00:01
In this question it is given that during a constant volume process, refrigerant r -40010a is heated from a temperature of 20 degrees centigrade to 200 degrees centigrade and the initial pressure was 300 kioskal.
00:21
We are required to find the change in specific entropy by using table b .4 and then we need to find change in specific entropy by using table b .4 and then we need to find change in specific entropy by using ideal gas with cv is equal to 0 .695 kilojoules per kilogram kelvin.
00:42
So let's see how to solve this question.
00:45
Zaffer table b .4 .2.
00:53
So from this table the value of entropy s1 corresponding to temperature 20 degree centigrade and pressure 300 kilo -pascal is equals to 1 .24.
01:05
85 kilojoule per kilogram kelvin and the value of specific volume v1 is equals to 0 .10720 2020 meter cube per kilogram.
01:20
Now we can observe that the specific volume at a state 1 is close to the specific volume of refrigerant r410a at the pressure 500 kilo -pascal and at temperature 200 degrees centigrade.
01:36
Therefore, from the same table, the value of entropy s2 corresponding to temperature 200 centigrade and 500 kylpacal pressure is equals to 1 .6413 kilojoules per kilogram kelvin and the value of specific volume veto is equals to 0 .10714 meter cube per kilogram.
02:08
And now let's find the change in specific entropy delta s which is equals to s2 minus s1.
02:18
Now substitute all the value so we will have change in specific entropy delta s is equal to s2 that means 1 .6413 minus s1 that means 1 .2485.
02:32
So when we further calculate we get change in specific entropy delta as is equal to 0 .6 .13...