Question and Answes

Question and Answes

wherein are answered questions relating to current problems in the fire protection

Distributor Problem

To the Editor:

In our fire department we use an Elkhart distributor nozzle. This nozzle has six openings, three of which are 9/16 inches in diameter and the other three are 5/8 inches in diameter.

What is the approximate equivalent of the combined nozzle outlet, the discharge, and the value of “K”?

Would appreciate your working out the following problem for me: A pumper is delivering water through a 500-foot line of 21/2-inch hose, to which is attached an Elkhart distributor. Thirty pounds pressure is needed at the distributor. What engine pressure is needed, and what will be the discharge?

F. F.

Answer: To find the size of a plain nozzle equivalent to a distributor, we must find the total port area of the distributor and then find a nozzle which has an equivalent area.

The distributor which you cite has three 9/16-inch ports and three %-inch ports.

The area of a port 9/16 inches in diameter is .2484 square inches.

The area of a port of %-inch diameter is .3068 square inches.

Three 9/16-inch ports have a total area of .7452 square inches, and three 5/8-inch diameter ports have a total area of .9204 square inches.

Together the six ports have a total area of 1.6656 square inches.

The plain nozzle having an area nearest to 1.6656 square inches is the 1-7/16inch nozzle which has an area of 1.623 square inches.

The value of K used in the discharge formula, for a 1-7/16-inch nozzle on 21/2-inch hose is 0.427.

And now to solve the example you give:

A pumper is delivering water through 500 feet of 21/2-inch hose equipped with the distributor.

Engine pressure = nozzle pressure X (1.1 + KL)

Nozzle pressure is 30 pounds.

K is, as given above, .427.

L, the number of 50-foot lengths of hose in the line, is 500 / 50, or 10.

Then E. P. = 30 X (1.1 + .427 X 10)

= 30 X 5.37 = 161.1 pounds

Discharge = 29.7 X d2 X √P, where “d” is the nozzle diameter in inches and P, the pressure in pounds per square inch.

Then Discharge = 29.7 X 1-7/16 X √30 = 29.7 X 1.4375 X 1.4375 X5.447 = 336 gallons per minute approximately.

Friction Loss

To the Editor:

What would be the friction loss in a line of 21/2-inch hose 100 feet in length wyed into two lines of 11/2-inch hose each 150 feet in length and equipped with a 1/2-inch nozzle?

If there is a “ride of thumb’’ for figuring friction loss in 11/2-inch hose wyed from 21/2-inch hose, I would appreciate your giving it also.

U. L. B.

Answer: As neither nozzle nor engine pressures are given in the problem you present, we will have to make assumption of either one. We will assume that the engine pressure is in the neighborhood of 58 pounds.

The usual procedure in solving a problem involving a line branched into two is as follows:

Change each line of 11/2-inch to 21/2inch.

Combine these two lines into a single line of 21/2-inch.

Combine two 1/2-inch nozzles into a single equivalent nozzle.

Calculate the problem as consisting of a single line of 21/2-inch hose with a single nozzle.

The nozzle pressure so found will be the nozzle pressure at each of the 1/2-inch nozzles.

Where the layout is placed horizontally, without incline, engine pressure minus nozzle pressure gives friction loss.

To find the friction loss in either branched line, change the 11/2-inch to 21/2-inch and use the friction loss formula for 21/2-inch hose, or we can employ the formula for finding friction loss in small diameter hose, making sure to use the proper constant C for the particular hose diameter.

Or we can find the friction loss in the 21/2-inch line from the pumper to the wye connection, and using the pressure at the wye connection (pumper pressure minus friction loss in 21/2-inch hose) solve for the friction loss in each branch line.

But where we are solving for friction loss by using the friction loss formula, we must know the discharge from the nozzle on the line which we are considering.

The calculations for the problem you give are as follows:

Change 11/2-inch hose to 21/2-inch hose, the conversion factor being 0.074.

150/0.074=2027 feet of 21/2-inch hose.

Combine the two parallel lines so found into a single line by use of the conversion factor 3.6.

