Ch1-Ch3solutions

Ch1-Ch3solutions - 1-3C There is no truth to his claim It...

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1-3C 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 10-0 z) In our case, W = 0.99Ws = 0.99mgs = 0.99(m)(9.807) Substiluting. 0.99(9.807) = (9.807-3.32xl0-6z) --+ z = 29,539m 0 Sea level 1.30 A lOOO-MW 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 uranium-235 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 uranium-235 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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1-46E T(K) = 18°C+273=291K T(F) = (1.8)(18)f32=64.4°F T(OF) = 64.4+460=524.4R 1-57E 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 A-3E). 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 =12.7-1.26=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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This note was uploaded on 01/13/2011 for the course MMAE EGN3343 taught by Professor Quanfangchen during the Spring '10 term at University of Central Florida.

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Ch1-Ch3solutions - 1-3C There is no truth to his claim It...

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