# Thermodynamics problem - Compressing the air. Energy Balance

1. Jan 21, 2015

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Dear!
I have a problem with one, complex exercise. I’ve divided it into three parts just to simplify finding the answer. So let’s start.
The air is sucked from the atmosphere by a compressor. Atmospheric pressure is 0,1MPa and the temperature is 283K, the suction pressure in the compressor is 1.5 MPa, whereas the discharge 4,5MPa and its temperature is 323K. Capacity of the compressor is 60000m3 /h. I need to analyse the important aspects related to the profitability of the operation of the installation (?) and first of all the energy balance. I do not know whether they are here given all the information necessary to solve the problem, but in case of any doubts my job is to ask for a specific data.

I divided this task into three parts,:
1. Compression of air (isentropic? I suppose).
2. Air Transport pipeline (adiabatically, with friction)
3. The expansion ((isentropic? I suppose).

1. Compression of air
In order to determine the technical work required to compress the air used the dependence of transformation polytropy:
Lt = (1 / (m-1)) * (p1V1-p2V2) + p1V1-p2V2
Just do not know how do I get m. Any suggestions?

2
I skip the phenomena which are taking place in the pipeline e.x pressure loss due to friction and due to local resistance. I assume that the temperature of the air in the pipe at the inlet = temperature of the air in the pipe at the outlet and the inlet pressure equal to the pressure of the outlet.

3. The expansion
The temperature in the nozzle outlet cross-section is:
T2 = T1-ƞd (T1-T2S)
The temperature of the T2S the isentropic expansion may be determined by the formula
T1 / T2 = (V2 / V1) χ-1 = (p1 / p2) ^ ((χ-1) / χ), assuming the air getting χ = 1.4
T2S = 323 * 0.1 ^ (0.286) = 323 * 0.52 = 168K
thus
T2 = 323-0,90 * (323-168) = 183,5K
Is it the right solution? What else I need in order to develop the energy balance?

Last edited by a moderator: Jan 21, 2015
2. Jan 21, 2015

### Staff: Mentor

This doesn't explain very well what the process looks like. Here's what I got out of what you wrote: You have a compressor, followed by a pipe, followed by a nozzle. Somehow the air to the compressor gets from 0.1 MPa to 1.5 MPa, which represents its suction pressure. Coming out of the compressor, the air is at 4.5 MPa and 323 K.

Questions:
How does the air get from 0.1 MPa to 1.5 MPa before the compressor?

Are you studying the open system steady flow version of the first law? What is the equation for that? Do you think that this might apply to the compressor?

Chet

3. Jan 22, 2015

Firts of all. Thank you very much for the answer!
That's right.

I feel dumb, that I haven't notice that. The pressure of air is raised in first compresson which is follwed by second compressor which get the air from 1.5MPa to 4.5MPa.

I'm not sure if it is right but :
$Q_{in}=m*(h_e-h_i)$ ?
I think that this may apply to the compressor, but i don't know how to find the h. Any advices?

4. Jan 22, 2015

### Staff: Mentor

So, the compression ratio for the first compressor is 15? (Sounds very high). Is the first compressor something that's part of your problem, or are you you just assigned to solve the second compressor?

You're missing a term:
$Q_{in}-W_s=m*(h_e-h_i)$ where Ws is the shaft work done. If the compressor is adiabatic, the Qin is zero. Do you think that adiabatic is a reasonable assumption? If it is, then your equation gives you the shaft work done by the compressor on the surroundings (minus the work required to run the compressor). You have air coming in at 15 Bar and 283 and leaving at 45 bar and 323. If it were an ideal gas (it's not), how would you determine the change in h? Have you learned about how to determine the effect of pressure on enthalpy? or, Do you have "air tables?"

Chet[/QUOTE]

Last edited: Jan 22, 2015
5. Jan 23, 2015

Thank you for the answer!
Yes, I have to analyse also the first compressor. But I suppose that it will be similar to the second compressor.
What's more I got the information that the temperature between first and second compressor is not important. (is it indeed?)

I think that it is quite reasonable assumption, but I have question. May it be a polytrophic compressor?

Is there a chance to calculate not ideal gas, but a real one?

$H=U+pV$
while U is internal energy, p is pressure and V is Volume.

