Part 9
A. 300 × 110 = 33,000 foot-pounds.
Q. Would that equal one horse-power?
A. Yes, if done in one minute.
Q. Suppose it took two minutes?
A. Then there would be only half a horse-power, or 33,000 ÷ 2 = 16,500 foot-pounds per minute.
Q. Is it correct to say “horse-power per minute” or “horse-power per hour”?
A. No. When an engine is working at the rate of 10 horse-power, it is doing 10 horse-power all the time. It is an error to assume that such an engine is doing 10 horse-power per minute, and 10 × 60 equals 600 horse-power per hour. When it is said that an engine uses 20 pounds of steam per horse-power per hour, it is meant that this amount of steam is used per hour for each horse-power developed.
Q. How is the foot-pounds of work done by a steam engine, found?
A. Multiply the average pressure per square inch during the stroke by the number of square inches in the piston, and by the number of feet through which the piston has moved.
Q. What do you understand by the “mean effective pressure”?
A. The mean pressure is the average pressure pushing the piston through the stroke, which is about one-third the pressure in the boiler. There is generally some back pressure working against it, therefore the “effective” pressure is only the difference between the two. It can only be determined accurately by measurements from an indicator diagram.
Q. What is a single acting engine?
A. An engine in which the steam acts on one side of the piston only.
Q. How do you find the “piston’s speed”?
A. On double acting engines, multiply the stroke in inches by two and by the number of revolutions per minute and divide by 12.
Q. Why multiply the stroke in inches by 2?
A. Because in double acting engines there are two working strokes to each revolution.
Q. Why do you divide by 12?
A. To reduce the inches to feet.
Q. How is the “piston’s speed” of a single acting engine found?
A. Multiply the stroke in inches by the revolutions per minute and divide by 12.
Q. What is the horse-power developed by an engine, say 12 × 24 inch, running 125 revolutions per minute, with 40 pounds mean effective pressure?
A. Area = 12 × 12 × .7854 = 113.0976 sq. ins.
Piston speed = 24 × 2 × 125 ÷ 12 = 500 feet per minute.
M. E. P. = 40 lbs.
Then (113.0976 × 500 × 40) ————————————————————— = 68.544 H. P. 33000
Q. What is a single valve engine?
A. It is an engine in which a single valve controls the admission and distribution of steam to both ends of the cylinder, or a common slide valve engine.
Q. What is a four valve engine?
A. An engine which has separate steam and exhaust valves for each end both top and bottom of cylinder, such as a Corliss engine.
Q. Into what three classes are engines divided with reference to the manner in which they are governed?
A. Throttling engines, Automatic cut-off and Governor engines.
Q. What is an Automatic cut-off engine?
A. An engine in which the amount of steam supplied is automatically cut off at various points in the stroke, in accordance with the load and pressure. In Throttling engines the volume admitted is constant and the pressure varied. In Automatic cut-off engines, steam is admitted at the highest available pressure, and the volume is varied to meet the requirements of the load. In Governor engines, the steam is admitted and cut off by the governor.
Q. What is a throttle governed engine?
A. An engine in which the amount of steam supplied is regulated by changing the pressure at which it enters the cylinder in accordance with the variation of the load.
Q. What is a Governor engine?
A. An engine in which the supply of steam is regulated by the governor.
Q. Into what classes may the Automatic cut-off engine be divided?
A. Into two classes: The four valve engine, in which the cut-off is usually effected by a detaching mechanism or trip under the control of the governor; the single valve engine, in which the point of cut-off is varied by changing the amount of travel of the valve.
