QUESTIONS and ANSWERS

QUESTIONS and ANSWERS

Wherein are answered questions relating to current problems in Fire Protection.

Pressure and Discharge of Large Nozzles

To the Editor:

A 500-gallon pumper is pumping through two 200-foot lines of 2 1/2-inch hose each equipped with a 1 3/4-inch nozzle. Engine pressure is 180 pounds. Stream from each nozzle is good, but short, reaching only about 100 feet.

What is the friction loss in each line? What is the nozzle pressure at each nozzle and the discharge from each nozzle?

J. F. I.

Answer: The solution to your problem is as follows:

N. P. = E. P. / (1.1 + KL), where E. P., the engine pressure, is 180 pounds.

K, a factor depending upon hose and nozzle diameters, is .907.

L is the number of fifty-foot lengths of hose in the line, and is 200 / 50 = 4.

Then N. P. = 180 / (1.1 + .907 x 4) = 180 / (1.1 + 3.62)

= 180 / 4.728

= 38.1 pounds.

This nozzle pressure is not sufficient to give a stream of satisfactory range or stiffness

Discharge — 29.7 X d X d X √P, where d is the diameter of the nozzle in inches and P, the pressure at the nozzle.

Discharge = 29.7 X 1 3/4 X 144 X √38.1 = 561 gallons per minute

The friction loss in the 200-foot line is 180—38.1, or 141.9 pounds.

The friction loss in 100 feet is 141.9 / 2 = 70.95 pounds.

Branched Lines of Small Diameter

To the Editor:

Five hundred feet of 2 1/2-inch hose from a pumper is branched into two 150-foot lines of 1 1/2-inch hose. Both the 1 1/2-inch lines are being used in the second story of a building using 1/2-inch tips and a nozzle pressure of 40 pounds per square inch. If the pump elevation is 36 feet below that of the ground floor of the building, what is the approximate pump pressure needed? N. D. F.

Answer: The solution to the problem you give is ns follows:

First, the elevation of the nozzle above the engine will be 36 feet plus the height of one story, or 12 feet, making a total of 48 feet.

First change each line of 1 1/2-inch to 2 1/2-inch. This is done by dividing by the factor .074.

150 / .074 = 2027 feet of 2 1/2-inch hose.

Combine these two parallel lines of 2 1/2-inch hose, each 2027 feet in length, into a single line.

The conversion factor is 3.6.

2027 / 3.6 = 563 feet of single line 2 1/2-inch hose.

Then the layout is equivalent to a single line of 2 1/2-inch hose 563 + 500, or 1063 feet in length.

Next combine the two 1/2-inch nozzles into a single nozzle. Two 1/2-inch nozzles are equivalent approximately to a 3/4-inch nozzle.

Now solve for engine pressure.

E. P. = N. P. X (1.0 + KL) = 40 X d + .068 X 21.26), where K for 2 1/2-inch hose and 3/4-inch nozzle is .068. = 40 X 2.44568 = 97.827, or 98 pounds approximately

Add the back pressure due to the elevation of the nozzle above the engine to this figure.

Back pressure = 48 X .434, or 20.822, or 21 pounds approximately.

Then the corrected engine pressure is 98 + 21 = 119 pounds.

Effect of Hydrant Pressure Upon Pump Performance, etc.

To the Editor:

Does the piston or rotary gear pump use any or part of the hydrant pressure fed into it?

When a pumper is taking suction, it is said that it is pumping from the source of supply. We can see where the pump would have to work to clear the line, or suction, of air, but when the vacuum is created, atmospheric pressure should push the water to the pump. Yet it is claimed that the pump is pumping from the source of supply. If this is true, what happens to the atmospheric pressure?

T. J. L.

Answer: Both piston and rotary gear pumps utilize a part of the pressure furnished by the hydrant, but not as much as a centrifugal pump.

Just what percentage of the hydrant pressure they can use, we are unable to say, as this depends largely upon the design of the pump.

When a pumper is taking from suction it first creates a vacuum in the suction hose, and the atmospheric pressure forces the water up to the pump. From then on, the suction line is filled with water. However, the back pressure, or tendency to create a vacuum, remains. The pump in lifting water from the source of supply up to the level of the suction inlet does just as much work as it would do in raising water by pressure a like distance.

When a pump is taking water from suction, a check of the compound gauge will show that there is a negative head. The amount of negative head (which for all practical purposes is the same as vacuum) will depend upon the height the water is being lifted. As you probably know, the compound gauge records either vacuum (negative head) or pressure.

Pump Discharge at Reduced Pressure, etc.

To the Editor:

In “Practical Hydraulics for Firemen,” where examples are given using Siamese stretches of 2 1/2-inch and 3-inch hose, why don’t you use the factor 6.1 instead of reducing the line of 3-inch to 2 1/2-inch? Where and when is the 6.1 used?

First size pumpers are rated at 1000 gallons per minute at 160 pounds pressure. As the pressure is increased, the discharge is lessened. What would be the result upon the discharge of decreasing the pressure below 160 pounds.

A. J. T.

Answer: In solving problems involving lines of 2 1/2-inch and 3-inch in the same layout, the 3-inch line is commonly reduced to 2 1/2-inch line to save the student the trouble of remembering another factor, namely, 6.1.

Furthermore, the factor 6.1 can only be used where the two lines are of equal length. The other method can be used no matter what the lengths of the two lines.

If the lines are of equal length, however. it is simpler and quicker to use the factor 6.1 to reduce the two of them to a single line of 2 1/2-inch hose.

Answering your second question, if a layout of hose and nozzle is so changed as to reduce the presssure at the pump, the discharge should be greater than its rated discharge at “capacity.”

(Continued on page 185)

(Continued from page 150)

However, there is a limit beyond which the increase cannot go, due to the fact that the motor driving the pump cannot be speeded up indefinitely.

Theoretically if the pressure at the pump you describe was reduced to 80 pounds, the discharge should be 2,000 gallons per minute. Actually, however, such a performance could not be obtained safely due to the great increase of speed that would be necessary on the part of the motor driving the pump.

The actual maximum discharge of a pump at lower than rated discharge pressure depends upon many factors such as the pump characteristics, the motor characteristics, etc.

Freezing Points of Extinguishers

To the Editor:

What are the limitations and freezing points, it any, to carbon dioxide and carbon tetrachloride extinguishers? M. C.

Answer: Pure carbon tetrachloride freezes at approximately 19 degrees below zero F.

However, an anti-freeze agent is commonly added to the carbon tetrachloride in extinguishers to depress the freezing point so low as to make the use of these extinguishers practical in almost any climate.

Carbon dioxide freezes at —86 degrees F., and hence is satisfactory to use anywhere on the face of the globe.

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