CHAPTER XI
HORSE POWER AND THE INDICATOR DIAGRAM
A book on “The Petrol Engine” would hardly be complete without some reference to horse-power and the indicator diagram. The following definitions must be carefully studied.
=Work.=—A force is said to do mechanical work when it overcomes a resistance in its own _line of action_. The line of action of a force is a line indicating the direction in which the force acts. Engineers _measure_ work in foot-pound units. The product obtained when we multiply the magnitude of the force or resistance (in pounds) by the distance through which it has acted or been overcome (expressed in feet) gives the quantity of work done in foot-pounds.
_Example_:—A force of 50 lbs. is exerted in overcoming a resistance through a distance of 12 feet. Find the work done.
Work done = Force (in lbs.) × Distance (in ft.) = 50 × 12 = 600 ft. lbs.
=Power.=—The _rate_ at which work is done is a measure of the power exerted. One horse-power is exerted when 33,000 foot-pounds of work are done in one minute. The work done per minute (in ft. lbs.) divided by 33,000 gives the horse-power expended.
_Example_:—To propel a motor-car along a level road at a speed of 30 miles an hour requires a tractive effort or pull of 70 lbs. if the vehicle weighs one ton. Find the horse-power required, at the road surface.
Work done per minute in ft. lbs. Horse-power = ——————————————————————————————— 33,000
Force (in lbs.) × Distance through which it acts per minute (in ft). = —————————————————————————————————— 33,000
70 × 30 × 5280 —————————————— 60 7 × 264 = —————————————— = ———-——— = 5·6 33,000 330
_Example_:—If the car in the preceding example had to climb a gradient which rose one foot for every four feet traversed by the car, find the additional horse-power needed to keep up a speed of 30 miles an hour while climbing the gradient.
Here we have to raise a weight of 1 ton vertically upwards through a height equal to one-fourth of the road surface covered, every minute.
Additional Horse-power required
(30 × 5280 1) 2240 (lbs.) × (————————- × -) ft. per min. ( 60 4) = ——————————————————————————————————————————— 33,000
2240 × 660 = —————————— = 44·8 33,000
Total Horse-power to climb the gradient of 1 in 4 at 30 miles an hour = 5·6 + 44·8 = =50·4=
=Brake Horse-Power.=—The length of the circumference or boundary line of a circle is 6·28 times the length of the radius of the circle or 3·14 times the length of its diameter. Hence, if an engine exerts a pull of P lbs. at the end of a brake arm of length R feet when it is maintaining a speed of N revolutions per minute (we may imagine the brake to be fitted round the rim of the flywheel), we can calculate the =brake horse-power= thus:—
Brake Horse-Power or B.H.P. = Work done on the brake per minute in ft. lbs./33,000
Hence B.H.P = (Pull at the end of the brake arm (in lbs.)) × (6·28 times the radius of the arm (in feet)) × (the number of revolutions made by the engine (in one minute))/33,000
= {P × 6·28 × R × N}/33,000
_Example_:—An engine being tested by a brake applied to the flywheel as shown in the sketch (Fig. 67) exerts a pull of 50 lbs. at a speed of 2,000 revolutions per minute. If the length of brake arm is 30 inches, calculate the brake horse-power developed.
Work done per minute = 50 × 6·28 × 30/12 × 2000 ft. lbs.
B.H.P. = {50 × 6·28 × 30/12 × 2000}/33,000 = 47·5
=Rated Horse-Power.=—For taxation purposes the Treasury makes use of a formula for the rating of petrol engines according to their probable horse-power. This formula is based on a certain speed of the piston which was regarded as a limiting value some years ago (when the formula was first proposed) and on the attainment of a certain effective pressure in the cylinder.
Horse-power from the Treasury formula = 0·4 d^{2}n.
Where d = diameter of cylinder in inches, n = number of cylinders.
