QUESTIONS and ANSWER
To the Editor:
- A length of 2 1/2-inch hose, with a given nozzle and nozzle pressure, requires 100 pounds engine pressure Would a second single line of the same length and size, equipped with the same size nozzle require twice the engine pressure in order to yield the same flow of water and nozzle pressure from both the single lines?
- What would he the maximum possibilities for a 500 gallon pumper in the following layouts?
- Two single 2 1/2-inch lines, one Equipped with a 3/4-inch nozzle and the other with a 1 1/8-inch nozzle at 25. 50 and 75 pounds nozzle pressure.
- Two 2 1/2-inch lines siamesed into one 2 1/2-inch line, one line equipped with a 3/4-inch nozzle and the other equipped with a 1 1/8-inch nozzle, at 25, 50 and 75 pounds nozzle pressure.
- One 2 1/2-inch line wyed into two 2 1/2-inch lines with the same size nozzles and at the same pressures as in the preceding layouts.
- One 2 1/2-inch line wyed into two 1 1/2-inch lines with a 1/2-inch nozzle at 25 and 50 pounds nozzle pressure. How many of these lines could be run off the pumper?
- Two lines of 2 1/2-inch hose siamesed into a deluge set with a 1 1/4-inch nozzle, a 1 1/2-inch nozzle and a 1 3/4-inch nozzle at 75 pounds nozzle pressure.
- Three lines of 2 1/2-inch hose siamesed into a deluge set equipped with a 1 1/4-inch nozzle, a 1 1/2-inch nozzle and a 1 3/4-inch nozzle at 75 pounds nozzle pressure.
S. C. H.
Answer: If the second duplicate line were attached to a pumper, the engine pressure would not have to be increased, but the speed of the pump would have to be stepped up to give the additional flow.
For identical layouts, identical engine pressures are required to give the same nozzle pressures.
The answers to question 2 (a) are as follows: with a single line with a 3/4inch nozzle and a single line with a 1 1/8inch nozzle, both 2 1/2-inches in diameter, at 25 pounds nozzle pressure, 1600 feet of hose could be used. At 50 pounds pressure, 600 feet, and at 75 pounds pressure, 200 feet. These figures are approximate.
Answers to (b), (c) and (d) cannot be given, because it is not stated what percentage of the line is single line and what percentage parallel line.
Answers to (e) are as follows: With 1 1/4-inch nozzle at 75 pounds nozzle pressure, about 700 feet of hose could be used. Using a 1 1/2-inch nozzle, about 100 feet of hose, and using a 1 3/4-inch nozzle, the engine could not make this nozzle pressure for it would have to discharge 787 gallons per minute which is more than its rated discharge.
(f) Where three parallel lines of 2 1/2inch hose are used, and 1 1/4-inch nozzle at 75 pounds nozzle pressure, the stretch would be 1,500 feet. If 1 1/2-inch nozzle is used at 75 pounds, the stretch would be 200 feet, whereas it could not maintain 75 pounds from 1 3/4-inch nozzle.
Foam from Animal Blood
To the Editor:
I recently read in a magazine article that the Navy was using foam made of animal blood. Is this true?
W. P. B.
Answer: Commander Harold J. Burke of the Bureau of Ships, U. S. Navy, Washington, D. C., writes as follows:
“I am afraid that this is largely another case of the ‘wayward press’ getting ‘slightly off the beam.’
“The facts briefly are as follows: As you well know, the Navy has for some time past been using a liquid foaming agent in producing the so-called mechanical types of foam. In this respect, we have been quite successful. The rapid and continued expansion of the Navy put a very heavy drain on the production facilities of our available source of supply of this material. As a consequence, we started to investigate another protein that could be used for manufacturing foam liquid. In the course of our investigation, Armour and Company submitted a protein made from animal blood. We tested this at one of the fire schools and determined that we could make foam, but the present state of the art had not been developed to a point where we were willing to accept it as comparable to our present product. Other than the test specimens referred to above, the Navy does not use any foam solutions made from animal blood.’’
Overcoming Back Pressure, Etc.
To the Editor:
- In comparing the back pressure as illustrated on Page 15 of “Fire Department Hydraulics” with the hack pressure of the Woolworth Building in question, we assume that three engines are siamesed into the standpi|ie When they are ready to pump does each have to overcome the Kick pressure of 558 pounds, or would each have to pump 115 pounds in order to cause water to flow at a rate of 1 pound, neglecting, of course, the standpipe friction loss? In either case, why ?
- In relay pumping, how can we estimate the cajKicity at which the first pumper must operate, or the pressure, in order to know at what rate it will deliver water to the second. Does it depend on the entire length of both lines and the nozzle tip used on the end? F. K. K.
Answer: Answering Question 1, the back pressure at each engine will be 888 pounds, where the two engines are independently siamesed into the standpipe.
The pressure at the Siamese connection is 338 pounds due to the elevation of the water in the standpipe, and this pressure would he exerted against any or all engines which attempted to force into the standpipe.
2. The capacity at which the first pumper must operate is only determined by the nozzle tip, and the desired nozzle pressure. The first engine can only give what water passes through the second engine on to and through the nozzle. Hence, the quantity of water that the second engine can discharge through the layout is the same quantity of water that the first engine must supply to it; neither more nor less will answer.
S. A. E. Horsepower and Brake Horsepower
To the Editor:
Should there he a variance in the actual horse-power as stamped on the motor by the manufacturer and the S. A. E. horse-power as determined by the formula: Bore x Bore x No. of Cylinders / 2.5 ? J. E. S.
Answer: The horse-power stamped on the pumper is probably the brake horsepower. This differs very much from the horse-power of a motor as determined by the S. A. E. formula.
The S. A. E. formula was originally developed by the Association of Licensed Automobile Manufacturers. When this organization went out of the picture, the Society of Automotive Engineers took over and endorsed the formula. It is, as you state: H. P. = D2 X N / 2.5, where D is the diameter, or bore, of the cylinder; N, the number of cylinders, and 2.5 a constant.
This formula is based upon an assumed piston travel of 1,000 feet per minute, and a mean effective pressure of 90 pounds per square inch in the cylinders.
With the change in design to high speed, high compression motors, the S. A. E. formula’s accuracy, of course, drops.
Brake horse-power of a motor is determined by the use of a prony brake, or a dynamometer, which actually indicates the horse-power delivered by the motor, no matter how many cylinders it has nor what the dimensions of the cylinders.