Modern Engineering Thermodynamics

10.11 Energy Efficiency Based on the Second Law 341
The second law efficiency is given by Eq. (10.29) as
ε =
A desirable result
A initial or net input
where
A desirable result = A 2 − A 1 = mu 2 − u 1 + p 0 ðv 2 − v 1 Þ − T 0 ðs 2 − s 1 Þ + ðV2 2 − V2 1 Þ/2g
c + gðZ 2 − Z 1 Þ/g c
since V 1 = V 2 , Z 1 = Z 2 , and v 1 = v 2 here, the this equation reduces to
A desirable result = A 2 − A 1 = mu ½ 2 − u 1 − T 0 ðs 2 − s 1 ÞŠ
For an incompressible liquid, u 2 − u 1 = c(T 2 − T 1 ) = c(ΔT ), and s 2 − s 1 = c ln(T 2 /T 1 ), then
A desirable result = mc T 2 − T 1 − T 0 ln T
2
T 1
h
= mc ΔT − T 0 ln T + ΔT
T
i
so
2
ΔT − T 0 ln T + ΔT 3
0
ΔT − T
6
ε = mc4
T 7
0 ln 1 + ΔT 1
B
5 = mc@
T C
A
Q in
Q in
But, from the first law efficiency defined earlier, Q in = mc (ΔT )/η T ,so
h
ε = η T 1 − T i
0 ΔT
ln 1 +
ΔT T
For example, if T 0 = 70.0°F = 530. R and T = 120.°F = 580. R, then
h
ε = η T 1 − 530: i
50:0
× ln 1 + = 0:124 × η
50:0 580:
T
so that, even if the first law thermal efficiency of the heating process were 100%, the corresponding second law efficiency
would be only 12.4%. This is due to the fact that ΔT is very small in comparison with T and T 0 .
Exercises
28. Determine the first and second law efficiencies for the closed liquid heater in Example 10.10 when 300. lbm of liquid
water at 50.0°F is heated to 100.°F using 16.5 × 10 3 Btu. The temperature of the local environment is 50.0°F.
Answer: η T = 90.9%, and ε = 4.20%.
29. If the closed liquid heater in Example 10.10 contains 150. kg of liquid water at 15.0°C and is heated to 40.0°C,
determine the first and second law efficiencies of this process. The amount of heat added to the heater is 23.0 × 10 3 kJ,
and the local environmental temperature is 10.0°C. Answer: η T = 68.3%, and ε = 6.20%.
30. Show that, when ΔT ≪ T, the second law efficiency defined in Example 10.10 for a closed liquid heating system can be
written as ε = η T (1 − T 0 /T). (Hint: If ΔT is small, i.e., ΔT ≪ T, then the logarithm term can be expanded as ln(1 + ΔT/T) ≈
ΔT/T.) Note that as T → T 0 (or as ΔT → 0) in this case, then ε → 0. This is due to the fact that there is no available energy
in the desired result (i.e., A 2 → A 1 ).
Example 10.10 shows that heating a liquid in a closed container is not a very efficient use of energy. What
would happen if the heating were done in an open system, such as a domestic hot water heater? Would that be
any better? Example 10.11 answers this question and shows that the process of heating a liquid has a poor
second law efficiency no matter how it is done.
EXAMPLE 10.11
In this example, we determine the first and second law efficiencies for heating a liquid from temperature T to temperature
T + ΔT in an open, uninsulated heating tank with a negligible pressure drop and compare the results to those obtained in
Example 10.10 for heating in a closed tank. As in Example 10.10, the temperature change ΔT is due to the external heat
transfer rate _Q in: , and since the tank is not insulated there is also a heat loss rate to the local environment, _Q loss: Again, take
the ground state temperature to be T 0 .
(Continued )