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

[Chestermiller]

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.

 

[MCTachyon]

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.

 

[Chestermiller]

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.

 

[MCTachyon]

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.

 

[MCTachyon]

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:

The work input to isothermal compression is given by:

W = p₁V₁In(p₂/p₁)

Is this the equation to use as we now know V₁?

 

[Chestermiller]

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.

 

[MCTachyon]

Thanks again Chester.

All the best.