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13C There is no truth to his claim It violates the second law of thennodynamics.
1.9 The variation of gravitational acceleration above the sea level is given as a function of altitude. The
height at which the weight of a body wiLl. decrease by 1 % is to be determined.
Analysis The weight of a body at the elevation z can be expressed
as
z
W
=
mg
=
m(9.807  3.31 x 100 z)
In our case,
W = 0.99Ws = 0.99mgs = 0.99(m)(9.807)
Substiluting.
0.99(9.807)
=
(9.8073.32xl06z)
+
z
=
29,539m
0
Sea
level
1.30 A lOOOMW power plant is powered by nuclear fuel. The amount of nuclear fuel consumed per year is
to be determined
Assumptions 1 The power plant operates continuously. 2 The conversion efficiency of the power plant
remains constant. 3 The nuclear fuel is uranium. 4 The uranium undergoes complete fission in the plant (this
is not the case in practice).
Properties The complete fission of 1 kg of uranium235 releases 6. 73x 1 010 kJ/kg of heat (given in text).
Analysis Noting that the conversion efficiency is 30%, the amount of energy consumed by the power plant
1S
Energy consumption rate
=
Power production/Efficiency
=
(1000 MW)/0.3
=
3333 MW
=
3.333 xl 0.6 kJ/s
Annual energy consumption
=
( Energy consumtion rate)( I year)
=
(3.333 xl 06 kJ/s)(365 x 24 x 3600 s/year)
=
1.051)(1014 kJ/year
Noting that the complete fission of uranium235
releases 6.73xlO10 kJ/kg of heat, the amount of uranium
that need~
to be supplied
to the power pl:'1nt
per year is
Annual
fuel consumption = Annual energy consumption
 1.051
x 1014 kJ/year
Heating
value
of fuel
 6.73
xl 010
kJ/kg
= 1562 kg/year
Therefore. this power plant will consume about one and a half tons of nuclear fuel per year.
{f
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View Full Document146E T(K) = 18°C+273=291K
T(F) = (1.8)(18)f32=64.4°F
T(OF) = 64.4+460=524.4R
157E The pressure in a tank is measured with a manometer by measuring the differential hl\ight of the
manometer fluid. The absolute pressure inthe'tank is to be determined for the cases of the manometer arm
with the higher and lower fluid level being attached to the tank.
Assumptions The fluid in the manometer is incompressible.
Properties The specific gravity of the fluid is given to be Ps
=
1.25. The density of water at 32°F is 62.4
Ibm/ft3 (Table A3E).
Analysis The density of the fluid is obtained by multiplying its specific gravity by the density of water,
P = (Ps )(PHzO) = (1.25)(62.4Ibm/ft3) = 78.0Ibm/ft3
The pressure difference corresponding
to a differential height of 28 in between the two arms of the
manometer
is
(
Ilbf
X
Ift2
)
.
L\P=pgh=(78Ibm/ft3)(32.174ft/s2)(28/12ft)
2"
=1.26psla
32.174Ibm.ft/s
144in"
Then the absolute pressures
in the tank for lhe two cases become:
(a) The fluid level in lhe arm altached
lo lhe tank is higher (vacuum):
Pabs
=
Patm Pvac =12.71.26=tl.44psia
Air
(b) Thc fluid Icvel in lhe arm allachcd lo lhe lank is lower:
Pabs
=
P
l(.1gc
+ Palin = 12.7 + 1.26 = 13.96 psia
Discu.~sion Nolc lhal wc can delerminc whcthcr lhc pressurc in a
k
.
b
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 Spring '10
 QuanfangChen

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