PRACTICAL HYDRAULICS FOR FIREMEN

PRACTICAL HYDRAULICS FOR FIREMEN

(Continued from page 64.)

Pump Discharge at Different Pressures.

From the previous discussion it is quite evident that the greater the discharge pressure on a pump is, the greater will be the amount of work required to discharge each pound of water. For example, to pump a quantity of water against a 200-foot head requires just twice as much work as it docs to pump a like quantity against a 100-foot head. For this reason the capacity of fire engines or other pumping engines is always less for high pressures than for low pressures.

Theoretically, a pumping engine should be able to produce the same water horsepower at any pressure. But this is not so, for the greater the pressure becomes the greater will be frictional resistances and slippage so that pumps seldom give the water horsepower at very high pressures that they do at low pressures. Nevertheless, for rough calculations for practical application in the fire service It may be assumed that pumping engines, including fire engines, can produce a constant water horsepower throughout the entire range of pressures at which they are capable of operating.

This assumption then gives a means of determining the capacity of a pumping engine at any pressure within working limits when the discharge it is capable of giving at a certain pressure is known. It is only necessary to solve for the water horsepower at the known pressure and discharge and consider that the horsepower so found will be constant throughout the entire range of discharge of the engine. With the water horsepower determined it is a simple matter to find the discharge for any given discharge pressure.

Example 1. A pumping engine discharges a maximum of 700 gallons per minute at 120 pounds pressure. What will be its maximum discharge at 200 pounds pressure? Solution: The water horsepower of the engine is (700X120) /1715=49, approximately. For convenience let Q=discharge at 200 pounds pressure. Then (QX200)/1715=49 for the horsepower will be the same at the higher pressure as stated above. Solving this equation, Q=(49X1715)/200=420 gallons per minute. Ans.

•Copyright, 1917, by F. Shepperd.

This value is somewhat high, and must necessarily be so, for no account is taken of the additional losses due to friction and slippage when the engine is working under high pressure. A good specification, given in the “Red Book,” and applying equally well to steam fire engines and automobile fire engines, is that the engine should deliver its full rated capacity at 120 pounds net pressure and 50 per cent of its rated capacity at 200 pounds net pressure. This will assure sufficient boiler capacity in steam fire engines and motors of high enough power in automobile fire engines. The method outlined in Example 1 may, however, be followed in the solution of problems wherein accurate results are not required.

Example 2. A motor pumping engine is rated at 800 gallons per minute at 100 pounds. How many effective 7/8-inch streams can it supply if each line is 1,000 feet in length? Solution: This example, given to bring out a definite point, is one which would probably never be met with in service.

Before solving for the number of streams the engine will supply, it is essential that the engine discharge pressure be known.

As mentioned in the chapter treating the subject of nozzles, 40 pounds is the minimum which could be considered as producing an effective stream where a l 1/8-inch nozzle is employed. This same figure may also be considered the minimum for a 7/8-inch nozzle. The discharge from a 7/8-inch nozzle under 40 pounds pressure=29.7X7/8X7/8X7/8X√40=144 gallons, approximately.

The friction loss in the 1,000-foot line of hose is determined by the usual formula, and is 10X(2Xl.44Xl.44+1.44)= 56 pounds. The discharge pressure of the pump must then be 40+56=96 pounds, approximately.

The water horsepower of the pump is (800X100)/1715=46.65. Substituting the new pressure in the formula and letting discharge at 96 pounds equal Q, (QX96)-4-1715=46.65. From this, Q= (46.65X1715)-4-96=833 gallons per minute.

As each stream requires 144 gallons per minute, the engine is capable of supplying 833/144=5.78, or six streams, approximately. Ans.

Example 3. A fire boat with a capacity of 9,000 gallons per minute at 150 pounds has a turret pipe with a 2-inch nozzle in action, and also supplies five 2 1/2-inch lines each 300 feet in length and equipped with l 1/4-inch nozzles. How many more similar lines can be added without overloading the pump? The pumps are working under 200 pounds pressure. Solution: The first step is to determine the maximum discharge at 200 pounds pump pressure. The water horsepower is (9000Xl50)+1715=787 horsepower, and the discharge at 200 pounds pump pressure can be found as before by substituting in the equation, (QX200)/1715=787, from which Q= 6,750 gallons per minute at 200 pounds.