Q. Give examples of the single valve type.
A. High speed, self-contained engines which have shaft governors.
Q. What are their advantages?
A. High rotative speed, compactness, portability, light weight and simplicity.
Q. Are they more economical than the four valve engine?
A. No; the four valve engines are the more economical.
Q. Give a prominent example of the four valve engine.
A. The Reynold’s Corliss.
Q. What is meant by an engine running “over”?
A. The top of the drive wheel running away from the cylinder.
Q. What is meant by an engine running “under”?
A. The top of the drive wheel running towards the cylinder.
Q. Which way are engines generally run?
A. “Over.”
Q. What advantages do engines have in running “over”?
A. The pressure of the cross-head on engines running over, is always downward upon the guides; for when the pressure is on the head end of the piston, the pressure against the connecting rod which is pointing upward, reacts by pressing the cross-head down upon the lower guide, and when the pressure is on the crank end of the cylinder, the cross-head will be dragging the crank, and as the crank is below the center line, it will pull the cross-head down upon the lower guide, while if the engine is running under, the pressure of the cross-head will be upon the top guide, both on the outward and inward strokes, and unless the cross-head is nicely adjusted to its guides and the guides are perfectly parallel, the cross-head will be lifted when subjected to thrust, and will fall on the center by its own weight, causing the engine to pound.
Q. At what point in the stroke is the pressure on the cross-head greatest with a uniform pressure in the cylinder?
A. When the crank is at right angles to the guide.
Q. How does the relative length of the connecting rod affect this pressure?
A. The longer the connecting rod as compared with the crank, the less will be the pressure on the guides.
Q. What is the usual ratio of connecting rod to crank?
A. The connecting rod is from four to six times the length of the crank.
Q. Are there any objections to a long connecting rod?
A. A long connecting rod makes a long engine, and makes extra cost in the bed or frame and the room occupied. The longer rod is heavier, and brings extra weight upon the cross-head, guides and crank-pin. The long rod also lacks stiffness unless excessively heavy.
Q. What determines the length of the crank?
A. The stroke.
Q. What limits the stroke?
A. The piston’s speed limits the length of stroke allowable with a given rotative speed, or the number of revolutions per minute with a given stroke.
Q. What is the practical limit of piston’s speed?
A. Engines of from four to six foot stroke can run at from seven to eight hundred feet piston’s speed per minute. Those of shorter stroke should not run over six hundred feet.
Q. Why do high speed engines have a short stroke in comparison with the diameter of their cylinders?
A. So that they can run at a high rate of speed without exceeding the limit of piston’s travel.
Q. What is the office of the fly-wheel?
A. It maintains a uniformity of motion of the crank, notwithstanding the unequal moving force upon the crank-pin.
Q. Is the force upon the crank-pin unequal, even when the pressure from the cylinder is uniform throughout the stroke?
A. Yes. No matter what the pressure on the piston is, it has no effect in turning the engine when the crank is in line with the guides, which is termed “on the center.” As the crank gets away from the centers, the effect of a given pressure becomes greater, and reaches its maximum when the crank is nearly at right angles with the guides.
Q. How does the fly-wheel counteract the jerky motion of the crank which would result from this?
A. By its tendency to resist an excessive moving force, and by its momentum keeps the engine in motion when the moving force is deficient.
Q. What would you do if the cylinder gets worn or cut from too tight rings or lack of oil?
A. Rebore the cylinder.
Q. What would you do if the crank-pin heats, gets worn or cut?
A. If bent it should be turned true again; if not bent it can be filed and polished perfectly true by hand.
Q. What would you do if the main bearings heat?
A. Loosen the caps and apply plenty of good oil. If this does not stop it take off the caps, examine the oil holes to ascertain why the oil does not reach the bearing. If the bearings have become rough and cut, the shaft will have to be smoothed again.
