If I had a prop that allowed the engine to hit 2700 rpm how much difference in speed should I expect. Thank you, Trent
About 2 knots.
I see about a 7 knot/100 rpm change at or around cruise settings ~ 2450 rpm/157 knots.
Kyle, I assume you are using the throttle to vary RPM. If so, you're changing torque as well as RPM, meaning a significant change in power.
That's not the same as Trent's proposal, in which the sole change is RPM. As a practical matter, we might consider his torque to be constant; the full throttle Lycoming torque curve is quite flat in that region.
A crude approach might be to assume Trent's 320 can make full rated power, 160HP, at 2700. If so, the usual HP equation says torque is 311 lbf-ft. If we then recalculate using 311 and 2600 RPM, current power is 154. All else equal, the RPM change alone is worth six HP.
So what will he get for six additional HP? Power required is proportional to velocity cubed, so..
New speed = 150*(160/154)^0.333333 = roughly 2 knots
Remember, this assumes the exact same prop, repitched for 100 more RPM. Changing to a more efficient prop might net better results, just because better prop efficiency means more of the available HP is applied to thrust.
BREAK
I wrote the above about 5:30 this AM. Then (given a second cup of coffee) it dawned on me I was up early to fly to TN and see a friend, and I have both props. Well, not quite, but same thing, a constant speed. Hmm...time for a test!
See photos below.
2600 and 2700 RPM, full throttle, mixture adjusted for essentially the same EGT. No other changes. We'll use the indicated power percentage:
210*0.68 = 142.8 HP @ 2600
210*0.71 = 149.1 HP @ 2700
New speed = Previous speed * (new power / old power)^0.3333333
so
New speed = 191 * (149.1 / 142.8)^0.3333333
New speed = 193.7.
I guess 194 indicated is close enough.