LARGE-DIAMETER HOSE HYDRAULICS IN YOUR HEAD

LARGE-DIAMETER HOSE HYDRAULICS IN YOUR HEAD

A glance at most hydraulics books convinces the average reader that he needs a computer for hydraulics calculation. However, there is a simple way to calculate gpm flowing through large diameter supply hose in your head.

HYDRANT PRESSURE, INLET PRESSURE, AND FRICTION LOSS

Because of friction loss (FL) in supply hose, the pressure at a hydrant (HP) decreases uniformly along the hose to the pump inlet. Since 10 pounds per square inch (psi) is needed at the pump inlet (maintaining at least 10 psi prevents cavitation, which will damage the pump impeller) the pressure available to overcome friction loss in the supply hose is HP — 10 psi. That is, to determine the pressure available to overcome friction loss in the supply hose, simply subtract 10 psi from the hydrant pressure when flowing. For example, if 1,000 feet of supply hose is laid from a 60-psi hydrant, there is 50 psi available to overcome friction loss. To determine the rate that friction loss reduces pressure, divide the pressure available by the length of hose. In this example pressure is lost at the rate of 50 psi/ 1,000 feet = 0.05 psi per foot. Because the decimal is inconvenient, friction loss is expressed in psi per 100 feet. In this example friction loss is 5 psi per 100 feet.

This means that 100 feet from the hydrant, the pressure drops from 60 psi to 55 psi. Halfway along the hose it drops by (5 psi/100 feet) x 500 feet = 25 psi, to give a pressure of 60 — 25 psi = 35 psi. These pressures can be measured with pressure gauges connected between hose couplings.

Remember that hydrant pressure is not static pressure; it is the pressure while the hydrant is flowing the full gpm required, bearing in mind that nearby hydrants may also be fully utilized.

To calculate flow and friction loss in supply hose, we tackle the mathematics head on. Table 17-7H of the NFPA Fire Protection Handbook, 16th edition, gives a friction loss equation for each of 19 types and diameters of hose. All of these can be reduced to one equation—the one for 2Vi*inch rubber-lined hose. With an understanding of friction loss and flow’ in a 2’Hnch hoseline you w ill be able to easily calculate the flow of other sizes of hose by converting that of 2 ½-inch hose for a given situation.

FRICTION LOSS AND FLOW FOR 2′-INCH HOSELINE

The general friction loss equation that appears in the NFPA handbook is: FL = C x q x L, where FL = friction loss in psi; C = a constant depending on hose diameter and type; q = flow rate in hundreds of gpm (gpm100); and L = hose length in hundreds of feet. As listed in the NFPA handbook, C = 2 for 2’/2-inch rubberlined hose, so the friction loss equation becomes FL = 2 x q x L. Substituting q = gpm/100 and rearranging this equation, we find gpm = 100 X FIV(2 X L). Phis shows that the gpm varies as the square root of the friction loss in a hose.

To calculate the flow for a convenient friction loss of 10 psi per 100 feet of 2‘/2-inch rubber-lined hose, substitute FL = 10 and L = 1 into the above equation: gpm = 100 X /10/2 224. Since 10 percent accuracy is adequate for the fireground hydraulics, simply remember that when a 2 ½-inch rubber-lined hose is flowing 200 gpm, then 10 psi is needed for every 100 feet of hose length to overcome the friction. This figure will he used for mental calculation in the rest of this article. (For 2 percent accuracy with 2’/2-inch hose, simply add 10 percent to the final gpm —not to the friction loss. For full accuracy add another 2 percent to the final gpm.)

For example, if 500 feet of 2‘/2-inch hose is flowing 200 gpm, the friction loss will he (10 psi/100 feet) x 500 feet = 50 psi.

