aldawsari (ma28683) – HW26 – Betancourt – (16856)
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20.320.7
001
10.0 points
The efficiency of a 905 MW nuclear power
plant is 21
.
5%.
If a river of flow rate 3
.
19
×
10
6
kg
/
s were
used to transport the excess thermal energy
away, what would be the average temperature
increase of the river?
Correct answer: 0
.
247451
◦
C.
Explanation:
Given:
P
output
= 905 MW = 9
.
05
×
10
8
W
and
e
= 21
.
5% = 0
.
215
.
The excess thermal energy transported per
second by the river is
P
excess
=
P
input
(1

e
)
=
parenleftbigg
P
output
e
parenrightbigg
(1

e
)
=
parenleftbigg
9
.
05
×
10
8
W
0
.
215
parenrightbigg
(1

0
.
215)
= 3
.
3043
×
10
9
W
where
e
is the efficiency and
P
output
the power
output of the plant. Thus the temperature of
the river is increased (per second) by
d
m
d
t
c
Δ
T
=
d
Q
d
t
=
P
excess
where
c
is the heat capacity of water and
d
m
d
t
is the flow rate of the water. Thus
Δ
T
=
P
excess
d
m
d
t
c
=
3
.
3043
×
10
9
W
(3
.
19
×
10
6
kg
/
s) (4186 J
/
kg
·
◦
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 Spring '10
 Gavrine
 Thermodynamics, Energy, Power, Heat, Heat engine, Carnot cycle

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