Calculating parameters of centrifugal compressor

  • #1
MysticDream
80
3
TL;DR Summary
Trying to understand Euler's turbine equation and velocity triangles.
I'm having a difficult time understanding Euler's turbine equation as it relates to designing a centrifugal air compressor. I need to calculate the impeller size if I know the inlet pressure, outlet pressure, rpm, and mass flow rate. The velocity triangles and definitions have me confused. It's the exit velocity of the air that I need to determine.

I have this equation for rate of work (power) from Euler's turbine equation:
$$P = \dot m(u_2c_2-u_1c_1)$$
where
u = tangential velocity of the impeller at inlet and outlet
c = absolute velocity of the fluid at inlet and outlet

I know I can use the formula for adiabatic compression to determine the power if I know the inlet volumetric flow rate, inlet pressure, outlet pressure, and RPM:
$$P = \left( \frac{(V_1P_1-V_2P_2)}{(\gamma-1)}\right)\left(\frac{RPM}{60}\right)$$

That leaves me with two variables in the first equation that I don't know, the absolute velocity of the air at the inlet and outlet. If the cross sectional area between the blades was constant (in say a tapered impeller), wouldn't the absolute velocity of the gas be the same at inlet and outlet? I know I can make variables "u" whatever I want by changing the internal and external diameter of the impeller until I get the desired exit gas velocity.

If there is another strategy to solve this problem, I'd like to know. If anyone has any insight to offer, I'd greatly appreciate it.
 
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  • #2
The inlet diameter of the impeller is normally determined by the diameter of the inlet duct, which corresponds with certain mass flow rate and inlet velocity.
For a compressor, it is desired the absolute velocity of the gas at the outlet to be smaller than at the inlet.
In that way, the achieved outlet static pressure is greater.

Please, see:
https://www.engineeringtoolbox.com/air-compressor-inlet-pipe-air-flow-d_1193.html

https://www.engineeringtoolbox.com/specific-work-turbo-machines-d_629.html

https://en.wikipedia.org/wiki/Velocity_triangle

https://en.wikipedia.org/wiki/Euler's_pump_and_turbine_equation
 
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  • #3
Lnewqban said:
The inlet diameter of the impeller is normally determined by the diameter of the inlet duct, which corresponds with certain mass flow rate and inlet velocity.
For a compressor, it is desired the absolute velocity of the gas at the outlet to be smaller than at the outlet.
In that way, the achieved outlet static pressure is greater.

Please, see:
https://www.engineeringtoolbox.com/air-compressor-inlet-pipe-air-flow-d_1193.html

https://www.engineeringtoolbox.com/specific-work-turbo-machines-d_629.html

https://en.wikipedia.org/wiki/Velocity_triangle

https://en.wikipedia.org/wiki/Euler's_pump_and_turbine_equation
Thanks, I've read the wiki articles but the information is lacking. It seems often variables aren't described thoroughly.
 
  • #4
MysticDream said:
TL;DR Summary: Trying to understand Euler's turbine equation and velocity triangles.

I'm having a difficult time understanding Euler's turbine equation as it relates to designing a centrifugal air compressor. I need to calculate the impeller size if I know the inlet pressure, outlet pressure, rpm, and mass flow rate. The velocity triangles and definitions have me confused. It's the exit velocity of the air that I need to determine.

I have this equation for rate of work (power) from Euler's turbine equation:
$$P = \dot m(u_2c_2-u_1c_1)$$
where
u = tangential velocity of the impeller at inlet and outlet
c = absolute velocity of the fluid at inlet and outlet

I know I can use the formula for adiabatic compression to determine the power if I know the inlet volumetric flow rate, inlet pressure, outlet pressure, and RPM:
$$P = \left( \frac{(V_1P_1-V_2P_2)}{(\gamma-1)}\right)\left(\frac{RPM}{60}\right)$$

That leaves me with two variables in the first equation that I don't know, the absolute velocity of the air at the inlet and outlet. If the cross sectional area between the blades was constant (in say a tapered impeller), wouldn't the absolute velocity of the gas be the same at inlet and outlet? I know I can make variables "u" whatever I want by changing the internal and external diameter of the impeller until I get the desired exit gas velocity.