The pressure on the turret pipe nozzle may be taken at 200 pounds, and the discharge will be 29.7X2X2X√200=1,680 gallons per minute. The pressure at the 1 1/4-inch nozzle on the 300-foot line of hose will be

200

N. P.=-=77 pounds, and with

1.1+.248X6

this pressure the discharge is 29.7X1 1/4 X 1 1/4 X √77=407 gallons per minute. The five lines discharge 5X407 =2,035 gallons per minute. Hence the total discharge thus far accounted for is 1680+2035=3715 gallons at 200 pounds pressure. The boat is still capable of giving 6750—3715=3,035 gallons per minute, or is capable of supplying 3035/407 =7.45, or 7, additional 300-foot lines of 2 1/2-inch hose equipped with l 1/4-inch nozzles.

Fire Engine Ratings.

Before starting the next example a few words relative to pumping engine-capacities may prove useful. Pumping engines, in fire departments, either of the steamer or automobile type, have been classified according to their capacities into sizes as follows:

But it would be impractical to design a pump with definite capacity, and the discharges of different engines of the same rated capacity are apt to vary considerably. The following method for rating fire engines may, therefore, appear more suitable:

All of the above discharges are at 120 pounds pump pressure.

Example 4. A building 100 feet high, called A, stands at the southeast corner of N street and Avenue R. Opposite A, and across the avenue, which is 90 feet wide, is a building 88 feet high called B, and across N street, 60 feet wide, is a building 70 feet high, called C. There is a hydrant 70 feet from B and another 100 feet from C, as shown in the accompanying sketch. Each building has its appropriate standpipe. A fire is raging in A. You wish to fight the fire from the tops of B and C. A double extra first size engine stands at each hydrant. Using two 3-inch lines to each standpipe, state what pressure you will order at either engine that you may get effective results. Show how you would figure loss of pressure from one engine to top of building. Is the double extra first size engine of 1,200 gallons capacity at 120 pounds capable of supplying the l 1/2-inch nozzle in each case? By effective results are meant effective streams across the streets to the center of the burning building. Building A fronts on Avenue R a distance of 80 feet and on N street, a distance of 50 feet. C fronts on Avenue R 90 feet, and B, on N street, 100 feet. The standpipe in each case has only a single 2 1/2-inch connection available.

Solution: The first step is to find what nozzle pressures would be required on buildings B and C. A stream from C would have to carry across N street and half the depth of the building, 40 feet, to be effective; or 60+40=100 feet. Likewise the stream from B must go 90+25 =115 feet feet. To find what pressure would be required to throw a l 1/2-inch Stream 115 feet, the formula for horizontal range is employed:

115=1/2N. P.X53, from which P.= 62, and N. P.=124 pounds.

The pressure in the nozzzle at C would be less than this, having a smaller range, so it may be neglected in the solution of the problem. The best layout of hose from the standpipe would be 2 2 1/2-inch lines 100 feet in length from the one connection and siamesed into the nozzle. To find the friction loss in the layout from nozzle to engine it is best to reduce all lines to equivalent of 2 1/2-inch hose by use of friction factors. The two 2 1/2-inch lines, 100 feet in length= 100/3.6=28 feet of 2 1/2-inch hose; 88 feet of standpipe, which for a building over 75 feet in height must be 6 inches in diameter, plus 100 feet for loss in valves, etc., is equivalent to 188/52=4 feet of 2 1/2-inch hose; two 3-inch lines, 100 feet in length=100=9.35=ll feet of 2 1/2-inch hose. Or the entire layout is equivalent to a single line of 2 1/2-inch hose 28+4+11=43 feet long.

The discharge per minute from the 1 1/2inch nozzle under 124 pounds pressure is 29.7 X 1 1/2 X 1 1/2 X√124=743 gallons, approximately. With this flow the friction loss in 43 feet of 2 1/2-inch hose would be

43

(2X7.43X7.43+7.43)X-=50.7 pounds.

100

The pressure loss in each of the two Siamese connections is 5 pounds, or a total of 10 pounds. The pressure loss in the 2 1/2-inch standpipe outlet is approximately the same as that experienced in hydrant outlet, or 8 pounds. The total pressure loss in the system is then 50.7+10+8=68.7 pounds. The nozzle pressure is 124 pounds, and the engine pressure must be 124+68.7=192.7 pounds.