And I count the Specific Enthalpy for 15 bar and 283K and it is 279.83 [ kJ / kg ], while for 45 bar and 323 is 315.68 [ kJ / kg ].
So now using the open system steady flow version of the first law I can calculate $Q_{in}-W_s=m*(h_e-h_i)$
Mass of air is 1,293 g per liter , so during one hour it is 60 000 000liters per hour * 1,293g=77580kg
So:
$Q_{in}-W_s=m*(h_e-h_i)$
$Q_{in}$ is 0 as you said.
$-W_s=m*(h_e-h_i)=77580*(315.68- 279.83)=2781242kJ.$
So
$W_s=-2781242kJ$
That's right? ;)

Last edited: Jan 23, 2015
6. Jan 23, 2015

### Staff: Mentor

You haven't provided enough information for me to address this. I feel like there are additional details that I need to know that are missing. In your engineering judgement, is the temperature entering the second compressor important in analyzing the second compressor? (Up to now, you've assumed it's 283, but what if it's not).

Are you saying that you would like to assume that the gas behaves polytropic and analyze the problem that way? How would you go about doing that? You still need to know the enthalpy change in the compressor, and you already have that (presumably, if 283 is the inlet temperature). Also, how would you apply polytropic to a real gas beyond the ideal gas region?
Apparently you already are. I assume the values of enthalpy you are getting in the calculation below are from tables.

Did these values come out of tables? If so, the people who prepared the tables already did the calculations for you, and took into account the difference between ideal gas behavior, and the real gas behavior that exists at 15 - 50 bars.
Did you assume that the 60000 m^3/hr were standard m^3 and is the 1.293 kg/m^3 the density at standard conditions? The final units should be kg/hr.

The units should be kJ/hr. What is that in kW?

Tell me more about the tables you are using. Do the tables also give entropy s? If so, how does the entropy at the inlet conditions compare with the entropy at the outlet conditions?

Chet

7. Jan 23, 2015

Oh, I understand. So could we leave this part now? :) I have to ask about more details of this part.
So let's start once again. We have only one compressor the suction pressure in the compressor is 1.5 MPa, temperature is 283K whereas the discharge 4,5MPa and its temperature is 323K.

Sorry for that. I suggested the polytropic because I heard that it is the most similar to the real conditions. I have to admit that have no idea how to to calculate this totally, not even for a polytropic reaction. So you suggest not to go into polytrophic reaction?

I used this site to calculate the enthalpy http://www.peacesoftware.de/einigewerte/calc_luft.php5

And obviously I did a mistake in units, it should be kg/hr.
Beacuse $Ws=−2781242kJ/hour$ it is $772567,2Watts$ so finally it is $772kW$
The entrophy of the outlet conditions is 5.831481 [ kJ / kg K ], while the inlet is 6.040657 [ kJ / kg K ]. But what I can do with it?

8. Jan 23, 2015

### Staff: Mentor

It can't really be treated polytropically, because the gas is beyond the ideal gas region (i.e., high pressure). So the equation of state is not the ideal gas equation. Besides, I have the feeling that you were supposed to solve this problem using the tables, correct? These tables were developed taking into account the non-ideal gas behavior of air.
I was asking because, if the compressor operated perfectly reversibly and adiabatically, the change in entropy would be zero. I just wanted to check. This looks pretty close to zero change. What would the final temperature have been if there were no change in entropy?

I suggest you try solving the compressor assuming that the air behaves like an ideal gas and see how the answer compares.

Chet

Last edited: Jan 23, 2015
9. Jan 25, 2015

Ok, I understand.
Nobody said to me that I am supposed to solve this problem by using the tables, but I think that is necessary to do this. Thanks for the suggestion!
The temperature won't change if there were no change in etropy, yes?

Tomorrow I will do it, and show the results here ;)
Thanks!

10. Jan 25, 2015

### Staff: Mentor

No. For the entropy to remain constant, the temperature would have to increase to compensate for the entropy decrease resulting from the increase in gas density. Do you know the equation for the entropy change of an ideal gas when its temperature and pressure both change? Look it up in your thermo book.

Chet

11. Jan 28, 2015

Let's start one again. I spend sometime on reading and understanding the basis and I want to ask whether I think in wrong or a right way. But first let's summarize what I have.

The air is sucked from the atmosphere by a compressor. Atmospheric pressure is 0,1MPa and the temperature is 283K, whereas the discharge 4,5MPa and its temperature is 323K. Capacity of the compressor is 60000m3 /h. I need to analyse the important aspects related to the profitability of the operation of the installation (?) and first of all the energy balance. There is also a Chance that two compressors are operating instead of one. In this way the first compress the air to 1,5MPa, and the second one to 4,5MPa.

