Therefore,
()
L
HL
1
=.
fi
F
dm
PT
dt
T
T
cT T
L
§·
¨¸
−
−+
©¹
Now,
P
= 100
×
10
6
W,
T
L
= 0 + 273 = 273 K,
T
H
= 800 + 273 = 1073 K,
T
i
= –40 + 273
= 233 K,
T
f
= 0 + 273 = 273 K,
c
= 2220 J/kg·K, and
L
F
= 333
×
10
3
J/kg, so
(
)
6
3
100 10 J/s 273 K
1
=
1073 K 273 K
2220 J/kg K
273 K 233 K +333 10 J/kg
82kg/s.
dm
dt
ªº
×
«»
−⋅
−
×
¬¼
=
We note that the engine is now operated between 0°C and 800°C.
51. (a) If
T
H
is the temperature of the hightemperature reservoir and
T
L
is the
temperature of the lowtemperature reservoir, then the maximum efficiency of the engine
is
H
800 + 40 K
=
=
= 0.78
or
78%.
800 + 273 K
TT
T
ε
−
(b) The efficiency is defined by
= 
W
/
Q
H
, where
W
is the work done by the engine and
Q
H
is the heat input.
W
is positive. Over a complete cycle,
Q
H
=
W
+ 
Q
L
, where
Q
L
is the
heat output, so
=
W
/(
W
+ 
Q
L
) and 
Q
L
 =
W
[(1/
) – 1]. Now
= (
T
H
–
T
L
)/
T
H
, where
T
H
is the temperature of the hightemperature heat reservoir and
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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