Modern Engineering Thermodynamics

10.11 Energy Efficiency Based on the Second Law 343
so that, even if the first law thermal efficiency of a domestic hot water heater were 100%, the corresponding second law
efficiency would be only 12.4%. This is again due to the fact that the ΔT here is very small in comparison with the values
of T and T 0 .
Exercises
31. Determine the first and second law efficiencies for the open liquid heater in Example 10.11 when 3.00 lbm/s of liquid
water at 50.0°F is heater to 100.°F with a heat input of 174 Btu/s. The temperature of the local environment is 50.0°F.
Answer: η T = 86.2%, and ε = 4.0%.
32. If the open liquid heater in Example 10.11 has a flow rate of 1.50 kg/s of liquid water at 15.0°C that is heated to 40.0°C,
determine the first and second law efficiencies of this process. The rate of heat added to the heater is 180. kJ/s, and the
local environmental temperature is 10.0°C. Answer: η T = 87.2%, and ε = 7.9%.
33. Show that, when ΔT is small (i.e., ΔT ≪ T ), the second law efficiency derived for the open liquid heating system
in Example 10.11 can be written as ε ≈ η T (1 − T 0 /T ). (Hint: When ΔT ≪ T, the logarithm can be expanded as ln
(1 + ΔT/T ) ≈ ΔT/T.)
Consider the operation of a heat engine. The “desired result” of the operation of a heat engine is the net power
output of the engine _W, and the availability of the desired result is again just the net power output _W: The
availability input rate to the engine is that due to the net heat transfer rate, which is the difference between the
heat transfer availability rate input from the high temperature source at T H and the heat transfer availability rate
loss to the sink at T L ,or
and
_A net = 1 − T
0
input T H
_A desired = _W
result
_Q H − 1 − T
0
j _Q
T L j
L
and the resulting second law availability efficiency of a heat engine is
ε HE =
1 − T
0
T H
_W
_Q H − 1 − T 0
T L
j _Q L j
If T 0 = T L , the second law efficiency of a heat engine that rejects heat to the local environment reduces to
Second law efficiency of a heat engine that rejects heat to the local environment
When T 0 = T L , ε HE =
where, from Eq. (7.16), we use η Carnot = 1 − T L /T H .
(10.30)
W/ _ _Q H
1 − T =
η T
(10.31)
L η Carnot
T H
A power plant is a type of modern heat engine. The following example illustrates the application of this material
to a power plant.
EXAMPLE 10.12
The electric power plant at Mount Etna has a heat input to the boiler of 1.00 × 10 6 kJ/s at 700.ºC and rejects 7.00 × 10 5 kJ/s of
heat to the condenser at 40.0ºC while producing 3.00 × 10 5 kJ/s of electrical power. The local environment (ground state) is at
0.101 MPa and 5.00ºC. Determine
a. The first law thermal efficiency of the power plant.
b. The rate at which available energy enters the boiler.
c. The rate at which available energy enters the condenser.
d. The second law efficiency of the power plant.
Solution
First, draw a sketch of the system (Figure 10.17).
(Continued )