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Thread: Aeronautical/Mechanical or engine Engineer help needed

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    Aeronautical/Mechanical or engine Engineer help needed

    Folks,


    I'm looking for some help from any of our resident aeronautical engineers as I don't seem to understand how the db601 power curve is happening.

    My problem starts with the following chart.


    Notice I have extended blue lines to simulate the output power as if the fluid coupling was removed by extending the upper one and dragging it down to FTH power.


    By looking at the blue lines, my logic tells me that the fluid coupling slip bellow 2100 meters, is saving the engine around 125 sPS

    Ok assuming all the above is correct, let's start with some math...

    I'm using the following formula to calculate the power consumed by the supercharger:
    Gamma / (Gamma - 1) * Q * P0 / Efficiency * (( (P2/P1)^ ((Gamma - 1) / Gamma)) - 1)

    Where:
    Gamma = 1.41
    Q = Engine displacement in M^3 * RPM /60 /2 (33.9 Liters displacement in this case)
    Efficiency = Guesstimaged at 0.7

    With this formula I calculated the following values power consumption values:
    Altitude Power sPS Power Watts Temperature Kelvin Pressure Pascals Compression Ratio
    to get to 1.3 ATA
    0 40.3 29640.74 288.16 101300 1.28
    2100 65.56 48226 274.52 78514 1.65
    4500 81.26 59770.18 258.87 57751 2.25






    So, if at FTH I need 59770.18 Watts to compress 2.25 times and at 2100 I need 48226.00, that means that the fluid coupling would be limiting to 80.68% the power transmitted.
    The same math but done in compression ratio would be 73.33%

    Going back to the chart, if the difference between the fluid coupling at full transfer and at reduced transfer is around 125 sPS, that would mean that I would need the supercharger to consume around 678 sPS at FTH.
    Where does this number come from?
    If we said that we need 73.33% of the compression at fluid coupling end, that means that the 125 sPS shown on the chart correspond to the difference of 26.67% (I know these numbers are not exact as I'm not inverting the formula above but should be good enough to illustrate my point)

    So, according to the chart I would need the supercharger to consume 680 sPS at FTH but my math tells me is only using 81.26.

    This is where I scream HELP! What am I doing so wrong????







    db601 edited.png

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    Re: Aeronautical Engineer help needed

    Quote Originally Posted by ATAG_Colander View Post
    Folks,


    I'm looking for some help from any of our resident aeronautical engineers as I don't seem to understand how the db601 power curve is happening.

    My problem starts with the following chart.


    Notice I have extended blue lines to simulate the output power as if the fluid coupling was removed by extending the upper one and dragging it down to FTH power.


    By looking at the blue lines, my logic tells me that the fluid coupling slip bellow 2100 meters, is saving the engine around 125 sPS

    Ok assuming all the above is correct, let's start with some math...

    I'm using the following formula to calculate the power consumed by the supercharger:
    Gamma / (Gamma - 1) * Q * P0 / Efficiency * (( (P2/P1)^ ((Gamma - 1) / Gamma)) - 1)

    Where:
    Gamma = 1.41
    Q = Engine displacement in M^3 * RPM /60 /2 (33.9 Liters displacement in this case)
    Efficiency = Guesstimaged at 0.7

    With this formula I calculated the following values power consumption values:
    Altitude Power sPS Power Watts Temperature Kelvin Pressure Pascals Compression Ratio
    to get to 1.3 ATA
    0 40.3 29640.74 288.16 101300 1.28
    2100 65.56 48226 274.52 78514 1.65
    4500 81.26 59770.18 258.87 57751 2.25






    So, if at FTH I need 59770.18 Watts to compress 2.25 times and at 2100 I need 48226.00, that means that the fluid coupling would be limiting to 80.68% the power transmitted.
    The same math but done in compression ratio would be 73.33%

    Going back to the chart, if the difference between the fluid coupling at full transfer and at reduced transfer is around 125 sPS, that would mean that I would need the supercharger to consume around 678 sPS at FTH.
    Where does this number come from?
    If we said that we need 73.33% of the compression at fluid coupling end, that means that the 125 sPS shown on the chart correspond to the difference of 26.67% (I know these numbers are not exact as I'm not inverting the formula above but should be good enough to illustrate my point)

    So, according to the chart I would need the supercharger to consume 680 sPS at FTH but my math tells me is only using 81.26.

    This is where I scream HELP! What am I doing so wrong????







    db601 edited.png
    Howdy Colander,

    Think I can clear this up for you.....

    What! First I need to know what language that is and what you've been drinking. I want a double! (Then again after seeing what it did to you maybe I'll stick to my Mother-in-law's moonshine).

    Are you saying the game is actually this deep?

    Hat's off to you!

    Gaidin

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    Re: Aeronautical Engineer help needed

    Quote Originally Posted by JG27-Gaidin View Post
    Are you saying the game is actually this deep?
    Hi Gaidin,

    The question is not that complicated, I simplified it to get to the point
    Now, the answer might be a lot more complicated or as simple as "you fool!! you forgot to multiply by X"

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    Re: Aeronautical Engineer help needed

    Believe it or not... this is what we, (our code guys) have to understand to fix the game errors and have the game systems replicate correctly the actual mechanical systems.

    The Daimler Benz DB601, (and later engines) used a fluid coupling to drive their supercharger... with two pumps driving the fluid, one which operated at sea level, the 2nd kicks in at a certain altitude. (2100 meters) This system in effect provides the advantages of a two speed supercharger, but with a characteristic and unique power curve.

