Problem 4.9 - A Home Air Conditioner & Hot Water Heater
We wish to do a preliminary thermodynamic evaluation
of a 500W input power home heat pump system as applied to summertime
use for both hot water heating to 50°C, and space cooling
(air conditioning), and thus maintain the inside home temperature
at a comfortable 20°C..
.
This unique combined air conditioning / hot
water heating system is designed to absorb heat from the air flowing
through the insulated duct in order to pump heat into the hot
water heating tank. The fan provides enough air flow over the
evaporator to cool the air by 10°C as it passes through the
duct, and the hot water is heated to a maximum of 50°C. In
this analysis we neglect the power provided to the fan. We also
assume that both the duct and the hot water tank are externally
adiabatic.
Carefully draw the complete cycle above on
the pressure-enthalpy [P-h] diagram provided below, showing
clearly all five processes (1) - (2) - (3) -(4) - (5) -(1). Using
the conditions shown on the diagram above and values obtained
from the R134a tables:
- Determine the enthalpy values at all five
stations [kJ/kg], and indicate these values on the P-h
diagram.
- Determine the mass flow rate of the refrigerant
R134a [0.0133 kg/s]
- Determine the heat rejected by the condenser
[-2.09 kW]. Assuming that all this heat is absorbed by the
water in the hot water tank, determine the time taken for 150
liters of water at 30°C to reach the required temperature
of 50°C [1 hr 40 min]. (Note:
The hot water heater is not a flow system, thus we need to first
evaluate the energy required to heat the water [12540 kJ]. This
section is solved as shown below)
- Determine the heat power absorbed by the
refrigerant in the evaporator [2.04
kW]. Assuming that all this heat is absorbed
from the air in the duct and neglecting the fan power, determine
the required mass flow rate of the in order reduce the air temperature
by 10°C while passing through the duct [0.204 kg/s].
- Determine the Coefficient of Performance
of the hot water heater (COPHW) (defined as the heat
rejected by the condenser divided by the work done on the compressor)
[COPHW = 4.17].
- Determine the Coefficient of Performance
of the air conditioner (COPAC) (defined as the heat
absorbed by the evaporator divided by the work done on the compressor)
[COPAC = 4.07].
- If we bypass the outside subcooler (State
(4) becomes saturated liquid as in State (3)) determine the change
in Coefficient of Performance of the air conditioner evaluated
above. Indicate this change on the P-h diagram and discuss
the relevance of the outside subcooling section in this system.
[COPAC reduced to 3.17]
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Engineering Thermodynamics by Israel Urieli is licensed under a
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