Question

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Refrigerant R134a is entering a condenser and is being cooled by water. The refrigerant enters the...

  1. Refrigerant R134a is entering a condenser and is being cooled by water. The refrigerant enters the condenser with a mass flow rate of 6 kg/min at 1 MPa and 70 0C and leaves at 35 0C. The cooling water enters at 300 KPa and 15 0C and leaves at 25 0C. Neglecting any pressure drop, determine:
  1. The mass flow rate of the cooling water
  2. The heat transfer rate from the refrigerant to the water.
  3. The rate of entropy production within the heat exchanger in kW/K

Solutions

Expert Solution

Refrigerant Entering mass flow rate mr = 6 kg/min

entering temperature T1 = 70 C and exit temperature T2 = 35 C at 1 MPa.

Water entering temperature tw1= 15 C and exit temperature tw2 = 25 C at constant pressure P = 300kPa

For refrigeration R-134a table :

At inlet condition (70C and 1 MPa) : specific enthalpy h1 = 452kJ/kg

Specific entropy s1 = 1.8265 kJ/kg.k

Exit condition (35 C and 1 MPa) : specific enthalpy h2 =249.05 kJ/kg

Specific entropy s2 = 1.1665 kJ/kg.k

For water table :

At inlet condition (15 C and 300kPa) : specific enthalpy hw1 = 63.269 kJ/kg

Specific entropy sw1 = 0.2244 kJ/kg.k

At final condition (25C and 300kPa) : specific enthalpy hw2 = 105.11 kJ/kg

specific entropy sw2 = 0.3671 kj/kg.k

a) applying energy balance for condenser: at steady state,

heat transfer by refrigerant = heat transfer by water

mr*(h1 - h2) = mw*(hw2 - hw1)

6 kg/min * (452 - 249.05)kJ/kg = mw*(105.11 - 63.269) kJ/kg

mw = 29 kg/min

mass flow rate of cooling water mw = 29 kg/min

b)  heat transfer rate from refrigerant to the water,

Q = mr*(h1 - h2) = 6 kg/min * (452 - 249.5) = 1217.7 kJ/min * (1min/60s) = 20.295 kJ/s = 20.295 kW

Q = 20.295 kW

heat transfer from refrigerant to water Q = 20.295 kW

c) rate of entropy production within heat exchanger :

mass flow rate mw = 6 kg/min = 0.1 kg/s

Entropy change due to refrigerant R-134a, (ΔS)r= mw*(s2 - s1)

(ΔS)r = 0.1kg/s * (1.1665 - 1.8265)kj/kg.k = -0.066 kW/K

Entropy change due to water ΔSw = mw*(sw2 - sw1) = (29kg/min)*(1min/60s) * ( 0.3671 - 0.2244)kJ/kg.K = 0.0689 kW/K

Rate of entropy production within heat exchanger,

ΔSgen = ΔSr + ΔSw = - 0.066 + 0.0689 = 0.0029 kW/K

entropy generation ΔSgen = 0.0029 kW/K


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