Question

An R-134 heat pump system obtains heat from a geothermal source and transfers 90,000 Btu/hr to an office building for space heating. The refrigeration cycle operates at pressure limits of 50 and 120 psia. If the compressor efficiency is 75%, find the power input (in hp and in Watts). (b) Find the COP for this heating application. If the building used electrical resistive heating rather than a heat pump, how much electrical power (in W) would be required? What is the percent increase in electrical power required (above that required by the heat pump) for the case of electrical resistive heating? (c) If the cycle operates 16 hrs/day and 30 days/month during the winter, and electricity costs 12 cents/kWH, how much money is saved per month by using the heat pump? (d) What are the working fluid temperatures in the two heat exchangers ? (For the condenser, we are mostly interested in the working fluid temperature during condensation because that is where most of the heat transfer occurs.)

Answer 1

For R-134, Latent heat of evaporation = 500 KJ/Kg

Pb, Total Heat absorbed in building =90000 BTU/hr = 94955 KJ/hr=26376 J/s

=> Q,Flowrate of R-134 in Office building = 94955/500=190 kg/hr=190/3600=.05 kg/s

For Compressure, Power, P

Given, P1= 50 psia = 50/14.5=3.44 bar =3.44*10^5 Pa

(14.5 psi=1bar, 1bar=10^5 Pa)

P2=120 psia = 120/14.5=8.27 bar

for R-134, Y=1.23

=>P=1.23*.05*3.44*10^5/(1.23-1)*((8.27/3.44)^((1.23-1)/1.23)-1)=16395 J/s

Compresor efficiency,n=.75

a)

=> Compressor electrical power, Pc=P/n=16395/.7=23421 J/s

Enery balance on R-134,

Heat absorbed from Geothermal(Pg)+Heat absorbed from compressor=heat evolved in building

=>Pg+P=Pb

=>Pg=26376-16395=9981 J/s

b) COP=Heat evolved/(Compressor electrical power+Geothermal Power)=Pb/(Pc+Pg)=26376/(23421+9981)=0.78

If electrical resistive heating given, Total ideal electrical resistve heating required = Pb=26376 J/s

Assuming, efficiency of electrical resistive heating = 100%

In Heat pump, Total electrical Power required = Pc=23421 J/s

For electrical resistive heating, Total power required = Pb= 26376 J/s

=> % increase in electrical heating=(26376-23421)/23421*100=12%

c) Power saved using Heat pump,Ps= Pb-Pc=26376-23421=2944 J/s

for 16 hrs a day, 30 day month, Total seconds,n = 16*3600*30=172800 seconds

=> Total energy saved using Heat pump=Ps*n=2944*1728000=508732000 Joule=508732000*2.7*10^-7=137 KWH

Energy cost given, E=12 cents/ KWH

=> Total Money saved = 137*12=1644 cents

d) @ 50 psia vapor pressure, R-134a saturation temperature = 3.33 C

@ 120 psia vapor pressure, R-134a saturation temperature = 32.2 C

Reference: http://www.csgnetwork.com/r134apresstempconv.html

=> 1st exchanger temperature = 3.33 C

=> 2nd exchanger temperature = 32.2 C