If there is another strategy to solve this problem, I'd like to know. If anyone has any insight to offer, I'd greatly appreciate it.
This is from my fluids text.

Theoretical adiabatic power of a centrifugal compressor is given by:

$$ P_{theo} = \frac{k}{k-1} Q_1 p_1 \left[ \left( \frac{p_2}{p_1} \right)^{(k-1)/k} - 1 \right] $$

Where;

##k## is the ratio of specific heats
##Q_1## inlet volumetric flowrate
##p_1## inlet pressure
##p_2## outlet pressure

So with the information you have (can find) you should be able to compute ##P_{theo}##.EDIT: Upon examining the "Euler Turbine Equation", I see this is not what you are after specifically. Whoops!
 
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  • #5
erobz said:
This is from my fluids text.

Theoretical adiabatic power of a centrifugal compressor is given by:

$$ P_{theo} = \frac{k}{k-1} Q_1 p_1 \left[ \left( \frac{p_2}{p_1} \right)^{(k-1)/k} - 1 \right] $$

Where;

##k## is the ratio of specific heats
##Q_1## inlet volumetric flowrate
##p_1## inlet pressure
##p_2## outlet pressure

So with the information you have (can find) you should be able to compute ##P_{theo}##.EDIT: Upon examining the "Euler Turbine Equation", I see this is not what you are after specifically. Whoops!
Yeah, but this is helpful as it's another way to double check if my power calculation is correct which I'm now doubting. If you think about it, a compressor not only has the work to do to adiabatically compress the gas, but it then has to push that reduced volume of gas into the reservoir which is more isobaric work. This equation unfortunately can't give me the impeller size but again, thanks anyway.

I'm surprised this equation doesn't include a mass flow rate or density factor. It includes the volumetric flow rate and pressure difference but wouldn't it take more power to compress 200 to 300 psi than it would to compress 100 to 200 psi? In each case the volumetric flow rate and pressure difference would be the same, but in the 200 to 300 psi case, more mass is flowing through the blades so you'd think more work would need to be done.
 
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  • #6
MysticDream said:
Yeah, but this is helpful as it's another way to double check if my power calculation is correct which I'm now doubting. If you think about it, a compressor not only has the work to do to adiabatically compress the gas, but it then has to push that reduced volume of gas into the reservoir which is more isobaric work. This equation unfortunately can't give me the impeller size, but again, thanks anyway.

I'm surprised this equation doesn't include a mass flow rate or density factor. It includes the volumetric flow rate and pressure difference but wouldn't it take more power to compress 200 to 300 psi than it would to compress 100 to 200 psi? In each case the volumetric flow rate and pressure difference would be the same, but in the 200 to 300 psi case, more mass is flowing through the blades so you'd think more work would need to be done.
Mass flow rate is implicitly there in ##Q_1## inlet volumetric flowrate( assuming we have inlet pressure and temp)?

However, it would seem that the adiabatic theoretical power of the compressor is fully constrained independent of the impeller design with the information you have as "known", so ##P_{theo}## should be one of the parameters in your equation, not a variable?Maybe this paper describes the variables of the reaction turbine better?

https://ceng.tu.edu.iq/med/images/محاضرات_1-10-2021/Ch.8_Reaction_Turbine.pdf

I also have a diagram in my text book if you are interested.
 
  • #7
erobz said:
Mass flow rate is implicitly there in ##Q_1## inlet volumetric flowrate( assuming we have inlet pressure and temp)?

However, it would seem that the adiabatic theoretical power of the compressor is fully constrained independent of the impeller design, so ##P_{theo## } should be one of the parameters in your equation, not a variable?https://ceng.tu.edu.iq/med/images/محاضرات_1-10-2021/Ch.8_Reaction_Turbine.pdf
I think you'd be correct IF the temperature was known. Volume and pressure alone can't give mass unless temperature is known. Temp and pressure can give you density. Density and volume can give you mass. EDIT: I see you caught that before I posted my reply.

