
Operational Theory
Talk Torque
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Torque is the name
given to a force applied through a radius to produce a turning
moment. It also can be defined as the tendency of a force
to rotate an object around a pivot.
If you have a spanner
0.1m long and apply a force of 10N at the end of the spanner to
tighten a nut you are applying a torque of 0.1 × 10 = 1Nm. Or if
you apply a force of 5N to the end of a spanner 0.2m long again
you apply a torque of 1Nm.
Or in imperial units
if you have a spanner 1ft long and apply a force of 10lb to the
end of the spanner to tighten the nut, you are applying a torque
of 10ft lb_{f}, and if you apply a force of 5lb to the
end of a spanner 2ft long again you apply a torque of 10ft lb_{f}.
Let us return to our
horse turning the mill stone (see
horsepower)

If we assume that Dobbin is
moving at 1rpm, then the radius of the bar rotating the
mill wheel is 181/2p =
28.81feet.
So if dobbin is exerting a
force of 180lb, the torque is 28.81 × 180 =
5185.8ftlb_{f}.
If we know the torque and the
RPM we can caculate the power.
Power = force × distance/time
and distance/time is velocity.
Angular velocity (ie rotation)
is measured in radians/sec. and there are 2p
radians in 1 revolution. (360°)
Angular Power = force ×
distance (ie radius) × angular velocity
But force × distance (ie
radius) = Torque
So Power = Torque × angular
velocity
angular velocity = RPM × 2p/60
(radians/sec)
Power = Torque × RPM × 2p/60
(ftlb_{f}/sec)
There are 550 ftlb_{f}/sec
in 1hp.
So hp = Torque × RPM × 2p/60
×550
hp = Torque × RPM/5252
Dobbin is producing 5185.8
×1/5252 = 0.987hp


If the situation is reversed so
that the output of a prime mover (steam engine, diesel
engine etc) is pulling a load around at 26rpm in a
circle of radius 10 feet so that the reading on the
spring balance is 202lb then the torque produced by the
engine is 202 × 10 = 2020ftlb_{f}. and the power
developed is 2020 × 26/5252 = 10hp.
If the speed was increased to
52rpm, then the load would have to be reduced to 101lb,
or the radius reduced to 5 feet for the same power
output.
The SI units are simpler to calculate.
Power (Watts) = Torque (Nm) × RPM ×2p/60
So if the reading on the spring balance was 900N and the radius was 3.05m. The
power would equal 900 × 3.05 × 26 × 2p/60 = 7474
Watts or approximately 7.5 KW
If we consider 2 diesel engines
of the same power output, one turning at 100 rpm and the
other at 600rpm. Then the torque produced by the faster
engine is only 1/6 of the torque produced by the slow
speed engine. However, if the speed of the medium speed
engine is geared down by a 6:1 ratio, then (ignoring
frictional losses), it's torque would be the same as the
slow speed engine.

The use of reduction gearing to increase the torque of an engine is easily
demonstrated in a car. It is almost impossible to pull away smoothly from a
standstill in a high gear, because the torque is low, and the engine will stall.
Using first gear gives a high torque but at the sacrifice of a low speed.
On a marine diesel engine used for propulsion, then there must be sufficient
torque to turn the propeller at a low enough speed to prevent excessive
cavitation and maintain efficiency. This is why medium speed engines rotating at
400 RPM + are geared down using reduction gears.
Torque also comes into play when we consider the crank of the engine. At TDC
there is no turning moment, and though the gas load on the piston is high, there
is no torque and thus no power is being transmitted.
As the crank turns past TDC there is now a turning lever (the distance OA
in the diagram below). If this is multiplied by thrust in the con rod (the
turning force) the torque at that particular crank angle can be calculated. When
the piston is moving up the cylinder on compression the torque is applied by the
rotating crank so for the purpose of this explanation is given a negative value.
Because the torque is varying infinitely throughout the engine cycle depending
on crank angle, so the power being developed is constantly varying. However if
the torque is averaged for the whole of one engine cycle, then the power
developed can be calculated.
To illustrate this, open this
exel spreadsheet, for which a screenshot of the first few rows is produced
below
Torque and Power Spreadsheet for a 980 Bore x 2.44m
Stroke 2 Stroke Engine 










RPM 
100.5 

Total Torque kNm 
173397 



Bore 
0.98 

Average Torque kNm 
481.7 


Stroke 
2.44 

Power kW 
5069.1 


Con rod length 
2.85 



























crank angle 
cyl press 
Gas load 
Sin f 
f 
q + f 
dist OA 
turn force 
Torque 
Power 
degrees q 
bar 
kN (P) 

degrees 
degrees 
metres 
kN 
KNm 
kW 
0 
106.7 
8048.343 
0 
0 
0 
0 
8048.343 
0 
0 
1 
108.9 
8214.288 
0.007471 
0.428052 
1.428052 
0.030404 
8214.517 
249.7574 
2628.531 
2 
111.2 
8387.776 
0.014939 
0.855998 
2.855998 
0.060788 
8388.712 
509.93 
5366.675 
3 
114.3 
8621.608 
0.022403 
1.283731 
4.283731 
0.091129 
8623.772 
785.8724 
8270.784 
4 
117.9 
8893.155 
0.029861 
1.711144 
5.711144 
0.121406 
8897.122 
1080.166 
11368.03 
5 
120.9 
9119.443 
0.037309 
2.138132 
7.138132 
0.151599 
9125.797 
1383.466 
14560.06 
6 
123.8 
9338.189 
0.044746 
2.564586 
8.564586 
0.181687 
9347.552 
1698.333 
17873.83 
7 
127.2 
9594.65 
0.052169 
2.9904 
9.9904 
0.211649 
9607.733 
2033.472 
21400.94 
8 
130.6 
9851.111 
0.059576 
3.415467 
11.41547 
0.241465 
9868.64 
2382.929 
25078.75 
It can be seen that whilst the torque varies depending on the crank angle and
the cylinder pressure, the average torque, (total torque/360) is
481.7kNm. If the power is now calculated (Torque ×revs/sec × 2p)
The answer comes out as 5069.1kW
Now compare this with the computer generated pressure/crank angle indicator
diagram taken from the same engine. Looking at the figures in the table it gives
the indicated power (pind) as 6779 hp. This is 5069kW.
