A refrigerator uses refrigerant 134a as the working fluid and operates on a idea

Question

A refrigerator uses refrigerant 134a as the working fluid and operates on a ideal vapor-compression refrigeration cycle between 0.1 and 0.6 MPa. The mass flow rate of the refrigerant is 0.08 kg/s. Show the T-s diagram with respect to saturation lines. Determine: 1. The rate of heat removal from the refrigerated space and the power input to the compressor. 2. The rate of heat rejection to the environment. Can you please explain how to get h2 when you have P2 and S2 when its not on the table. Thank you!

Explanation / Answer

Get the software Engineering equation solver. Here is a code for it, which will solve the problem for you (with the condenser temperature of 60 F): Note: R134a is another name for tetrafluoroethane. "!---------Begin EES-------" "!----Unit settings: kJ/C/kg/kPa/degrees--------" "ASSUMPTIONS: 1. Compressor is adiabatic 2. TXV is adiabatic 3. Velocity effects are negligible everywhere" "Data" m_dot = 0.05 [kg/s] P[1] = 140 [kPa] P[2] = 800 [kPa] T[1] = -10 [C] T[2] = 50 [C] T[3] = 26 [C] P[3] = 720 [kPa] P[4] 140 [kPa] F$ = 'R134a' {This is the name of the fluid flavor, a string variable instead of a number} "Strategy" "Look up enthalpy at states 1 through 3" h[1] = enthalpy(F$, T=T[1], P=P[1]) h[2] = enthalpy(F$, T=T[2], P=P[2]) h[3] = enthalpy(F$, T=T[3], P=P[3]) "Use isenthalpic TXV condition to lock-down state 4" h[4] = h[3] "Rate of heat transfer from refrigerated space in to evaporator" Q_dot_in = m_dot*(h[1] - h[4]) "Power input to compressor" W_dot_in = m_dot*(h[2] - h[1]) "Use entropy to confirm that our compressor is actually possible as an adiabatic compressor" s[1] = entropy(F$, T=T[1], P=P[1]) s[2] = entropy(F$, T=T[2], P=P[2]) "If s[2] > s[1], you have a real compressor that is irreversible yet of course possible If s[2] = s[1], you know darn well what you are doing on compressor design, because you have a reversible isentropic compressor If s[2]

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