Power conversion?

[bada]

Something that i do not really undestand came out after the reading of the bmw801 and the merlin 61/65/70/75 manual.
Maybe you'll be able to help me with that one!

In the 801 manual, the power quoted is 1730PS ON THE PROPELLER, i will call that "final power".

In the Merlin manual,the power quoted for the M61 is 1400BHP.

Now, the pb lies in the fact that the definition of the BHP is (from Wiki, but seems to be the same everwhere):
Brake horsepower (bhp) is the measure of an engine's horsepower without the loss in power caused by the gearbox, generator, differential, water pump, and other auxiliary components such as alternator, power steering, and AC compressor. Thus the prefix "brake" refers to where the power is measured: at the engine's output shaft, as on an engine dynamometer. The actual horsepower delivered to the driving wheels is less. An engine would have to be retested to obtain a rating in another system. The term "brake" refers to the original use of a band brake to measure torque during the test (which is multiplied by the engine RPM and a scaling constant to give horsepower).

now, i suppose (and only suppose) that if the BHP definition is right, in this case, we have to replace the wheels and gearbox with the reductor and the propeller.
So, my pb here is: if the BHP definition is right, was was the Final power of the merlin61? (in this case ,only as exemple)
How much power loss was due to the reductor and propeller?

Thanks for helping me out with that one

(1PS=0.7355Kw)
( =0.98632Hp)
(1BHP=0.7457Kw)
( =1.01389PS)

[HoHun]

Hi Bada,

>now, i suppose (and only suppose) that if the BHP definition is right, in this case, we have to replace the wheels and gearbox with the reductor and the propeller.

I'm not sure of the exact details of the setup, but with aviation engines, it seems to have been standard operating procedure to use a specially designed "brake propeller" for power testing. It generated enough drag to dissipate the engine power and at the same time generated some cooling airflow.

As engines for most WW2 era combat aircraft had to use a reduction gear to harness their power to a propeller efficiently, the losses in this reduction gear were already figured into the brake losses. Reduction gear losses seem to be an important consideration for the designers of modern experimental aircraft because they have to compare the use of an ungeared engine without these losses, but a possibly less efficient propeller to the use of the same engine suffering some losses, but swinging efficient larger propeller more slowly for better efficiency. I don't think any main combat types of WW2 were equipped with engines for which a direct drive option existed - if they did, they certainly were obsolescent or at least low-powered.

Further considerations when discussing engine power:

- Are the figures given for the engine at rest, or for an engine as installed in an aircraft with a forward-facing air inlet under the assumption of a certain forward speed? The former case is considered for "static" pressure, the latter for "ram" pressure. The resulting curves for the same engine are similar, but the ram power curve will be shifted to higher altitudes compared to the static power curve.

- What kind of exhaust system is installed in the engine? The exhaust system determines the so-called backpressure that has an impact on the efficiency of the exhaust stroke. Jet exhaust nozzles tend to increase back pressure, decreasing brake power but increasing overall thrust from the engine.

- What is the effect of exhaust thrust on the actual thrust power in flight? In high-speed flight, as much as 25% of the available thrust might be provided by the exhaust system, which doesn't even register on the dynometer brake.

- According to which conventions is the power curve calculated? Difficult to find out - as a rule of thumb, early in WW2 the methods for calculating a power-over-altitude curve were imperfect, while later on they got better. Due to limitations of the test machinery, it was not possible to test engines on the test stand under realistic conditions, so some data points were measured and the complete power curve was calculated. Of course, there were also "flying test stands" - Jumo for example used a Ju 52 for this purpose, mounting the test engines in the nose while keeping the original outboard engines to sustain flight even if the test engine (often a prototype) failed.

- For which operating conditions (rpm, boost pressure) was the power curve valid? Most major engine types were re-rated now and then during WW2. You'll have to add the question for altitude if you're asking for a single power figure and not for the entire curve.

>So, my pb here is: if the BHP definition is right, was was the Final power of the merlin61? (in this case ,only as exemple)

Usually, you should be able to take the figure for rated power, check for the conditions supplied with it (if you're lucky enough to have them supplied), and take it as brake power figure. It's a good idea to check for the altitude though as rating systems could differ - the British engines were rated for optimum power, while the German engines are often quoted with their sea-level powers. Nothing wrong with either, but they're not directly comparable of course.

>How much power loss was due to the reductor and propeller?

I'm not sure about the reduction gear - maybe 1 or 2 %, I guess. The propeller of course was a major source of losses: even an ideal propeller inevitably has losses due to its operating principle. In the main operating range, it had maybe 10 to 20% losses, but at very low or very high speeds, it could be much more than that.

Regards,

Henning (HoHun)

[engguy]

Good question.
I think the way it worked is the complete engine is plugged into the dynamometer, the torque is read at what ever crankshaft rpm is noted, since the tachometer drive measures crankshaft rpm. And then the crankshaft torque is calculated from the measured quantity of torque and the gear ratio in the reduction unit plus what ever losses they determine exist in the reduction unit. HP is most always crankshaft out put. I'm sure it was not an exact science, what really is?
If the HP figure is what you call the final at the prop, the torque there has to be taken as well as the rpm at that point.
If there are no losses the HP at the crankshaft will be the same at the output shaft, no matter how low the reduction. Remember a reduction is a torque multiplier. And HP is a time dependant measurement.

[bada]

thanks,
so can we assume that the "final power" is correct concerning the merlins?