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Automotive Electric Turbocharger Questions

  1. Oct 23, 2016 #1
    Hello Everyone,
    I just registered here to try to get some answers on a question that has been bugging me for a few days.
    So back in June, Audi announced the upcoming SQ5, a medium sized SUV, but in the sport package. In Europe, the car will receive a 3.0L TDI engine, however it will be released with one large turbocharger, and one electric supercharger. I put the article below if you are interested:
    The electric supercharger is made by a French company called Valeo, they seem to make many OEM parts. The power rating is 7kW at 48 Volts. I believe that this is the product in the Audi:

    Anyway, they don't sell that to the public. (perhaps because the car won't even be released until next year...)
    The purpose of the electric supercharger is to provide instant boost pressure under hard, and sudden, acceleration -- to eliminate turbo lag. I would like to use a similar system in a car I am building. My engine will be a twin turbocharged Buick 350 (that equals 5.7L), hopefully capable of near 800 crank HP. So, I'm looking to mimic the system that Audi/Valeo developed. I was wondering how many HP/kW I need in an electric motor to push the turbo. I have found some BLDC motors, ranging from 3kW to 20kW.
    Some motors I've looked at: http://www.goldenmotor.com/frame-bldcmotor.htm
    I would spline the motor into a gear up box (I know it's not ideal, but I can't seem to find any 5-10kW 75,000 RPM motors) and then the output was splined onto an additional turbocharger's shaft in place of the turbine. This is essentially the setup of a ProCharger: https://www.procharger.com/auto-superchargers/models
    I would want to provide 5-10lbs of boost up from idle (600 RPM) through about 3000 RPM, while the normal turbos spool. The motor and compressor would only have to operate for about a three second interval.

    So, after all of that the question is essentially how many kW do I need in an electric motor? Is there an idealized way to calculate/estimate how much power is needed to push 1lb/min at a pressure ratio between 1.25 and 1.75?
    Thank you to anyone who can help. This math is far above my head.
    Last edited by a moderator: Apr 15, 2017
  2. jcsd
  3. Oct 23, 2016 #2


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  4. Oct 23, 2016 #3
    Thanks. That was a good article. "I found that flowing 780cfm at 15 psi required 80 to 100 hp."
  5. Oct 24, 2016 #4

    jack action

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    The equation is:
    [tex]P = \frac{\left( {PR}^{0.2857} - 1 \right) (T + 459.67) \dot{m}}{237.269\ \eta_c\ \eta_m}[/tex]
    • ##P## is the power required in kW;
    • ##PR## is the pressure ratio across the compressor;
    • ##T## is the inlet temperature in °F;
    • ##\dot{m}## is the mass flow rate in lb/min;
    • ##\eta_c## is the isentropic efficiency of the compressor (see efficiency islands of a compressor map);
    • ##\eta_m## is the mechanical efficiency of the drivetrain (gearbox, bearings, etc.).
    You can also use the volumetric flow rate (##\dot{V}##) in CFM instead:
    [tex]P = \frac{\left( {PR}^{0.2857} - 1 \right) (T + 459.67) \dot{V}}{3102.5472\ \eta_c\ \eta_m}[/tex]

    For example:
    • ##PR## = 1.5
    • ##T## = 70 °F
    • ##\dot{m}## = 1 lb/min
    • ##\eta_c## = 70%
    • ##\eta_m## = 95%
    [tex]P = \frac{\left( (1.5)^{0.2857} - 1 \right) (70 + 459.67) (1)}{237.269\ (0.70)\ (0.95)} = 0.4123\ kW[/tex]
    This is based on the definition of work for an isentropic compressor:

    ##P = \dot{m}W##
    ##P = \dot{m}C_p \left(T_{out} - T_{in}\right)##
    ##P = \dot{m}C_p T_{in}\left(\frac{T_{out}}{T_{in}} - 1\right)##
    ##P = \dot{m}C_p T_{in}\left(\left(\frac{p_{out}}{p_{in}}\right)^{\frac{k-1}{k}} - 1\right)##

    I just added some conversion factors for the units and the efficiency losses. More info here.
  6. Oct 24, 2016 #5


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    Based on some back-of-the-envelope calcs using isentropic compression and an engine flow rate calculator:
    • A 5.7l engine will flow about 243 CFM at 3000 rpm
    • At this flow rate with isentropic compression, it would take about 5.2 hp to compress air to 5 psi of boost, or 9.73 hp to compress air to 10 psi of boost
    • Assuming a 12 V power system, the motor(s) would need to draw about 323 amps for 5 psi, or 605 amps for 10 psi
    This makes it clear why you would need a 48 V power system, but even at higher voltage you're talking about a heavy power bank, alternator/generator, and motor to operate this. Maybe time to consider a centrifugal supercharger?


  7. Oct 25, 2016 #6
    Wow, what awesome answers! Thank you so much, that was all that I needed to know. jack action, thank you for taking the time to put those standard formulas into something I could use easily. I really appreciated the Imperial units too. I know it's not the accepted way, but I'm a little old fashioned.
    Thanks again guys, this is an awesome community.
  8. Oct 25, 2016 #7


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    If it's got to go from 0 to 75,000rpm in less than 1-2 seconds best check the power required to accelerate the mass/moment of inertia.
  9. Nov 2, 2016 #8
    I have looked into this before to eliminate turbo lag off of the line. The problem i found is it almost detrimental as the benefits are very small and power consumption very high, which the engine will have to put back into the batteries through the alternator, and then add into that the weight or the motors turbo and the extra pipework. For a project its not very efficient, but for commercial company i guess it must be viable... although just another part to go wrong.

    I came to the conclusion that a small belt driven supercharger was a lot more efficient, and maybe have it run on a clutch or ratcheting system so at higher rpm it isn't pulling the engine down.

    This was a fair few years ago but i would be very interested if you try it out the electric idea.

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