To Calculate Heating Load of an air compressor

[mistershark]

I need to calculate the heating load of an air compressor (powered with an electrical engine 500kWe for a flow rate of 83 sm3/min, 6bar pressure) for an application in compressed air for a pneumatic conveyor.

Anyone can help me?

thanks in advance

[mistershark]

Hello GUYS
I really need your help. HELP ME, HELP ME

[Yuri B.]

Hello. I am an electrician, am sure somebody, more competent than me, will help you. Can you explain to me, what do you mean by "heating load" ? The engine, I guess, is 50 Kw - not 500.

[Sandro Baptista]

I need to calculate the heating load of an air compressor (powered with an electrical engine 500kWe for a flow rate of 83 sm3/min, 6bar pressure) for an application in compressed air for a pneumatic conveyor.

Anyone can help me?

thanks in advance

Assuming that the compressor sucks air at atmospherica pressure 1 bar (a) at 293 K and the compressed air is cooled and delivery to the pneumatic conveyer at 300 K. For 83 m3/min for a pressure ratio of 6 (» 6 bar (g)) then I would say that compressor + air cooler will disspate about 450 kW of heating load. This value will also depend of the isentropic efficiency which depends of each compressor and is own conception.

I hope this would help.

Regards

[Sandro Baptista]

I just want to add something that I remember

Because the electric motor (500 kW of mechanical power) doesn't have an electric efficiency of 100% so we have to count with some heat generated by it. So I would add about more 30 kW.

[Yuri B.]

My God! How you handle such monsters!

[Yuri B.]

Surely, elephant's size.

[mistershark]

Thanks to all for the answers, at this i would to explain my problem but i'm busy with another one!
as soon as I can I will post my considerations

thanks very much again

P.S.
@Yuti B
for "heating load" I mean the emission of heat from the compressor, caused by your electrical motor

see you

[Sandro Baptista]

P.S.
@Yuti B
for "heating load" I mean the emission of heat from the compressor, caused by your electrical motor

see you

Mistershark,

I think is more correct you say: emission of heat from the compressor (compressor itself, air cooler, oil cooler) plus heat load from the motor casing, due the fact the electric efficiency not be 100%.

Oil cooler should apply if it is a screw compressor (for this capacity it should be).

Regards

[Yuri B.]

Mistershark, tell me please, are such powerfull motors driven not by 400 V or 660 V but by much higher voltages ? (to produce less heat to dissipate) I suppose, it be at least 3 kV.

[mistershark]

Ok, good morning at all.
I can provide more information for clarify my question.

the machine is a compressor three stage, integrally geared and packaged centrifugal compressor, oil free, cooling water. This machines (in number of 3, in the same industrial structure of 140 square meters of surface) serving a pneumatic system for conveying fly ash. The coincidence factor is 2/3 of the compressors working at the same time.

at the worst condition the power absorption of the motor is 464 kW.
assuming an efficiency of 0.8, can I consider a formula that I find on an italian manual:

(0,735/efficiency)x0,86xpower

that for my case means a load of 367kW per compressor

@Yuri B
the motor has a voltage of 6kV

[Sandro Baptista]

Hi,

Just a few questions and comments:

1) Power absorption of the motor is 464 kW. I think you meant active electric power absorbed (not apparent) by the motor, right?

2) An efficiency, electrical I suppose of 80% is very low for this kind of motor even if it's a part load of 75% aprox. I would say about 90% - 95% as a better approach.

Regards

[mistershark]

Hi,

Just a few questions and comments:

1) Power absorption of the motor is 464 kW. I think you meant active electric power absorbed (not apparent) by the motor, right?

2) An efficiency, electrical I suppose of 80% is very low for this kind of motor even if it's a part load of 75% aprox. I would say about 90% - 95% as a better approach.

Regards

1) generally the suppliers, for rotary machines, supply the active electric power absorbed (i know that for the power transformer is the opposite)

2) i considered 0.8 for safety calculations.

do you agree with the formula that i write in the preceding post?

[Sandro Baptista]

do you agree with the formula that i write in the preceding post?

I don't know from where those numbers (0,735 ; 0,86) came from...but I guess that:

0,735 mean 0,735 kW / CV
0,86 maybe is a rough value for the power factor (cos phi)

So I would said that maybe you pick the wrong formula and even so it following my thoughts it should have an error (it should be eff/0,735 and not 0,735/eff).

If the "power" that appears in the formula is the apparent power (which includes the reactive electrical power) and if you seek the shaft power absorbed by the compressor the formula should be:

efficiency / 0,735 x 0,86 x power = P_mechanical

You told me that the motor absorb 464 electrical power (which is active power so don't consider 0,86) so considering an efficiency of 80% (as you told me as safety calculation) I would said that the compressor should absorb (mechanically) about

0,8 / 0,735 x 464 kW = 505 CV or 0,8 x 464 kW = 371 kW.

