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 lbf, and if you apply a force of 5lb to the end of a spanner 2ft long again you apply a torque of 10ft lbf.

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 = 5185.8ftlbf.   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 (ftlbf/sec)   There are 550 ftlbf/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 = 2020ftlbf. 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. 