    The game does not replicate this correctly.

    Modeling a standard two speed gear driven supercharger is simple, because the lines are straight. (see red line below)



    Not a spurious request.
    Last edited by RAF74_Buzzsaw; Oct-27-2015 at 16:06.

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    Re: Aeronautical Engineer help needed

    The issue is not the curved line but the delta in power that does not add up to my numbers.

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    Re: Aeronautical Engineer help needed

    Hi Colander,

    Sent you a PM reply. I haven't been checking ATAG as much as I normal do but I think I figured out potentially what the issue is. I'll be double checking with some more people soonish as I am better with aerodynamics rather than actual engine qualities.

    If there are any more questions, feel free to PM me.

    Cheers,
    Eagle

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    Re: Aeronautical Engineer help needed

    Just FYI all, the question is still open.

    Also, to make my question clearer in case it get's lost in the math...
    If at FTH I loose 80 sPS to the supercharger, how can I get 125 back because the fluid coupling is only at 74% power transmission?
    Last edited by ATAG_Colander; Oct-27-2015 at 17:37.

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    Re: Aeronautical Engineer help needed

    Where is Cuck_Owl? Doesn't he study crazy stuff like this?
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    Re: Aeronautical Engineer help needed

    I've deleted the last couple of posts.

    Let's keep this thread clean please.

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    Re: Aeronautical Engineer help needed

    I was with you upto 'Folks, I'm looking for some help'....
    "The trouble with the world is that the stupid are cocksure and the intelligent are full of doubt.'' - Bertrand Russell
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    Re: Aeronautical Engineer help needed

    Here my opinion,

    I believe that the numbers on chart are experimental and net values, that is; the engine gives 1000 PS at sea level, at 2400 rpm, at 1.3 ata. Lets assume that supercharger consumes 80 PS (hypothetically). Meaning of this, the engine has 1080 PS potential if the supercharger operates by an outer power source at same conditions.

    Lets look first part of chart (0-2100m). Shows us the supercharger power draw is nearly constant because of the linear and parallel nature of graphs(at different ata and rpm).

    Second part of chart (2100-4500m) the supercharger reaches max. limit, fight between low tempeature vs low air density creates mildly lowering curves.

    After then if we look lower blue line, we can not reach from beginning of line to 1.3 ata point, unless without having an "super" supercharger actually DB 601 doesn't have.

    So, my opinion there is no any kind of power saving here.

    Hope it helps...

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    Re: Aeronautical Engineer help needed

    Since I took the time to write this in a PM I thought I would copy/paste it here as it explains a bit better:

    A fluid coupling varies the transmission of power due to slip. The more slip, the less torque is transmitted. On the other hand, the more slip, the more heat is generated which is transferred to the engine as in this case, they share the same oil.

    So, what the fluid coupling is doing is restricting the engine transfer of power to the supercharger since, at lower altitudes, you need less compression hence less power. If you transmit 100% of the power, then you are loosing unnecessary power compressing air that you don't need.

    So, theoretically, the power used by the supercharger at FTH (where the fluid coupling is minimum) is around 86 sPS. This means that at lower altitudes, the power requirements are less.

    Now jump to the chart, see the blue lines I drew and count the power difference between them. It will be around 150 sPS.

    If the max power it can loose due to the supercharger is 86, I do not understand how the heck it is gaining 150.
    If the supercharger was not needed at all bellow 2100 meters, you would be able to add 86 HP (not exactly right but you get my point) to the engine power. This would still leave us with a difference of 150-86 = 64 sPS to reach the upper blue line.

    Now, add to that the fact that you are still loosing power compressing air bellow 2100 so that difference would be bigger.

    Hope this explains.

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    Re: Aeronautical Engineer help needed

    Quote Originally Posted by Continu0 View Post
    Where is Cuck_Owl? Doesn't he study crazy stuff like this?
    CAE working at other problem

    o7
    It takes an airplane to bring out the worst in a pilot.
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    Re: Aeronautical/Mechanical or engine Engineer help needed

    Changed the thread title to broaden the request for help.

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    Re: Aeronautical/Mechanical or engine Engineer help needed

    Could Mystic post that on Facebook as well?
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    Re: Aeronautical/Mechanical or engine Engineer help needed

    Bumpy Bump...

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    Re: Aeronautical/Mechanical or engine Engineer help needed

    Quote Originally Posted by ATAG_Colander View Post
    Bumpy Bump...
    It looks conventional up to 2100m.

    from 2100 to FTH is where it gets interesting.

    what I think may be happening is that the second altitude stage is robbing the engine of more power than the first, due to the fact that the rpm is going up on the fluid coupling (think less dense atmosphere requires more spinning of s/c to generate pressure).

    This is something that is solved in torque converters on cars by mechanically locking up above some low-medium rpm, because as RPMs increase the fluid transfer becomes less and less efficient. Fluid dynamics. I think you are thinking linear power losses across the alt range like on a geared a/c, wheras it is a sigmoidal increase in power loss due to parasitic drag after 2100m because it is a fluid coupling.

    I'm not an engineer so I'm not going to attempt to calculate the maths - except to say I think there are WAY too many variables to get a reasonable chance of a good number.
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    Re: Aeronautical Engineer help needed

    Quote Originally Posted by JG27-Gaidin View Post
    <snip> ... Are you saying the game is actually this deep? .... <snip>
    Yes it is! and TF strives to create as realistic a simulation of the flight & engine dynamics as possible on a PC.

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