Q. Would any harm result from starting an engine with the drip cocks closed?
A. Yes, the condensed steam filling the space would smash the cylinder or piston head.
Q. What do you mean by atmospheric pressure?
A. The weight of the atmosphere, which is 14.7 lbs. per square inch at sea level.
Q. How hot can you get water with exhaust steam under atmospheric pressure?
A. 212° Fahr.
Q. Does atmospheric pressure have any influence on the boiling point?
A. It does.
Q. Would you run an engine with throttle wide open, or partly open?
A. Wide open on governor engines, as it is more economical.
Q. How many pounds of water required per horse-power for the best engines?
A. From 25 to 30 pounds.
Q. At what temperature has iron the greatest tensile strength?
A. About 600 degrees.
Q. How much water is consumed (in pounds) per hour per indicated horse power?
A. From 25 to 60 pounds.
Q. How much steam will be evaporated from one cubic inch of water under atmospheric pressure?
A. About one cubic foot, approximately.
Q. How much coal is consumed per hour per indicated horse-power?
A. From two to seven pounds.
Q. How much does one cubic foot of fresh water weigh?
A. 62½ pounds.
Q. How much does one cubic foot of iron weigh?
A. 486⁶/₁₀ pounds.
Q. What does one square foot of half inch boiler iron weigh?
A. Twenty pounds.
Q. For steam purposes, how much wood is required to equal one ton of coal?
A. About 4000 pounds of wood.
Q. Of what does coal consist?
A. Carbon, nitrogen, sulphur, hydrogen, oxygen and ash.
Q. What are their relative proportions?
A. There are different proportions in different specimens of coal. The average per cent is carbon eighty, nitrogen one, sulphur two, hydrogen five, oxygen seven, ash five.
Q. Of what is air composed?
A. It is composed of nitrogen and oxygen in the proportion of seventy-seven of nitrogen and 23 of oxygen.
Q. Of what does water consist?
A. Hydrogen and oxygen in the proportion of one of hydrogen to eight of oxygen by weight.
Q. What are the different kinds of heat?
A. Latent heat, sensible heat, and sometimes total heat.
Q. What is meant by latent heat?
A. Heat that does not affect the thermometer and which expends itself in changing the nature of a body, such as turning ice into water or water into steam.
Q. Under what circumstances do bodies get latent heat?
A. When they are passing from a solid to a liquid state, or from a liquid to a gaseous state.
Q. How can latent heat be recovered?
A. By bringing the body back from a state of gas to a liquid, or from a liquid to a solid.
Q. If the power is in coal, why should we use steam?
A. Because steam has some properties which make it an invaluable agent for applying the energy of the heat to the engine.
Q. What is steam?
A. It is an invisible elastic fluid generated from water by the application of heat.
Q. What are its properties which make it so valuable to us?
A. First. The ease with which we can condense it.
Second. The small space which it occupies when condensed.
Third. Its great expansive power.
Q. What do you understand by the term “horse-power”?
A. A horse-power is equivalent to raising 33,000 pounds one foot per minute.
Q. What do you understand by “lead” on an engine valve?
A. Lead on a valve is the admission of steam into the cylinder before the piston completes its stroke.
Q. What are considered the greatest improvements on the stationary engine in the past forty years?
A. The Corliss valve gear, the governor, the compound and triple expansion.
Q. What is meant by triple expansion engine?
A. A triple expansion engine has three cylinders using the same steam expansively in each one.
Q. What is the clearance of an engine as the term is applied at the present time?
A. Clearance is the space between the cylinder head and the piston head with the ports included.
Q. What is the principal which distinguishes a non-condensing from a condensing engine?
A. Where no condenser is used, and the exhaust steam is open to the atmosphere, it is a non-condensing engine.
Q. Why do you condense steam?
A. To form a vacuum and thus remove the atmospheric and back pressure that would otherwise be on the piston, thereby getting more useful work out of the steam.
Q. What is meant by vacuum?
A. A space void of all pressure.
Q. How can you maintain a vacuum?
A. By the steam used being constantly condensed by the cold water or cold tubes, and the air pump constantly clearing the condenser.
Q. Why does condensing the used steam form a vacuum?
A. Because a cubic foot of steam at atmospheric pressure shrinks into about one cubic inch of water.
Q. What is a condenser as applied to an engine?
A. The condenser is that part of an engine into which the exhaust steam enters and is condensed.
Q. About how much gain is there by using the condenser?
A. Seventeen to twenty-five per cent. where cost of water is not figured.
Q. What do you understand by the use of steam expansively?
A. Where steam admitted at a certain pressure is cut off and allowed to expand to a lower pressure.
Q. How many inches of vacuum gives the best result in a condensing engine?
A. About 25 inches.
Q. What is meant by a horizontal compound tandem engine?
A. One cylinder being back of the other with two pistons on the same rod.
Q. What do you understand by lap?
A. Outside lap is that portion of the valve which extends beyond the ports when valve is placed on the center of its travel; inside lap is that portion of valve which projects over the ports on inside or toward the middle of the valve.
