Calculate compressor efficiency

  • Thread starter varnish
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  • #1
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Hello

I am struggling with how a compressor's efficiency is calculated.

This is the set up:

I have a Garrett T3 compressor housing being spun by an electric motor.
I can measure the airflow in, the airflow out, the pressure and temperature on both sides of the compressor, the power being used by the motor and the rpm the thing is spinning at.

Can someone give me a nice simple equation to calculate the efficiency and explain please?
Please consider I am not a physics or engineering major and do not not have a degree in mathematics.

Laymen's terms would be appreciated.

Thanks guys
 

Answers and Replies

  • #2
minger
Science Advisor
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There are various definitions of efficiency when it comes to compressors and turbines. Any efficiency is going to a function of what you put in divided by what you "get out". A common definition is something like
[tex]
\eta = \frac{ \dot{W}_{in}}{\Delta h_{in}}
[/tex]
Or basically, the ratio of the work we're inputting divided by the enthalpy rise in the stage. We can write this as:
[tex]
\eta = \frac{ T\omega}{\dot{m}c_p \Delta h}
[/tex]
Where T is the input torque and [tex]\omega[/tex] is the rotational speed, combined they are the work per unit time (power) of the motor. On the bottom we have total enthalpy.

Certainly one doesn't measure enthalpy, so typically I've seen the enthalpy term written out in terms of total temperature and change in total pressure, which you should be able to measure with what you have.

edit: OK, I actually read your post and seen you're not an engineer, so some of the terms may not come easy. So, we can use isotropic definitions for pressure and temperature to convert the previous equation to
[tex]
\eta = \frac{T\omega}{ \dot{m}c_p T_{T1} \left[ \left( \frac{P_{T2}}{P_{T1}}\right)^{\frac{\gamma-1}{\gamma}} - 1 \right] }
[/tex]
Where the total pressure is defined as:
[tex]
P_T = p + \frac{\rho V^2}{2}
[/tex]
And the total temperature defined as:
[tex]
T_T = T + \frac{V^2}{2c_p}
[/tex]
So, you will need to use the temperature at the inlet and outlet planes to compute a gamma (ratio of specific heats). You will also need to compute density. This can be done from tabular lookup or from an empirical relation. From density, area, and the measured airflow you can compute velocity. From velocity and measure pressure/temperature you can compute total properties (you'll also need to look up specific heat at constant pressure cp similar to gamma).

Once you have total pressure and temperature at both inlet and outlet planes, you can compute an efficiency given your mass flow and input from your motor. Note that you will need a few conversion factors depending on the units of your power from the motor.
 
Last edited:
  • #3
jack action
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Gold Member
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If you are talking about the efficiency on a http://www.turbobygarrett.com/turbobygarrett/tech_center/turbo_tech103.html" [Broken], they are called isentropic (or adiabatic) efficiencies. They are calculated this way:

[tex]\eta_c = \frac{T_{out\ th} - T_{in}}{T_{out} - T_{in}}[/tex]

Where:

[tex]\eta_c[/tex] = compressor efficiency
[tex]T_{in}[/tex] = inlet temperature
[tex]T_{out}[/tex] = outlet temperature
[tex]T_{out\ th}[/tex] = theoretical outlet temperature

The actual outlet temperature will aways be greater than the theoretical one. The energy taken to heat up the air is consider energy lost that you have to supply by putting more work at the input shaft (Temperature is a measure of the energy «stored» in the air).

The theoretical outlet temperature depends on the pressure ratio and inlet temperature. For air:

[tex]T_{out\ th} = T_{in} \left( \frac{P_{out}}{P_{in}} \right)^{0.2857}[/tex]

Mixing both equations, the efficiency equation can be rewritten this way:

[tex]\eta_c = \frac{\left( \frac{P_{out}}{P_{in}} \right)^{0.2857} - 1}{\frac{T_{out}}{T_{in}} - 1}[/tex]

You have to use temperature in the proper unit: Kelvin (K). (http://en.wikipedia.org/wiki/Unit_conversion#Temperature")

[tex]\frac{P_{out}}{P_{in}}[/tex] is the pressure ratio, so you have to use absolute pressure and NOT gauge pressure. (http://www.turbobygarrett.com/turbobygarrett/tech_center/turbo_tech103.html?#t103_1" [Broken])

If you need another source, you can read http://www.not2fast.com/turbo/glossary/turbo_glossary.shtml#compressor_efficiency" which basically says the same thing I wrote (the equation is the same even if it is written slightly differently).
 
Last edited by a moderator:
  • #4
549
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Very very much simplified, if you start with your gas going into the compressor at a certain state, pressure, temperature, density and velocity and you measure the outlet pressure, density and velocity, you can calculate what the temperature should be, any amount that the actual outlet temperature is above the calculated is a measure of the compressor inefficiency.
ASME PTC 10 will give you all the details
 
Last edited:
  • #5
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Hello

I am struggling with the difference between isentropic efficiencies and adiabatic efficiencies.
It kinds like have a same meaning. But why they need different requirements?



thanks guys
 
  • #6
jack action
Science Advisor
Insights Author
Gold Member
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From Wikipedia:

Adiabatic:

An adiabatic process is any process occurring without input or output of heat within a system.

(...)

An adiabatic process that is reversible is also called an isentropic process. Conversely, an adiabatic process that is irreversible and extracts no work is in an isenthalpic process
Isentropic:

an isentropic process (...) is one in which for purposes of engineering analysis and calculation, one may assume that the process takes place from initiation to completion without an increase or decrease in the entropy of the system, i.e., the entropy of the system remains constant. It can be proven that any reversible adiabatic process is an isentropic process.
So, if an adiabatic process is reversible it is also isentropic. This is why the two terms are often used without distinction, although we should really use the term reversible in conjunction with adiabatic to truly say it is isentropic (but our laziness makes it always implied).
 

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