aljabr (faa335) – Hw30 – Ross – (89251)
1
This
printout
should
have
6
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
The efficiency of a 846 MW nuclear power
plant is 18
.
5%.
If a river of flow rate 2
.
8
×
10
5
kg
/
s were
used to transport the excess thermal energy
away, what would be the average temperature
increase of the river?
Correct answer: 3
.
17979
◦
C.
Explanation:
Given:
P
output
= 846 MW = 8
.
46
×
10
8
W
and
e
= 18
.
5% = 0
.
185
.
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
8
.
46
×
10
8
W
0
.
185
parenrightbigg
(1

0
.
185)
= 3
.
72697
×
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
.
72697
×
10
9
W
(2
.
8
×
10
5
kg
/
s) (4186 J
/
kg
·
◦
C)
=
3
.
17979
◦
C
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 Gavrine
 Energy, Power, Heat, Thermal Energy, Heat engine, Carnot cycle

Click to edit the document details