Superchargers?

[ThomasP]

Hey Snowygrouch,

Please see my 2nd paragraph post#105 and 1st paragraph post#112 re work.

I think the only difference we see re the Merlin 2-stage air flow is that I read the route from the 1st stage outlet to the 2nd stage inlet as a spiral instead of a U?

 

[tomo pauk]

This is a WW2 German set of compressor maps for a 2 stage supercharger (centirifugal) showing the performance of each stage
and the combined performance. Theses are in Had, adiabatic pressure head (work) in which the total compressor work can be
represented as the work of each stage ADDED. which you can see as its basically 6000+6000=12,000 m.kg/kg

This does not work for pressure ratio, and needs to be caculated as per the equation I previously posted

(look for "314" which is the impeller tip speed in m/s)
View attachment 535124

I appreciate your posts Callum, I just wish I could understand them. Great stuff.

The 1st diagram (Bild 4), says roughly: "supercharger map of the 1st stage of 2-stage blower, calculated per Diagram 1 (Bild 1)".
2nd diagram (Bild 5) says roughly: "supercharger map of the 2nd stage of a 2-stage blower, calculated per Diagram 3 (Bild 3)".
3rd diagram: "supercharger map of a 2-stage blower, calculated per Diagrams 4 and Diagram 5"
(impeller tip speed in m/sec; VL in m^3/sec is the weight of air the blower is 'moving', Had in m.kg/kg is blower's work, efficiency is noted being, for example, 0,75 = 75%, or 0.65 = 65%)

 

[wuzak]

The actual formula for multiple stage compression is similar to the formula I used to show the pattern of heat rise in the compressed air in my post#94 earlier in this thread:

Ta + T1 - 25 + T2 - TAC = MT

If you substitute P for each T increase , and substitute i for each temperature decrease you get:

Pa + P1p - P1i + P2p - P2i = MP

simplified to

P1e + P2e = MP

Your direct correlation between pressure and temperature presupposes that the volume remains constant.

In fact pressure temperature and volume are interconnected.

In a 2 stage system, the second stage pressure ratio is determined by the pressure, temperature and mass flow of the air fed by the first stage.

The second stage and feed ducts cause a back pressure for the first stage, changing its performance.

Once all those factors are considered, the overall pressure ratio is the product of each stage's ratio.

 

[wuzak]

And regarding calculating the pressure ratio of axial compressor, I believe the Navier-Stokes equations were applied.

 

[pbehn]

The 1st diagram (Bild 4), says roughly: "supercharger map of the 1st stage of 2-stage blower, calculated per Diagram 1 (Bild 1)".
2nd diagram (Bild 5) says roughly: "supercharger map of the 2nd stage of a 2-stage blower, calculated per Diagram 3 (Bild 3)".
3rd diagram: "supercharger map of a 2-stage blower, calculated per Diagrams 4 and Diagram 5"
(impeller tip speed in m/sec; VL in m^3/sec is the weight of air the blower is 'moving', Had in m.kg/kg is blower's work, efficiency is noted being, for example, 0,75 = 75%, or 0.65 = 65%)

Thanks Tomo but my German is good enough to understand the German but my knowledge of fluid dynamics and all the other laws of physics is overloaded, I can however follow the basic principles and find it all interesting.

 

[ThomasP]

Hey Wuzak,

re: from your post#115. "My point is that in order for the method espoused by Thomas of adding the pressure ratios together to get the overall ratio, since it would require ratios less than 1:1"

This illustrates some of the confusion over additive work (physics term) processes.

If we use the MetroVick F2/2 9-stage compressor as an example, and it is capable of achieving a 3.5:1 pressure ratio at sea level, the actual pressure rise is from 14.7 lb/in^2 to 51.45 lb/in^2, a difference of 36.75 lb/in^2. If each of the stages achieved the same amount of work (physics term), then (as you said) each stage would add ~.4, not as a ratio, but as 4 lb/in^2 (4.083 to be more precise) to that stage's intake pressure. So we get:

14.7 + 4.083 + 4.083 + 4.083 + 4.083 + 4.083 + 4.083 + 4.083 + 4.083 + 4.083 = 51.45

However, since we are starting at an ambient air pressure that is greater than 0 (in this case 14.7 lb/in^2) none of the stages would have a ratio of less than 1:1. So we get the following ratios for the individual stages:

1st stage ratio = (14.7 + 4.083) / 14.7 = 18.783 / 14.7 = 1.278:1
2nd stage ratio = (18.783 + 4.083) / 18.783 = 22.866 / 18.783 = 1.217:1
3rd stage ratio = (22.866 + 4.083) / 22.866 = 26.949 / 22.866 = 1.179:1
4th stage ratio = (26.949 + 4.083) / 26.949 = 31.032 / 26.949 = 1.151:1
5th stage ratio = (31.032 + 4.083) / 31.032 = 35.115 / 31.032 = 1.132:1
6th stage ratio = (35.115 + 4.083) / 35.115 = 39.198 / 35.115 = 1.116:1
7th stage ratio = (39.198 + 4.083) / 39.198 = 43.281 / 39.198 = 1.104:1
8th stage ratio = (43.281 + 4.083) / 43.281 = 47.364 / 43.281 = 1.094:1
9th stage ratio = (47.364 + 4.083) / 47.364 = 51.447 / 47.364 = 1.086:1

If you then multiply the ratios you get:

1.278 x 1.217 x 1.179 x 1.151 x 1.132 x 1.116 x 1.104 x 1.094 x 1.086 = 3.497

3.497 = 3.5 (almost )

 

[Shortround6]

My knowledge of fluid dynamics and all the other laws of physics may be somewhat on the elementary side so let's try a little example and perhaps people can point out the flaws in my thinking.

