Calculate kW of Compressor through enthalpy 1

[eltjo]

Hi all,
In order to check a compressor offer I want to calculate the kW needed to raise the pressure.
In short, I do not trust the stated 480 kW of Horsepower.

The compressor needs to compress 10.000 kg/hour CO2 @ 55 C saturated with water (about 670 kg/hour of water --> Total flow is 10.680 kg/hour) from 1 bar(a) to 2,75 bar(a) @ 180 C (Compressor 1)

Then the flow is cooled to 50 C so approx. 470 kg/hour of water is knocked-out making the total flow 10.200 kg/hour.


And then compression from 2,5 bar(a) to 7,2 bar(a) @ 180 C (Compressor 2)

For Compressor 1:
If I see it as a blackbox with input and output the Enthalpy is reased and to raise the Enthalpy 365 kW is needed. (1.314.360 kJ/hour devided by 3600 --> 365 kW)

For Compressor 2: 345 kW

In total the Enthalpy raise is 710 kW, which is far more than the stated 480 kW. Can anyone tell me what I am missing here.

I do also have a centifigal compressor calculation sheet, but this is a new type of compressor which claims to have less energy needs so I wanted to use a complete other approach.

Greetz,
Eltjo (From the Netherlands)

 

[dcasto]

I ran numbers here and I get the same 710 KW for the system. When I back into the polytropic efficency I get 66% for the first stage and 60% for the second.

To achieve 480 KW, the polytropic Eff would be about 89%. That would only be achievable with a very tightly build, huge valve area reciprocating compressor.

I might call foul on this myself.

 

[25362]

I didn't take the time to do a full calculation, but it seems to me the outlet temperatures for an adiabatic compression are too high, probably by some +/- 50 deg C. Correct me if I'm wrong.

 

[eltjo]

Thanks for the quick answers!

@25362: As it is a new design for a compressor I can't tell if the outlet temperatures are too high. That's why I am threating it as a black box and use data provided to me.

@dcasto: Since compressor calculations are not my specialty this might seem a stupid question...(Excuse in advance) but if the system is reased 710 kW, this energy must come from somewhere, how can I get 480 kW with polytropic Eff 89%? Where does the other 230 kW comes from?

 

[25362]

By adjusting the temperatures to values obtained by adiabatic compression, the kW obtained from enthalpies drop by about 60%.

 

[25362]

I suggest you recheck discharge temperatures. They may be the result of a typing error.

 

[eltjo]

@ 25362 You mean that if the outlet temperatures are adjusted downwards the enthalpy will also drops?

I just did a "micro-calculation" for an isothermal compression (n=1) (This is the most optimistic calculation right?)

Pisoth = Qme * R * T1 * LN (P2/P1)
Where:
Qme = 10.000 kg/hour --> 2,78 kg/s
R = 287 J/(kg * K)
T1 = 323, 15 K
P1 = 1 bar(a)
P2 = 7,2 bar(a)

Pisoth = 508 kW

This is without correction to CO2 which is heavier and without the saturated water.

 

[25362]

Yes, by dropping the temperatures to, say 140 deg C the enthalpy differences drop to about 60%.

In your formula R = 8314.5/44 = 189, not 287 J/(kg*K).

 

[25362]

Apparently, you have used R for air, not for carbon dioxide.

 

[eltjo]

Yes, thank you for the notification.

Ofcourse the R is not correct because there needs to be a correction for the water too.

I am going to email the supplier. For information I will post their response here.

 

[CJKruger]

eltjo, there are at least 3 methods in frequent use nowadays:

Method 1 - Use a simulator. They do the calc in two steps.
Step 1. Starting at known inlet conditions, they do an isentropic flash to the outlet pressure. Then they calc the outlet enthalpy based on the outlet pressure and entropy.
Step 2. They now calc the actual outlet enthalpy by dividing the enthalpy change in step 1 with the adiabatic efficiency. Lastly they recalc all outlet properties (including work required), based on the new outlet enthalpy and pressure.

Method 2 - Use the well known compressure head/power equation based on adiabatic efficiencies.

Method 3 - Use the well known compressor head/power equation based on polytropic efficiencies.

For recips you have to check the outlet temp (per API std) and for centrifugals you have to make sure your head is realistic.

In all cases you motor/turbine will supply all the power requires, including efficiencies not used above (seal losses, gear losses, etc).

 

[eltjo]

Dear people,

The answer was indeed as stated above the discharge temperature. They say it is 150 C instead of 180 C. This wil drop the enhalpy from 710 kW to 415 kW.

The overall efficiency must then be 92% to get the 450 kW. But I will look into that.

Thanx for al the answers!

 

[dcasto]

Yes, 92% is to good to be true.

 

[25362]

I wouldn't be surprised if the correct again their discharge temperature to 140 deg C.

 

[25362]

Using the conventional formula for adiabatic compression:

kW_ad=[k/(k-1)] WRT₁[p^(k-1)/k – 1]

gives us an idea of the kW consumed in compression.

R=8314÷MW, and taking MW = 40 for the 1st stage and 43 for the 2nd stage, assuming k =1.3 for both cases, W=kg/s, T₁=328 and 323 K, respectively, one gets:

For the 1st stage: kW_ad = 230
For the 2nd stage: kW_ad = 212

 

[eltjo]

This is correct, I have made some calculations with my old schoolbooks and it uses the same formula (and answers).

But when I make that calculation complete with the discharge temperature:

T2/T1 = P2/P1^((n-1)/n)

The discharge temperatures are lower than the stated 150 C.

With your numbers approx. 135 C.

And this calculation doesn't use the efficiency of the motor, right?

 

[25362]

The formula refers to the work of compression on the gas, nothing on the mover.

 

[eltjo]

Yes, I thought so.
Thanx for all the replies. If I have news from the manufacturer I'll share the info.