The discussion focuses on calculating the volumetric flow rate of free air at 1 bar and 20°C during a two-stage compression process. The polytropic index of compression is specified as 1.28, and the input power per stage is 2 kW. The correct volumetric flow rate was determined to be 0.0148 m³/s using the isothermal compression equation W = p1V1In(p2/p1), where p1 is the inlet pressure and p2 is the outlet pressure. The participants clarified the concept of a two-stage compressor and the importance of intercooling in the compression process.
PREREQUISITESMechanical engineers, thermodynamics students, and professionals involved in HVAC and compressor design will benefit from this discussion, particularly those focusing on air compression systems and energy efficiency.
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 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?
Oh. Missed that. Then you're done. Do you want to try to prove that the 2nd compressor will operate the same?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?
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 said:Is it a similar equation?
But subbing in (p3/p2) for compressor ratio?
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: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.
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.
MCTachyon said:The work input to isothermal compression is given by:
W = p1V1In(p2/p1)
This equation can be used.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?