It often seems that
people are confused about the relationship between POWER and TORQUE. For
example, we have heard engine builders,
camshaft consultants, and
other technical experts ask
customers:
And the question is posed
in a tone which strongly suggests that these experts believe power and torque
are somehow mutually exclusive.
In fact, the opposite is
true, and you should be clear on these facts:
(At the
bottom of this page, the derivation of that equation is shown, for anyone
interested.)
An engine produces POWER
by providing a ROTATING SHAFT which can exert a given amount of TORQUE
on a load at a given RPM. The amount of TORQUE the
engine can exert usually varies with RPM.
A dynamometer determines the POWER an
engine produces by applying a load
to the engine output shaft by means of a water brake, a generator, an
eddy-current absorber, or any other controllable device capable of absorbing
power. The dynamometer control system causes the absorber to exactly match the
amount of torque the engine is producing at that instant, then measures that TORQUE as well as the RPM
of the engine shaft, and from those two measurements, it 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).
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.
Figure 1
Referring to Figure 1, assume that the handle is
attached to the crank-arm so that it 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.
Because the shaft is
fixed to the wall, 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 sketch was twice as long (i.e. 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).
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. The definition of one
horsepower is 33,000 foot-pounds per
minute. The power which the guy produced by 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 following
change to the handle-and-crank-arm sketch above. The handle is still
12" from the center of the shaft, but now, instead of being fixed to the
wall, the shaft now goes through the wall, supported by frictionless bearings,
and is attached to a generator behind the wall.
Suppose, as illustrated
in Figure 2, that a
constant force of 100 lbs. is somehow applied to the handle so that the force
is always perpendicular to both the handle and the crank-arm as the crank
turns. In other words, the "arrow" rotates with the handle and
remains in the same position relative to the crank and handle, as shown in the
sequence below. (That is called a "tangential force").
Figure 2
If that constant 100 lb.
tangential force applied to the 12" handle (100 lb-ft of torque) causes
the shaft to rotate at 2000 RPM, then the power the shaft is
transmitting to the generator behind the wall is 38 HP, calculated as follows:
100 lb-ft of
torque (100 lb. x 1 foot) times 2000 RPM divided by 5252 is 38 HP.
Example 1: How much TORQUE is required
to produce 300 HP at 2700 RPM?
since
HP = TORQUE x RPM ÷ 5252
then by
rearranging the equation:
TORQUE = HP x 5252 ÷ RPM
Example 2: How much TORQUE is required
to produce 300 HP at 4600 RPM?
Example 3: How much TORQUE is required
to produce 300 HP at 8000 RPM?
Example 4: How much TORQUE does the
41,000 RPM turbine section of a 300 HP gas turbine engine produce?
Example 5: The output shaft of the
gearbox of the engine in Example 4 above turns at 1591 RPM. How much TORQUE is
available on that shaft?
The point to be taken
from those numbers is that a given amount of horsepower can be made from an
infinite number of combinations of torque and RPM.
Think of it another way: In
cars of equal weight, a 2-liter twin-cam engine that makes 300 HP at 8000 RPM
and 400 HP at 10,000 RPM will get you out of a corner just as well as a 5-liter
engine that makes 300 HP at 4000 RPM and 400 HP at 5000 RPM. (In fact, in cars
of equal weight, the smaller engine will probably race BETTER because it's much
lighter, therefore puts less weight on the front end.)
In order to design an
engine for a particular application, it is helpful to plot out the optimal
power curve for that specific application, then from that design information,
determine the torque curve which is required to produce the desired power
curve. By evaluating the torque requirements against realistic BMEP values you
can determine the reasonableness of the target power curve.
Typically, the torque
peak will occur at a substantially lower RPM than the power peak. The reason is
that, in general, the torque curve does not drop off (%-wise) as rapidly as the
RPM is increasing (%-wise). For a race engine, it is often beneficial (within
the boundary conditions of the application) to operate the engine well beyond
the power peak, in order to produce the maximum power within a required RPM
band.