In this case the value that you get is close to my but the concept of your calculation is wrong...if I understood the meaning of the 0,735 and 0,86.

Regards

[mistershark]

I don't know from where those numbers (0,735 ; 0,86) came from...but I guess that:

0,735 mean 0,735 kW / CV
0,86 maybe is a rough value for the power factor (cos phi)

So I would said that maybe you pick the wrong formula and even so it following my thoughts it should have an error (it should be eff/0,735 and not 0,735/eff).

If the "power" that appears in the formula is the apparent power (which includes the reactive electrical power) and if you seek the shaft power absorbed by the compressor the formula should be:

efficiency / 0,735 x 0,86 x power = P_mechanical

You told me that the motor absorb 464 electrical power (which is active power so don't consider 0,86) so considering an efficiency of 80% (as you told me as safety calculation) I would said that the compressor should absorb (mechanically) about

0,8 / 0,735 x 464 kW = 505 CV or 0,8 x 464 kW = 371 kW.

In this case the value that you get is close to my but the concept of your calculation is wrong...if I understood the meaning of the 0,735 and 0,86.

Regards

I report you the main paragraph, concerning this formula:
“the absorbed power by an electrical engine is : HP/eff where ‘eff’ is a value 0,8-0,9 inclusive, so the amount of heat for the a conditioning treatment is: (735/eff)*0,86 kCal/h per each HP”
Analysing this formula I think that 0,86 means the conversion from kCal/h to kW (and not the classic cos (phi) ) and 735 (as you suggest) is the conversion factor from kW to HP.
So you think properly to consider as amount of heat passed through a electric motor:
(Active Electric Power) X eff ?

[Sandro Baptista]

I report you the main paragraph, concerning this formula:
“the absorbed power by an electrical engine is : HP/eff where ‘eff’ is a value 0,8-0,9 inclusive, so the amount of heat for the a conditioning treatment is: (735/eff)*0,86 kCal/h per each HP”
Analysing this formula I think that 0,86 means the conversion from kCal/h to kW (and not the classic cos (phi) ) and 735 (as you suggest) is the conversion factor from kW to HP.
So you think properly to consider as amount of heat passed through a electric motor:
(Active Electric Power) X eff ?

If 0,86 is 0,86 kCal/h and not 0,86 kCal/h the things make sense. But in this case the formula means:
The mechanical shaft of the compressor absorves 505 HP. So if the motor has an eff of 0,8 the heat load gain is 735/0,8*0,86*50 = 399.013 kCal/h = 464,0 kW. 464 kW is the value that you had gave in a previous email about the electric power absorved by the motor and 505 cv was the value that I have calculated for the 464 kW in a previous email.

ATTENTION: 1 HP » 745W, and 1 CV » 735W


So you think properly to consider as amount of heat passed through a electric motor:
(Active Electric Power) X eff
Active electric power x eff would be the mechanical shaft absorved by the compressor, which is only a part the other part is the heat load produced by the motor due to the non efficiency, so we can say that the active electric power should be the total heat load. Of course this is only true if you consider that the air at exit of the compressor is more or less equal at the inlet (293 K » 300 K by suppose in one of the previous post).

Regards

[dougheret0]

I report you the main paragraph, concerning this formula:
“the absorbed power by an electrical engine is : HP/eff where ‘eff’ is a value 0,8-0,9 inclusive, so the amount of heat for the a conditioning treatment is: (735/eff)*0,86 kCal/h per each HP”
Analysing this formula I think that 0,86 means the conversion from kCal/h to kW (and not the classic cos (phi) ) and 735 (as you suggest) is the conversion factor from kW to HP.
So you think properly to consider as amount of heat passed through a electric motor:
(Active Electric Power) X eff ?

Correct. Another way to put it is that if both the motor and the load are in the space, then all of the electrical energy delivered to the motor when it is running appears in the space as heat. So you only have to determine what percentage of time the motor actually runs, and what percentage of nameplate amps it pulls when it is running.

If the load is outside the conditioned space, then only the heat resulting from the motor inefficiency appears as cooling load in the conditioned space.

[mistershark]

Correct. Another way to put it is that if both the motor and the load are in the space, then all of the electrical energy delivered to the motor when it is running appears in the space as heat. So you only have to determine what percentage of time the motor actually runs, and what percentage of nameplate amps it pulls when it is running.

If the load is outside the conditioned space, then only the heat resulting from the motor inefficiency appears as cooling load in the conditioned space.

the electrical motor is covered and the compressor box not. The equipment (motor+compressor) in number of 3 is on a structure of 140 mq of surface.