Lets say we have a 2 stage supercharger that operates on a 3 to ratio in the first stage and a 2 to ratio in the second stage. The first stage (in theory) will raise the pressure from 15lb/sq/in to 45 lb/sq/in and the 2nd stage raises it from 45 lb/sq/in to 90lb/sq/in.

Now is the 2nd stage "work" result added to the 1st stage or has it multiplied it?

Now please note I have made no mention of power required in each stage or temperature rise or any other factors.

lets also assume, as a variation on this that this was a design goal and for some reason the 1st stage didn't achieve it's goal of 45lb/sq/in output and only made 42 lb/sq/in, what happens?

My theory/line of thinking says the 2nd stage multiples the 42 lb/sq/in by 2 to 84lb/sq/in.
Other, less likely lines of thinking may call for an "additive" (second stage miraculously still adds 45lbs/sq/in to the total for 87 psi total?) or some other total?

My working knowledge of fluid dynamics (such as it is) is from working with water which is not compressible so a lot my experience may not apply.

In my example the 2nd stage may be doing more actual "work" since it is compressing the air by 45psi vs the 30psi of the first stage but then it is only reducing the volume to 1/2 the starting volume vs 1/3 like the 2st stage.

 

[pbehn]

I find the discussion interesting, from the theoretical physics then to the practical application which throws up news laws of physics to be got around, then the whole lot has to be mounted in an aircraft and perform at all sorts of altitudes and speed. Finally when talking about temperature pressure and flow rates how do you measure and control them with 1940s tech.

 

[Shortround6]

They could measure them in test houses/on test stands but in operational aircraft everything had to be on the conservative side (as much as possible) to take care of unknowns/weather-temperature variations.

Pilots didn't care about most of the technical details and the pressure gauge was simple, cheap, and worked close enough to get by with. In a test house they could rig pressure gauges at various points in the intake tract and measure the exhaust pressure. It was possible (although not easy) to measure the amount of air (at least the cubic ft) per minute goring through the system.

For flying there were often test aircraft

with one or more positions for engineers/technicians to monitor banks of gauges.

 

[ThomasP]

Hey wuzak,

re your post#124 "And regarding calculating the pressure ratio of axial compressor, I believe the Navier-Stokes equations were applied."

The Navier-Stokes series of equations can and have been used in the development of compressive systems, like the airflow around an airfoil (i.e. a turbine blade), compression in a confined space (i.e. within the confines of a cylinder), and resulting momentum of the air flow (i.e. the mass of the air being moved times the velocity of the air), which can then be used to calculate the Kinetic Energy (KE) of the airflow (i.e. the energy contained in the exhaust), which when applied to a real world mechanism (such as a jet engine) can give you the theoretical thrust. In and of themselves the Navier-Stokes equations will not give you an exact real world value, but said value (if you substitute solvable sub-formula for the variables) should get pretty close to the real world values after refinement. A couple of the equations that can be used to solve the problems of motion of/in a system are below.

I am sure you have heard the term Reynolds number used in the calculation of the modeling of airflow around an airfoil (i.e. usable to help figure the pressure decrease/increase and overall lift of a wing, or the equivalent effect relative to a turbine blade)

and this set of equations used in the modeling of 3-dimensional movement of a body/system (i.e. perhaps a spiral vortex flow of compressible gas with changing radii and varying radial and axial velocities, aka the airflow through a jet engine).

Please note all the adding and subtracting going on.

 

[wuzak]

What if the inlet condition is 1/2 atmosphere?

In your example of the MetroVick F.2/2 does each stage give a 4psi rise still?

 

[ThomasP]

Hey wuzak,

re: "What if the inlet condition is 1/2 atmosphere?"

If you are talking about having 1/2 standard pressure at SL but with the same air temperature as the standard day (15ºC), you would add 1/2 the pressure gain listed above for each stage( i.e. 4.083 / 2 = 2.0415 lb/in^2, so +2.0415) and the resulting CR of 3.5:1 would be the same. If, however, you are asking what the compression ratio would be at 18,000 ft (where the ambient pressure is 1/2 the standard pressure at SL and the standard temperature is -20.6ºC) the pressure gain would be 2.196 lb/in^2, with resulting CR of 3.76:1. The achievable CR at 33,800 ft (where the pressure is 1/4 SL and the temperature is -51.8ºC) would be 4.04:1. From 36,000 ft on up the temperature remains the same (-56.5ºC) and the achievable CR would remain the same at 4.08:1. This is assuming the engine is not moving and the turbine rpm remains the same at all heights.

Altitude______CR
36,000 ft+___4.08:1
33,800 ft____4.04:1
18,000 ft____3.76:1
S.L._________3.5:1

 

[wuzak]

I must admit that I had been under the impression that you were summing the pressure ratios, not the pressure deltas.

I wrote down your addition method in terms of pressure ratios.

That is:
dP1 = P0 x R1 - P0 = P0(R1 - 1)

And you do that for all stages, it comes out as:

P = P0.R1.R2.....Rn

So if you do know the individual pressure ratios you can multiply them to get the overall ratio.

You are correct that you can't directly calculate the pressure ratio of a stage. You have to go through the process of calculating the pressure at each stage.

 

[Buster01]

Here is a link to a table of characteristics. Here is a link to an article on turbochargers with a lot of historical data.

 

[ThomasP]

Hey Buster01,

Great article, thanks for the link. I have visited the Engine History website but had not noticed the article.