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

[MCTachyon]

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 = mR_specificT
V = (mR_specificT)/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?

 

[scottdave]

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).

 

[MCTachyon]

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?

 

[scottdave]

What value did you use for W in this formula? Is p1 1bar and p2 is 15bar?

 

[MCTachyon]

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?

 

[scottdave]

Oh OK. I had misread your n-1/n as n - (1/n) rather than (n-1)/n

 

[Chestermiller]

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.

 

[MCTachyon]

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.

 

[MCTachyon]

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?

 

[Chestermiller]

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?

 

[MCTachyon]

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?

 

[Chestermiller]

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.

 

[MCTachyon]

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 P_in = 2kW, T₁= 20°C and p₂ = 15 bar?

 

[Chestermiller]

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 P_in = 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?

 

[MCTachyon]

In first compressor it is (p₂/p₁) and in the second it is (p₃/p₂)?

 

[Chestermiller]

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?

 

[MCTachyon]

(p₂/p₁) = (p₃/p₂)?

 

[Chestermiller]

(p₂/p₁) = (p₃/p₂)?

Why don't you tentatively assume that is correct and see how it plays out.

 

[MCTachyon]

Working out for p₂:

(p₂/p₁) = (p₃/p₂)

p₂² = p₁ x p₃

p₂ = √p₁ x p₃

Getting there?

 

[Chestermiller]

Working out for p₂:

(p₂/p₁) = (p₃/p₂)

p₂² = p₁ x p₃

p₂ = √p₁ x p₃

Getting there?

Yes. How about a number?

 

[MCTachyon]

Oops, that would be a good start.

p₂ = √1 x 15

p₂ = 3.873 bar

 

[Chestermiller]

Oops, that would be a good start.

p₂ = √1 x 15

p₂ = 3.873 bar

OK

 

[MCTachyon]

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?

 

[MCTachyon]

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?

 

[Chestermiller]

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?

 

[MCTachyon]

It's the equation I believe that is used to work out the work input to polytropic compression..

 

[Chestermiller]

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).

 

[MCTachyon]

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.

 

[Chestermiller]

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?

 

[MCTachyon]

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)?

 

[MCTachyon]

The work input to isothermal compression is given by:

W = p₁V₁In(p₂/p₁)

2000 = 10⁵ * V₁ * In(3.873)

V₁ = 2000 / 1.354x10⁵

V₁ = 0.0148 m³ s^-1

Better?

 

[Chestermiller]

The work input to isothermal compression is given by:

W = p₁V₁In(p₂/p₁)

2000 = 10⁵ * V₁ * In(3.873)

V₁ = 2000 / 1.354x10⁵

V₁ = 0.0148 m³ s^-1

Better?

Actually, your results in post #24 look correct to me, if you did the arithmetic correctly. The equation for the shaft work with the n in the numerator is, I believe, for a continuous flow system. So you have determined the volumetric flow rate into the first compressor. But the problem statement asks for the mass flow rate. So, what is the mass flow rate?

 

[MCTachyon]

The problem asked to..

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

Have we not done that in #24?

Is or more work needed?

 

[Chestermiller]

The problem asked to..

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

Have we not done that in #24?

Is or more work needed?

Oh. Missed that. Then you're done. Do you want to try to prove that the 2nd compressor will operate the same?

 

[MCTachyon]

Is it a similar equation?

But subbing in (p₃/p₂) for compressor ratio?