I read also a thread on eng-tips forums:

<<ASHRAE Classifies heat gain into (3) as follows:

A = Motor in Driven Equip in. Btu/hr Heat gain = (BHP x 2545)/Eff

B = Motor out, Driven Equip in. Btu/hr Heat Gain = BHP x 2545

C = Motor in, Driven Equip out. Btu/hr Heat Gain = (1/Eff -1) x BHP x 2545>>

that it gives interessant solutions with the ASHRAE specifications

[Sandro Baptista]

Hi Mistershark,

Let's go by parts:

- What is the delivery temperature of the compressedo air to the process. Is the air cooled in a air cooler heat exchanger? Where is the air cooler mounted (I suppose outside of the machine room)

- Does this compressor have oil cooler? Where is oil cooler mounted? Is it air cooled? Is the oil cooler mounted outside of the machine room?

- Do you want to know the heat load of the motorcompressor inside of the machine room or do you want to know the total heat load?
If you only want to know the heat load inside the machine room and you have all the heat exchangers (air and oil) outside the machine room then you should only count with the non efficiency of the compressor motor.

[mistershark]

Let's go by parts:

- What is the delivery temperature of the compressedo air to the process. Is the air cooled in a air cooler heat exchanger? Where is the air cooler mounted (I suppose outside of the machine room)

> I don't have informations about the temperature of the air at the outlet compressor but i can say that it has a removable slide-in-and-out intercooler and aftercooler bundles with water in the tubes that cool the air stream between compression stages. Tthe intercooler and after cooler are of the plate fin extended surface type featuring tubes with no internal obstructing fins for easy inspection and cleaning
Shaft power at coupling (guaranteed point): 447kW
Driver Rating: 550kW
so the calculated ratio is 0,81 (is this value the efficiency?)

- Does this compressor have oil cooler? Where is oil cooler mounted? Is it air cooled? Is the oil cooler mounted outside of the machine room?

> no, it'is an oil free so it has a mecanism of lubrification only for the shaft and dynamic parts.

- Do you want to know the heat load of the motorcompressor inside of the machine room or do you want to know the total heat load?
If you only want to know the heat load inside the machine room and you have all the heat exchangers (air and oil) outside the machine room then you should only count with the non efficiency of the compressor motor.

> I want to know the max heat that could be released from motor surface to atmosphere in room if we can consider that the electric motor is the only cause of heat emission for the entire equipment motor-compressor

[Sandro Baptista]

By what you just say the electric motor is the only part that could release heat. The intercooler and after cooler are water cooled and it doesn't matter if they are indoors or not.

If you teel me that the shaft power at coupling is 447 kW and in a previous email you said that the elictric power absorved was 464 kW so heat load release is 17 kW. The electric efficiency of that motor wich is not at 100% loaded would be 96%.

[mistershark]

By what you just say the electric motor is the only part that could release heat. The intercooler and after cooler are water cooled and it doesn't matter if they are indoors or not.

If you teel me that the shaft power at coupling is 447 kW and in a previous email you said that the elictric power absorved was 464 kW so heat load release is 17 kW. The electric efficiency of that motor wich is not at 100% loaded would be 96%.

You don't consider the previous value 464kW. We have changed the compressor supplier. So I can link an image of the compressor and I ask you what's the influence of the motor cover on the calculation of heat emitted by the motor surface.



And after the compressor, in the same area, we also have a dryer for air treatment because the end user of the compressed air need air dryed.
So the drying machine is like this:



and with these technical caracteristics:

Compressed air flow rate
4830 Nm³/h

Working pressure
6,15 bar g.

Dryer air inlet temperature / outlet temperature
45 / 38,4 °C

Ambient temperature / Suction
40 / 40 °C

Ambient pressure
0 bar g.

Relative humidity
95 %

Ambient water content
59,02 g/Nm³

Water content inlet / outlet dryer
10,77 / 0,84 g/Nm³

Moisture flow rate
47,97 kg/h

Pressure dew point / Atmospheric dew point
3 / -19,6 °C

Compressed air pressure drop
0,1 bar

Absorbed power by the refrigerating compressor
19,36 kW

Maximum allowable pressure drop
0,5 bar

so what do you think about the heat emission from the dryer too.

thanks very much in advance

[Sandro Baptista]

Dryer air inlet temperature / outlet temperature 45 / 38,4 °C

Hi,

I think you meant the mositure air inlets at +45ºC after the aftercooler and after take out water vapour from the air and be reheated then the air outs at +38,4ºC


Well, concerning the motor supposing that it dissipates xxx kW it shouldn't matter if it is cover or not...it disspates the same. What would happens is that the thermal equilibrium with the cover outside air would be reached at a bigger temperature inside the cover.

