Calc Volumetric Flow Rate of Free Air: 0.841 m3/s

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Discussion Overview

The discussion revolves around calculating the volumetric flow rate of free air in a two-stage compressor system. Participants explore the implications of pressure, temperature, and power input on the flow rate, while addressing the complexities of the compression process and the equations involved.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents an initial calculation for volumetric flow rate but expresses uncertainty about unit consistency.
  • Another participant suggests incorporating power and the effects of compression into the calculations.
  • A different participant introduces a new equation from their notes, leading to a revised volumetric flow rate calculation.
  • There is a discussion about the definition and functioning of a two-stage compressor, with differing understandings among participants.
  • Participants clarify the relationship between pressures in the two stages and the implications for the compression ratio.
  • One participant proposes an equation for isothermal compression and calculates a new volumetric flow rate, while another questions the appropriateness of the equations used.
  • There is an ongoing exploration of whether the equations used are suitable for continuous flow systems versus closed systems.
  • Participants discuss the need to determine mass flow rate in addition to volumetric flow rate, with some confusion about the problem requirements.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correctness of the equations used for the calculations. There are multiple competing views regarding the definitions and implications of the two-stage compressor and the appropriate equations for the calculations.

Contextual Notes

Participants express uncertainty about the assumptions underlying their calculations, particularly regarding the type of system (open vs. closed) and the nature of the compression process (isothermal vs. polytropic). There are unresolved questions about the relationships between pressures and the implications for flow rates.

MCTachyon
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Homework Statement



Air is to be compressed from a pressure of 1 bar and a temperature of
20°C to a pressure of 15 bar in a two stage compressor with intercooling.
After intercooling the air temperature returns to 20°C and the polytropic
index of compression is 1.28. If the input power per stage of compression
is 2 kW, calculate:

The volumetric flow rate of free air (at 1 bar and 20°C) in m3 s–1.

Homework Equations


PV = mRspecificT
V = (mRspecificT)/P

The Attempt at a Solution


[/B]
V = (1 x 287 x (20+273))/1x105

V = 84091/1*105

V = 0.841

This attempt is using all the information I have. But I don't think this is the correct solution as the units don't balance to leave m3 s-1 (I think).

Any direction?
 
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In the formula that you use V is volume (meters3). This is what you calculated. Somehow, you need to incorporate the power (2 kW) which is a rate of energy per time. Also, you should use the fact that you are compressing the gas to a higher pressure (which takes energy).
 
Thank you Scott for this input.

I have found an equation in my theory notes which I missed completely when trying to solve this.

W = n/(n-1) p1V1 [(p2/p1)(n-1/n) - 1]

Rearranging for V1 gives me an answer of 5.416 x 10-3.

Subbing in SI base units leaves me with m3 s-1.

This look better?
 
Last edited:
What value did you use for W in this formula? Is p1 1bar and p2 is 15bar?
 
W = 2000 J s-1, p1 = 1x105Pa & p2 = 15x105Pa

Subbing into: W = n/(n-1) x p1V1 x [(p2/p1)(n-1/n) - 1]

2000 = (1.28/0.28) x 105 x V1 x [(15/1)(0.28/1.28) - 1]

2000 = 4.57 x 105 x V1 x 0.808

V1 = 2000 / (4.57 x 105 x 0.808)

V1 = 2000 / 369.26x103

V1 = 5.416x10-3 m3 s-1In the ball park?
 
Oh OK. I had misread your n-1/n as n - (1/n) rather than (n-1)/n
 
What you have done so far is mostly incorrect. Do you know what a two stage compressor is.? If so, please define it for us so that we know you understand.
 
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Hi Chester,

My understanding of a two stage compressor is to take in air from the atmosphere and outputs this air at a pressure substantially higher.

This is achieved by intercooling between stages which causes a contraction in volume causing the pressure to rise sharply in the second stage.
 
scottdave said:
Oh OK. I had misread your n-1/n as n - (1/n) rather than (n-1)/n

Ah ok, anything you can add about my direction with this question?
 
  • #10
MCTachyon said:
Hi Chester,

My understanding of a two stage compressor is to take in air from the atmosphere and outputs this air at a pressure substantially higher.

This is achieved by intercooling between stages which causes a contraction in volume causing the pressure to rise sharply in the second stage.
This is not correct. A two stage compressor is actually two compressors in series, with the output from the first compressor first cooled (inter-cooling) and then fed as the input of the second compressor. Were you aware of this?
 
