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

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The discussion revolves around calculating the volumetric flow rate of free air in a two-stage compressor system, where air is compressed from 1 bar to 15 bar with intercooling. Initial attempts to solve the problem involved using the wrong equations and assumptions about the compressor's operation. After clarifying the concept of a two-stage compressor and correcting the equations, the participants derived the volumetric flow rate to be approximately 0.0148 m³/s using the isothermal compression formula. The conversation highlights the importance of understanding thermodynamic principles and the correct application of formulas in solving engineering problems. The participants gained a clearer understanding of the compressor's operation and the calculations involved.
  • #31
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?
 
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  • #32
MCTachyon said:
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?
 
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  • #33
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?
 
  • #34
MCTachyon said:
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?
 
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  • #35
Is it a similar equation?

But subbing in (p3/p2) for compressor ratio?
 
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  • #36
MCTachyon said:
Is it a similar equation?

But subbing in (p3/p2) for compressor ratio?
It's the exact same equation. Because the inlet temperatures are the same, the product of specific volume and pressure going into the 2nd compressor as that going into the 1st compressor. And, as you pointed out, the compression ratios are the same.
 
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  • #37
Thanks for your help with this. I have a much greater understanding of 2 stage compressors now and hopefully have picked up enough here to answer the question thoroughly.

Thanks again.
 
  • #38
MCTachyon said:
Thanks for your help with this. I have a much greater understanding of 2 stage compressors now and hopefully have picked up enough here to answer the question thoroughly.

Thanks again.
I have one more thing to add. If the molar flow rate of air through the compressor is ##\dot{m}##, then your equation for the power becomes: $$\dot{W}=\dot{m}\left[\frac{n}{(n-1)}P_{in}v_{in}\left(R^{\frac{(n-1)}{n}}-1\right)\right]$$where ##P_{in}## is the inlet pressure to the stage, ##v_{in}## is the inlet molar volume of the air to the stage, and R is the compression (pressure) ratio. The inlet molar volume is related to the inlet pressure and inlet temperature ##T_{in}## by$$v=\frac{RT_{in}}{P_{in}}$$. Substituting this, we get:
$$\dot{W}=\dot{m}\left[\frac{n}{(n-1)}RT_{in}\left(R^{\frac{(n-1)}{n}}-1\right)\right]$$Since both stages have the same inlet temperature, molar flow rate, and compression ratio, the power for the two stages must be the same.
 
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  • #39
Chestermiller said:
I have one more thing to add. If the molar flow rate of air through the compressor is ##\dot{m}##, then your equation for the power becomes: $$\dot{W}=\dot{m}\left[\frac{n}{(n-1)}P_{in}v_{in}\left(R^{\frac{(n-1)}{n}}-1\right)\right]$$where ##P_{in}## is the inlet pressure to the stage, ##v_{in}## is the inlet molar volume of the air to the stage, and R is the compression (pressure) ratio. The inlet molar volume is related to the inlet pressure and inlet temperature ##T_{in}## by$$v=\frac{RT_{in}}{P_{in}}$$. Substituting this, we get:
$$\dot{W}=\dot{m}\left[\frac{n}{(n-1)}RT_{in}\left(R^{\frac{(n-1)}{n}}-1\right)\right]$$Since both stages have the same inlet temperature, molar flow rate, and compression ratio, the power for the two stages must be the same.

This is something I've been reading about to further my understanding of the Thermodynamics course.

I have printed out this thread to add to my notes for this particular area.

Thanks again Chester.
 
  • #40
Just a quick one while this thread is still active:

Further on in my reading (on the same compressor by the way) there is a question that askes:

"Calculate; The power required if the same compression was carried out in one stage isothermally."

I believe earlier in this post I state that:

MCTachyon said:
The work input to isothermal compression is given by:

W = p1V1In(p2/p1)

Is this the equation to use as we now know V1?
 
  • #41
MCTachyon said:
Just a quick one while this thread is still active:

Further on in my reading (on the same compressor by the way) there is a question that askes:

"Calculate; The power required if the same compression was carried out in one stage isothermally."

I believe earlier in this post I state that:
Is this the equation to use as we now know V1?
This equation can be used.
 
  • #42
Thanks again Chester.

All the best.
 

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