Compressor Work in Brayton Cycle

Thats maybe the Brayton factor. What does the book tell you about Gamma?

thanks for your reply..

the book doesnt tell anything. But alot of problems.. i tried to obtain work using different methods like using efficiency and enthalpy.. which implied the modified equation

gives the correct result. But the actual derived equation

gives wrong results for work in compression or expansion in brayton cycle.
And i dont know anything about brayton factor... what is it?

"...the book doesnt tell anything...", means that you didn't really look, so look again or get a better book, gamma is the Heat Capacity Ratio. You can start your education process here, not by relying on random uneducated guesses from anonymous strangers.

Gas simply undergoes non ideal compression or at polytropic process where gamma>1 because control mass is insulated.

If it in the case that compression or expansion is done in such a way that temperature is maintained at a fixed value or constant, gamma is equated to 1.

W = ∫Pdv ; Pv^δ = const, δ=1 for isothermal process, for adiabatic δ>1 ; P= const./v

W = mRT ∫dv/v

W= mRT ln(v₂/v₁)

Try deriving it when gamma is greater than 1, and you'll see your posted equations.

---This means another bottle of beer

Re Noudge 79 #2

Can you get a gamma of 1?

If gamma equates to 1, then you would get a 'divide by zero'

Can you elaborate? Did #6 answer your question?

Re #2, #3 and #6. Is there a situation where λ (or k) = 1, or near 1.

Because then W → ∞.

Perpetual motion ??

No. As k → 1, n → 0 but

W = (P₁*V₁)/n*((P₂/P₁)ⁿ - 1) does not → ∞. It becomes

W = (P₁*V₁)*ln(P₂/P₁) corresponding to k = 1 for isthermal compression.

You can show that limit of (xⁿ - 1)/n as n → 0 is ln(x) using L'Hospital's rule for indeterminate forms, or easier, let Mathcad do it.

To Codemaster #12

GA and thanks for taking the trouble to reply in detail. It is appreciated.

No problem! Interesting questions.

I think I remember seeing this as well, but I do not recall the reason for the additional factor of heat capacity ratio. Do you think the first equation applies when there only one simple process defining the compression or expansion? Perhaps then the need for the extra factor is due to a Brayton (efficient) cycle requiring multiple stages of compression and expansion to arrive at the operating points. Or is this due to the inclusion into the calculation the changes in velocity of the gas as it moves off the blading onto the diaphrams, etc.?

Should not Mach number come into this somewhere?

I was hoping nobody would ask that because I don't know the answer!

Doing the integration for compressing a fixed volume gives the formula without the extra factor. I can see that in a continuous process like a compressor the power could be higher, but I haven't seen a calc to give the factor and I can't think how to work it out. I got it from a book by Hick Hargeaves, Vacuum/pressure producing machines and associated equipment, which is a technical guide, not a textbook. But I'm sure it's correct as the powers calculated agree with figures in literature from various suppliers, e.g. Aerzen.

I think your second explanation is nearer the mark as the formula applies to various types of compressor - Roots, screw, piston, 1-stage high-speed centrif, multi-stage centrif etc.

Not sure about the Mach number. For "ordinary" compressors gas velocity doesn't appear explicitly. Maybe high velocity would give higher losses hence lower efficiency (which obviously has to be taken into account to get compressor shaft power) but doesn't alter the basic formula.

http://web.mit.edu/16.unified/www/FALL/thermodynamics/thermo_6.htm

The above is a pretty well thought out derivation of the work flow equation.

Good link, I'll get into it and see if I can follow it.

I had a look at your link but found it heavy going. I found link below which was a bit easier.

http://www.see.ed.ac.uk/~johnc/teaching/plantengineering2/2005-06/ohs/compressors.pdf

Apparently for flow work we need to consider the change of enthalpy so integrate V.dP instead of P.dV as in non-flow compression. Not entirely sure why, my thermodynamics is somewhat rusty, but it works. I think my link overcomplicates it, if you just substitute for V from P.V^k = constant, the integration comes out quite easily and with n = (k - 1)/k gives the formula

W = (P₁*V₁)/n*((P₂/P₁)ⁿ - 1) directly.

I think that's the more useful form as usually P₁, V₁ and P₂ are known.

Formula in my link

w_s = n*R*T₁/(n - 1)*((P₂/P₁)ⁿ - 1) (with n in place of γ) doesn't include flow or volume, so I can only assume it means work per unit mass (hence lower-case w_s).

Thanks for the good link, better than my sorry one.

James - when discussing this we only looked at work for compressible flow, in non-flow and flow conditions. I don't know whether you read thread below, about power for liquid flow, but it raises an interesting point (interesting to me, anyway).

With n = (k - 1)/k, in non-compressible flow n = 1, k = ∞ so for non-flow work

W = 1/(k - 1)*P₁*V₁*((P₂/P₁)^(k-1)/k - 1) = 0.

But flow work W = k/(k - 1)*P₁*V₁*((P₂/P₁)^(k-1)/k - 1) = P₁*V₁/n*((P₂/P₁)ⁿ - 1)

= P₁*V₁*(P₂/P₁ - 1) = V₁*(P₂ - P₁) = V₁*ΔP as for liquid flow.

So for non-compressible, difference between non-flow and flow work goes all the way from zero to greater than adiabatic compressible flow, with k = γ.

http://cr4.globalspec.com/thread/100127/Comparing-Work-Required-At-Compressing-Gas-Pumping-Liquid

Looks like a good general solution that applies to all cases. Good work!