
of the integration. Then Eq. 7–51 simplifies to

(7–54)

For the steady flow of a liquid through a device that involves no work inter-
actions (such as a nozzle or a pipe section), the work term is zero, and the
equation above can be expressed as

(7–55)

which is known as the Bernoulli equationin fluid mechanics. It is devel-
oped for an internally reversible process and thus is applicable to incom-
pressible fluids that involve no irreversibilities such as friction or shock
waves. This equation can be modified, however, to incorporate these effects.
Equation 7–52 has far-reaching implications in engineering regarding
devices that produce or consume work steadily such as turbines, compres-
sors, and pumps. It is obvious from this equation that the reversible steady-
flow work is closely associated with the specific volume of the fluid flowing
through the device. The larger the specific volume, the larger the reversible
work produced or consumed by the steady-flow device(Fig. 7–42). This
conclusion is equally valid for actual steady-flow devices. Therefore, every
effort should be made to keep the specific volume of a fluid as small as pos-
sible during a compression process to minimize the work input and as large
as possible during an expansion process to maximize the work output.
In steam or gas power plants, the pressure rise in the pump or compressor
is equal to the pressure drop in the turbine if we disregard the pressure
losses in various other components. In steam power plants, the pump han-
dles liquid, which has a very small specific volume, and the turbine handles
vapor, whose specific volume is many times larger. Therefore, the work out-
put of the turbine is much larger than the work input to the pump. This is
one of the reasons for the wide-spread use of steam power plants in electric
power generation.
If we were to compress the steam exiting the turbine back to the turbine
inlet pressure before cooling it first in the condenser in order to “save” the
heat rejected, we would have to supply all the work produced by the turbine
back to the compressor. In reality, the required work input would be even
greater than the work output of the turbine because of the irreversibilities
present in both processes.

v 1 P 2 P 12 

V 22 V 11

2

g 1 z 2 z 12  0

wrevv 1 P 2 P 12 ¢ke¢pe¬¬ 1 kJ>kg 2

wrev,in

2

1

v¬dP
¢ke
¢pe

Chapter 7 | 363

wrev

1

2

wrev P dv

wrev

= –∫ 1

2

wrev v dP

(a) Steady-flow system

(b) Closed system

= ∫

FIGURE 7–41

Reversible work relations for steady-

flow and closed systems.

= –∫ 1

2

w v dP

= –∫ 1

2

w v dP

= –∫

1

2

w v dP

FIGURE 7–42

The larger the specific volume, the

greater the work produced (or

consumed) by a steady-flow device.