2027/3.6=563 feet of 21/2-inch hose, single line.

Add to this the 100 feet of 21/2-inch hose given in the problem. The total length of line is then 100 + 563, or 663 feet.

A 1/2-inch nozzle has an area of .19635 square inches. Two such nozzles have a combined area of 2 x.19635, or .3927 square inches.

The nearest size single nozzle having this area would be an 11/16-inch nozzle.

K for an 11/16-inch nozzle on 21/2-inch hose is equal to 11/16×11/16×11/16×11/16/ 10, or 0.0223 approximately.

The layout is thus equivalent to a single line of 21/2-inch hose 663 feet in length, equipped with an 11/16-inch nozzle.

Nozzle Pressure = Engine Pressure (1.1 + KL)

L, the number of 50-foot lengths of hose in the line, is equal to 663/50 or 13.26.

K, as given above, is .0223. Engine pressure is 58

Then E. P.= 58/(1.1+ 0.223x 13.26) = 58 / (1.1 + .2957)

= 58/1.3957

= 42 pounds approximately

This is the pressure at each nozzle. The friction loss from the engine to the nozzle would then be equal to engine pressure, 58 pounds, minus nozzle pressure, 42 pounds, or 14 pounds.

Friction Loss in Small Lines

To the Editor:

We carry two skid loads of 150 feet each of 144 -inch hose. We either use a gated wye or run them directly from the truck. We use Elkhart S-O-S fog nozzles which have a nozzle pressure of from 100 to 125 pounds for proper fog.

What is the proper way to figure friction loss in 11/2-inch hose, single line? In two 150-foot lines from a wye that is supplied from a 21/2-inch line?

What is the difference in friction loss between one 21/2-inch line and two parallel 21/2-inch lines?

How do you figure friction loss in 3/4inch rubber booster hose?

B. P.

Answer: Before we can solve for friction loss in a line of hose, it is necessary to know the flow in the line, or the discharge through the nozzle which that line supplies.

Ordinarily the manufacturers tell you what the discharge is, for the various pressures, from a fog nozzle, and these are the figures you would normally employ.

To solve for friction loss we proceed as follows:

The friction loss per 100 feet of 1 Viinch hose equals 40 x (2Q2 + 1/6Q) x C, where C is the coefficient, and for 11/2inch hose it is .4.

“Q” is the flow in hundreds of gallons per minute.

Suppose the flow through the line is 30 gallons per minute. Then the friction loss per 100 feet is 40 x (2 x .30 x .30 + 1/6 X .30) x .4, or

40 x .23 x .4

= 3.68 pounds per 100 feet

For 150 feet, the friction loss would be 1 1/2 times as much, or 1.5 x 3.68, or 5.52 pounds.

Where a 2 1/2-inch line of hose is supplying two 1 1/2-inch lines through a wye connection, we solve for friction loss in either of the two 1 1/2-inch lines precisely as was done above. Of course, we must know the discharge from the nozzle for the nozzle pressure in consideration.

Where a 2 1/2-inch line is supplying two 1 1/2-inch lines, we first find the friction loss in each of the 1 1/2-inch lines. Then we add the discharges from the two nozzles, and the total thereof is the flow through the 21/2-inch line. Using this combined flow, we solve for friction loss in the 21/2-ineh hose as follows:

F. L. per 100 feet = 2Q2+ Q (if the flow is greater than 100 gallons per minute). If the flow is less than 100 gallons per minute, then the friction loss per 100 feet of 21/2-inch hose is 2Q2 + 1/2Q.

In the above, “Q” is the flow in hundreds of gallons per minute.

Having the friction loss per 100 feet of 2 1/2-inch hose, we multiply by the number of 100-foot sections in the line to get the total friction loss.

Friction loss for a given flow in parallel lines of 2 1/2-inch hose is 28 per cent of the friction loss in a single line at the same flow.

To solve for friction loss in the 3/4-inch booster line, the formula is as follows:

F. L. in 100 ft. of %-inch hose= 40 X (2Q2 + 1/6Q) x 11.7.

In the above, “Q” is the flow in hundreds of gallons per minute.