log p/h question, flow rate

[Peter_1]

Cycle R134a, TE =-10°C, TC = 40°C, no SH nor SC, needed 2 kW cooling capacity.

Q: what flow do we need?

-10°C liquid = 186.93 kJ/kg
-10°C gas = 391.32 kJ/Kg
Both saturated from tables

Added in evaporator = 391.32 - 186.93 = 204.39 kJ/kg
Isentropic compression gives us 425 kJ/kg, so 33.68 (425-391.32) was added during compression.

Flow for 2 kW = 2000 W = 2 kJ/s, so we need 2kJ/s
Evaporator takes 204,39 kJ/kg
kJ/s x kg/s = kg/s
2 x 1/204.39 = 0.009785 kg/s or 9.785 g/s flow

Coolpack gives me 14.73 g/s and this varies with altering TC

Where's my fault ?

Then flow rate x heat of compression= needed compression power (kW) kg/s x kJ/kg = kJ/s = kW
0.009785 x 33.68 = 0.329 kW
Coolpack 0.496 kW (due to my erratic flowrate of course).

[mad fridgie]

Cycle R134a, TE =-10°C, TC = 40°C, no SH nor SC, needed 2 kW cooling capacity.

Q: what flow do we need?

-10°C liquid = 186.93 kJ/kg
-10°C gas = 391.32 kJ/Kg
Both saturated from tables

Added in evaporator = 391.32 - 186.93 = 204.39 kJ/kg
Isentropic compression gives us 425 kJ/kg, so 33.68 (425-391.32) was added during compression.

Flow 2 kW = 2000 W = 2 kJ/s, so we need 2kJ/s
Evaporator takes 204,39 kJ/kg
kJ/s x kg/s = kg/s
2 x 1/204.39 = 0.009785 kg/s or 9.785 g/s flow

Coolpack gives me 14.73 g/s and this varies with altering TC

Where's my fault ?

Then flow rate x heat of compression= needed compression power (kW) kg/s x kJ/kg = kJ/s = kW
0.009785 x 33.68 = 0.329 kW
Coolpack 0.496 kW (due to my erratic flowrate of course).

Flash Gas!

[mad fridgie]

your liquid is at 40C, needs to be -10C (so a percentage of flow is used to cool itself) then your calcs are right, for evap duty
give those scales a nudge, lol

[desA]

Hope this helps, Peter.

x roundoff, or slight table differences could account for the tiny difference in computed result.

[Chef]

-10°C liquid = 186.93 kJ/kg
-10°C gas = 391.32 kJ/Kg
Both saturated from tables

Added in evaporator = 391.32 - 186.93 = 204.39 kJ/kg
.

Peter - If you use the sat tables for -10C you get h4=186.9 but you should use h4=h3 which is at 40C then h4=256Kj/Kg

That gives you 391-256=135

Then 2/135=O.148Kg/s which seems to agree with coolpack.

Probaly better to use a PH diagram than a sat table.

Chef

[Peter_1]

Thanks to all, clarified now. At least 4 schoolbooks needs to be rewritten after this thread.
3 used a fault from the 1st in their books and so this fault is spread widely.
Many, me included has learn it the wrong way. I became confused because I found that something had to be wrong but the books stated something different.
Thanks guys, never too old to learn something.

[desA]

Great, Peter. Glad to see that you're on your way forwards again.

[Peter_1]

Continuing on this thread, another problem..here we go again, hope others can benefit from this

Cycle, R404a,
TE-30°C
TC 40°C
SH 7K
SC 5K
Isentropic compression 0.8
No SGHX
2 kW

Cycle specification in Coolpack Statepoints gives for point 6 (entrance evaporator) 104.4 kJ/kg and for point (or 8 or 1) 207.4 kJ/kg

If you draw this cycle on a log p/h under Refrigeration Utilities, you then get 254 kJ/kg and 355 kJ/kg along the enthalpie axis.

Where these differences comes from?

Calculating the massflow via Coolpack gives 0.01942 kg/s where the calculation via the drawn log p/h and the values read in it gives 0.019924 kg/s

Also, value x in Statepoints is 0.49 where in the drawn cycle it is very close to 0.5 (enlarged view)
Using this different figures gives also not forgiving errors when calculating compressor performance and condenser capacity.

My question, where's my fault regarding mass flow?

[desA]

Very interesting observation. I suspect that Coolpack is using different refrigerant tables in the different modules, for some odd reason.
With different tables, sometimes the zero reference points may differ.

I would however expect the value (hsup-h2phase) = h1-h6 to be of similar value using either method. You have found a slight difference here as well. Perhaps Coolpack has a few bugs?

You may want to send your findings to the developer of Coolpack v1.49 & find out exactly what gives.

[Peter_1]

Well, I calculated it once the other way.

QEvaporator = m.(1-x).(hg-hf)

If I use the mass flow found in Refrigeration Utilities, 0.01942 kg/s, for a capacity of 2 kW and use for hg 350.26 kJ/kg and for hl 159.17 kJ/kg (tables), then I can extract x which gives me 0.46.
Coolpack gives a calculated x of 0.49

I know now - I hope it in fact - that this is a small fault in Coolpack and that it is not me.

Hey DesA, thanks to you, I was yesterday evening better prepared in classes. Thanks.

[desA]

^ An absolute pleasure Peter. Any time.

[Chef]

Very interesting observation. I suspect that Coolpack is using different refrigerant tables in the different modules, for some odd reason.
With different tables, sometimes the zero reference points may differ.

desA You are right that coolpack uses different references and the IIR value is the one normally used by the refrigeration industry. It is correct in the utilities package but in coolpack it somehow reverts back to the default value and so gives the wrong enthalpy. R134a and R22 and other gases are correct but some blends are reset to the default.

Care must be taken it seems when using coolpack.

This is from the manual.
IIR - the value of specific enthalpy is set to 200 kJ/kg and the value of specific entropy is set to 1.0 kJ/kg-K for saturated liquid at 0°C (273.15 K). This is the standard reference state for the International Institute of Refrigeration. Note that this option is not applicable to fluids for which the critical temperature is less than 0°C. In this case, the reference will be reset to the default value.


Chef

[desA]

Thanks Chef. An excellent review.

[Peter_1]

Send a message to Coolpack, will post their reaction here

[NoNickName]

Peter, would you like to try the cycle simulation by Solvay (software solkane 7.0).

[Peter_1]

Didn't know they had something similar.
Will have a look. Just back (1:30 midnight) after a concert of U2 in Brussels. Very good concert