Problem 4.9 - A Home Air Conditioner & Hot Water Heater
We wish to do a preliminary thermodynamic evaluation
of a 500W input power home air conditioner/hot water heater system
using refrigerant R134a. Consider the following system flow diagram
This unique combined air conditioning / hot
water heating system is designed to absorb heat from the 32°C
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 to 15°C at the outlet of 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 adiabatic.
Plot the four processes on the P-h diagram
provided below and use the R134a refrigerant property tables in
order to determine the following:
- Determine the mass flow rate of the refrigerant
R134a [0.0128 kg/s]
- Determine the heat rejected by the condenser
[-2.02 kW]. Assuming that all this heat is absorbed by the
water in the hot water tank, determine the time taken for 100
liters of water at 32°C to reach the required temperature
of 50°C [62 minutes].
- Determine the heat absorbed by the evaporator
[1.516 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 inlet air at station (5) in order reduce
the air temperature from 32°C to 15°C at the outlet of
the duct [0.089 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 = 4.03].
- 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 = 3.03].
- Is this a suitable and practical system for
hot climates? Discuss.
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Engineering Thermodynamics by Israel Urieli is licensed under a
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