Concerning with the drying system: the condenser unit is attached to the rest of the unit dryer? It appears so. If so do you have any ducts to transport the hot air to outside?

[mistershark]

Hi,

I think you meant the mositure air inlets at +45ºC after the aftercooler and after take out water vapour from the air and be reheated then the air outs at +38,4ºC

yes


Well, concerning the motor supposing that it dissipates xxx kW it shouldn't matter if it is cover or not...it disspates the same. What would happens is that the thermal equilibrium with the cover outside air would be reached at a bigger temperature inside the cover.

I agree

Concerning with the drying system: the condenser unit is attached to the rest of the unit dryer? It appears so. If so do you have any ducts to transport the hot air to outside?

the condenser is part of the dryer. in the same case. we don't have some duct to pull off the room compressor the hot air

So dear Sandro, can you provide the exact, for you, value of heat load (or emission heat, as you prefer) of the compressor and the dryer too? I need to write a technical report for this calculation

thanks

[Sandro Baptista]

Mistershark,

We have seen previous that the heat release by the electric motor (no efficient part) would be around 20 kW.

Concernig the air drier unit here's what I get:

4.830 Nm3/h » N is for the following conditions p=1 bar(a) and 273K/0ºC
Density = 1,291 kg/m3 at 273K 1 bar(a)

At 7 bar(a) it represents 690 m3/h at 273K

Inlet mass flow at 318 K (+45ºC) = 765 kg/h
Inlet mass water vapour = 44.658 g/h
Outlet mass water vapour = 3.557 g/h

0,84 g/Nm3 represents 0,651 g/Nkg. At 7 bar(a) the dewpoint is about +1ºC

» Sensible Cold capacity to cold the air from +45ºC to +1ºC = 9,4 kW
» Latent Capacity to remove 44,658 kg/h to 3,557 kg/h of water = 28,5 kW

TOTAL COLD CAPACITY OF THE AIR DRIER UNIT = 37,9 kW

» Reheating the air from +1ºC to +38,4ºC = 8,0 kW


Total heat load dissipated by the air dry unit indoors would be = COLD CAPACITY + ELECTRIC POWER ABSORBED OF THE COMPRESSOR - REHEATING POWER OF THE AIR = 37,9 + 19,4 - 8 = 49,3 kW
(remember you told me before that the condenser was indoors and duct was putting out the hot air from the condenser)


So in total we should count with a heating load in the machine room of about 70 kW FOR EACH AIRCOMPRESSOR UNIT PLUS AIR DRIER UNIT.

I hope this help...and I hope this subject been closed

P.S. what's the purpose of the water in the air drier unit? Is it removing any heat? If it does we must discount that value on the total heating load.

Regards

[mistershark]

Mistershark,

We have seen previous that the heat release by the electric motor (no efficient part) would be around 20 kW.

<< sorry Sandro, but it needs only to consider this values for the electrical motor:
Shaft power at coupling (guaranteed point): 447kW
Driver Rating: 550kW
So the constructor of the machine assure me that the efficiency of the compressor is 95%. 5% is used for compression of the air, the rest is heat losses.
So, what number I can consider for a calculation of an air traitment to digest compressor heat load in the room?>>


Concernig the air drier unit here's what I get:

4.830 Nm3/h » N is for the following conditions p=1 bar(a) and 273K/0ºC
Density = 1,291 kg/m3 at 273K 1 bar(a)

At 7 bar(a) it represents 690 m3/h at 273K

Inlet mass flow at 318 K (+45ºC) = 765 kg/h
Inlet mass water vapour = 44.658 g/h
Outlet mass water vapour = 3.557 g/h

<< can you explain me how you have fish this number out?>>

0,84 g/Nm3 represents 0,651 g/Nkg. At 7 bar(a) the dewpoint is about +1ºC

» Sensible Cold capacity to cold the air from +45ºC to +1ºC = 9,4 kW
» Latent Capacity to remove 44,658 kg/h to 3,557 kg/h of water = 28,5 kW

<< and these one too >>

TOTAL COLD CAPACITY OF THE AIR DRIER UNIT = 37,9 kW

» Reheating the air from +1ºC to +38,4ºC = 8,0 kW

P.S. what's the purpose of the water in the air drier unit? Is it removing any heat? If it does we must discount that value on the total heating load.

<< the water serves for cooling the condenser in the refrigerating dryer part as well as you can see in the above posted image, the condenser is near the two blowers >>

thanks a lot Sandro

[Sandro Baptista]

Have your "superiors" accepted the calculations?

[mistershark]

Have your "superiors" accepted the calculations?

Yes, for sure! Thanks again

I will offer a beer to you

[Sandro Baptista]

Come to Portugal-Lisbon and buy me a lunch