Q. Of what use is lap?
A. It gives expansion to the steam in the cylinder.
Q. What is the dead center of an engine?
A. The point where the center of shaft, center of wrist-pin and center of piston rod are in the same straight line.
Q. From what cause do belts have power to drive shafting?
A. By friction and cohesion.
Q. When would you oil an engine?
A. Before starting it and as often while running as is necessary.
Q. What is the tensile strength of American boiler iron?
A. 40,000 to 60,000 pounds per square inch.
Q. What are the principal defects found in boiler iron?
A. Imperfect welding, brittleness, low ductility.
Q. What is the advantage of steel as a material for boiler plate?
A. Tensile strength, ductility, homogeneity, malleability and freedom from laminations and blisters.
Q. What are the disadvantages of steel as a material for boiler plate?
A. It requires greater care in working than iron and is subject to flaws induced by the pressure of gas bubbles in the ingots from which the plates are made.
Q. How far apart should stay bolts be put in a boiler?
A. They vary from 4 to 6 inches apart, depending on thickness of plates, size of stay bolts and amount of steam pressure to be carried.
J. I. CASE PORTABLE SKID ENGINE.
The engine frame is of the girder pattern, cast in one piece, and contains the guides for cross-head and pillow block bearing. It also forms front head of cylinder to which it is firmly bolted. The cylinder is supported and firmly attached to the large feed water heater, and both cylinder and frame are entirely disconnected from the boiler. The heater forms a support for front end of boiler. The engine is placed diagonally with the boiler, allowing the crank shaft to pass over the fire box end. This permits the use of a very large band wheel. The outer end of shaft is supported by a pillow block attached to a large bracket bolted to the boiler.
The locomotive style of boiler has a large steam dome in center and an ash pan under fire box. It is supported upon the long wooden skids by brackets bolted to the sides of boiler, the heater being also bolted to the skids.
The independent pump is connected to the heater, and bolted to skid.
The engine has a Plain Slide Valve, which receives its motion from the rocker arm, operated by the eccentric and rod. It has Governor, Steam Gauge, Pop Valve and all necessary fittings. These engines range in size from 20 to 30 horse-power, and are extensively used for driving portable saw mills.
RULES AND TABLES.
_To find the steam pressure_ on slide valve, multiply the unbalanced area of valve in inches by pounds pressure of steam per square inch, add weight of valve in pounds, and multiply the sum by 0.15.
_Safety boiler pressure_ according to the United States Government rule is as follows: Multiply ⅛ of the lowest tensile strength found on any plate in the cylindrical shell by the thickness expressed in inches or part of an inch of the thinnest plate in the same cylindrical shell, and divide by the radius or half the diameter, also expressed in inches, and the sum will be the pressure allowable per square inch of surface for single riveting, to which add 20 per centum for double riveting.
_To find the water pressure_ on steam pipes leading from boiler to steam gauge, divide the difference in height between the highest point of pipe and the center of steam gauge by 2³/₁₀; the result will be the pressure exerted by the water in the pipe in pounds upon the gauge.
_Area of a Circle._—To find the area of a circle when the diameter is given, multiply the diameter by itself, or in other words square the diameter and multiply the result by .7854.
Ex. Diameter 5 inches, 5 × 5 = 25 × .7854 = 29.635 area.
_Circumference of a Circle._—To find the circumference of a circle when the diameter is given, multiply the diameter by 3.1416.
Ex. Diameter is 5 inches. 5 × 3.1416 = 15.708 inches circumference.
_Diameter of a Circle._—To find the diameter of a circle when the circumference is given, multiply the circumference by .31831.
