A swimming pool hall is heated by the system shown below A v

A swimming pool hall is heated by the system shown below. A volume flow rate of air of 15,000 cubic meters perhour leaves the pool hall at 28 degrees Celcius, with a percentage saturation of 70%, passes across a plate heatexchanger and then the coils of the heat pump evaporator, leaving at 17degrees Celcius. The fresh air enters at 5 degrees Celcius,percentage saturation 80%, passes across the plate heat exchanger and then the coils of the heatpump condenser, entering the pool hall at 31 degrees Celcius. Using the data given, calculate:

(i) the mass flow rate of refrigerant required in the heat pump circuit;

(ii) the power input required to the electric motor;

(iii) the percentage cost saving in using the energy recovery system instead of electric heating of the fresh air to the pool hall temperature;

(iv) Sketch on the psychrometric chart with all the salient points marked.

Data
Refrigerant R-12: evaporator pressure, 4.233 bar; condenser pressure, 8.477 bar; vapor leavesevaporator dry saturated; vapor leaves compressor superheated at 40 degrees Celcius; liquid leaves condenserin the saturated state; combined mechanical and electrical efficiency of motor, 90%. Effectivenessof plate heat exchanger, 0.8; air enters evaporator coils with a percentage saturation of 90%.

Air Air out Evaporator Plate heat 3 exchanger Throttle valve Compressor Air extract Swimming pool hall X Condenser Supply air

Solution

i) From psychrometric chart, h₁ = 73.2kJ/kg, h₆ = 44.5kJ/kg, h₅ = 90kJ/kg, h₃ = 16 kJ/kg

Total heat lost by air = 73.2-44.5 = 28.7 kJ/kg

Total heat gained by air = 90-16 = 74 kJ/kg

74- 28.7 = (h_b-h_c)+(h_a-h_d)

Where, h_a=Enthalpy of R12 at compressor inlet = 198.762 kJ/kg

h_b=Enthalpy of R12 at compressor outlet=205.924 kJ/kg

h_c=Enthalpy of R12 at condenser outlet

h_d=Enthalpy of R12 at evaporator inlet

h_c=h_d

Therefore, 45.3 = 205.924+198.762-2*h_c

h_c = -19.069 kJ/kg

Let x be heat echanged in plate heat exchanger

74 = 0.8x -(205.924 -(-19.069))

x = 373.74125 kJ/kg

Now, 5.10417*(90-16) = m_r*0.4055*(205.924 -(-19.069) - (5.10417*373.74125) (m_r = mass of refrigerant)

m_r = 25.049 kg/s

ii) power input required to motor = h_b - h_a =25.049*( 205.924-198.762 )= 1179.40175/0.9 = 199.335 KW

iii) The percentage cost saving in using the energy recovery system instead of electric heating of the fresh air to the pool hall temperature = 90.24%