QUESTIONS and ANSWERS

QUESTIONS and ANSWERS

Formula for Small Nozzles

To the Editor:

In answering the question submitted by J. P. T. in the June issue of FIRE ENGINEERING, why do you use the equation N. P. = E. P. / (1 + KL)?

I would also appreciate your solving the following problem for me: An engine at 120 pounds pressure is pumping through a 300-foot line of 2 1/2-inch hose branched into two lines of 1 1/2-inch hose, each 300 feet in length and equipped with 1/2-inch nozzles.

What is the nozzle pressure on each nozzle?

What is the friction loss in this line of hose?

What will be the discharge?

Answer: In the equation for solving nozzle and engine pressures on lines involving small nozzles, we find that the formula N. P. = E. P. / (1 + KL) gives more accurate results than the formula N. P. = E. P. / (1.1 + KL).

As you may realize, all formulas for nozzle and engine pressures, as well as friction loss, are based upon experimental results.

The formula which gives results closest to experimental results is the one which is considered the most suitable.

For lines of hose with large nozzles, the original formula still holds.

In the problem you give for solution, the first named formula is used, for we have a nozzle of less than 1-inch in diameter.

The solution is as follows:

First reduce each line of 1 1/2-inch hose to a 254-inch. The conversion factor is .074.

300 / .074 = 4054 feet of 2 1/2-inch hose.

Now combine the two parallel lines of 2 1/2-inch so secured into a single line of 2 1/2-inch. The conversion factor is 3.6.

4054 / 3.6 = 1126 feet of 2 1/2-inch hose (single line).

Next combine the nozzles into a single nozzle.

1/2 X 1/2 = .25 1/2 X 1/2 = .25 .25_± .25 = .50 √.50 = .707

The nearest size nozzle to one of .707 inches diameter is 3/4-inch.

For small diameter nozzles, the formula is:

N. P. = E. P. / (1 +KL) E. P. = 120 pounds K = .042 I. = 22.5 Then N. P. = 120 / (1 + .042 X 22.5) = 61.7 pounds.

The nozzle pressure at each of the 1/2-inch nozzles is thus 61.7 pounds.

The discharge from a 1/2-inch nozzle at 61.7 pounds is equal to 29.7 X 1/2 X 1/2 X √61.7 = 58 gallons per minute approximately.

The discharge from the two nozzles is twice as much, or 116 gallons per minute.

The friction loss in 100 feet of 1 1/2-inch hose carrying 58 gallons per minute is 12.2 pounds.

The friction loss in 200 feet will be two times 12.2, or 24.4 pounds.

The pressure at the wye will then be 61.7 + 24.4, or 86.1 pounds per square inch.

The friction loss in the single line of 2 1/2-inch hose from the pumper to the wye will be 120 — 86.1, or 33.9 pounds.

Use of Factors

To the Editor:

In “Fire Service Hydraulics,” for solving friction loss for siamesed lines, you use the value of K as .135. The Underwriters use 3.6 as the factor. In one of your problems on Page 73, Example 83, using 3.6, there’s a difference of 20 pounds.

P.J.F.

Answer: The figure .135 is the value of K for two parallel lines of hose siamesed into a 1 1/2-inch nozzle.

The factor 3.6 is the factor for reducing two parallel lines of 2 1/2-inch hose to a single line of 2 1/2-inch hose.

Thus the factors are for different purposes and are not to be used interchangeably.

In solving Example 83 on Page 73 of “Fire Service Hydraulics,” by first reducing the two parallel lines of 2 1/2-inch hose to a single line, and then solving as a single line, we get the following:

600 / 3.6 = 166.67

E. P. = N. P. X (1.1 + KL)

Nozzle pressure is 60 pounds.

K for a single line of 2 1/2-inch hose with 1 1/2-inch nozzle is .505.

L, the number of 50-foot lengths of hose in the line, is 166.67 / 50, or 3.33 lengths.

E. P. = 60 X (1.1 + .505 X 3.333)

= 166.98, or 167 pounds pressure. You will note that this answer is but 4 pounds off the answer secured by solving for nozzle pressure using the factor of K for parallel lines siamesed into the nozzle.

Changing 1 1/2-Inch to 2 1/2-Inch Hose

To the Editor:

The ever increasing use of 1 1/2-inch hose in fire fighting and salvage work has made it necessary to become familiar with the formulas used to solve problems involving the use of 2 1/2-inch lines wyed into two or more lines of 1 1/2-inch hose.

What are the factors for changing two, three and four siamesed lines of 1 1/2-inch hose into an equivalent length of 2 1/2-inch? J. L. H.

Answer: Values for reducing various layouts of 1 1/2-inch hose to a single line of 2 1/2-inch hose may be secured by use of the conversion factor for to 2 1/2-inch and the factors for the various layouts of 2 1/2-inch hose.

For example, the factor for converting a single line of 1 1/2-inch to a single line of 2 1/2-inch hose is .074; this factor is also employed for changing 2 1/2-inch to 1 1/2-inch by multiplication.

The factor for changing two lines of 2 1/2-inch to a single line of 2 1/2-inch is 3.6; the factor for changing three lines of 2 1/2-inch to a single line of 2 1/2-inch is 7.75.

The factor for changing four lines of 2 1/2-inch to a single line of 2 1/2-inch is 12.4.

Then to find the factors for changing 1 1/2-inch hose, in various layouts, to a single line of 2 1/2-inch hose, we merely multiply the factor for the corresponding layout of 2 1/2-inch by .074 and we get the factor for changing 1 1/2-inch hose to 2 1/2-inch for the various layouts.

From the above the factor for two parallel lines of 1 1/2-inch hose for conversion to single line of 2 1/2-inch is .266.

The factor for converting three parallel lines of 1 1/2-inch hose to a single line of 2 1/2-inch is .57.

The factor for converting four parallel lines of 1 1/2-inch hose to a single line of 2 1/2-inch is .92.

Opening Glass Doors

To the Editor:

I enjoy and find very helpful your magazine, but I would like to see something more on the subject of heavy glass doors, such as the experiences of various departments with the same.

E. P. McC.

Answer: So would the editors! There is a dearth of sound information on the question of opening the modern type thick glass and imitation (plastic) glass doors. Have any of our other readers run into this problem? And if so, what did they do about it?

Correction

In the July issue of this journal, under the heading “Carbon Dioxide in Extinguishers” on Page 468, the statement was made that the sealing of the valve in a squeeze-type carbon dioxide extinguisher was brought about by “expansion of a compressed spring.”

It should be pointed out, however, that the popular squeeze grip valve in common use today utilizes the pressure from within the cylinder to prevent the discharge of liquid carbon dioxide when the extinguisher is not being used.

A spring is provided in the valve mechanism, whose purpose is to return the valve to its seat after using, but not to keep the valve sealed.

The editor is indebted to Captain Charles A. Key, US-TVA Public Safety Service, Tennessee Valley Authority, Knoxville, Tenn.; K. D. Wanless, District Manager, American-LaFranceFoamite Corporation, Spokane, Wash., and Albert T. Hall, McAlester, Okla., for directing our attention to this discrepancy.

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