With modern engines much greater horse-power is obtained, and a near approximation to the true output is obtained by using what is now known as the Joint Committee’s formula.
Brake Horse-Power = 0·46 n (d + s) (d - 1·18)
Where d = diameter of cylinder in inches. s = length of piston’s stroke in inches.
This formula is only to be used in an attempt to predict the _probable_ maximum horse-power which any engine will give. It must not be confused with the ordinary brake horse-power formula.
_Example_:—Find the probable maximum horse-power of an engine having four cylinders each 3 in. bore and a piston stroke of 4 in. What would be its horse-power for taxation purposes?
_By Joint Committee’s formula_—
B.H.P. = 0·46 × 4 (3 + 4)(3 - 1·18) = 1·84 × 7 × 1·82 = 23·35
_By Treasury formula_—
B.H.P = 0·4 × 3^{2} × 4 = 0·4 × 9 × 4 = 14·4
=Indicated Horse-Power.=—The horse-power which an indicator would show as being developed inside the cylinder of a petrol engine, above the piston, would be called the _indicated_ horse-power, and should always work out a greater number than the brake horse-power or power available at the engine flywheel, because some of the power liberated from the combustion of the petrol within the cylinder is lost in friction of the piston and bearings.
The Indicated Horse-Power or I.H.P. = P_{e} × A × L x N_{e}/33,000.
Where P_{e} = mean effective pressure from the diagram, in lbs. per sq. inch.
A = area of piston in square inches = 0·7854 (diameter of cylinder)^{2}
L = length of stroke of piston, in feet.
N_{e} = number of power impulses per minute delivered to the crankshaft.
Since a four-stroke engine gives one power impulse to the crankshaft in every two revolutions, it follows that N_{e} is equal to _half_ the number of revolutions per minute for a single-cylinder engine of that type, and _twice_ the number of revolutions for a four-cylinder engine. A four-cylinder two-stroke engine might be arranged to give either _two or four_ impulses per revolution of the crankshaft—depending upon the arrangement of the cranks.
_Example_:—A four-cylinder four-stroke engine runs at a speed of 2,000 revolutions per minute and the mean-effective pressure in the cylinders is 75 lbs. per square inch. Calculate the indicated horse-power if the cylinders are 4 in. × 4 in.
I.H.P = P_{e} × A × L × N_{e}/33,000
= 75 × 0·7854 × 4^{2} × (4/12) × 4000/33,000
= {75 × 12·56 × 4000}/99,000 = 38
=The Indicator Diagram.=—At the commencement of this chapter we explained that the work done by a force was measured by multiplying the number representing the magnitude of the force (in pounds) by the distance through which it had acted (measured in feet). This product gave us the quantity of work done in foot-pound units. Thus “_quantity of work done_” is really the product of two numbers, just as the area of a rectangular floor space is measured by length times breadth. In symbols we write W = F × S where F is the magnitude of the force or resistance in pounds and S the distance through which it has acted, in feet. It is interesting to contemplate this symbolical expression W = F × S together with the expression Area = Length × Breadth, because it gives us a new idea for measuring work. Imagine a diagram of the kind shown in Fig. 68, in which the curved line AB has been obtained by _plotting_ values of F and S for any imaginary case. The diagram is supposed to represent pictorially how the particular force under consideration has varied in magnitude as it has traversed a space represented, to some scale, by the length DC. It is clearly seen that the force has been _decreasing_ in an _irregular_ manner from some large value represented by the height DA to a small value represented by the height CB. We now proceed to show that the shaded area ABCD measures the total amount of work done by this force.
Considering for a moment just the small strip _efdc_ of the diagram we see that it is easy to find a rectangle _abcd_ equal in area to it. Now the _height_ of this rectangle will be the _average value_ of the force while it traversed the space _cd_, and hence the area of the rectangle _abcd_ gives the =work done= by the force in passing from _c_ to _d_. Similarly by dividing up the whole diagram we would obtain a number of little rectangles each equal in area to the magnitude of the work done from point to point. Thus the whole area ABCD gives the whole work done. To measure the work done in an engine cylinder we must use some form of _indicator_. An indicator is an instrument which traces out a diagram on which _abscissæ_ (or horizontal distances) represent displacements of the piston and _ordinates_ (or vertical distances) represent the pressures acting on the piston.