Square root law. For other gpm flow rates remember the square root law, which states that the gpm through a hose varies as the square root of the friction loss: gpm a FL. This means that if gpm is doubled, the friction loss increases by four times. If the friction loss doubles, the flow causing the friction must increase by a factor of2 ⅞ 1.4 or 40 percent. This method allows you to mentally calculate flows and the corresponding friction loss in 2‘/2-inch hose. The following table covers the useful range.

Flow and Friction Loss (psi/100 feet) in 2‘/>-Inch Hose

To solve for the flow and friction loss values, remember that for 100 feet of 2½-inch hose flow ing 200 gpm the friction loss is 10 psi. According to the square root law, 200 a10. To calculate gpm for other friction losses, and vice versa, you must do to one side of the equation what is done on the other. For example, if we double the friction loss to 10 psi x 2 = 20 psi per 100 feet, then in the friction loss-side of the square root law equation we get10 x 2, which is equal to 10 x J2. We must then multiply the gpm by J2 to find the new flow: 1.4 X 200 gpm = 280 gpm. If we were to double the original flow to 200 gpm x 2 = 400 gpm, then we must double the friction loss side of the equation: /10 x 2, which is equal to v/10 x 2= ⅛ 10 x 4 = 40. The friction loss at 400 gpm is 40 psi per 100 feet.

Supply Hose Flow Relative to 2’/2-Inch Hose

Any intermediate values of flow or friction loss can be estimated with sufficient accuracy for fire service use.

For example, what is the friction loss through a 2’/2-inch hose flowing 300 gpm? We remember that for 200 gpm, the friction loss is 10 psi per 100 feet. It is 20 psi per 100 feet for 1.4 X 200 = 280 gpm. Remembering that this estimate is 10 percent low, we add 10 percent, to 280, which comes to 308 gpm. Close enough. The friction loss for 300 gpm will be just under 20 psi per 100 feet.

What about other sizes of supply hose? Do we have to go through the whole procedure again? Mercifully, no. The hard part is over.

FLOW IN LARGE-DIAMETER HOSELINES

The only factor that changes with hose size and type in the NFPA’s general friction loss equation is the constant C. To determine flow’ in hoses larger than 2‘/2-inches, all we need to remember is how much more efficient a large-diameter hose is than 2‘/2-inch hose. This is shown in the fourth column of the chart above. For example, for the same friction loss, five-inch hose flows roughly six times as much water as 2‘/2-inch hose. (Note: The flow values in the second column correspond to a friction loss of 10 psi per 100 feet. The 12 percent underestimate for 2‘/2-inch hose is included in the gpm factor for other hose sizes, so the accuracy of gpm determination is limited only by the accuracy of constant C; the tabulated NFPA figures are a worthwhile guide. Friction loss also depends on the type and condition of the hose lining and how straight the hose is).

LDH example 1. How much water will flow through 800 feet of 4-inch supply hose fed from a 50-psi hydrant? To solve, first calculate the friction loss for 800 feet of hose supplied by a 50-psi hydrant. Since 10 psi is needed at the pump inlet, there is 40 psi available to overcome friction loss in the hose. The 40 psi is to be distributed over 800 feet, so friction loss in 100 feet is (40 psi/800 feet) x 100 feet = 5 psi. With this friction loss, determine how much water would flow’ through 2‘/>-inch hoseline. Since 5 psi is half of 10 psi (the friction loss that corresponds to 200 gpm), we must divide 200 gpm by J2. Since /2 = 1.4, the hose would flow 200 gpm/ 1.4 = 200 gpm x 0.7 = 140 gpm. From the fourth column of Table 1, four-inch hose flows roughly four times as much water as 2‘/2-inch hose for the same friction loss. The fourinch hose will therefore flow 4×140 gpm = 560 gpm —only enough to fully supply a 500-gpm pumper.

