1. 1 ISAT 413 - Module III: Building Energy Efficiency Topic 9:Advanced Heat Pumps  Heat Pump Water Heater  Heat Recovery from a Swimming Pool Using a Heat Pump  Mechanical Vapor Re-Compression  Demonstration of a 2-RT Heat Pump

2. 2  Heat Pump Water Heater — Ground Coupled A small nursing home is to be heated by a low-thermal mass under-floor hot water heating system. Two ways of providing the heat input are being considered. (a) An electric heat pump with a low temperature source provided by ground water; the ground water is pump to evaporator and returned by gravity to a rubble-filled soak way. (b) A gas-fired condensing boiler. The required design heating load is 100 kW and the under- floor system has an average water temperature of 35 o C. The annual hours of use of the heating system at the average conditions may be taken as 4000. Using the data below, neglecting heat losses and pumping power, make an initial estimate of the time after which the heat pump system will start to make a net saving neglecting all annual costs other than the fuel costs.

3. 3 Ground Coupled HPWH — (continued) Data: Heat Pump = Vapor compression with refrigerant R-12 Minimum Temperature Difference for HXs = 10 K Overall Compressor Efficiency = 0.86 Average Temperature of Ground Water = 10 o C Compressor Inlet = Dry Saturated Vapor Condenser Sub-cooling = 0 K Compressor Isentropic efficiency = 1.0 Capital Cost of the Complete System = L 21,000 Cost of Electricity = 3.7 p/kW Condensing Boiler Efficiency = 90% capital Cost of Boiler = L 5,500 Cost of gas = 1.2 p/kW

4. 4 Ground Coupled HPWH — (continued)

5. 5 {Heating load & energy balance} Q_dot_h=100 Q_dot_h=m_dot_r*(h_2-h_3) Q_dot_c=m_dot_r*(h_1-h_4) W_dot_isn=Q_dot_h-Q_dot_c eta=0.86 W_dot_c=W_dot_isn/eta {Savings} C_run_hp=W_dot_c*3.7/100 C_run_Boiler=(Q_dot_h*1.2/100)/0.9 21000+C_run_hp*t=5500+C_run_Boiler*t y=t/(4000)

6. 6 Ground Coupled HPWH — (continued) {Solutions} C_run_Boiler=1.333 [EP/hr] C_run_hp=0.737 [EP/hr] h_1=187.53 [kJ/kg]h_2=209.81 [kJ/kg] h_3=79.70 [kJ/kg]h_4=79.70 [kJ/kg] m_dot_r=0.76853 [kg/s]P_1=3.083 [bar] P_2=10.835 [bar]P_3=10.835 [bar] Q_dot_c=83 [W]Q_dot_h=100 [W] s_1=0.696 [kJ/kg-K]s_2=0.696 [kJ/kg-K] t=25991 [hr]T_1=0.00 [C] T_3=45.00 [C]W_dot_c=20 [W] W_dot_isn=17 [W]y=6.50 [yr]

7. 7  Heat Recovery from a Swimming Pool Using a Heat Pump

8. 8  Mechanical Vapor Re-Compression z The use of heat pump for heat recovery in cases where vapor is continuously evaporated is becoming well-established. The vapor evaporated from the process is compressed to a higher pressure and then condensed providing a heating effect; the system therefore is known as Mechanical Vapor Re- compression, MVR. Examples of MVR applied to evaporation, distillation, and drying are shown below:

9. 9 An Example of MVR (Mechanical Vapor Re-Compression) z A rotary steam drier similar to that shown in the following Figure operates with 1 kg/s of steam entering the drier at 6 bar and condensing with no undercooling. The condensate is throttled to 0.7 bar before passing through the heat exchanger where it is evaporated leaving as a dry saturated vapor. Assuming that the isentropic efficiency of the steam compressor is 0.65 and that the combined mechanical and electrical efficiency is 0.9, calculate: (i) the overall coefficient of performance of the heat pump; (ii) the rate of cost saving compared with providing the heat for the drier with a gas boiler of 80% efficiency; take the cost of electricity as 4 p/kWh and of gas as 1.3 p/kW-h.

10. 10 An Example of MVR (continued)

11. 11 {Power required and system COP} W_c=m_dot_s*(h_2-h_1) W_mech=W_c/0.9 {Combined mechanical and electrical efficiency is 0.9} COP_MVR=Q_dot_h/W_mech {Costs/Savings} C_run_MVR=W_mech*4/100 C_run_boiler=(Q_dot_h/0.8)*1.3/100 {Gas boiler efficiency is 80%} S_%=(C_run_boiler-C_run_MVR)/C_run_boiler*100 An Example of MVR (continued)

12. 12 {Solutions} COP_MVR=3.402 C_run_boiler=43.945 [NP/hr] C_run_MVR=31.795 [NP/hr] eta_isn=0.650h_1=2660 [kJ/kg] h_2=3375 [kJ/kg]h_2s=3125 [kJ/kg] h_3=670.618 [kJ/kg]m_dot_s=1.0 [kg/s] P_1=0.7 [bar]P_2=6.0 [bar] Q_dot_h=2704.3 [kW]S_%=27.6 [%] s_1=7.479 [kJ/kg-K]s_2s=7.479 [kJ/kg-K] W_c=715.4 [kW]W_mech=794.9 [kW] An Example of MVR (conclusion)