How much horsepower was required to run the supercharger impeller in the R-2800 engine at cruise conditions, as used in the DC-6?

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    $\begingroup$ There may be a way to roughly calculate it based on fuel consumption and/or engine rpm. Interesting that superchargers increased Horsepower output of an engine, but at the cost of higher fuel use. The higher efficiencies (higher compression ratios) may have made it close to a "free lunch". $\endgroup$ Commented Sep 4, 2023 at 23:08
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    $\begingroup$ This is not an answer, just a comment. The answer can be reasonably determined for 65% power (assuming 100% of the sea-level rated power is available from boost at altitude) to be about 450 hp, with equivalent power also being provided to the shaft by the compounding turbine. Compounding increased engine thermal efficiency to about 36 pct; normal would be ~28-30 pct. The assumed full rated hp at sea level is 2800 hp. The fuel savings from compounding is about 25 pct. $\endgroup$ Commented Sep 4, 2023 at 23:10
  • $\begingroup$ @nielsnielsen Are we good here? Wow - talk about good fortune finding the source by Graham White noted farther below. Maybe my previous impression was not a DC-6, but a DC-7, of which more sense can be made. I made an appropriate edit to the originally provided answer. Evidently, although somewhat off base, the answer was apparently informative. Best of regards. -TP $\endgroup$ Commented Sep 5, 2023 at 21:16
  • $\begingroup$ @ThomasPerry, all good, thanks for your input, I appreciate it, answer accepted!- NN $\endgroup$ Commented Sep 5, 2023 at 23:37

1 Answer 1


For compounding see below. The R-2800 was not compounded. From information available in print, the power absorbed by the supercharger turbine was about 400 horsepower. This is comparable to the power required by the compounding supercharger. A comment below gives the source in print for the noted supercharger power requirement of 400 hp. This was a very interesting question!

Figure 8:8 from Dommasch

The caption explains these diagrams. On the right, (b) shows the equivalent work produced by the supercharger and compounding turbine. The excess hatched area is equivalent to power produced at the propeller. Since compounding will realize equivalent power at the propeller with 25 pct less fuel, the portion of the power at the propeller produced by the compounding is equivalent to the total shaft power from the turbine less that portion of the shaft power consumed by the compressor. If we make the rough assumption that the area of the turbine curve (hatched area in b) is about twice that of the compressor curve, then the power consumed by the compressor would be equivalent to the half the power produced by the turbine. The other half of the power from the turbine is given to the propeller. In other words, if the engine is operated at 65 pct power, approximately one-fourth of that power is produced at the propeller by the turbine, and equivalent power is consumed by the compressor in boosting the engine.

If we do the math we will find the turbine produces about 910 hp, of which half is given to the propeller and half is given to the compressor. These do not seem to be unreasonable approximations for the engine operating at 65 pct of its rated 2800 horsepower.

Source of the illustration -

Dommasch, et. al, Airplane Aerodynamics, 3rd ed. 1961, or 4th ed. 1967, Pitman Publishing, New York.

  • $\begingroup$ But the R-2800 was not compounded, It had a single-stage, 2-speed supercharger driven off the crankshaft. It is the power consumed by that supercharger that I am interested in. -NN $\endgroup$ Commented Sep 5, 2023 at 5:44
  • $\begingroup$ Sorry, Niels. If the answer is wrong, I'll take it down. I do remember asking about that first time I saw a DC-6, what kind of engines were on that plane, and was told they were compound. Ummm... yeh. I know; 68 yrs ago, now... Otherwise, I do not know the power absorbed by the direct-drive compressor. $\endgroup$ Commented Sep 5, 2023 at 13:32
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    $\begingroup$ OK - I found your answer! Up to 400 hp is necessary to drive the supercharger in order to produce the necessary mass air flow and manifold pressure. source: Graham White, 2001. R-2800: Pratt and Whitney's Dependable Masterpiece. Society of Automotive Engineers, Warrendale, PA. p. 89. $\endgroup$ Commented Sep 5, 2023 at 17:58
  • $\begingroup$ Hum BABY! 400HP to spin that puppy! Time for my next question! -NN $\endgroup$ Commented Sep 5, 2023 at 23:41
  • $\begingroup$ As compounding is not relevant to this question, I think it should be removed from this answer. $\endgroup$
    – Jpe61
    Commented Sep 6, 2023 at 11:44

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