Yes, it is true that 5.52 KW equals 7.40 HP however, that assumes 100% efficiency, which is not true. If we use 80% for the motor efficiency we get 5.92 HP; the rest of the electricity goes to waste heat. Since your motor is listed as 5 HP the 5.92 number would suggest a SF (service factor of 1.2) for a short term overload condition.
Including the SF for your motor we need to find a gasoline engine which has 6 HP at 3450 RPM at its rated (not maximum output) output. This engine should be equivalent to your current electric motor.
HP = [Torque x RPM] divided by 5252 and we want to solve for torque. Therefore we have, Torque = [6 x 5252] divide by 3450 = 9.1 ft-lbs. Now, we can look at engine performance curves for a engine that makes approximately 9.1 ft-lbs torque at its rated output.
I have performance curves for Kawasaki engine so I will use Kawasaki egines for an example.
Case I: Kawasaki Model FE170D - 5.5 HP at 3600 RPM, maximum output. This engine has 8.0 ft-lbs torque at its maximum output so there is no point in looking at the rated output torque.
Case II: Kawasaki Model FE250D - 8.0 HP at 3600 RPM, maximum output. Checking the "recommended" maximum output (rated output) performance curve at 3450 RPM, I read 6.6 HP which calculates to a torque output of 10.0 ft-lbs. Therefore, this engine should replace your current electric motor when run at 3450 RPM and using the current pulley size.
This engine is a single cylinder, air cooled, horizonal shaft and runs on gasoline. Kawasaki also makes similar engines in 9.5, 11.5 and 13 HP versions. The point of this reply is not so much as to dictate this is the size engine you must purchase but rather to show the sizing method. Using a larger engine simply means it will not have to work as hard and provides for additional over load capacity. Finally, I used Kawasaki because I have the data, there are many other engine manufactures to choose from.
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