Well, the power isn't actually constrained because I know the other parameters: mass flow rate, inlet and outlet pressures. For these parameters there will be a power requirement. I surely have to calculate this power requirement correctly, however. Then after having the power requirement, I want to be able to calculate the impeller geometry and size to get the job done at a specified RPM. You know, typical induction motors turn at 3600 rpm due to the grid frequency, so the RPM could be 3600 or some multiple of it with a transimission. Either way, the RPM options are limited so it is a defining factor.
 
  • #8
MysticDream said:
I think you'd be correct IF the temperature was known. Volume and pressure alone can't give mass unless temperature is known. Temp and pressure can give you density. Density and volume can give you mass.

Well, the power isn't actually constrained because I know the other parameters: mass flow rate, inlet and outlet pressures. For these parameters there will be a power requirement. I surely have to calculate this power requirement correctly, however. Then after having the power requirement, I want to be able to calculate the impeller geometry and size to get the job done at a specified RPM. You know, typical induction motors turn at 3600 rpm due to the grid frequency, so the RPM could be 3600 or some multiple of it with a transimission. Either way, the RPM options are limited so it is a defining factor.
But you have defined thermodynamic end states and process? If you know inlet temperature and pressure ##p_1,T_1##, and it undergoes an adiabatic process to ##p_2##, then ##T_2## is fixed. And that's why the equation I show doesn't require "compressor design" factors. I truly think it becomes an input ##P## to the equation you wish you use to design the impeller?

I'm probably missing something.
 
  • #9
erobz said:
But you have defined thermodynamic end states and process? If you know inlet temperature and pressure ##p_1,T_1##, and it undergoes an adiabatic process to ##p_2##, then ##T_2## is fixed. And that's why the equation I show doesn't require "compressor design" factors. I truly think it becomes an input ##P## to the equation you wish you use to design the impeller?

I'm probably missing something.
Well I think equation you showed determines the power requirement to do the work on the gas itself in an ideal theoretical case. It's similar to how we do calculations assuming a massless piston. In reality, there could never be a massless piston.
 
  • #11
MysticDream said:
I'm surprised this equation doesn't include a mass flow rate or density factor.
It does because the density ratio is linked to the pressure ratio:
$$ \left( \frac{p_2}{p_1} \right)^{(k-1)/k} = \left( \frac{\rho_2}{\rho_1} \right)^{(k-1)}$$
Also, note that the exponential pressure ratio in the equation is really just a simple temperature ratio:
$$ \left( \frac{p_2}{p_1} \right)^{(k-1)/k} = \frac{T_2}{T_1}$$
 
  • #12
jack action said:
It does because the density ratio is linked to the pressure ratio:
$$ \left( \frac{p_2}{p_1} \right)^{(k-1)/k} = \left( \frac{\rho_2}{\rho_1} \right)^{(k-1)}$$
Also, note that the exponential pressure ratio in the equation is really just a simple temperature ratio:
$$ \left( \frac{p_2}{p_1} \right)^{(k-1)/k} = \frac{T_2}{T_1}$$
Ah, good point. It's not the pressure difference but the pressure ratio. I'm still not understanding though. Are you saying the pressure ratio is all that matters; that a constant volume process of 400/200 psi requires the same power as 200/100 psi? This doesn't make sense to me because even a simple fan would require more power to move more mass and a higher starting density would be more mass per volume. Could you please clarify?
 
  • #13
MysticDream said:
that a constant volume process
The outlet density is different from the inlet density so it is not an isochoric process; it is an isentropic one.
 
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  • #14
jack action said:
The outlet density is different from the inlet density so it is not an isochoric process; it is an isentropic one.
Sorry, I used the wrong term. I meant a constant inlet volumetric flow rate. That is what the formula had as a variable, the inlet volumetric flow rate.
 
  • #15
MysticDream said:
even a simple fan would require more power to move more mass and a higher starting density would be more mass per volume.
A compression process doesn't imply a moving mass.

