**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 **for**mula, 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.