jamesk
04-08-2013, 09:28 PM
Good day chaps, :)
I am using the Coolpak software to help in creating a cycle for me to analyse.
I have come up with this set of temps. press. and enthalpy values:
10609
In the picture at State point 1: temperature= 0 degs C; pressure= 292.9KPa; Enthalpy= 246.6 KJ/Kg.
My question:
If I look at the saturation tables for R-134a,for 0 degs C, and a corresponding pressure of 292. KPa I get an enthalpy of 250.45 KJ/KG, (saturated vapour condition) and not 246.9 KJ/Kg as in the diagram above.
Can anyone figure out why? I am taking the State point as a saturated refrigeratant, and I have set all variables to either 1 for efficiencies, or 0 for losses, to make the cycle as ideal as possible.
Anyone good with CoolPak who is kind enough to help?
Thanks,
James
I am using the Coolpak software to help in creating a cycle for me to analyse.
I have come up with this set of temps. press. and enthalpy values:
10609
In the picture at State point 1: temperature= 0 degs C; pressure= 292.9KPa; Enthalpy= 246.6 KJ/Kg.
My question:
If I look at the saturation tables for R-134a,for 0 degs C, and a corresponding pressure of 292. KPa I get an enthalpy of 250.45 KJ/KG, (saturated vapour condition) and not 246.9 KJ/Kg as in the diagram above.
Can anyone figure out why? I am taking the State point as a saturated refrigeratant, and I have set all variables to either 1 for efficiencies, or 0 for losses, to make the cycle as ideal as possible.
Anyone good with CoolPak who is kind enough to help?
Thanks,
James