Question

An inventor reports having a compressor that receives R-134a at -20C and delivers vapour at 1MPa and 40C. If the compression is adiabatic, does this process violate the second law?

Answer 1

@Jemurray3 can you explain this please

Answer 2

im thinking since its adiabatic, there is no heat lost, but what does that mean about entropy does that mean it has to be isotropic, but then temperature cannot go up?

Answer 3

isentropic*

Answer 4

the entropy i got from tables for 40C and 1MPA = 0.9179
and from the tables again 20 C and 0.5mpa = entropy = 0.9383 kj/kg.K
and the next pressure up at 0.6MPA the sat temperature is at 21.55 degrees

Answer 5

not sure if that helps though

Answer 6

Adiabatic processes are not necessarily isentropic. What tables are you using?

Answer 7

superheated refrigerant 134-a table

Answer 8

and yes that what i was thinking too, it didnt say *reversible* and adiabatic this time so

Answer 9

how can we know if a process violates the 2nd law or not

Answer 10

1 more method i was thinking about is finding the highest entropy possible if the 134 -a can have at 20 degrees at the same volume of the 1 MPA , 40 degrees

Answer 11

Well, what does the 2nd law say?

Answer 12

that entropy always increases or is 0 between state 1 and 2

Answer 13

or S = k Ln (N)
where N is the total number of possible microstates

Answer 14

derived from S= Q/T

Answer 15

Okay, so what you need to do is find the entropy of the system before and after the compression. I suggest you try this: http://www.peacesoftware.de/einigewerte/r134a_e.html

Answer 16

how can i find the entropy of state 1 though, they only give u temperature and u just know that there is no heat flow in the system

Answer 17

I'm not sure. You need more information to determine the entropy so perhaps you should assume it's at atmospheric pressure.

Answer 18

its adiabatic so there is no interaction between environment or anything no heat lost no heat gained that is the hint

Answer 19

ok forget this question ill comeback to it ill work on the calculation questions