CoolPak software Help

[jamesk]

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

[xxargs]

I think coolpack not using exactly same algorithm and reference point as your papers.

ASHRAEA seems using reference point 0 for enthalpy and entropy for liquid boiling R134a on -40 degree Celcius (-40 F) and giving value is close to your 250.45 kJ/kg at saturated vapor and 0 degree Celsius

coolpack seems have reference point around -37 degree Celsius in some reason

Absolute value is not important, important in calculation is differences on enthalpy (and entropy) in each step on cooling cycle and so far i not seeing big differences compare if you make same cycle on ex. refprop from NIST.

check:

0 C, 0.2929 MPa -> enthalpy (vapor) = 246.6 kJ/kg for coolpack
60.4 C, 1.696 MPa -> enthalpy (liquid) = 136.6 kJ/kg for coolpack

differences are 246,6 - 136,6 = 110 kJ/kg

Refprop - standard setting R134a (reference point is _not_ -40 degree Celsius) :

0 C, 0.2928 MPa -> enthalpy (vapor) = 398.6 kJ/kg for refprop
60.4 C, 1.6978 MPa -> enthalpy (liquid) = 288.16 kJ/kg for refprop

differences 398.6 - 288.16 = 110.4 kJ/kg

You see some differences but not to much to ruin calculate even if programs using complete different reference points and not exact same algorithm to calculate values and possible rounding error for pressure, temperature, enthalpy, entropy etc. on R134a.

[jamesk]

Thank you very very much for your time.

I have carried out the same calculation with my R-134a tables and recieved the same difference of 0.4 kJ/kg, which made me confident to continue.

My tables state: "enthalpy and entropy values of saturated liq. are set to zero at -40deg C (and -40def F)."