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Prob 8.3
PROBLEM 8.3
PROBLEM STATEMENT:
An electric utility company has a plant with a capacity of
1000 MW of electricity for sale to its customers. The overall thermal efficiency of
the plant is 25 percent and toil (the fuel) costs 25 cents a gallon. The heating
value of the oil is 4.65 x 10
6
kJ/kg and there are 3.4 kg of oil per gallon. What is
the cost of running at maximum capacity for one week?
GIVEN:
NET
ΗΕ
W
=1000 MW,
= 25 %
η
&
Fuel cost
=
25¢/gallon, 1 gallon
=
3.4 kg,
×
6
HV = 4.65
10 kJ/kg
FIND:
Total operational cost for one week ($).
ASSUMPTIONS:
GOVERNING RELATIONS:
Thermal efficiency,
HE
NET
IN
=W
Q
&
&
η
QUANTITATIVE SOLUTION:
From the thermal efficiency relation, the total power input into the plant is
NET
IN
HE
6
W
1000
Q
=
=
= 4000 MW
0.25
=4 10 kJ/s
η
×
&
&
One week is equivalent to 6.048
x
10
5
s. Thus, the total energy consumed is
()
( ) ( )( )
65
IN
IN
12
Q
= Q
time = 4
10
kJ/s
6.048
10
s
=2.4192 10 ±kJ
×× ×
×
×
&
The required mass and volume to produce this energy is
( )
×
×
×
××
×
12
5
IN
6
5
5
2.4192
10
kJ
Q
m=
=
=5.203 10 kg
HV
4.65
1 g a l
V
V = m
= 5.203
10 kg
m3
.
4
k
g
o
=1.530
10 gal
i
l
The total cost is computed as follows:
( ) ( )
×
5
cost = volume
oil price =1.530
10
gal
0.25 $/gal
=$38,250
DISCUSSION OF RESULTS:
Thermal efficiency is not just of thermodynamic interest; it is directly related to the
economics of the power plant. The power company produces kilowatthours (W
NET
)
for which it charges its customers, and one of the biggest factors in the cost of
production is fuel (Q
IN
). Thus the efficiency, which is the ratio of W
NET
to Q
IN
, can be
thought of as the power plant’s return on its energy investment.
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View Full Document Prob 8.3
PROBLEM 8.4
PROBLEM STATEMENT:
Plot the Pv, T
s
, Ph, and hs cycle diagrams for the air
standard Carnot cycle operating with 1 lbm of working fluid between 60°F and
400 °F, if the maximum volume reached in the cycle is 1 ft
3
, and the maximum
pressure in the cycle is 2000 psia. Assume that the specific heat of air can be
taken as temperatureindependent.
DIAGRAM DEFINING SYSTEM AND PROCESS:
W
IN
Q
H
Q
C
3
1
2
4
Adiabatic
Reversible
Compression
W
OUT
W
IN
Isothermal Cooling & Compression
Isothermal Heating & Expansion
Adiabatic
Reversible
Expansion
W
OUT
GIVEN:
State 1:
T
1
= 60
o
F
State 2:
T
2
= 400
o
F, P
2
=2000 psia
State 3:
T
3
= 400
o
F
State 4:
T
4
= 60
o
F, V
4
= 1 ft
3
m = 1 lbm
FIND:
Specific enthalpies (h), pressures (P), and specific entropies (s) for each state
and plot Pv, Ts, Ph, and hs diagrams
ASSUMPTIONS:
1. Air Standard Carnot cycle
2. Constant mass
3. Constant specific heats (evaluate at 60
°
F for air standard)
4. Nonflow processes
GOVERNING RELATIONS:
1. PV = mRT
Prob 8.3
2. For adiabatic and reversible (isentropic) process, constant specific heat:
−
=
k1
k
bb
aa
TP
PROPERTY DATA:
From Table 5e at 60
°
F
k = 1.4
c
P
= 0.24 Btu/lbm
o
R
h
1
(60
o
F) = h
4
= 124.5 Btu/lbm
h
2
(400
o
F) = h
3
= 206.74 Btu/lbm
QUANTITATIVE SOLUTION:
−
−−
=⇒
= =
⇒
=
k
22
2
11
.
4
1
.
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This homework help was uploaded on 03/19/2008 for the course ME 326 taught by Professor Schmidt during the Spring '07 term at University of Texas at Austin.
 Spring '07
 Schmidt

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