If the air has velocities at the inlet and outlet, you used the stagnation values which do take into account the velocity, namely:
$$T_0 = T + \frac{V^2}{2C_p}$$
Or:
$$\frac{T_0}{T} = 1+\frac{k-1}{2}M^2$$
Keep in mind that the original equation comes from a difference in [total] enthalpy between the inlet and outlet:
$$P = \dot{m}\left(h_{0_2} - h_{0_1}\right)$$
$$P = \dot{m}C_p T_{0_1} \left(\frac{T_{0_2}}{T_{0_1}} - 1\right)$$
$$P = \dot{m}\frac{k}{k-1} RT_{0_1} \left(\frac{T_{0_2}}{T_{0_1}} - 1\right)$$
$$P = \dot{m}\frac{k}{k-1} \frac{p_{0_1}}{\rho_{0_1}} \left(\frac{T_{0_2}}{T_{0_1}} - 1\right)$$
$$P = \dot{m}\frac{k}{k-1} \frac{p_{0_1}}{\rho_{0_1}} \left[ \left(\frac{p_{0_2}}{p_{0_1}} \right)^{(k-1)/k} - 1\right]$$

Here I did not convert ##\dot{m} = \rho \dot{V}## as I don't think it would simplify that easily with stagnation properties. I would have to think about it. Anyway, keeping everything with total temperature is a lot cleaner and less error-prone:

$$P = \dot{m}\frac{k}{k-1} RT_{0_1} \left(\frac{T_{0_2}}{T_{0_1}} - 1\right)$$
 
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  • #16
jack action said:
Here I did not convert ##\dot{m} = \rho \dot{V}## as I don't think it would simplify that easily with stagnation properties. I would have to think about it. Anyway, keeping everything with total temperature is a lot cleaner and less error-prone:

$$P = \dot{m}\frac{k}{k-1} RT_{0_1} \left(\frac{T_{0_2}}{T_{0_1}} - 1\right)$$

Yes, makes sense to approach it that way.

I plugged the initial Power equation into excel and it turns out the power does increase for a higher initial pressure, even if the compression ratio and volumetric flow rate (at inlet) is the same. It must the P1 in the equation that accounts for this. I'm still surprised there isn't a density or temperature factor, because a higher temp could be a higher pressure but not more mass per volume.

Did this equation not imply stagnation pressures? If so, I'm assuming the power would be the same as your equation above.

$$ P_{theo} = \frac{k}{k-1} Q_1 p_1 \left[ \left( \frac{p_2}{p_1} \right)^{(k-1)/k} - 1 \right] $$
 
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  • #17
jack action said:
Keep in mind that the original equation comes from a difference in [total] enthalpy between the inlet and outlet:
You're saying I'd have to use an equation that incorporates the mass flow rate to get an accurate power value first. Then the initial equation still holds, even though that particular equation did not include density or mass flow rate. Is that correct?
 
  • #18
You don't seem to appreciate the information given by the equations for the ideal gas law and the isentropic process.

Ideal gas law:
$$\frac{p}{\rho} = RT$$
isentropic process characteristic:
$$\frac{p}{\rho^k} = \text{constant}$$
Assuming you know the constant and ##R##, you have two equations and 3 unknowns. This means that if you know one of those unknowns - ##\rho##, ##p## or ##T## - you necessarily know the other two, For instance, if you know the temperature ##T## then:
$$\frac{\rho RT}{\rho^k} = \text{constant}$$
$$\frac{RT}{\rho^{k-1}} = \text{constant}$$
$$\rho = \left(\frac{RT}{\text{constant}}\right)^\frac{1}{k-1}$$
And the pressure ##p## is naturally:
$$p= \rho RT$$
$$p = \left(\frac{RT}{\text{constant}}\right)^\frac{1}{k-1} RT$$
$$p = \left(\frac{(RT)^k}{\text{constant}}\right)^\frac{1}{k-1}$$
Coming back to your original equation:
$$P = \left( \frac{(V_1P_1-V_2P_2)}{(\gamma-1)}\right)\left(\frac{RPM}{60}\right)$$
it represents the work done per revolution times the number of revolutions per second. The work done per revolution can be rewritten as (because of the ideal gas law):
$$\frac{(V_1P_1-V_2P_2)}{(\gamma-1)} = \frac{(mRT_1-mRT_2)}{(\gamma-1)} = \frac{mR}{(\gamma-1)}(T_1-T_2)$$
Where ##m## is the mass of air processed by the compressor per revolution.
 
