In order to discuss powerplants in any depth, it is essential to understand the concepts of POWER and TORQUE.

Sometimes it seems people are confused about both entities. People often discuss them as if they were mutually exclusive. For example, we have heard engine builders ask a customer:

"Do you want your engine to make TORQUE or HORSEPOWER?"

You should be clear on the concept that POWER is dependent on torque. TORQUE and RPM are the MEASURED quantities of engine output. POWER is CALCULATED from torque and RPM, by the following equation :

HP = Torque x RPM ÷ 5252

An engine produces POWER by applying a TORQUE on a rotating shaft.

A dynamometer measures torque and RPM, and from those two values calculates observed power. Then it applies various factors (air temperature, barometric pressure, relative humidity) in order to correct the observed power to the value it would have been if it had been measured at standard atmospheric conditions (corrected power).

A few basic definitions will help. First, TORQUE is defined as a FORCE around a given point, applied at a RADIUS from that point.

Note that the unit of TORQUE is one pound-foot (often misstated), while the unit of WORK is one foot-pound.

Referring to this picture, assume that the handle is parallel to the supported shaft and is located at a radius of 12" from the center of the shaft. In this example, consider the shaft to be fixed to the wall. Let the arrow represent a 100 lb. force, applied in a direction perpendicular to both the handle and the crank arm, as shown.  The shaft does not turn, but there is a torque of 100 pounds-feet (100 pounds times 1 foot) applied to the shaft. Note that if the crank arm in the picture was twice as long and the handle was located 24" from the center of the shaft, the same 100 pound force applied to the handle would produce 200 lb-ft of torque (100 pounds times 2 feet).

Next, POWER is the measure of how much WORK can be done in a specified TIME. In the example on the Work and Energy page, the guy pushing the car did 16,500 foot-pounds of WORK. If he did that work in two minutes, he would have produced 8250 foot-pounds per minute of POWER (165 feet x 100 pounds ÷ 2 minutes).

In the same way that one ton is a large amount of weight (by definition, 2000 pounds), one horsepower is a large amount of power (by definition, 33,000 foot-pounds per minute). The power the guy produced pushing his car across the lot (8250 foot-pounds-per-minute) equals ¼ horsepower (8,250 ÷ 33,000).

OK, all that’s fine, but how does pushing a car across a parking lot relate to rotating machinery?

Consider the handle-and-crank sketch again. The handle is 12" from the center of the shaft, and the shaft is now supported by frictionless bearings, and is attached to some rotational load behind the wall.

Suppose that a constant force of 100 lbs. is applied to the handle so that the force is always perpendicular to both the handle and the crank as the crank turns. (The "arrow" rotates with the handle and remains in the same position relative to the crank and handle).

If that constant 100 lb. tangential force causes the shaft to rotate at 2000 RPM, then the power the shaft is transmitting to the load behind the wall is shaft is 100 lb-ft of torque (100 lb. x 1 foot) times 2000 RPM divided by 5252, which is 38 HP.

The following four examples illustrate the relationship.

Example 1: How much torque is required to produce 300 HP at 2700 RPM?

HP = TORQUE x RPM ÷ 5252

so

TORQUE = HP x 5252 ÷ RPM

TORQUE = 300 x 5252 ÷ 2700 = 584 lb-ft.

Example 2: How much torque is required to produce 300 HP at 4600 RPM?

TORQUE = 300 x 5252 ÷ 4600 = 343 lb-ft.

Example 3: How much torque is required to produce 300 HP at 8000 RPM?

TORQUE = 300 x 5252 ÷ 8000 = 197 lb-ft.

Example 4: How much torque does the turbine section of a 600 HP Garrett TPE-331 produce? The turbine of that engine turns at 41,000 RPM, so the torque is:

TORQUE = 600 x 5252 ÷ 41,000  = 76.86 lb-ft.

Example 5: After the 600 HP (in Example 4 above) goes through the TPE-331 propeller gearbox, it emerges at the prop flange at 1591 RPM. How much torque is applied to the prop?

