Question

Refrigerant 134a at 1MPa and 90℃ is to be cooled to 1 MPa and 30℃ in a condenser by air. The air enters at 100kPa and 27 ℃ with a volume flow rate of 600m³/min and leaves at 95kPa and 60℃. Determine the mass flow rate of the refrigerant.

Answer 1

hR-134a,in = hR(1 MPa, 90oC) = 324.64kj/kg
hR-134a,out = h(1 MPa, 30oC) is the enthalpy of a compressed liquid
hR-134a,out = h(1 MPa, 30oC) ≈ hf(30oC) = 93.58 kJ/kg
gas constant for air: R = 0.2870 kJ/kg∙K
v,in = RTin/Pin.

air ar600 m3 kg.K 100 kPa mar air 0.287 _ 300.15Km"7-696 5 min Ava arJa

We can how substitute Dh = c_pDT for air into our first law and solve for the refrigerant flow rate

airair.out airp.aiair out ai RR.inRout

In using this equation, we can compute the temperature difference in degrees Celsius or in kelvins and get the same result

6965kg1.005 k(60-27)K min kg K 32464k/ kg mair p.air atr,oud ar 9538k/ kg