Let's treat the air as an ideal gas has $CP* = (7/2)R$ . (It's okay?)
An initial state of P = 1 bar and T = 283 K to a final state of P = 45 bar and T = 323 K.
First option is:

The gas is compressed to P = 45 bar in an adiabatic reversible compressor, and then enters a heat exchanger in which it is cooled to T = 323 K.

The second option is:

The gas is compressed in two separate adiabatic reversible compressors; the first has inlet P = 1 bar and outlet P = 15 bar, and the second compresses the gas from P = 15 bar to P = 45 bar. After first and second compressor there is a heat exchanger that cools the air to T =323 K.
I found the work added and heat removed per mole of gas for each of the following processes because it is easy to compare two ways of compressing the air.\

This is an adiabatic, steady state compressor, so the energy balance:
$Ws/n=Hout - Hin$

The entropy balance for a steady state, adiabatic, reversible compressor is:
$Sout-Sin=0$

The gas in this problem is an ideal gas with constant heat capacity so:
Sout - Sin=0=Cpln(Tout/Tin)+Rln(pin/pout)
And I count the Tout
which is about 840K.
For any ideal gas, $dH = CP*dT.$ Because CP* is constant, the change in enthalpy is:
WSA/n=Hout -Hin=Cp(Tout-Tin)
and finally the WSA/n=16120 J/mole
in case of count it into J/hour I count the numer of moled during one hour by using n=pV/RT but should I use to this the pressure of 45 barthe volume of 60000m^3/hour and the temperature of 837K?

The energy balance for one side of a heat exchanger
Q/n=Hout -Hin=Cp(Tout-Tin) and I received about -14956,88J/mole

Now the second option

The temperature of air which leaves the first compressor is Sout - Sin=0=Cpln(Tout/Tin)+Rln(pin/pout)
And it is about 610 K

The temperature of air which leaves the second compressor is Sout - Sin=0=Cpln(Tout/Tin)+Rln(pin/pout)
And it is about 442K.
And now I counted the amount of energy taken by first and secong compressor by using an aquation
WS/n=Hout -Hin=Cp(Tout-Tin)
and for first compressor I got 9500 J/mole and for second one it is 3500J/mole
Then I count "Heat removal required" for the heat exchanger by using an aquation

Q/n=Cp(Tout/Tin)
and for first It is -8460J/mole and for second one it is -3500 J/mole

So for first option:
Shaft work required (J/mol)
16120 J/mole
Heat removal required (J/mol)
-14956,88J/mole

And for second option:
Shaft work required (J/mol)
9500 J/mole and for second one it is 3500J/mole = 13000J/mole

Heat removal required (J/mol)
-8460J/mole and for second one it is -3500 J/mole = 11960 J/mole
Is it energy balance for that compressors? Thank you for reading

12. Jan 29, 2015

### Staff: Mentor

I just have some comments. The problem statement is very imprecise and flexible, so you have to use your best engineering judgement.

You are already considering using 2 compressors in series with intercooling. I would like you to consider using 3 compressors in series. In practice, if there is a large overall compression ratio required, engineers typically use compressors in series, with intercooling. This approach reduces the amount of mechanical work that needs to be done, because each compression is carried out at a lower temperature. Typical compression ratios for industrial compressors are 3 to 4.

Your analyses of the individual compressors look very competent. You might also consider using your thermodynamic air tables to determine the outlet temperatures by looking up the final temperature required for the final entropy to match the inlet entropy. These tables take into account the non-ideal gas behavior of air. You could then compare the results, and see whether working with the ideal gas law is adequate for your purposes.

Regarding the 60000 cubic meters per hour, that is typically specified at the very initial temperature and pressure (283 and 1 atm), or at standard conditions (273 and 1 atm). You don't use the conditions coming out of the compressors or the heat exchangers.

Chet

13. Feb 2, 2015

Lastly I was without the Internet so that's why I was quiet.
Thanks for advice I try to do it today.

The same, I will do it today.

I do not understand what you mean by:
Could you please explain this?

What's more what should I do to finalize something what is called "energy balance"?

14. Feb 2, 2015

### Staff: Mentor

If the mass flow rate through the system is constant, the volume flow rate has to be changing because the gas volume is decreasing as you increase the pressure. So you need to know what that 60000m^3/sec refers to. You need to use this number to calculate the mass flow rate. Usually, when you are given a volume flow rate like this, it is referring to the volume flow rate that would exist if the gas were at Standard Conditions of temperature and pressure.

You are expected to use your own personal engineering judgement to provide this. My advice is to "do your best."

Chet