The engine would be capable of supplying the nozzle, for its water horsepower is (1200X120)/ 1715=84, approximately, and 743 gallons at 192.4 pounds= (743 X 192.4) / 1715=83.4 water horsepower, approximately.

Pump Slippage.

When the piston in a pump moves back and forward under pressure a certain amount of water either gets back into the supply pipe as the direction of motion of the piston changes, or leaks to the opposite side of the piston. This leakage is termed slippage and is usually expressed in per cent. For example, the volume swept out by a piston in a pump may be 1,000 cubic inches but only 950 cubic inches of water is discharged. The slippage is therefore 1000—950=50 inches, or 50 / 1000=5 per cent.

In new engines the slippage is between 3 and 5 per cent. After an engine has been in use for a few months slip will generally increase about 1 per cent; thereafter, if valves and packings are properly cared for, the increase will be very slight. A slip of 10 per cent, or over, according to the engineers of the National Board of Fire Underwriters, usually indicates broken or displaced valve springs, and more than this, a badly worn plunger or pumps barrel, or possibly a leaky suction.

Example: The discharge from an engine, calculated from the water cylinder dimensions, and number of strokes, is 755 gallons per minute. Accurate measurements, however, show that only 743 gallons are discharged per minute. What is the slip? Solution: The leakage or slip is 755—743=12 gallons. This is equivalent to 12 / 755=1.59 per cent.,— an unusually small slip.

Brake Horsepower, Indicated Horsepower, and A.L.A.M. Horsepower. Brake horsepower is to the power plant of a pumping engine what water horsepower is to the pump. It is the power output of the engine, and is usually measured by a device known as the Prony brake.

Indicated horsepower is the horsepower of an engine determined by taking the average pressure in the power cylinders multiplied by the travel of the pistons in feet. The pressure in the cylinders is secured by a device called an indicator, while the travel of the piston can be figured from the r.p.m. of the engine and length of stroke.

By A.L.A.M. horsepower is meant the horsepower of an automobile engine determined by the formula of the Association of Licensed Automobile Manufacturers. This formula does not take into consideration the work actually performed by an engine, but gives an engine a horsepower rating based upon the number of cylinders and the bdre thereof. The formula is

Bore X Bore X No. of Cylinders

Horsepower=–

2.5

Example: A motor pumping engine has a 6-cylinder motor, and the cylinder bore is 5 inches. What is the A.L.A.M. horsepower of the engine? Solution:

5 X 5 X 6

Horsepower=-=60 horsepower.

2.5

Ans.

(The End.)

District Forester of Pennsylvania has issued a notice to fire wardens to be vigilant at this season. He says the forest fire season will soon be with us and acres of timberland and much game and game birds will be in danger of destruction. As a forest fire warden and an officer of the State of Pennsylvania, it will be your duty. to promptly take measures to extinguish all forest fires in your locality. Let me urj?e upon you the necessity of quick action whenever you see a fire or one is reported to you. Prompt action at the start of a fire will save thousands of dollars of damage. If you have a large fire and you and your men cannot control it, call on your nearest warden for assistance. In this case, let one warden make up the reports and accounts for the fire.

PRACTICAL HYDRAULICS FOR FIREMEN

3

PRACTICAL HYDRAULICS FOR FIREMEN

(Continued from page 250)

PUMPS.

Piston or Reciprocating Pump.

The first water pump developed and which still predominates from the standpoint of numbers in use is the piston pump. It is shown in its simplest form in Figure 49, which is a sketch of a singleacting pump. A is the cylinder; B, the piston rod; C, the piston; D, the inlet valve; E, the outlet valve; F, the water space. The operation is as follows; When the piston travels out, the back pressure of the water in the discharge pipe at E forces the valve E to close. The partial vacuum produced in the cylinder by the piston sweeping out the space causes the atmosphere without to force water up through valve D, which is opened by the water pressure. Thus the cylinder is filled with water on the outward stroke of the piston. As the piston returns it puts the water in the cylinder under pressure, forcing the valve D to close, and stay closed during the stroke. In the meantime the pressure of the water in the cylinder continues to increase until it overbalances or exceeds the back pressure in the discharge pipe when the valve E reopens and the water in the cylinder escapes through the discharge pipe. This complete operation is repeated on every double stroke of the piston. The above describes the simple single acting pump of the piston, or reciprocating type.

