Gas Turbine compressor efficiency calculation

[mikebon111]

Hello
I need to create a PI calculation that can trend compressor degradation. I have found an old calculation in xl from another plant that looks like this
A = RATIO OF PRESSURES AT INLET AND DISCHARGE =((BAROMETER + CDP) DIVIDED BY BAROMETER)RAISED TO THE EXP (0.28627)
B = RATIO OF TEMPERATURE AT INLET AND DISCHARGE = (INLET TEMP + 460) DIVIDED BY (COMP DISCHARGE TEMP+460)
FINAL EFFICIENCY CALC = (1- A)/(1-B)*100

What is EXP (.28627) and what is the 460 added to T1 and T2

Thank you for in advance

 

[zdas04]

You are an Engineer and you don't recognize the conversion from Fahrenheit to Rankine? You really might think of having someone sit down with you for a month or two for a review of high school physics.

I don't have a clue what language you cut and pasted, but it is no programming language I'm familiar with. I'm going to assume it is program documentation and the phrase "RAISED TO THE EXP" means "^" so it would be:

[(P(bara)+P(disch))/P(bara)]^0.28627

If I misinterpreted and it is really "RAISED TO" exp(0.28627) why wouldn't they have just said "RAISED TO 1.331". In case you ditched algebra while you were avoiding physics "exp(x)" usually taken as e^x and "e" is called Euler's Constant and has a value of 2.71828.

So this is an air compressor and P(bara) is actually local atmospheric pressure and the term is generally called "compression ratios".

The part that makes no sense is that raising compression ratios >1 to .28627 is greater than 1. Ratio of temperatures is less than 1 so this equation results in "efficiency" around -150%. Negative efficiency is a bad thing. Efficiency greater than 100 is also pretty tough.

David

 

[msquared48]

"EXP" is the experience factor.

"460" is the interest the bank charges per year in eurodollars per 1000 milliliters of displacement.

Mike McCann
MMC Engineering

http://mmcengineering.tripod.com

 

[mikebon111]

Take it easy, I am not an engineer, I am an operator that was asked to create a PI trend. I stumbled upon this forum and figured I'd ask for help. I guess that was a mistake....

 

[msquared48]

Sorry. I couldn't help it. It's Friday. No offense intended.

FYI, this is a site where one engineering professional helps another and it is posted as such on the home page.

Not being an engineer, you should really engage the help of a mechanical engineer to answer your question, although the question already has been answered by zdas04, a registered ME.

Mike McCann
MMC Engineering

http://mmcengineering.tripod.com

 

[DLite30]

I don't recognize that form of the equation...the one I would use for adiabatic efficiency would be:

((P2/P1)^0.28627-1) / ((T2-T1)/T1)

T1=inlet temp in Rankine
T2=discharge temp
P1=inlet presure in psia
P2=disharge pressure in psia
0.28672 = (k-1)/k, where k is the ratio of specific heats for air, which is somewhere around 1.4 depending on your average temperature.

 

[zdas04]

DLiteE30,
Thanks for describing the constant, I've been thinking about where it might have come from since the original post. So since:

T(adiabatic)=T1*(P2/P1)^((k-1)/k)

and your equation is

T1*[(P2/P1)^((k-1)/k)-1]/[Δ]T

That is the same as

(T(adiabatic)-T1)/(T2-T1)

It all makes sense as long as T2>T(adiabatic) which is generally true except when you have leaking suction valves (which cause the early introduction of gas and lower compression ratios and less heat of compression within the cylinder). This equation could easily show a malfunctioning machine as >100% efficient.

David