An air-conditioner operates on the vapor-compression refrigeration cycle with re

Question

An air-conditioner operates on the vapor-compression refrigeration cycle with refrigerant R-134a. The evaporator is inside an air handler of a building. The air flowing through the air handler enters the air hander at 27 oC and is limited to a 20 oC temperature drop. The waste heat is rejected to the ambient air at 37 oC. The refrigerant enters the compressor at 180 kPa superheated by 2.7 oC at a rate of 0.06 kg/s and leaves the compressor at 1200 kPa and 60 oC. R-134a is subcooled by 6.3 oC at the exit of the condenser. Determine (a) the rate of cooling provided, in Btu/h, and the COP, (b) the isentropic efficiency of the compressor, (c) the ratio of volume flow rate of air entering the air handler (m3/min) to the mass flow rate of refrigerant (kg/s) through the air handler, in (m3air/min)/(kgR134/s), and (d) the power input.

Explanation / Answer

From R134a properties at P1 = 180 kPa we get T_saturation = -12.7 deg C

Since compressor inlet condition is 2.7 degrees superheated, T1 = -12.7 + 2.7 = -10 deg C

At P1, T1 we get h1 = 245 kJ/kg, and s1 = 0.948 kJ/kg-K

At P2 = 1200 kPa and T2 = 60 deg C we get h2 = 290 kJ/kg

Assuming isentropic compression, for P2_is = P2 = 1200 kPa and s2_is = s1 = 0.948 kJ/kg-K we get h2_is = 285 kJ/kg

At P3 = P2 = 1200 kPa we get T_saturation = 46.3 deg C

Since state is subcooled at condensor exit, we get T3 = 46.3 - 6.3 = 40 deg C

At P3, T3 we get h3 = 108 kJ/kg

Also, during throttling h4 = h3 = 108 kJ/kg

a)

Rate of cooling = m*(h1 - h4)

= 0.06*(245 - 108)

= 8.22 kW

Compressor power input = m*(h2 - h1)

= 0.06*(290 - 245)

= 2.7 kW

COP = Rate of cooling / Power input

= 8.22 / 2.7

= 3.04

b)

Compressor eff = (h2_is - h1) / (h2 - h1)

= (285 - 245) / (290 - 245)

= 0.889 or 88.9 %

c)

For air handler, 8.22 = m_air*Cp*delta T

8.22 = m_air*1.005*20

m_air = 0.409 kg/s

Volume flow rate of air = 0.409 / 1.2 = 0.341 m^3/s

= 0.341*60 m^3/min

= 20.45 m^3/min

Ratio = 20.45 / 0.06

= 340.8

d)

Power input = 2.7 kW as found in part a.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.