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  • #19
jack action said:
You don't seem to appreciate the information given by the equations for the ideal gas law and the isentropic process.

Coming back to your original equation:
$$P = \left( \frac{(V_1P_1-V_2P_2)}{(\gamma-1)}\right)\left(\frac{RPM}{60}\right)$$
it represents the work done per revolution times the number of revolutions per second. The work done per revolution can be rewritten as (because of the ideal gas law):
$$\frac{(V_1P_1-V_2P_2)}{(\gamma-1)} = \frac{(mRT_1-mRT_2)}{(\gamma-1)} = \frac{mR}{(\gamma-1)}(T_1-T_2)$$
Where ##m## is the mass of air processed by the compressor per revolution.
Sorry, I had planned to study your response. You currently have a much better understanding of the subject than I do. I appreciate your insight.

In my last comment I was referring to the initial equation that the 2nd commenter posted which only had P1, P2, and volumetric flow rate as variables to calculate power. I was still stuck on that, not understanding how power could be derived without a mass flow rate. It seems though, that it matters not what the density of the fluid is as the enthalpy is still a product of it's pressure and volume. Therefore, if there was a greater density at a lower temp (at the inlet) it would have the same enthalpy at a lesser density but a higher temp.

Aside from that equation, you've provided other equations that will allow me to calculate power. I will study them.
 
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  • #20
They are both basically the same equation, at least for the work part. So the equations for ideal gas law and isentropic process apply in both cases, meaning if you know one characteristic - ##p##, ##\rho## or ##T## - you know the other two.
MysticDream said:
I was still stuck on that, not understanding how power could be derived without a mass flow rate.
Sorry, maybe I went too fast in my previous post. Here is where the mass flow rate and density are "hidden" in the original equation:
$$P_{theo} = \frac{k}{k-1} Q_1 p_1 \left[ \left( \frac{p_2}{p_1} \right)^{(k-1)/k} - 1 \right]$$
$$P_{theo} = \frac{k}{k-1} \left(\frac{\dot{m_1}}{\rho_1}\right) p_1 \left[ \left( \frac{p_2}{p_1} \right)^{(k-1)/k} - 1 \right]$$
$$P_{theo} = \frac{k}{k-1} \dot{m_1} \frac{p_1}{\rho_1} \left[ \left( \frac{p_2}{p_1} \right)^{(k-1)/k} - 1 \right]$$
Which is the second-to-last equation in post #15
 
  • #21
jack action said:
They are both basically the same equation, at least for the work part. So the equations for ideal gas law and isentropic process apply in both cases, meaning if you know one characteristic - ##p##, ##\rho## or ##T## - you know the other two.

Sorry, maybe I went too fast in my previous post. Here is where the mass flow rate and density are "hidden" in the original equation:
$$P_{theo} = \frac{k}{k-1} Q_1 p_1 \left[ \left( \frac{p_2}{p_1} \right)^{(k-1)/k} - 1 \right]$$
$$P_{theo} = \frac{k}{k-1} \left(\frac{\dot{m_1}}{\rho_1}\right) p_1 \left[ \left( \frac{p_2}{p_1} \right)^{(k-1)/k} - 1 \right]$$
$$P_{theo} = \frac{k}{k-1} \dot{m_1} \frac{p_1}{\rho_1} \left[ \left( \frac{p_2}{p_1} \right)^{(k-1)/k} - 1 \right]$$
Which is the second-to-last equation in post #15
Wow, thanks a lot! This is a lot of information. I couldn't find any source online with all these equations.
 

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