However, for an engine
which operates in a relatively narrow RPM band, such as an aircraft engine, it
is generally a requirement that the engine produce maximum power at the maximum
RPM. That requires the torque peak to be fairly close to the maximum RPM. For
an aircraft engine, you typically design the torque curve to peak at the normal
cruise setting and stay flat up to maximum RPM. That positioning of the torque
curve would allow the engine to produce significantly more power if it could
operate at a higher RPM, but the goal is to optimize the performance within the
operating range.
An example of that
concept is shown Figure 3 below. The three dashed lines represent three torque
curves having exactly the same shape and torque values, but with the peak
torque values located at different RPM values. The solid lines show the power
produced by the torque curves of the same color.
Figure 3
Note that, with a torque
peak of 587 lb-ft at 3000 RPM, the pink power line peaks at about 375 HP between
3500 and 3750 RPM. With the same torque curve moved to the right by 1500
RPM (black, 587 lb-ft torque peak at 4500 RPM), the peak power jumps to about
535 HP at 5000 RPM. Again, moving the same torque curve to the right another
1500 RPM (blue, 587 lb-ft torque peak at 6000 RPM) causes the power to peak at
about 696 HP at 6500 RPM
Using the black curves as
an example, note that the engine produces 500 HP at both 4500 and 5400 RPM,
which means the engine can do the same amount of work
per unit time (power) at 4500 as it can at 5400. HOWEVER, it will burn less
fuel to produce 450 HP at 4500 RPM than at 5400 RPM, because the parasitic
power losses (power consumed to turn the crankshaft, reciprocating components, valvetrain) increases as the
square of the crankshaft speed.
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.
An example of that is
shown in Figure 4 below. It is the same as the graph above, EXCEPT, the blue
torque curve has been altered (as shown by the green line) so that it doesn't
drop off as quickly. Note how that causes the green power
line to increase well beyond the torque peak. Such a change to the
torque curve can be achieved by altering various components (lobe profiles,
lobe separation, runner length and / or cross section, etc.). Alterations
intended to broaden the torque peak will inevitable reduce the peak torque
value, but the desirability is determined by the application.
Figure 4
This part might not be of
interest to most readers, but several people have asked:
"OK,
if HP = RPM x TORQUE ÷ 5252, then where
does the 5252 come from?"
Here is the answer.
By
definition, POWER = FORCE x DISTANCE ÷ TIME
Using the example in
Figure 2 above, where a constant tangential force of 100 pounds was applied to
the 12" handle rotating at 2000 RPM, we know the force involved, so to calculate
power, we need the distance
the handle travels per unit time,
expressed as:
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 π x radius
DISTANCE per
revolution. = 2 x 3.1416 x 1 ft = 6.283 ft.
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 2000 rev. per min. = 12,566 feet per minute
Now we know enough to
calculate the power, defined as:
POWER =
FORCE x DISTANCE ÷ TIME
so
Power = 100 lb x
12,566 ft. per minute = 1,256,600 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. We have already calculated that the power
being applied to the crank-wheel above is 1,256,600 ft-lb per minute.How many HP is that?
HP =
(1,256,600 ÷ 33,000) = 38.1 HP.
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
of that equation by RADIUS, we get:
(a) FORCE = TORQUE ÷
RADIUS
Now, if DISTANCE per
revolution = RADIUS x 2 x π, then
(b) DISTANCE per minute =
RADIUS x 2 x π x RPM
We already know
(c) POWER = FORCE x
DISTANCE per minute
So if we plug the
equivalent for FORCE from equation (a)
and distance per minute from equation (b) into equation (c), we get:
POWER =
(TORQUE ÷ RADIUS) x (RPM x RADIUS x 2 x π)
Dividing both sides by
33,000 to find HP,
HP = TORQUE
÷ RADIUS x RPM x RADIUS x 2 x π ÷ 33,000
By reducing, we get
HP = TORQUE
x RPM x 6.28 ÷ 33,000
Since
33,000 ÷
6.2832 = 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.