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Methods of handling Supply air in HVAC
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### 18) R-134a is compressed by an adiabatic compressor from the saturated.docx

1. 18) R-134a is compressed by an adiabatic compressor from the saturated vapor state at 0.12 MPa to 1.2 MPa and 70 Solution 18) Pi = 0.12 Mpa (saturated) Pe = 1.2 Mpa Tt = 70 deg. C m_dot = 0.108 kg/s hi = h(Pi) = h(0.12 MPa) = 233.86 kJ/kg he = h(Pe,Te) = 298.96 kJ/kg Energy balance for compressor at steady state, dEcv/dt = 0 = Q - W + m_dot(hi-he), Q = 0 W = m_dot(hi-he) W = 0.108*(233.86 - 298.96) W = - 7.0308 kW 19) Twi = 16 deg. C Two = 50 deg. C m_dot = 0.2 kg/min = 3.33*10^-3 kg/s COP = 2.8 Solve for heat transfer at hot reservoir, Qh = m_dot*c*(Two-Twi) Qh = 3.33*10^-3*4.18*34 Qh = 0.473733 kW Power input, Cop = Qh/W W = Qh/COP
2. W = 0.473733/2.8 W = 0.1692 kW ~ 0.17 kW 20) COP=Qh/W (for heat pump) Qc-heat absorbed...not needed in this calculation Qh-heat rejected Qh= 30000kJ/h W=3kW Qh=30 000kJ/h =(25/3) kJ/s =(25/3) kW COP=(25/3)/(3) = 2.7777....=2.78 u cannot have COP's like 0.36 or even 1.18...these values are very low....usually COP's range from 2-4.5 where 4.5 is the COP for very efficient heat pumps
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