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Solution HW 6
W
net
= 535 MW
Thermal efficiency =
η
thermal
= 0.40
Heat in = Q
H
= W
net
/
η
thermal
= 535 MW / 0.4 = 1337.5 MW or MJ/s
Î
Heat out = Q
L
= Q
H
– W
net
= 1337.5 – 535 = 802.5 MJ/s
W
net
= 175,000 MW
Option 1:
η
thermal
= 0.32
Cost at $1,150/kW = (1.75 x 10
8
kW)($1,150/kW) = $ 2.0125 x 10
11
Q
H
= W
net
/
η
thermal
= 175,000 MW / 0.4 = 546875 MW
Coal required in 5 years
= (5.46875 x 10
8
kJ/s) (5x365x24x3600 s) / (2.8 x 10
7
kJ/T)
= 3.0979 x 10
9
T
Option 2:
η
thermal
= 0.47
Cost at $1, 520/kW = (1.75 x 10
8
kW)($1, 250/kW) = $ 2.66 x 10
11
Q
H
= W
net
/
η
thermal
= 175,000 MW / 0.47 = 372340.4 MW
Coal required in 5 years
= (3.723404 x 10
8
kJ/s) (5x365x24x3600 s) / (2.8 x 10
7
kJ/T)
= 2.0968 x 10
9
T
Excess cost of more efficient plants = ($2.66  $2.0125) x 10
11
= $6.475 x 10
10
Savings in coal over 5 years = 3.0979 x 10
9
 2.0968 x 10
9
= 9.8288 x 10
8
T
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View Full Document If the extra cost is to be recovered from fuel savings in 5 years, the cost of coal
should be $6.475 x 10
10
/ 9.8288 x 10
8
T = $ 65.88 per ton
Q
L
= 7.2 kW
COP = 1.21
COP
R
= Q
L
/ W
Î
W = Q
L
/ COP = 7.2 / 1.21 kW = 5.95 kW
Heat pump condenser
Fluid
R134a
Flow rate = m = 0.045 kg/s
Inlet
P = 800 kPa, T = 35
0
C = 308.15 K
From table A12
T
sat
= 31.31
0
C
> superheated state
h
in
= 267.29 + (276.45267.29) (35 – 31.31)/(40 – 31.31) kJ/kg
= 271.18 kJ/kg
Outlet P = 800 kPa
Saturated liquid
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This note was uploaded on 11/25/2009 for the course ENGR thermo taught by Professor Staff during the Spring '09 term at University of Louisiana at Lafayette.
 Spring '09
 Staff

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