Modern Engineering Thermodynamics

7.5 The Absolute Temperature Scale 215
The results of Example 7.1 are highly unrealistic since no real heat engine can ever be
reversible. The irreversibilities within modern heat engines limit their actual operating
thermal energy conversion efficiency to around 30%.
Exercises
1. Rework Example 7.1 for a flame temperature of 2500.°C and an environmental
temperature of 20.0°C. Answer: (η T ) max = 89.4%.
2. If the engine described in Example 7.1 has a maximum (reversible or Carnot) thermal
efficiency of 60.0% when the environmental temperature is 70.0°F, determine the
flame temperature of the combustion process. Answer: T flame = T H = 1324 R = 865°F.
Burning fuel
at 4000.°F
Heat engine
Environment
at 70.0°F
Heat from combustion
Exhaust
Work
v7 = ?
max
possible
FIGURE 7.7
Example 7.1.
EXAMPLE 7.2
A coal-fired electrical power plant produces 5.00 MW of electrical power while exhausting 8.00 MW of thermal energy to a
nearby river at 10.0°C. The power plant requires an input power of 100. kW to drive the boiler feed pump.
Determine
a. The actual thermal efficiency of the power plant.
b. The equivalent heat source temperature if the plant operated on a reversible Carnot cycle.
Solution
First, draw a sketch of the system (Figure 7.8).
Q H (from the combustion of coal)
Boiler
Boiler feed
pump
Turbine
Electrical
generator
W E = 5.00 MW
W P =−1.00 kW
Condenser
Q L (to river at 10.0°C) = 8.00 MW
FIGURE 7.8
Example 7.2.
The unknowns are the actual thermal efficiency of the power plant and the equivalent heat source temperature if the plant operates
on a reversible Carnot cycle. If we construct the system boundary as shown in the sketch, the power plant is a closed system.
a. The actual thermal efficiency of this system is given by Eq. (7.9) as
η T = ð _W out Þ net W
= _ E − j _W P j Q
= _ H − j _Q L j
_Q in
_Q H
_Q H
and the energy rate balance (ERB) for the steady state operation of this system is
or
_Q H − j _Q L j − ð _W E − j _W P jÞ = 0
_Q H = j _Q L j + ð _W E − j _W P jÞ
= j−8:00 MW j + ð5:00 MW − j− 0:100 MWjÞ
= 12:9MW
(Continued )