Modern Engineering Thermodynamics

10.11 Energy Efficiency Based on the Second Law 345
Exercises
34. Calculate the second law thermal efficiency of the power plant in Example 10.12 if the boiler outlet temperature is
increased from 700.°C to 1000.°C. Assume all the other variables remain unchanged. Answer: ε HE
= 42.7%.
35. Suppose we lower the condenser temperature in Example 10.12 from 40.0°C to 15.0°C. Assuming all the other variables
remain unchanged, determine the new second law thermal efficiency of the power plant. Answer: ε HE
= 43.5%.
36. Determine the second law thermal efficiency of the power plant in Example 10.12 when we set T L = T 0 = 5.00°C.
Answer: ε HE
= 42.0%.
37. Why is the value of ε HE calculated in Exercise 36 less than the value calculated in the Example 10.12? (Hint: Does it
include the heat transfer irreversibilities and the lost available energy in the heat transfer between the condenser and the
environment?)
CRITICAL THINKING
The combustion temperature of the burning fuel in the boiler of Example 10.12 is 1800.°C. Suppose this temperature is
taken as the heat source temperature T H . Would the value of the second law availability efficiency ε HE
calculated in Example
10.12 increase or decrease? Explain what additional irreversibilities are introduced by this change in system boundary that
would produce the change you observe in ε HE
.
In the case of a heat pump, the net available energy input rate is again just the pump work rate (power). The
available energy rate of the “desired result” (heating) is (1 − T 0 /T H ) H . Then, the second law availability efficiency
of a heat pump is
ε HP
=
1 − T
0 _Q
T H
H
j _W in j
= 1 − T
0
T H
_Q H
j _W in j
!
= 1 − T
0
COP actual
T H heat pump
(10.32)
and, if T 0 = T L , the second law efficiency of a heat pump that absorbs heat from the local environment reduces to
Second law efficiency of a heat pump that absorbs heat from the local environment
When T 0 = T L , ε HP
=
COP actual
heat pump
COP Carnot
heat pump
(10.33)
where the coefficient of performance (COP) of a Carnot heat pump given in Eq. (7.18) is used. The following
example illustrates the use of this material.
EXAMPLE 10.13
Aheatpumpisdesignedtoprovide30.0× 10 3 Btu/h of heat to a small house at 70.0°F when the outside temperature is
30.0°F. The electric motor driving the heat pump draws 1.50 hp. Determine
a. The first law thermal efficiency (i.e., the COP) of the heat pump.
b. The second law availability efficiency of the heat pump.
Solution
First, draw a sketch of the system (Figure 10.18).
The unknowns are the first law thermal efficiency (i.e., the COP) of the heat pump and the second law availability
efficiency of the heat pump.
a. The first law thermal efficiency of the heat pump is given by Eq. (4.70) as
η T = COP actual =
heat pump
Desired energy result
Required energy input =
Q _ H
=
j _W in j
30:0 × 10 3 Btu/h
ð1:50 hpÞð2545 Btu/hp.hÞ = 7:86 (Continued )