Figure 50 shows a simple double-acting pump. It is precisely the same as the single-acting pump with the exception that the cylinder is closed at both ends and there are a duplicate set of valves at the rod end (the right hand end in the figure) of the cylinder. Water is discharged from both ends of the cylinder, the water leaving the space PP and entering F as the piston moves to the right and entering PP and leaving F as the piston moves toward the left.

The quantity of water discharged from a pump cylinder is found as follows; Calculate the piston area, which is, of course, the internal cross sectional area of the cylinder. Knowing the number of strokes per minute, the discharge is equal to the piston area multiplied by the length of stroke (all in the same units of measurement) multiplied by the number of strokes in one direction per minute. This product gives the discharge per minute from the head end of the cylinder, that is, from the left hand end of the cylinder shown above—not from the end through which the piston rod travels. If the pump is double-acting, it then discharges from both ends of the cylinder and the piston strokes in both directions are discharge strokes. But the discharge from the side of the piston to which the rod is attached is less than that from the other side, for the rod takes up a certain amount of space which is not filled on the filling stroke or in-stroke.

In calculating the discharge from this end of the cylinder, find the piston area, and subtract from it the cross section area of the piston rod. This then is the effective piston area on the rod end. The discharge per minute from this end equals the effective area (piston area minus rod area) times the length of stroke times the number of strokes per minute in one direction. The sum of the discharges from both ends of the cylinder per minute gives the total discharge of the pump per minute working double-acting.

Example 1. Find the discharge from a single cylinder pump 5 inches diameter, stroke 6 inches, and running 150 r.p.m. The piston rod is ¾ (or 0.75) inch in diameter. Pump is working double-acting. Solution: Area of piston on head end side =5X5X0-7854=19.63 square inches. Discharge per stroke=6X19.63=117.78 cubic inches. Discharge per minute on head end side=150X117.78=17,667 cubic inches=76.5 gallons per minute. Area of piston rod= 3/4X3/4X0.7854=0.44 square inches. Corricted area of piston on rod end=19.63— 0.44=19.19 square. inches. Discharge per stroke=6X19-19=115.14 cubic inches. Discharge per minute on head end side= 150X115.14=17,271 cubic inches=74.6 gallons per minute. Total discharge of pump per minute=76.5+74.6=151.1 gallons per minute.

Reciprocating or piston pumps, as found on fire apparatus, have assumed quite a variety of designs, but all, of course, embody the fundamental principles of operation outlined in the above paragraphs. In Computing the displacement or discharge from fire apparatus pumps account must, of course, be taken of the total number of water cylinders. Figure 51 shows a sectional view of one of the types of Vshaped piston pumps found on motor fire apparatus. In this particular case the two pump cylinders employed are set at right angles to each other.

Rotary Gear Pump.

The second type of pump to come into general use on fire apparatus was the rotary gear pump. The interior mechanism of this type of pump in its present state is shown in Figure 52. All rotary gear pumps have two parallel gear shafts carrying impellers (i. e.gears) which mesh with each other. As shown in Figure 52 the water enters through the left hand side of the chamber, gets in the spaces between the gear teeth as the gears or impellers revolve in the direction shown by the arrows, and is pushed upward and out of the outlet. The gears fit the casing snugly and water is permanently prevented from leaking gack between the gears and casing by small plugs fitted in the ends of alternate gear teeth and which are replaceable; and as the water cannot return between the wheels in consequence of the cogs being there always in contact, it must necessarily rise in the discharge or forcing pipe.

•Copyright, 11117, by F. Shcpperd.

It is not feasible to accurately calculate the discharge of a rotary pump from the dimensions of the moving parts and their speed, and the proper method of measuring the flow is by use of either a tank with weir or nozzle with pressure recording gauge. The latter procedure is probably the more convenient for fire departments.

Centrifugal Pump.

When a bucket filled with water is swung around at full arm’s length, the water will not leave the bucket even though it be completely inverted during its course through the air.

