Sample Problem

Problem Statement:

A window air conditioner is used to cool a house in the summer, during a day when the outside temperature is a sizzling 98 degrees Fahrenheit. Refrigerant enters the evaporator as saturated liquid and leaves as a saturated vapor at 0.01MPa before it enters the compressor at 100⁰C, which causes a pressure change of 10-fold in the fluid. About 15 kJ of energy is then removed from the refrigerant as it passes through the condenser. The expansion valve reduces the pressure of the fluid to 0.01 MPa before the cycle is repeated. What is the estimated efficiency of this air conditioner and how does it compare to typical efficiencies?

Diagram:	Assumptions:
(Source: article on "Air Conditioners" taken from http://www.howstuffworks.com/)	Steady State process Molar basis No pressure change across the evaporator and the condenser The refrigerant evaporates from a saturated liquid to a saturated vapor The compressor is reversible and adiabatic The refrigerant is water (for ease of thermodynamic data acquisition) One hour operation for efficiency calculation

Equations:
	Compressor - , Evaporator - Thermal Efficiency -

Values (given & obtainable from steam tables):
	Pi,evap = Pf,evap = 0.01 MPa Pi,compressor = 0.01 MPa; Pf,compressor = 0.01*10 = 0.1MPa Ti,compressor = 1000C Tsat @ Pi = 45.81 0C DHvap = 2392.05 kJ/kg @ given Pi Si = 8.4489 kJ/kg-K @ given Pi and Ti of compressor Hi = 2687.5 kJ/kg @ given Pi and Ti of compressor

Solution:

Evaporator - = 2443.86 kJ/kg
Compressor - = 8.4489 kJ/kg-K

from interpolation of steam tables, Hf,compressor = 3216.18 kJ/kg

= -[Hf,compressor - Hi,compressor]
Ws = -(3216.18 - 2687.5) kJ/kg
Ws = -528.68 kJ/kg

thermal efficiency = -(-528.68 kJ/kg)/(2443.86 kJ/kg) = 0.22 = 22%

Comparison:

Given a window air conditioner unit's typical EER rating of 8.5,
Converting EER of 8.5 to thermal efficiency:
and
Thus EER of 8.5 is approximately 40%

Assessment:

Due to the many assumptions and "ideal cases" dealt with in this sample problem, it is unlikely that the 22% calculated is truly a valid value. However, it may be noted that coolants used in air conditioners are not water but instead hydroflurocarbons or freons that result in higher efficiencies due to a combination of their volatility and enthalpies of vaporization.

| | | | | | | |

Solution:
	Evaporator - = 2443.86 kJ/kg Compressor - = 8.4489 kJ/kg-K
		from interpolation of steam tables, Hf,compressor = 3216.18 kJ/kg
		= -[Hf,compressor - Hi,compressor] Ws = -(3216.18 - 2687.5) kJ/kg Ws = -528.68 kJ/kg
	thermal efficiency = -(-528.68 kJ/kg)/(2443.86 kJ/kg) = 0.22 = 22%

Comparison:
	Given a window air conditioner unit's typical EER rating of 8.5, Converting EER of 8.5 to thermal efficiency: and Thus EER of 8.5 is approximately 40%