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MAE204_F11_HW3-Sol

MAE204_F11_HW3-Sol - Q3 We queA—‘Im SkU-AA “AV...

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Unformatted text preview: Q3 We queA—‘Im SkU-AA “AV 3*‘468 MOLD MULL‘ 909557 ddeg H- "\‘JEL A“ 80 Q‘FW bentenL <25 3 Sq \‘bhc'\‘: D \? C10“) k °\ SquJpq-‘ra UQ‘PUA' AASW‘. LOQ\L «3T *4‘0‘1 A’S’ qugg all‘ TN Lt—AVA‘ HeA‘ as? qum‘impm 367+ Arab) 33W "571:5 6* mag/V, 1') ML] [CT/1%! MAL! W so me e‘ 3A,; block mi thcn M LQJTM'X“ kccfi‘ u‘Q‘ Q“ . s/un pluJ .34.; \CA'EO'F Kchcd‘ a}? UQ/bufi gcjruq' TIM; I3 becaugg ‘HNFK pol“ \OL q “E‘IM LAH’IN. H‘t bCAkmL '5 q) 5 L [C3 9 COA e \IQUS ‘UMA‘ fl€<c53 ”k X’L heQ'L-J |’$‘3>.§‘&°Q (1‘qu pp? VPr (96% m\~m bah 33‘ Locust / '7 Q90“ ; 90% U‘oL Cun5em¢:\—,‘y,\ «3? 5M7, / bEanfim‘Vl-JMM‘ / N hasmwkewpe M S 0 mafia wrfim‘ 1 ‘ C9 “A“. D0 r‘ \ c9 \- Egrgr‘ |\ C7¢nc7 :: ha :. FLT—1&3 2 0&39‘5 =1 :20 L: C9 2?) Pow Parka 51 wmhm :_ H00 L.) 2-28C The work done is the same, but the power is different. 2—40C No. This is the case for adiabatic systems only. 2-47 A classroom is to be air-conditioned using window air-conditioning units. The cooling load is due to people, lights, and heat transfer through the walls and the windows. The number of 5-kW window air conditioning units required is to be determined. Assumptions There are no heat dissipating equipment (such as computers, TVs, or ranges) in the room. Analysis The total cooling load of the room is determined from Qcooling = Qlights + Qpeople + Qheat gain where Qlights =10 x 100 w =1 kW Qpeople = 40x 360 kJ / h = 4 kW Room QMgain = 15,000 kJ / h = 4.17 kW 15,000 kJ/h 40 people le 10 bulbs Substituting, Qcooling =1+4+4.17 = 9.17 kW Thus the number of air-conditioning units required is 9.17 kW —. = 1.83—)2 units 5 kW/un1t 2-74 Water is pumped from a lower reservoir to a higher reservoir at a specified rate. For a specified shaft power input, the power that is converted to thermal energy is to be determined. Assumptions 1 The pump operates steadily. 2 The elevations of the reservoirs remain constant. 3 The changes in kinetic energy are negligible. Properties We take the density of water to be p = 1000 kg/m3. Analysis The elevation of water and thus its potential energy changes during pumping, but it experiences no changes in its velocity and pressure. Therefore, the change in the total mechanical energy of water is equal to the change in its potential energy, which is gz per unit mass, and rhgz for a given mass flow rate. That is, w AEmech = rhAemech = mApe = mgAZ = Pl’gAz 1N 1kW =1000k 3 0.03 3/ 9.81m/2 45 —[—)=13.2kw Then the mechanical power lost because of frictional effects becomes Wfrict = Vi/pumin — AEmech = 20 — 13.2 kW = 6.8 kW Discussion The 6.8 kW of power is used to overcome the friction in the piping system. The effect of frictional losses in a pump is always to convert mechanical energy to an equivalent amount of thermal energy, which results in a slight rise in fluid temperature. Note that this pumping process could be accomplished by a 13.2 kW pump (rather than 20 kW) if there were no frictional losses in the system. In this ideal case, the pump would function as a turbine when the water is allowed to flow from the upper reservoir to the lower reservoir and extract 13.2 kW of power from the water. 3-1C A liquid that is about to vaporize is saturated liquid; otherwise it is compressed liquid. 3-2C A vapor that is about to condense is saturated vapor; otherwise it is superheated vapor. ...
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