hw 6 thermo.pdf

# hw 6 thermo.pdf - 5.49 A large heat pump should upgrade 4...

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Unformatted text preview: 5.49 A large heat pump should upgrade 4 MW of heat at 65°C to be delivered as heat at 145°C. What is the minimum amount of work (power) input that will drive this? For the minimum work we assume a Camot heat pump and Q1. = 4 MW. Q H TH 273.15 +145 '3HP=. =TH-TL= 145-65 =5'227 Win Q BREF=BHP-l =—=4.227 Win Now we can solve for the work win = QL/ﬂREF = 4/4227 = 0.946 MW This is a domestic or small ofﬁce building size AI’C unit. much smaller than the 4 MW in this problem. 5.57 Calculate the amount of work input a refrigerator needs to make ice cubes out of a tray of 0.25 kg liquid water at 10°C. Assume the refrigerator works in a Carnot cycle between —8°C and 35°C with a motor-compressor of 660 W. How much time does it take if this is the only cooling load? Solution: (1V. Water in tray. We neglect tray mass. Energy 1341.: ml:u2 — u1)=1Q2 — 1W2 Process : P = constant + Po 1W2 =1 P ‘W = Pom(V2 — V1) 1% = 111012 — 1L11l+1wz = 111012 — 111) Tb]. 13.1.1 : h] = 41.99 lil-"lig, Tb]. B.1.5 :32 = - 333.5 3mg 1Q2 = 0.25 kg (-3334 — 41.99 } kl’kg = - 93.343 1:] Consider now reﬁ'igerator QL QL TL 233 — 3 B- w _QH-QL_ TH-TL_35-(-3)_ 6.1-5 QL 1Q: 93.343 [3 - [3 = 616 =15.24kJ For the motor to transfer that amount of energy the time is found as W=Jurdt=9lr3t w 15.24x1000 =—= 4s . 600 25° W At: Comment: We neglected a baseload of the refrigerator so not all the 650 W are available to make ice= also our coefﬁcient of perfomiance is very optimistic and ﬁnally the heat transfer is a transient process. All this means that it will take much more time to make ice-cubes. 5.83 A house is cooled by a heat pump driven by an electric motor using the inside as the low-temperature reservoir. The house gains energy in direct proportion to the temperature difference as anin = K(TH - TL). Determine the minimum electric power to drive the heat pmnp as a function of the two temperatures. Solution: Refrigerator CUP: [3- : {EL-"Win 3 TLJ'CI‘ H - TL} ; Heat gain must be removed: QL = anin = K[TH - TL) Solve for required work and substitute in for B o Wm = (ED-"ﬂ 2 KU‘H — TL) x (TH - TL).-“1'L . 2; WmEKCTH' L) ’TL 5.97 Hydrogen gas is used in a Carnot cycle having an efﬁciencyr of 60% with a low temperature of 300 K. During the heat rejection the pressure changes from 90 kPa to 120 kPa. Find the high and low temperature heat transfer and the net cycle work per unit mass of hydrogen. Solution: As the efﬁciencyr is known, the high temperature is found as TL 1] =05 =1—T— =>TH=TL.-='(1- 0.6): ?50K H Now the volume ratio needed for the heat transfer, T3 = T4 = TL, is 1.3 v4 = (RT3 .5 P3 ) 1" (RT4I' P4 ) = P4_-" P3 = 120 .r' 90 = 1.333 so from Eq.'}".9 we have with R = 4.1243 ﬁ'om Table A5 qL = RTL ln (v3.-"'v4 ) = 355.95 kakg Using the efﬁciency from Eq.7.4 then qH= qL (1 - 0.6) = 889.9 kakg The net work equals the net heat transfer w = qH - qL = 533.9 kakg ...
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