Modern Engineering Thermodynamics

4.9 Work Efficiency 125
In the case of work-absorbing systems, such as pumps or compressors, we can use an equation similar to
Eq. (4.70) to define a work transport energy conversion efficiency, orreversible efficiency, η W , as work efficiency for
work-absorbing systems:
Work efficiency for work-absorbing systems
η W ð%Þ = W rev W
× 100 =
_ rev
× 100
W act _W act
(4.71)
In the case of work-producing systems, such as engines or electrical generators, the reversible or work transport
energy conversion efficiency becomes:
Work efficiency for work-producing systems
η W ð%Þ = W act W
× 100 =
_ act
× 100
W rev _W rev
(4.72)
When these systems consist only of mechanical components, as, for example, in an internal combustion engine, the
work transport energy conversion efficiency is simply called the mechanical efficiency and η W is usually written as η m .
Even though work transport energy conversion efficiencies are always less than 100%, not all energy conversion
efficiencies are less than 100%. The value of the efficiency depends on the nature of the desired result in
Eq. (4.70). An electrical resistance can convert electrical energy (the energy input) into heat (the desired result)
with an energy conversion efficiency of 100%, but when this process is reversed, we find that the conversion of
heat into work occurs with a much lower efficiency (a consequence of the second law of thermodynamics). On
the other hand, refrigeration systems normally produce more “desired result” (cooling) than it actually costs in
required energy input. Such systems normally have energy conversion efficiencies far in excess of 100%, not
because they violate any law of physics, but simply because of the way their energy conversion efficiency is
defined. Because it seems paradoxical to most people to speak of efficiencies in excess of 100%, we call such
efficiencies coefficients of performance (COPs) instead. For example,
ðCOPÞ refrigerator
=
Refrigerator cooling rate
Refrigerator power input
EXAMPLE 4.11
The automobile engine shown in Figure 4.19 produces 150. hp on a test
stand while consuming fuel with a heat content of 20.0 × 10 3 Btu/lbm at
a rate of 1.10 lbm/min. A design engineer calculates the reversible power
output from the engine as 223 hp. Determine
a. The energy conversion efficiency of the engine.
b. The work efficiency of the engine.
1.10 lbm/min of fuel
W actual = 150. hp
Solution
a. The energy conversion efficiency is given by Eq. (4.70) as
W reversible = 233 hp
η E =
Desired energy result
Required energy input
FIGURE 4.19
Example 4.11.
The desired energy result here is the engine output power, 150. hp. The required energy input here is the energy coming from
the fuel, 20.0 × 10 3 Btu/lbm × 1.10 lbm/min × 60 min/h = 1320 × 10 3 Btu/h × (1 hp)/(2545 Btu/h) = 519 hp. Then,
η E
=
150: hp
519 hp
= 0:289 = 28:9%
b. Since an engine is a work producing machine, Eq. (4.72) gives the work efficiency as
η W =
_W actual 150: hp
× 100 =
_W reversible
223 hp × 100 = 67:3% (Continued )