The sparks from an emery wheel and the particles of metal which leave it when the wheel is in motion shoot off the wheel in straight lines and do not follow the curvature of the wheel nor the direction of its rotation.

On these facts is based the principle of operation of the centrifugal pump. To fully explain the application of these performances it is necessary’ to carefully consider the case of the revolving bucket. Figure 53 shows a bucket attached to a string and whirling around the nail A. If the string should break when the bucket is at position C, the bucket would not travel in a curved line, but, on the contrary would, like the particles from the emery wheel, shoot straight off the circle at right angles to the radius at the point where the bucket broke loose; in other words it would shoot off tangentially from the circle.

A law of physics states that a body continues in its state of rest or of uniform motion in a straight line, except in so far as it may be compelled, by outside force, to change that state. This means that if a body were thrown through the air it would travel forever in a straight line were there no outside effect. The reason why an object so projected does not so continue is due to the action on it of such outside forces as gravity and the retarding force of air friction. But for short spaces and for the purpose of demonstrating the action of the centrifugal pump these two forces may be neglected. In that case the whirling bucket continually wants to travel off tangentially from its path but the string keeps pulling it from its straight line course to follow around a circular path. In Figure 53, while the bucket would be at D if it could travel in the direction it would, were it free, the string deflects it and it is as the result at position E instead. The bucket exerts a tension or pull on the string in being deflected from its straight path. This pull is known as centrifugal force, and varies as the square of the velocity. It is this same centrifugal force which is the means of producing pressure by centrifugal pumps.

Figure 54 represents the simplest form of centrifugal pump. The water enters through the inlet at the center of the rotating impeller and is thrown outward by centrifugal force. As it passes out into the water passage it draws after it more water through the intake, so that the flow is solid and continuous. The arrows in the sketch show the course of the water. The pump in Figttre 54 is said to be a single stage pump, i. e., the water passes through but one set of impellers. Such a pump, while capable of giving a large flow is quite unsuitable for, and quite incapable of, producing much increase in pressure. In the fire service where high pressures especially are required such a pump would be unsatisfactory. But if the water from this first pump were discharged into a second similar pump under pressure, and from this to a third pump, by the time it had emerged from the last pump the pressure would have been boosted sufficiently to answer the needs of the fire department. This multiple raising of the pressure is exactly what takes place in three-stage pumps. By three-stage pumps are meant pumps having three sets of impellers through which the water passes one at a time. Each stage boosts the water pressure in turn.

Figure 55 shows the cross section of the first type of centrifugal pump giving successful performance on motor fire apparatus. This particular pump has a bronze casing in which are placed impellers on a single shaft. Each impeller represents a stage, and they are separated by suitable guides that convey the water from one impeller to another. The only moving part is the steel shaft to which the impellers are permanently keyed. The shaft is rotated, and the water, which lies in the first impeller, is thrown to its outer edge and through openings on this edge at high pressure by centrifugal force. From this point the water passes through vanes, which are nothing more than stationary guides to prevent the water from forming whirlpools and eddies. As it passes into the next stage or impeller this operation is repeated. The passing of the water out of the impeller, as mentioned above, draws more in to take its place. After passing through the last stage, the water is carried into the discharge pipe and from there to the different lines of hose. Each impeller stores a certain amount of pressure in the water as it passes through.

Pump Primer.

If a centrifugal pump is attached to a fire hydrant where there is pressure to force water into the impellers the pumps may be put into operation instantly.

But if water must be drafted, i. e., from a stream or cistern, the suction created by spinning the impellers idly in the confined air is not sufficient to draw water into the pump, and other means have to be provided for supplying the pump with water before it can be set in operation. In the pump described above provision is made for this by a separate arrangement, known as a vacuum pump. This is a small pump of the rotary type. It is mounted in conjunction with the pump transmission. It requires about twenty seconds to get the centrifugal pump in operation when this rotary primer is used.

Discharge.

As in the rotary pump, there is no accurate method of determining the displacement or discharge of the centrifugal pump outride of tank, weir, or nozzle measureFire engine pumps in good condition can lift water a vertical distance of twentyfive feet, as mentioned in a previous chapter treating the subject of atmospheric pressure. However, where the suction pipe is small and the quantity of water being drawn large this vertical lift is diminished by the friction loss in the suction pipe. Hence there is a different limit of suction for each different discharge of a pump. For instance, a pump can lift 500 gallons per minute through a vertical distance of 23 feet if 5-inch suction hose is employed, whereas it can lift the same amount only 12½ feet if 3 1/2-inch suction is used.

ment. The centrifugal is not a positive displacement type of pump.

