Understanding Multistage Compressors: Efficiency & Pressure Ratios Explained

[shaiqbashir]

Hi Guys!

well! I am having a problem in understanding the concept of compressors.

my teacher was explaining today the MULTISTAGE COMPRESSORS in which the initial pressure was "P1", the intermediate pressure was "Px" and the final pressure was "P2", my teacher said this statement:

" for higher efficiency we need to make the ratio P2/P1 as small as possible"

this thing just starts clicking in my mind. Because we know that we use compressors in order to increase the pressure of the fluid. It means that we want P2 i.e. final pressure to be as high as possible, higher the P2, higher will be the ratio P2/P1, so why the efficiency wouldn't be greater then?

why according to our sir, this ratio should be as small as possible in order to have a greater efficiency,

Plz help me

Thanks in advance

 

[Clausius2]

The more final pressure the more temperature, and so the more energy needed to reach such temperature... the more irreversibility generated, therefore the less efficiency. You can check it in a T-S diagram and see how the isobarics diverge as you advance up in Temperatures. Three main reasons come to my mind

i) Thermodynamic reasons: as I have said, the more compression ratio the more final temperature, the more irreversibility generated and so the more work needed for reaching such pressure jump compared with the isentropic work needed.

ii) Fluid Dynamics reasons: The more counter pressure, the more diffuculty of flowing to compressor outlet. A great jump of pressure might cause stalling inside compressor, decreasing its performance.

iii) Thermal reasons: the more final temperature the more heat transfer through walls, and so the more work needed for establishing the desired jump of pressures.

 

[FredGarvin]

It means that we want P2 i.e. final pressure to be as high as possible, higher the P2, higher will be the ratio P2/P1, so why the efficiency wouldn't be greater then?

For any application, you don't really want P2 to be as high as you can get it. You want it only as high as you need for the particular system. This is another engineering tradeoff that is always made. You want P2 to be the smallest amount your system can handle. You won't sacrifice the required output just to increase the compressor efficiency. Likewise you wouldn't want to sacrifice efficiency to provide more than required. I hope that made sense because I had to type it three times.

 

[Danger]

Clausius & Fred;
Hi guys. For some reason I always think of axial-flow compressors when in these forums, but I just got to wondering if the same 'law of diminishing returns' formulae apply to centrifugal or piston types. Is one better than the others at reaching maximum efficiency (regardless of weight or size trade-offs)? For instance, I've always much preferred a Roots blower to a turbocharger for vehicle applications even though it robs horsepower to run it. (Okay, I admit that the look and the noise have some influence there , but I still prefer positive-displacement for that application.)

 

[Q_Goest]

Consider an idealized compression cycle where a gas is taken from a pressure, P1, to a pressure P2. For the sake of visualization, let's consider a reciprocating compressor, though any compressor type is applicable. A recip has a piston that has a force on it equal to the area times the pressure. The work done by the piston is the force times the distance the piston travels through. As the piston compresses the gas, the molecules get closer together and energy is added to the molecules. Eventually, the pressure gets up to P2 and the gas is pushed out of the cylinder at that pressure. The total work then, is given by the pressure at each step, times the distance.

W = PdV

By integrating the pressure over the volume, we get the total work. Note that the force is Pressure times Area, which is then multiplied by the distance the piston moves through to give Pressure times Volume. Volume = Area * Length so this is really saying:

W = PAdL where A is area and L is length. So if it helps to visualize the work in small steps, consider how the pressure increases as the piston moves some small length L.

Compression is a reversible process if it is done adiabatically (ie: if there is no heat transferred into or out of the gas during the compression process). In this case, the compression follows a line of constant entropy*. It is an isentropic process. During an isentropic compression process, the gas heats up. The work that goes into the gas can be thought of as going into 1) increasing the pressure and 2) heating the gas. Remember that for any gas that heats up, the pressure is also increased simply because of the increased temperature. You can see how temperature affects pressure by remembering the ideal gas equation: PV=nRT. With only an increase in temperature, and if there is no increase in volume, the pressure must also rise. This isn't what happens, since the volume is also decreasing. But what this points out is that the pressure will rise more quickly if the energy added to the gas increases the temperature during compression. In other words, if we compared two processes, one in which some of the heat were removed so the temperature rose more slowly, and one in which none of the heat were removed, there would be less work needed to compress the gas in which heat was removed since the pressure would rise more slowly.

What we learn from this is that there are a two of ways of reducing the amount of work needed to compress a gas. The first way is to simply remove heat during the compression process. By doing so, the pressure will not rise as quickly and the piston will not be pushing as hard at each point in the stroke. The second way of reducing the amount of work is less obvious. By making a compressor in multiple stages, we cool the gas after each compression process. So even though we have the same flow rate through the compressor and the same final pressure, if we stop the compression process half way and cool the gas back down at constant pressure and then continue to compress it, we see there is an advantage to be gained by reducing the pressure half way though the cycle. If we do this an infinite number of times, then we're essentially taking out all the heat that enters the process and the process is equal to an isothermal process.

So the conclusion we have is that if we must compress a given mass flow rate of gas from a pressure P1 to a pressure P2, the more stages of compression we add, the more cooling we can obtain, and the closer to an isothermal process we can obtain which reduces the total amount of energy needed to compress a gas.

*For a compression process, the pressure and volume relationship can be calculated using the polytropic formula: PV^n = constant. For an adiabatic process n is equal to the ratio of specific heats. For an isothermal process n is equal to 1.

 

[Danger]

Thanks, Q. That's pretty much the maximum amount of math that I can handle, but it sure clarified things and will be useful in future. In fact, I copied your post into my Notepad files so I don't have to worry about trying to find it again.

 

[quark]

I think your teacher is giving an example for multistage compression as you alluded to in the subject line of your question. When P2/P1 is high, what you have to do is to split the process. Optimum power consumption of a multistage compressor is possible when the adiabatic power requirement for each stage is same. In simple terms, the itermediate pressure Px should equal (P1P2)^1/2 where all the pressures are in absolute terms.

 

[shaiqbashir]

Thanks a lot Guys for all ur precious Support