Question

A 10-ft3 tank contains a saturated mixture of refrigerant R-134a at a pressure of 40 psia. If the saturated liquid occupies 2% of the volume, determine the a) total mass (lbm) b) quality c) average internal energy (Btu lbm) d) temperature (℉).

Answer 1

a pressure of 40 psia = 2.72 bar

from R-134 thermoynamics tables, this correspond to a temperatue of 2 deg.c

From F= 1.8C+32= 1.8*2+32= 35.6 F

from the tables, volumes :Saturated liquid : 0.775*10-³ kg/m3 and vapor =0.0689 m3/kg

liquid occupies 2% of the tank volume

Volumes in the tank : liquid =2%= 10*2/100= 0.2 ft3 and vapor =10-0.2 =9.8 ft3

but 1ft =0.3048m

volumes of liquid and vapor in the tank : 0.2*(0.3048)³ =0.005663 m³ and vapor =9.8*(0.3048)³=0.277m³

mass : =Volume/specific volume

Mass : Liquid : : 0.005663/{0.775*10-3) kg=7.3 kg and vapor =0.277/0.0689=4.02 kg

total mass = 7.3+4.02=11.32 kg but 1 lb= 0.4535 kg , 11,32 kg =11.32/0.4535lb=24.96 lb

Quality of steam= amout of vapor/ total quantity=100* (4.02/11.32)=35.5%, liquid fraction= 100-35.5=64.5

internal energyof liquid ( from tables)= 52.01 kj/kg and that of vapor = 228.32 Kj/Kg

average internal energy= liquid mass fraction* Internal energy of liquid + vapor mass fraction* internal energy of vapor =0.645*52.01+0.355*228.32 Kj/Kg=114.6 Kj/Kg

1 Kj/Kg =0.429 But/lb

114.6 Kj/Kg= 114.6*0.429 Btu/lb=49.16 Btu/lb

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