Question #112255
A piston-cylinder device contains 6 kg of refrigerant-134a at 800 kPa
and 50oC. The refrigerant is now cooled at constant pressure until it
exist as a liquid at 24oC. Show the process on T-v diagram and
determine the heat loss from the system. State any assumption made.
1
Expert's answer
2020-04-28T09:47:09-0400

Solution.

P2=800kPa;P_2 = 800kPa;

T2=24oC;T2 = 24^oC;

P1=800kPa;P_1 = 800kPa;

T1=50oC;T_1 = 50^oC;

u1=261.62kJ/kg;v1=0.02846m3/kg;u_1 = 261.62 kJ/kg; v_1 = 0.02846 m^3/kg; Tabl A-12;

u2=82.37kj/kg;v2=0.0008257m3/kg;u_2 = 82.37kj/kg; v_2 = 0.0008257m^3/kg;

m=6kg;m = 6 kg;

ΔV=m(v2v1);ΔV=6(0.00082570.02846)=0.1658m3;\Delta V = m(v_2 - v_1) ; \Delta V = 6\sdot(0.0008257 - 0.02846) =-0.1658m^3;

ΔU=m(u2u1);ΔU=6(82.37261.62)=1075.5kJ;\Delta U = m( u_2 - u_1); \Delta U = 6\sdot(82.37 - 261.62) = -1075.5kJ;

ΔQ=W+ΔU;\Delta Q = W + \Delta U;

W=PΔV;W=8105(0.1658)=132kJ;W = P\sdot\Delta V; W = 8\sdot10^5\sdot(-0.1658) =- 132kJ;

ΔQ=1321075.5=1207.5kJ=1.2075MJ;\Delta Q = -132 -1075.5 = -1207.5kJ = -1.2075MJ;




Answer: ΔQ=1.2075MJ.\Delta Q = -1.2075MJ.



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