TORQUE = 600 x 5252 ÷ 1591= 1981 lb-ft.
(ignoring losses in the gearbox, of course).

General Observations

In order to design an engine to produce a particular value of HP, you must know the RPM at which you want that HP to occur, and design the engine to produce the required torque at that RPM. Typically, the torque peak will occur at a lower RPM than the power peak. The reason is that typically, the torque curve does not drop off (%) as rapidly as the RPM is increasing (%).

The RPM band within which the engine produces its peak torque is limited. You can tailor a high peak with a very narrow band, or a lower peak value with a wider band. Those characteristics are usually dictated by the parameters of the application for which the engine is intended.

If an engine is tailored to produce a peak torque of 500 lb-ft at 2500 RPM, then at that 2500 RPM it will produce 238 HP, and the torque will be declining as speed exceeds 2500 RPM. Suppose torque drops to 415 lb-ft at 3000 RPM. The HP is still 238, and the engine can do the same amount of work per unit time at 2500 as it can at 3000, but it will burn less fuel producing that power at 2500 RPM than at 3000 RPM. (That's because the power consumed to turn the crankshaft and the reciprocating machinery attached to it increases as the square of the crankshaft speed.)

If that same engine is tailored to produce the same peak torque (500 lb-ft) but with the torque peak at 6100 RPM, the power the engine would produce at 6100 is 581 HP. The peak HP depends on the shape of the torque curve after peak torque, that is, on how quickly the torque curve falls off above peak torque.

Derivation of the Equation

This part will probably not be of interest to most readers, but several readers have asked: "OK, where does the 5252 come from?" Here is the answer.

By definition, POWER = FORCE x DISTANCE ÷ TIME

Using the example above where a constant tangential force of 100 pounds was applied to the rotating handle at a 6" radius, we know the force, so to calculate power, we need the distance the handle travels per unit time.

Power = 100 pounds x distance per minute

OK, how far does the crank handle move in one minute? First, determine the distance it moves in one revolution:

DISTANCE per revolution = 2 x p x radius

DISTANCE = 2 x 3.1416 x 1 ft = 6.283 ft. per revolution.

Now we know how far the crank moves in one revolution. How far does the crank move in one minute?

DISTANCE per min. = 6.283 ft .per rev. x 60 rev. per min.
= 377 feet per minute

Now we know enough to calculate the power, defined as:

POWER = FORCE x DISTANCE ÷ TIME
so
Power = 100 lb. x 377 ft. per minute
or
Power = 100 x 377 = 37,700 ft-lb per minute

Swell, but how about HORSEPOWER?  

Remember that one HORSEPOWER is defined as 
33000 foot-pounds of work per minute. 

Therefore,
HP = POWER (ft-lb per min) ÷ 33,000

So the power being applied to the crank-wheel above is:
 37,700 ft-lb per minute, or 1.14 HP     (37,700 ÷ 33,000).

Now we combine some stuff we already know to produce the magic 5252.

We already know that TORQUE = FORCE x RADIUS.

If we divide both sides by RADIUS, we get:

FORCE = TORQUE ÷ RADIUS (a)

Now, if DISTANCE per revolution = RADIUS x 2 x p

then DISTANCE per minute = RADIUS x 2 x pi x RPM (b)

We already know POWER = FORCE x DISTANCE per minute (c) so if we plug the equivalent for FORCE from (a) and distance per minute from (b) into (c), we get:

POWER = (TORQUE ÷ RADIUS) x (RPM x RADIUS x 2 x p)

Dividing both sides by 33,000 to find HP, 

HP = TORQUE ÷ RADIUS x RPM x RADIUS x 2 x p ÷ 33,000

By reducing, we get

HP = TORQUE x RPM x 6.28 ÷ 33,000

Since

6.2832 ÷ 33000 = 1 ÷ 5252

therefore

HP = TORQUE x RPM ÷ 5252

Note that at 5252 RPM, torque and HP are equal. At any RPM below 5252, the value of torque is greater than the value of HP; Above 5252 RPM, the value of torque is less than the value of HP.

We thanks EPI, Inc. for those informations