  • #11
I wasn't.

So from the information given we taking in air at P1 and T1 which will come out the first compressor as P2 and T2 and pass into a inter-cooler? The inter-cooler will bring the temperature down so the air passes into the second compressor as P2 and T1?

Am I getting it in terms of what is happening to the air?
 
  • #12
MCTachyon said:
I wasn't.

So from the information given we taking in air at P1 and T1 which will come out the first compressor as P2 and T2 and pass into a inter-cooler? The inter-cooler will bring the temperature down so the air passes into the second compressor as P2 and T1?

Am I getting it in terms of what is happening to the air?
Yes.
 
  • #13
So following on from that.

The air leaves the second compressor as P3 and T3?

So what we know about the air as it enters the second compressor is Pin = 2kW, T1= 20°C and p2 = 15 bar?
 
  • #14
MCTachyon said:
So following on from that.

The air leaves the second compressor as P3 and T3?

So what we know about the air as it enters the second compressor is Pin = 2kW, T1= 20°C and p2 = 15 bar?
No. P2 is not 15 bars. P3 is 15 bars. P2 is unknown, and the mass flow rate is unknown. But we do know that the amount of work done in each of the compressors is 2 kW. What does that tell you about the compression ratio in each compressor?
 
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  • #15
In first compressor it is (p2/p1) and in the second it is (p3/p2)?
 
  • #16
MCTachyon said:
In first compressor it is (p2/p1) and in the second it is (p3/p2)?
How are these related, given that the inlet temperatures and the amounts of power are the same?
 
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  • #17
(p2/p1) = (p3/p2)?
 
  • #18
MCTachyon said:
(p2/p1) = (p3/p2)?
Why don't you tentatively assume that is correct and see how it plays out.
 
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  • #19
Working out for p2:

(p2/p1) = (p3/p2)

p22 = p1 x p3

p2 = √p1 x p3

Getting there?
 
  • #20
MCTachyon said:
Working out for p2:

(p2/p1) = (p3/p2)

p22 = p1 x p3

p2 = √p1 x p3

Getting there?
Yes. How about a number?
 
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  • #21
Oops, that would be a good start.

p2 = √1 x 15

p2 = 3.873 bar
 
  • #22
MCTachyon said:
Oops, that would be a good start.

p2 = √1 x 15

p2 = 3.873 bar
OK
 
  • #23
I guess the next step is to being it all in with an equation that'll give us the Volumetric Flow Rate? Or does T2 and T3 play a part?
 
  • #24
Right I've tried to tie this up here:

W = (n / n-1) * p1 * V1 * [(p2/p1)(n-1 / n) -1]

2000 = (1.28 / 0.28) * 105 * V1 * [(3.873/1)(0.28 / 1.28) -1]

2000 = 4.57 * 105 * V1 * 0.3447

V1 = 2000 / (4.57 * 105 * 0.3447)

V1 = 0.127 m3 s-1

Have I jumped the gun here? Got the right equation?
 
  • #25
MCTachyon said:
Right I've tried to tie this up here:

W = (n / n-1) * p1 * V1 * [(p2/p1)(n-1 / n) -1]

2000 = (1.28 / 0.28) * 105 * V1 * [(3.873/1)(0.28 / 1.28) -1]

2000 = 4.57 * 105 * V1 * 0.3447

V1 = 2000 / (4.57 * 105 * 0.3447)

V1 = 0.127 m3 s-1

Have I jumped the gun here? Got the right equation?
I'm not sure whether your first equation is correct. Is this correct equation for a continuous flow compressor?
 
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  • #26
It's the equation I believe that is used to work out the work input to polytropic compression..
 
  • #27
MCTachyon said:
It's the equation I believe that is used to work out the work input to polytropic compression..
Is it the equation for a closed system or the equation for a continuous flow system (shaft work).
 
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  • #28
From what I can gather in front of me its for a closed system. I can't find an equation for a continuous flow system in what I'm reading.
 
  • #29
Are you familiar with the open system (control volume) version of the first law of thermodynamics? You are aware that your compressor is an open system device (with gas flowing through), right?
 
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  • #30
Does this make it an isothermal process then? In the sense the system has some means of controlling the temperate (I.e. the inter-cooler)?
 

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