00:01
All right guys, in order to answer this question, it is required to determine the airflow in an automobile air conditioner.
00:08
The system uses r134a as a working fluid.
00:12
The evaporator operates at 100 kioskal gauge pressure and the condenser operate at 1 .5 megapascal gauge pressure.
00:21
The compressor consumes 6 kilowatts of electrical power operating at 85 % isontropic efficiency.
00:29
The air enters the system at 25 degrees centigrade at relative humidity of 60 % and leaves at 8 degrees centigrade with relative humidity of 90%.
00:42
So the absolute pressure of the elevators and the condenser is pe is equal to 100 kioskal plus 100 kilopascal and that is equal to 200 kilopascal.
00:58
So pc is equal to 100 kilopascal plus 1 .5 megapascal and that is equal to 1 ,600 kilo -pascal.
01:14
So the state just before the condenser is state 1 and can be read from figure a14, that is h1 is equal to 400 kilojoules per kg.
01:25
And s1 is equal to 1 .74 kjoules per kg.
01:33
Now the state 2 is the state the compressor would compress if the process was isentropic.
01:42
It has the same entropy as the 0 .1 and has the pressure pc from the same table as we read before.
01:50
So h2 dash is equal to 445 kilojoules per kg.
01:59
Now the real specific enthalpy added to the fluid that is h2 minus h1 is equal to h2 dash minus h1 divided by eta isentropic...