When to Use Temperature or Pressure for R134-a Tables?

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When using R134-a tables, both temperature and pressure are necessary to accurately determine the state of the refrigerant. For the given problem, the intersection of the temperature at -24ºC and pressure at 100kPa should be used to establish the initial state. The isentropic process can then be followed to find the final state at 800kPa. It is noted that R134-a at -24ºC and 100kPa is not technically saturated, so using temperature values directly is recommended. The mass flow rate can be calculated using the flow rate divided by the specific volume at the vapor stage.
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R134-a Tables Use T or P??

Using the Saturated R134-a tables, I'm a little confused as to when I should look up values according to temperature or pressure.

In this particular problem:
Refrigerant-134a enters a compressor at 100kPa and -24ºC with a flow rate of 1.35 cfm and leaves at 800kPa and 60ºC. Determine the mass flow rate of R-134a and the power input to the compressor.

The equations I plan to use are: mdot = Q/v. And I think that to find power I can use
P = mdot * (u2 - u1)

The problem is that I don't know if I should look at the values according to T1=-24C or P1 = 100kPa (same for T2). Can anyone shed any light on this?
 
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You need two properies to set the state of the fluid, so you need to use both temperature and pressure. In order to set the state, you have to find the intersection of two property lines that intersect somewhere in the T_S or P_V diagram; in your case for state (1) you can find the intersection of the T=-24 C line (or near to it) and 100 kPa line.

Once you've set the state, you would follow an isentropic line up to the higher pressure and this would give you State (2) (the intersection of an isentropic line through state (1) and the 800 kPa line). You have both temperature and pressure for state (2), which means you'll also be able to find the compressor's efficiency if state (2) is not isentropic w.r.t. state (1).
 


Mech_Engineer said:
You need two properies to set the state of the fluid, so you need to use both temperature and pressure. In order to set the state, you have to find the intersection of two property lines that intersect somewhere in the T_S or P_V diagram; in your case for state (1) you can find the intersection of the T=-24 C line (or near to it) and 100 kPa line.

Once you've set the state, you would follow an isentropic line up to the higher pressure and this would give you State (2) (the intersection of an isentropic line through state (1) and the 800 kPa line). You have both temperature and pressure for state (2), which means you'll also be able to find the compressor's efficiency if state (2) is not isentropic w.r.t. state (1).

I see what you mean when you say to find the intersection of T and P. It makes sense. But we are only given the tables. Would I have to interpolate the two values (vf @ T=-24C and vf @ P=100kPa)?

Also, when you say follow the isentropic line - would I use the liquid or vapor entropy value?
 


Ah, sorry I missed that you are using saturation tables. In the case of saturated fluid, the saturation state is technically the second property setting the state of your fluid. In a refrigeration system, the compressor compresses vapor rather than liquid (liquid would be a pump, and not a refrigeration cycle), so I would say you you need to look at saturated vapor properties.

In the case of R-134a, saturated vapor/liquid at -24 C is at 1.1160 bar (111.6 kPa) so you can really look up properties using either value, I'm guessing there was some rounding error in the problem statement because R-134a at -24 C and 100 kPa is not techically saturated fluid any more. I wouldn't bother interpolating, just use temperature.
 


Mech_Engineer said:
Ah, sorry I missed that you are using saturation tables. In the case of saturated fluid, the saturation state is technically the second property setting the state of your fluid. In a refrigeration system, the compressor compresses vapor rather than liquid (liquid would be a pump, and not a refrigeration cycle), so I would say you you need to look at saturated vapor properties.

In the case of R-134a, saturated vapor/liquid at -24 C is at 1.1160 bar (111.6 kPa) so you can really look up properties using either value, I'm guessing there was some rounding error in the problem statement because R-134a at -24 C and 100 kPa is not techically saturated fluid any more. I wouldn't bother interpolating, just use temperature.

Awesome! Thanks. Also, just to double check - I can find mdot by using:
mdot = Flow Rate / (Specific Volume-vapor stage 1), correct?
 


Correct, and you can see it in a units analysis:

Flow rate (SI): m^3/s
Specific Volume: m^3/kg

(Flow Rate) / (Specific Volume) = kg/s
 
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