Modern Engineering Thermodynamics

8.2 Systems Undergoing Reversible Processes 253
Solution
First, draw a sketch of the system.
Here, the unknown is _W electrical , and the system is the entire power plant as shown in Figure 8.2. You must now realize
max
that a system like this produces maximum work output when the internal losses (friction etc.) are a minimum, and the
absolute maximum occurs when the system is reversible. Therefore, what we wish to find here is _W electrical
rev :
200.°F
System boundary
Q solar
100. × 10 ³ Btu/h
W elect = ?
Heat
engine
Electrical
generator
Solar collector
River
Condenser
40.0°F
Q condenser
FIGURE 8.2
Example 8.2.
Since no clearly defined system states are given in the problem statement and all the given values are rate values, we recognize
that this problem requires a rate analysis. The energy rate balance (ERB) equation for this system is
_Q net − _W net = _U
+ _ KE + _ PE
0
ðsteady
stateÞ
0
ðassume the system
is stationaryÞ
or
_W electrical rev = _W net = _Q net
= _Q solar − j _Q j condenser
Note that there are two heat transfer surfaces in this system (the solar collector and the condenser), each at a different isothermal
temperature, and one work mode (electrical). The entropy rate balance (SRB) equation for this system is
_S =
_ms
0
ðsteady
stateÞ
Z
=
Σ
_q
T b
dA + _S P
rev
0
ðreversible
systemÞ
The system surface area Σ system is the sum of the surface areas of the collector, condenser, and the remaining surfaces where
no heat transfer occurs
Since the heat transfer surfaces are all isothermal, we can set
Z
_q
Σ
T b
∑ system = ∑ collector +∑ condenser +∑ no heat transfer sufaces
Z
dA =
Σcollector
Z
_q
dA +
T b
Σcondenser
_q Q
dA =
_ solar
−
T collector
T b
_Q condenser
T river
(Continued )