Ordinary steam engine indicators with pencil motion and paper drum are not suitable for use with fast running petrol engines. The moving parts of these indicators are too heavy and their springs too sluggish in action to keep correct time with these high speed engines. Again, there is too much friction between the pencil and the paper drum, as well as in the lever joints. Therefore special indicators have to be used, in which the diagram is traced out by a beam of light reflected from a mirror on to a ground glass screen or photographic plate. One corner of the mirror is tilted in time with the movement of the engine piston by means of a special reducing mechanism, and another corner of the mirror is tilted in a direction at right angles to the first by means of a very short thin rod kept in contact with a metal diaphragm subjected to the pressure of the gases in the engine cylinder. A beam of light is thrown on to the mirror from a lamp, and after reflection traces out the diagram on the screen or plate. Such an instrument would generally be described as a _manograph_. An indicator diagram from a four-stroke engine is shown in Fig. 69. The line ABC represents the suction stroke of the piston during which the pressure of the gases in the cylinder falls a little below that of the atmosphere. Atmospheric pressure is shown by the height of the line LL above the base, or line of zero pressure (perfect vacuum). The inlet valve can be opened at B and closed at D after the crank has turned the bottom dead-centre and begun the compression stroke. The line CDE represents the compression stroke of the engine, during which the gases are compressed and their pressure rises. The height of the point E above the line LL gives the _compression pressure_ to the scale of the diagram. Ignition occurs at E, and results in an instantaneous rise of pressure to F due to the explosion, which is, however, quickly followed by expansion to G. The exhaust valve opens at G, the gases are _released_, and the pressure falls still further to point H. The line HA represents the exhaust stroke of the piston, and the exhaust valve would be closed after the crank had passed its upper dead-centre and commenced the suction stroke. The distance marked (x) on the diagram measures the _clearance_ volume (or volume of the space above the piston containing the valves and referred to as the combustion chamber) to the same scale that the length of the diagram measures the volumetric displacement of the piston. The volume traced out by the piston during any working stroke is measured by multiplying the area
of the piston in square { centimetres} by the length { inches } of the stroke in { centimetres} the product giving us {inches } the capacity of the cylinder in cubic { centimetres}. The { inches }
area of the diagram HEFG gives the work done during one cycle of operations, and the area of the small diagram ABCD gives the work lost in taking in and expelling the charge. The small area should be subtracted from the large one to get the useful work done per cycle of operations. The area of the diagram HEFG may readily be obtained by finding its vertical height at a number of equidistant points, and from these measurements ascertaining the average or _mean_ height of the diagram. The average height of the diagram (in inches) multiplied by its length (also in inches) gives the area in square inches.
The average or mean height of the diagram also gives what we term the _mean effective pressure_ acting on the piston, and constitutes the P_{e} of the indicated horse-power formula above. The area ABCD is always small and generally neglected with four-stroke engines. There are _two separate diagrams_ for a two-stroke engine. The diagram for the working cylinder is A_{1}B_{1}C_{1}D_{1} in Fig. 70, and that for the crankchamber is E_{1}F_{1}G_{1}H_{1}. The effective work done per cycle is measured by the difference in the area of these two diagrams. The piston uncovers the exhaust port at B_{1} and closes it again at C_{1}; it uncovers the inlet port at F_{1} and covers it again at G_{1}. From F_{1} to G_{1} the charge is being delivered from the crankchamber to the working cylinder. The area of the loop E_{1}F_{1}G_{1}H_{1} is larger than the corresponding portion of the four-stroke diagram and should not be neglected.