Required Sizes for Suction Pipes.

The following table, prepared by the engineers of the National Board of Fire Underwriters, well merits careful study. The figures are based on the ability of the pump to maintain a vacuum of 23 inches, or in other words, to lift water approximately 26 feet.

TABLE SHOWING MAXIMUM LIFT, IN FEET, WHEN DRAFTING VARIOUS QUANTITIES OF WATER WITH A FIRE ENGINE IN GOOD CONDITION.

From “Fire Engine Tests and Fire Stream Tables” (copyrighted), and reprinted by courtesy of the National Board of Fire Underwriters.

Example: An engine in good condition is required to lift 500 gallons of water per minute from a stream a vertical distance of 20 feet. What is the minimum sized suction that can be used? Answer: Refer to the table above, find 500 in the first column and follow’ the line across the table to 20½. Look at the top of this column, and the 4 1/2-inch suction is therefore the minimum that can be used.

Pump Horsepower,

When a one-pound weight is lifted vertically a distance of one foot the work done is termed a foot pound. If 10 pounds are lifted one foot vertically, ten foot pounds of work are done. If 100 pounds are lifted vertically a distance of 10 feet, 10X100=1,000 foot pounds of work are done.

Power is the rate of doing work. For instance, if a machine is capable of lifting 33,000 pounds through a vertical distance of 1 foot in a minute, that machine possesses 1 horsepower. In other words, one horsepower is that power which a machine possesses when it is able to do 33,000 foot pounds of work a minute, or 550 foot pounds of work a second.

Example 1: A pumping engine is able to raise 7,000 pounds of water a vertical distance of 250 feet per minute. What is its horsepower? Solution: The work it does per minute is 7,000X250=1,750,000 foot pounds. 33,000 foot pounds of work per minute equals one horsepower, therefore the horsepower of the engine is 1,750,000-4-33,000=53 horsepower, approximately.

This horsepower is what is known as the water horsepower of an engine, for it is the horsepower equivalent of the work done by the discharge from the engine. In other words, it is the output horsepower of the pump, and does not include the power required to overcome friction in the power plant, pump, and connections between, and also slippage. To get the water horsepower of a pump the discharge pressure must be taken at the pump, and not at the nozzle, for the friction in fire hose connected to a pump has no relation to the pump performance.

Example 2: A pumping engine is capable of discharging 700 gallons per minute against a pressure of 120 pounds per square inch. What is its horsepower? Solution: A gallon of water weighs 8.35 pounds, so that a discharge of 700 gallons per minute is equivalent to a discharge of 700X8.35= 5,845 pounds per minute. A pressure of 120 pounds is equivalent to a head of 120X2.304=276.48 feet. In other words, the pressure of 120 pounds per square inch is equivalent to the weight of a column of water 276.48 feet in height and one square inch in area. Hence to pump one pound of water against the -120 pounds pressure is equivalent to pumping it a vertical distance of 276.48 feet, and the work required would be 276.48 foot pounds (per pound of water). In this connection it should be remembered that to lift a 1 foot column of water a vertical distance of 276.48 feet is exactly equivalent in work to lifting a 276.48-foot column of water one foot vertically.

The work done, therefore, in pumping the 700 gallons per minute against a pressure of 120 pounds is equal to 5,845X276.48 =1,616,025 foot pounds per minute. This is equivalent to 1,616,025-4-33,000=49 horsepower, approximately.

Where it is necessary to make frequent calculations for the water horsepower of a pump, the method outlined above would prove too lengthy for practical use. A shorter method has been devised for securing the same results by combining those terms in the solution which are constant for all problems. This condensed solution consists of multiplying the discharge pressure in pounds per square inch by the discharge in gallons per minute and dividing the product by 1,715. The answer is the water horsepower.

Example 3: Find the water horsepower of the pump in Example 2. Solution: (700X120)/1715=29 horsepower, approxmately.

(To be continued.)