written 6.3 years ago by
teamques10
★ 69k
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modified 6.3 years ago
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TE=TsatPe=−15℃,TC=TsatPc=35℃
From P-H chart of R-12,
We know that at point 1,
h1=181kjkg;h2=210kJkg;h3=h4=68kJkg
COP=R.EWc=181−68(210−181=3.896
Also,
Qa=ṁr×(h1−h4)
For 1 TR capacity,
1×3.5=ṁr(181−68)
ṁr=0.0309kgsec=1.8584kg/min
Also
m ̇_r×v_1=piston \ displacement \ per \ TR ×η_vol
v_1=0.076 m^3⁄kg \ from \ p-h \ chart
Therefore,
piston displacement per TR=0.1765 m^3⁄min
Rate at which heat is rejected in condenser per TR
Q_r=m_R×(h_2-h_3 )
=0.0309×(210-181)
=0.899\frac{ kJ}{sec}
We know, ideal COP
=\frac{T_E}{T_C-T_E}
=\frac{258}{308-258}
=5.16