LDH example 2. Suppose 1,500 gpm is needed for an attack pumper from a 2,000-foot relay with five-inch supply hose. Is this practical —meaning, will the required upstream engine pressure be reasonable? From the table, 2‘/2-inch hose flows roughly six times less water than five-inch hose, meaning that the 2‘/2-inch hose would flow 1,500/6 = 250 gpm if used in this relay. For 2’/2-inch hose at 200 gpm the friction loss is 10 psi per 100 feet. At double this friction loss (20 psi^per 100 feet), the flow would be /2 times as much, or 280 gpm. Since 250 is roughly halfway between 200 and 280, splitting the difference, the friction loss will be about 15 psi per 100 feet.

Since 2,000 feet of hose is needed, the total friction loss will be (15 psi/ 100 feet) x 2,000 feet = 300 psi. Since 10 psi is needed at the downstream pumper inlet, the upstream engine pressure needs to be 300 + 10 = 310 psi. This is not practical, since a 1,500-gpm pump is rated to pump only 750 gpm at 250 psi, and 310 psi may burst a hose.

The best solution is to use another 1,500 gpm pumper in the middle of the 2,000-foot hose run, so each feeds 1,000 feet of hose. The friction loss through 1,000 feet of hose will be 1,000 feet x (15 psi/100 feet) = 150 psi. Adding 10 psi to keep the hose stiff at the inlet of the downstream pumper requires an engine pressure of 150 + 10 = 160 psi. A well maintained 1,500-gpm pumper should be able to handle this.

With practice, such calculations can be done in your head faster than they can be done with paper and pencil. While you are getting used to mental calculations it is useful to write down the steps.

LDH example J. To see if six-inch hose in the above relay would avoid the need for an intermediate relaypumper, you might jot down notes such as:

• 1,500 gpm needed
• 2,000 feet of six-inch hose
• FL = 140 psi (since 1,500-gpm pumps are rated at 150 psi)
• FL per 100 feet = 140/20 = 7 psi per 100 feet
• Flow in 2‘/2-inch hose with a friction loss at 7 psi per 100 feet = 180-
• gpm (estimated)
• Since six-inch hose flows 7 times as much as a 2‘/2-inch hose, 7 x 180 gpm = 1,300 gpm (not quite enough).

To determine the engine pressure required to flow 1,500 gpm 2,000 feet with an intermediate relay pumper:

• 1,000 feet of six-inch hose
• 1,500 gpm needed
• Flow in 2’/2-inch hose for same FI. is 1,500/7 ~ 210 gpm (since a 2 ‘ inch hose will flow one-seventh that of a six-inch hose)
• FL in 2’/2-inch hose at 210 gpm ⅞ 10 psi per 100 feet
• FL in 1,000 feet of six-inch hose is 1,000 feet X (10 psi/100 feet) = 100 psi
• Engine pressure = 100 psi + 10 (at downstream inlet)
• Conclusion: Hngines should run cool delivering 1,500 gpm.

APPLICATIONS

What practical difference does understanding hydraulics make? Consider traditional departments with 1,500-gpm pumps, 50-psi hydrants, and no understanding of hydraulics. For “big water” perhaps 500 feet of traditional dual 2’/2-inch hose is laid. The pressure available to overcome friction loss will be 50 — 10 = 40 psi. The friction loss is therefore 40 psi/ 500 feet = 8 psi per 100 feet. If it were 5 psi per 100 feet the flow would be 0.7 of the 200 gpm at 10 psi per 100 feet, or 140 gpm. At 8 psi per 100 feet, the flow is roughly 180 gpm. Dual 2’/2-inch lines would double the flow to 360 gpm.

If, for a large fire three 2‘/2-inch attack lines are advanced from this pumper, they could only flow an average of 1 20 gpm each. With a big 1 ‘/4-inch tip, (wanting to flow 300 gpm at 50 psi), the tip pressure would be about 7 psi giving a limp stream with a 20-foot reach. A single five-inch hose would flow six times as much as one 2‘/2*inch line, or 1,080 gpm — enough to supply the three attack lines.

Learn large-diameter hose hydraulics to save bucks and buildings.