Although power is a basic unit, in agriculture the more common unit is horsepower. Different measures of horsepower are used. For example, when discussing tractors, you may use engine horsepower, brake horsepower, drawbar horsepower, or power take-off horsepower. In the following section we will investigate the principles of horsepower. Horsepower is an arbitrary unit that was developed by James Watt to promote his early steam engines. He watched horses pulling loads of water out of mine shafts and concluded that one horsepower was equal to performing work at the rate of 33,000 foot-pounds per minute. Expressed algebraically the conversion from power to horsepower is:
Power min lb-ft 1 min
33 000 lbft min 33,000 lb-ft min
Note that unless power is in the units of ft-lb/min, the use of the conversion factor 1hp = 33,000 ft-lb/min will not produce correct results. The common horsepower equation is:
T x 33,000
This is an example of an equation with a conversion constant. This means all of the variables must have the correct units. This equation requires distance (D) expressed in feet, force (F) in pounds, and time (T) in minutes.
Problem: How many horsepower are developed if a person loads six 60-lb bales onto a truck platform 4.0 ft high in 1.5 min?
Now consider what would happen if the time required to load the hay is measured as 90.0 sec instead of 1.5 min. Obviously 90.0 sec equals 1.5 min, but if the same equation is used either the seconds must be converted to minutes or a different conversion factor from power to horsepower must be used. If the 33,000 conversion unit is used with time measured as 90.0 sec, the answer will be incorrect. The problem can be solved in two different ways. The first way is to add a conversion unit for time to the equation. This would be the preferred method for just a few calculations.
The second way is to determine the appropriate conversion value for the units being used. In a situation where horsepower will be calculated several times and time is measured in seconds, it would be more efficient to use a conversion value from power to horsepower appropriate for when time is measured in seconds instead of minutes. This can be accomplished using units cancellation:
ft-lb 33,000 ft-lb 1 min ft-lb
sec 1 min 60 sec sec
Using this conversion value, the previous problem can be solved by:
Horsepower can also be calculated using torque and the speed of the shaft. This is called shaft horsepower. This equation was used to evaluate the horsepower being produced by early engines using a Prony Brake, Figure 3.1.
The early Prony Brake used the friction between a rotating flywheel and a stationary block of wood to produce a force on the lever. Once the force on the lever and the speed of rotation of the flywheel is known, the brake horsepower can be determined. Mathematically brake horsepower is:
where F = Force produced (lb); L = Length of the lever arm (ft); N = Rotary speed of the Prony brake shaft (rpm); 5252 = Units conversion constant. This equation is derived from the horsepower equation:
F 1 hp 5252
min where the length of the lever arm equals the radius of the circle and the distance the force is working through.
Problem: How many horsepower is an engine producing when 8 lb of force are measured at the end of an 18.0 inch Prony brake arm when rotating at 1,700 revolutions per minute?
FLN V 12 inj min
[Note: the answer was rounded to two significant figures because the significance of the zero's in 1,700 are ambiguous, but at least two significant figures should be used because 8 lb is a measurement and the uncertainty is +/- 0.5 lb.]
Brake horsepower contains a length and a force. We know from earlier discussions that if we have a force times a length, we are dealing with torque. Therefore, the brake horsepower equation can be rewritten to use torque:
A Prony brake can still be used to measure the horsepower of engines, but electrical and hydraulic dynamometers are more accurate than the Prony brake. The rotary power of modern engines is measured by connecting directly to the flywheel or by using the power take off (pto) shaft.
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