Power at Work: Back to Basics - Part II - What you should know about horsepower

Wednesday, December 19, 2007

Power is really the rate of doing work and horsepower is one important measure we need to understand as we go about our daily tasks

by RALPH WINFIELD

In the last issue, I covered some basics on electricity, heating air and air-flow rates. This month, I would like to deal with some questions that come up regularly about horsepower and hydraulic power.

What exactly is horsepower?

It doesn't matter whether you are dealing with a car, a tractor or a boat. A horsepower is still a horsepower.

In the late 19th and early 20th century, people like Henry Ford, Charles Hart, Charles Parr and other entrepreneurs were endeavouring to replace the horse on the road and in the field. In order to compare power capability, it was determined that one good horse could produce about 33,000 lb-ft of work per minute. Do remember that power introduces the element of time along with work. Thus, power is really the rate of doing work.

If we put a simple scale between a two-furrow plow and a small tractor and determine how many minutes it takes to cross a field, we could calculate the "drawbar horsepower" it developed. For example, if the scale showed 1,200 pounds of pull and the tractor moved the plow 600 feet in five minutes, we could say that the:
Drawbar horsepower = (1,200 x 600) /(33,000 x 5) = 4.86

In other words, the little tractor could effectively replace a four-horse hitch.

But early tractors were often used more extensively for "belt work" than for field or drawbar work, so a rotational horsepower rating was also required. Well, all we have to do is set up a rope and pulley system in order to wrap rope on a wide pulley in order to maintain a constant pulley diameter. Each wrap of rope will be equal to the circumference of the pulley, which can be written as: pi () times the diameter or, better still, pi times twice the radius.

Now, we can also see that the pull on the rope times the radius of the pulley will give us a torque value (T) in lb-ft.

This will give us the following rotational horsepower equation: Hp = 2 NT/33,000 which can be simplified to: Hp = NT/5255 where N is the shaft (pulley) rotational speed in revolutions per minute (rpm) and T is the torque applied in lb-ft.

This is how the dynamometer at your dealer works. The dynamometer is driven from the power-take-off (pto) at either 540 or 1,000 rpm and the torque load is created in the dynamometer by a hydraulic pump or other power-consuming device. That power is converted to heat and removed from the dynamometer as hot water.

Let's assume that we are load-testing a 100 pto horsepower tractor using the 540 and then the 1,000 rpm pto shaft into the dynamometer and compare the torque values.
T = Hp x 5,255/N
T@540 = 100 x 5,255/540 = 973 lb-ft
T@1,000 = 100 x 5,255/1,000 = 525 lb-ft

Do you see why we increased the power-take-off speed to 1,000 rpm as tractor horsepower increased? We reduced the torque or twisting action on the pto shaft, as well as the other drive line components, to just about half. You can also see why John Deere had to conform and give up the old slow-speed, two-cylinder tractors in 1960.

When you read the literature for the new car or pickup truck that you are considering, note the engine speed at which maximum horsepower is established. I'll bet it is well over the 2,000-2,500 rpm that is typical for all of our farm workhorses, such as tractors and combines.

But, please, torque is a force acting at a perpendicular distance - the moment arm - and should always be quoted as lb-ft or Nm, not ft-lb as we often see in print. If the metric term, newton-metre (Nm) for engine torque is used, the power will be given in watts, not horsepower. For your information, one horsepower is about 746 watts.

Now let's look at hydraulic horsepower. We are seeing more and more hydraulic drive systems for powering augers, drivelines and other rotating parts. The big new combines in particular are being driven hydraulically to eliminate many of the moving drive parts, such as belts and chains.

If you know the gallonage (flow rate) and pressure requirements of your work functions, you can easily calculate the "hydraulic horsepower" (hhp): hhp = P x Q/1,714

If, for example, the maximum working (relief) pressure is 3,000 psi (P) and the maximum flow rate (Q) in U.S. gallons per minute is 40, the hydraulic horsepower will be:
hhp = 3,000 x 40/1714 = 70.0

When this value has been calculated, do remember that all hydraulic systems will have power or energy losses as heat. Heat will be created whenever the hydraulic fluid is pushed through small clearances around valve spools, restrictor valves or passed pistons. Thus, you must add at least 20 per cent to the hhp to determine the engine hp needed to drive the hydraulic pump(s). For the above example, a 90-horsepower engine would be okay.

Note that I used U.S. gallons per minute (gpm) and pounds per square inch) psi. Virtually all hydraulic equipment is sourced in or through the United States. U.S. gallons have been and will be the units of hydraulic flow into the foreseeable future. If you prefer litres, one U.S. gallon is equal to 3.78 litres, not the 4.5 you might have used to convert from Imperial gallons.

Then we have the pressure issue. The United States uses pounds per square inch. In Canada we talk about (kilopascal) KPa and megapascal MPa,, but we are alone in this area. Europe, the other primary source of hydraulic components, uses "bars" and they are not likely to change any time soon.

So you should remember the following pressure comparisons:
One atmosphere = 14.5 psi = one bar

If you are checking your tire pressures in Europe and a taxi driver is in the lineup, you will be told very quickly to set the pressure regulator to two bars. I speak from experience! BF

Agricultural engineer Ralph Winfield farms at Belmont in Elgin County.