Modern Engineering Thermodynamics

10.11 Energy Efficiency Based on the Second Law 339
The entrance and exit specific flow availabilities are given by Eq. (10.20) (neglecting all flow stream kinetic and potential
energy terms as per the problem statement) as
so
a f = h − h 0 − T 0 ðs − s 0 Þ
and
a f 1 = h 1 − h 0 − ðT 0 Þðs 1 − s 0 Þ = ð3456:5 − 83:9 kJ/kgÞ− ð20:0 + 273:15 KÞð7:2346 − 0:2965 kJ/kg .KÞ = 1339 kJ/kg
a f 2 = h 2 − h 0 − ðT 0 Þðs 2 − s 0 Þ = ð2488:9 − 83:9 kJ/kgÞ− ð20:0 + 273:15 KÞð7:8509 − 0:2965 kJ/kg .KÞ = 190:4 kJ/kg
Then, Eq. (10.27), for a reversible process, gives the required heat transfer as
or
_Q =
_ W + _mða f 2 − a f 1 Þ
1 − T 0
T b
_Q = 20:0 × 103 kJ/s + ð18:0kg/sÞð190:4 − 1339 kJ/kgÞ
= −1260 kJ/s = −1260 kW
20:0 + 273:15 K
1 −
350: + 273:15 K
Exercises
25. Determine the heat loss from the turbine in Example 10.9 if it is not reversible but instead has a power output of
15.0 × 10 3 kW with an isentropic efficiency of 88.0%. Assume all the other variables remain unchanged.
Answer: _Q = 6840 kJ/s:
26. If we lower the surface temperature on the turbine in Example 10.9 from 350.°C to 100.°C, determine the new heat loss
rate from the turbine, assuming all the other variables remain unchanged. Answer: _Q = 3120 kJ/s.
27. An error was made in reporting the steam mass flow rate in Example 10.9. The actual mass flow rate is 20.0 kg/s not
18.0 kg/s. Determine the heat loss rate now, assuming all the other variables remain unchanged.
Answer: _Q = 5600 kJ/s.
10.11 ENERGY EFFICIENCY BASED ON THE SECOND LAW
The common work and thermal energy conversion efficiencies defined in Chapters 4 and 7 for fluid pumps and
compressors, heat engines, heat pumps, air conditioners, refrigerators, and so forth are based on energy transport
ratios for these technologies. Such efficiencies do not reveal the source of the losses within these devices, because
they have no term containing the irreversibilities within the system. Consequently, these efficiencies are often
called first law energy conversion efficiencies and are described by Eq. (4.70).
10.11.1 First Law (Energy) Efficiency
η E
Desired energy result
Required energy input
For example, in Chapter 7, we introduce the first law (thermal) energy efficiency of a heat engine as
(4.70)
η T
Net work output
Total heat input = ðW outÞ net
ðQ in Þ total
= ð _W out Þ net
ð _Q in Þ total
(7.5)
Equation (4.70) tells us that the energy conversion efficiency is the ratio of the magnitude of the energy that has
been converted divided by the magnitude of energy initially present that could be converted. Since energy is
conserved, if the energy conversion process is not 100%, then some of the energy initially present must have
been converted into a form different from that desired. This energy is said to be “lost” to the system, since it did
not end up in the proper energy form.
All technology can be categorized into four broad genres:
1. Those that output some form of energy (such as an engine) as their primary function.
2. Those that absorb energy (such as a pump) as their primary function.