Question

Air enters a counterflow heat exchanger operating at steady state at 27 C, 0.3 MPa and exits at 12 C. Refrigerant 134a enters at 0.4 MPa, a quality of 0.3, and a mass flow rate of 35 kg/h. Refrigerant exits at 10 C. Stray heat transfer is negligible and there is no significant change in pressure for either stream.(a) For the Refrigerant 134a stream, determine the rate of heat transfer, in kJ/h.(b) For each of the streams, evaluate the change in flow exergy rate, in kJ/h, and interpret its value and sign.Let T0 = 22 C, p0 = 0.1 MPa, and ignore the effects of motion and gravity.

Answer 1

A) 337.21 kj/h

b) -224.823 kj/h

Step-by-step explanation:

Steady state temperature = 27⁰C

Pressure = 0.3 MPa

exit temperature = 12⁰c

Refrigerant 134a :

entering pressure = 0.4 MPa

mass flow rate = 35 kg/h

quality = 0.3

inlet temp = 8.93⁰c

A) Determine the rate of heat transfer in kJ/h for Refrigerant 134a stream

calculate the specific enthalpy of refrigerant at the inlet

hm = hf + x ( hg - hf )

at 0.4 MPa : hf = 64 kj/kg , hg = 256 kj/kg

x = 0.3 ( quality )

hence : hm = 64 + 0.3 ( 256 - 64 ) = 64.3 + 57.6 = 121.9 kj/kg

next calculate the specific enthalpy at outlet

Tout = 10⁰c , T sat = 8.93⁰c

since the Tsat < Tout the refrigerant is super heated and from the super heated refrigerant table at P = 0.4 MPa, hout = 257 kj/kg

Heat gained by refrigerant ( Qin)

Qin = mref ( hout - hin )

= 35 ( 257 - 112) = 5075 kj/h

To determine the rate of heat transfer we have to apply the equation below

heat gained by refrigerant = heat gained by air

Qin = Qout

5075 kj/h = mair ( hout - hin ) ------------ 1

considering ideal conditions

at T = 300 k , hout = 300.19 kj/kg ( specific enthalpy )

at T = 285 k , hin = 285.14 kj/kg ( specific enthalpy )

back to equation 1

5075 kj/h = mair ( 300.19 - 285.14 )

mair ( rate of energy transfer ) = 5075 / 15.05 = 337.21 kj/h

B) evaluating the change in flow energy rate in kJ/h

attached below is the detailed solution