Ex. Circumference 20 inches. 20 × .31831 = 6.362 diameter.
_To find the pressure_ of pounds per square inch of a column of water, multiply the height of the column in feet by .434. Approximately, we generally call every foot elevation equal to one-half pound pressure per inch, this allows for ordinary friction.
_To find the diameter_ of a pump cylinder to move a given quantity of water per minute (100 feet of piston being the standard of speed), divide the number of gallons by 4, then extract the square root and the product will be the diameter in inches.
_To find the horse-power_ necessary to elevate water to a given height, multiply the total weight of water in pounds by the height in feet and divide the product by 33000 (an allowance of 25% should be added for friction, etc.).
_The area of the steam piston_ multiplied by the steam pressure, gives the total amount of pressure that can be exerted. The area of the water piston multiplied by the pressure of water per square inch gives the resistance. A margin must be made between the power and the resistance to move the piston at the required speed—say 50%.
_To find the capacity_ of a cylinder in gallons, multiply the area in inches by the height of stroke in inches, which gives the total number of cubic inches; divide this amount by 231 (which is the cubic contents of a gallon in inches), the product is the capacity in gallons.
RULES FOR CALCULATING THE SPEED OF GEARS AND PULLEYS.
In calculating for pulleys, multiply or divide by their diameter in inches.
In calculating for gears, multiply or divide by the number of teeth required.
The driving wheel is called the “driver,” and the driven wheel the “driven.”
PROBLEM 1.—To find the diameter of the driver when the revolutions of the driver and driven and the diameter of the driven are given.
RULE.—Multiply the diameter of driven by its number of revolutions, and divide by the number of revolutions of the driver.
PROBLEM 2.—To find the diameter of the driven to make the same number of revolutions in the same time as the driver, when the diameter and revolutions of driver are given.
RULE.—Multiply the diameter of the driver by its number of revolutions, and divide the product by the required number of revolutions.
PROBLEM 3.—To find the number of revolutions of the driven when the diameter or number of teeth and number of revolutions of the driver and the diameter or number of teeth of the driven are given.
RULE.—Multiply the diameter or number of teeth of the driver by its number of revolutions, and divide by the diameter or number of teeth of the driven.
PROBLEM 4.—To find the number of revolutions of the driver when the diameter of the driven and the number or revolutions of driven are given.
RULE.—Multiply the diameter of driven by its number of revolutions, and divide by the diameter of the driver.
SHAFTS AND PULLEYS.
To find the size of pulley needed to give the main shaft a certain number of revolutions, multiply the diameter of fly-wheel in inches by the number of revolutions of the engine, and divide by revolutions of main shaft.
To find the revolutions of main shaft when diameter of its pulley is given, multiply the diameter of fly-wheel in inches by its number of revolutions, and divide by the diameter of pulley in inches.
To find diameter of fly-wheel needed to drive a pulley at a given number of revolutions, multiply the diameter of pulley in inches by its number of revolutions, and divide by number of revolutions of the engine.
To find revolutions of an engine with a given fly-wheel to drive a pulley at a given number of revolutions, multiply the diameter of pulley in inches by its number of revolutions, and divide by the diameter of fly-wheel in inches.
Rule for finding the amount of heating surface in a locomotive boiler:
Find surface of flues by multiplying the diameter by 3.14 to get circumference, and multiply this product by the length of the flue, then multiply this result by the number of flues in the boiler, and divide by 144 to get number of square feet of surface in the flues.
Multiply length and width of fire box for number of square inches in crown sheet, then multiply the length and height of the fire box and the result by two, which gives the number of square inches in the sides; then multiply the width and height and multiply the result by two, which gives the number of square inches in the ends, from which subtract number of square inches in space left for door and flues—then add all these results together and divide by 144 to get the number of square feet in the fire box. Add same to number of square feet of flues, and the total will be the number of square feet in the boiler.
By dividing number of square feet by rated horse-power of boiler will give number of square feet of heating surface to each horse-power.
_=TABLES.=_
=ALLOYS.=