Modern Engineering Thermodynamics

254 CHAPTER 8: Second Law Closed System Applications
EXAMPLE 8.2 (Continued )
Therefore,
Q _
T
condenser = _Q river
solar
T collector
and the ERB becomes
_W electrical
rev = _Q solar 1 − T river
T collector
= ð24,300 Btu/hÞ
= ð100: × 10 3 40:0 + 459:67
Btu/hÞ 1 −
200: + 459:67
1kW
3412 Btu/h
= 7:11 kW
The positive sign indicates that the electrical power is out of the system.
This example could also have been solved by using the Carnot heat engine efficiency (which is defined only for
a reversible system) and the definition of the absolute temperature scale given in Eq. (7.15).
EXAMPLE 8.3
Determine an expression for the minimum isothermal system boundary temperature required by the second law of thermodynamics
as an incompressible material is heated or cooled from a temperature T 1 to a temperature T 2 in a closed system
with a constant heat flux.
Solution
First, draw a sketch of the system (Figure 8.3).
Here, we are asked to derive a formula (the unknown) for the minimum
value of an isothermal boundary temperature, (T b ) min =?Thesystemis
(T b ) min = ?
closed and made up exclusively of an incompressible material (solid or
T 1 T 2
liquid). The energy balance equation for this system is
Q
1Q 2 − 1 W 2 = mu ð 2 − u 1 Þ+ KE 2 − KE 1 + PE 2 − PE A = Constant
1
Incompressible
State 1
State 2
0 ðassume the system is stationaryÞ
material
For an incompressible Z substance, V = constant, so dV = 0, and
2
FIGURE 8.3
1W 2 = − pdV = 0: Since no other work modes are mentioned in Example 8.3.
1
the problem statement, we assume that there are none. The resulting
energy balance is then
1Q 2 = mu ð 2 − u 1 Þ
The entropy balance equation for this system is
Z Z
_q
ms ð 2 − s 1 Þ = dA dt + 1 ðS P Þ
τ Σ T 2
b
which, for a constant heat flux ð _q Þ and isothermal boundaries, reduces to
ms ð 2 − s 1 Þ = 1 Q 2
+ 1 ðS P Þ
T 2
b
Combining the energy balance and the entropy balance, we get
mu ð 2 − u 1 Þ
T b =
ms ð 2 − s 1 Þ− 1 ðS P Þ 2
Now, since 1 (S P ) 2 ≥ 0, clearly T b is at a minimum when 1 (S P ) 2 = 0 (a reversible process). Therefore,
ðT b Þ min
= u 2 − u 1
s 2 − s 1