**Straight Line Hydraulics**

**FIRE SCIENCE**

During a fire situation, one of the last things a pump operator needs is to hesitate over hydraulics. A new concept using a nomograph could serve to make pump operation calculations as quick and easy as drawing a straight line.

Hydraulic rules of thumb have long been popular with pump operators to enable them to quickly determine proper water pressures and flows for fireground evolutions. However, there are times when these rules are inadequate, and more rigorous mathematical calculations are required, such as:

- When traditional rule of thumb formulas do not apply,
- During pre-planning sessions of major hazards when accuracy counts,
- During classroom training situations or examinations,
- When testing rule of thumb solutions to determine their adequacy and/or shortcomings in situations.

The drawback to mathematical calculations is that their formulas are often difficult to use and even more difficult to remember and apply. Then there follows the need to transform your hydraulic calculations, which are always based on 2 1/2-inch hose sizes to the actual hose sizes being used. This requires the memorization of a variety of mathematical factor constants (see Table 1). There are even times when the complexity of the mathematics becomes such an obstacle that calculations are not even attempted.

To meet the needs of department members who find the textbook calculations burdensome and yet still use them to solve most hydraulics problems. the following graphical method has been developed. A number of typical problems and several contrived problems are presented, showing a side by side comparison of the common mathematical solutions and the new graphical solutions. This direct comparison should show how much quicker and simpler the graphical method can be than the mathematical method.

The graphical method of solving fireground hydraulics problems involves the use of the Fire Service Nomograph. The nomograph has five vertical scales, one each for the typical fireground variables:

- Nozzle pressure (graduated in pounds per square inch or psi)
- Nozzle size and type
- Flow (graduated in gallons per minute or gpm)
- Hose size and type
- Friction loss (graduated in psi loss per 100 feet of hose).

Scales 1, 2, and 3 are related and are always worked together; similarly, scales 3, 4, and 5 are related and always worked together.

The first three scales (nozzle pressure, nozzle size, and flow) are related according to the nozzle flow formula:

Q = 29.7 d^{2} √p

Q = flow in gpm

d = nozzle diameter in inches

p = nozzle pressure in psi.

This formula may be used to determine flow through a given nozzle if its diameter and the nozzle pressure are known. However, use of the formula requires squaring and finding square roots of numbers. This takes time, and requires mathematical steps in which many mistakes can be made. The graphical solution for solving the same problems requires no more than drawing a straight line through two known points on the graph. The point at which the line drawn intersects the third scale indicates the solution to the problem. The following examples should clarify the use of the nomograph and point out its inherent simplicity.

**Example 1**

**Determine the flow through a 1 1/8-inch smooth bore nozzle if the water pressure at the nozzle is 45 psi.**

*MATHEMATICAL METHOD*

**Flow is determined by the equation Q = 29.7 d p**

**where**

**d = 1 1/8 inches**

**p = 45 psi**

**Q = (29.7)(1 1/8) ^{2} x √45**

**= (29.7)(9/8) ^{2} √45**

**= (29.7)(81/64)(√45)**

**= (29.7)(1.26)(6.7)**

**= 250.7 gpm**

*GRAPHICAL METHOD*

**Flow is determined by drawing a line through the 1 1/8-inch mark on scale 2 (nozzle size) from the 45 psi mark on scale 1 (nozzle pressure).**

**The flow is read directly from scale 3 (flow) where the line down crosses that scale.**

As seen in this example, both methods produce the same answer. However, the mathematical solution might take five minutes to solve and check by hand, where the graphical solution can be solved as quickly as a straight line can be drawn, and with no mathematical skill required at all!

THIS NOMOGRAPH IS TO BE USED AS A WORKSHEET TO BE DETACHED AND CARRIED THROUGH THE READING OF THE ARTICLE SO THAT PROBLEMS CAN BE WORKED ALONG WITH THE EXAMPLES.

THE REVERSE SIDE CAN BE PRESERVED (LAMINATED) FOR FIRE DEPARTMENT USE AFTER THE VALUE OF USING STRAIGHT LINE HYDRAULICS ISACCEPTED.

**►Key:**

**A. Rockwood 1 1/2-inch L-11 spray nozzle**

**B. 60 gpm fog setting**

**C. 1 1/2-inch mystery nozzle**

**D. 125 gpm fog setting**

**E. 2 1/2-inch mystery nozzle**

**F. 2 1/2-inch 9 port Bressnan distributor**

**G. 750 gpm rated fog**

**H. 1,000 gpm rated fog**

**I. 3-inch hose with 2inch couplings**

**J. 3-inch hose**

**K. Two 2 1/2-inch lines siamesed**

**L. 3-inch and 2 1/2-inch lines siamesed**

**M. Two 3-inch lines siamesed**

**►Key:**

**A. Rockwood 1 1/2-inch L-11 spray nozzle**

**B. 60 gpm fog setting**

**C. 1 1/2-inch mystery nozzle**

**D. 125 gpm tog setting**

**E. 2 1/2-inch mystery nozzle**

**F. 2 1/2-inch 9 port Bressnan distributor**

**G. 750 gpm rated fog**

**H. 1,000 gpm rated fog**

**I. 3-inch hose with 2inch couplings**

**J. 3-inch hose**

**K. Two 2inch lines siamesed**

**L. 3-inch and 2inch lines siamesed**

**M. Two 3-inch lines siamesed**

Note also that the graphical solution remains identical no matter which two of the three variables (flow, nozzle size, or nozzle pressure) are given. If the mathematical solution is tried, the equation must be transposed (another potential source for error) if the original question is stated differently. For example, if the original question gave the flow (250 gpm) and the nozzle size (1 1/8 inches) and asked what nozzle pressure would be required to produce this flow, the mathematical solution would be:

√P = Q/29.7d^{2}

Q^{2}

p= (29.7)^{2} d^{4} (250)^{2}

p= (882)(1 1/8)^{4}

p = 44.5 psi

The graphical solution would remain the same as before by drawing a straight line through the 250 gpm mark and the 1 1/8-inch nozzle size mark and extending the line to intersect the nozzle pressure line. The answer (45 psi) is read where this line crosses the pressure scale.

The right side of the nomograph is used in an identical fashion to solve problems involving friction loss in a given hose size.

The most common mathematical method to determine friction loss in a hose is to determine what the friction loss would be in a standard 2 1/2-inch hose, and then multiplying that value by one of the factors listed in Table 1. The traditional formula to determine friction loss in 2 1/2-inch cotton jacketed hose is the Underwriters Formula:

FL = 2Q^{2} + Q

where

FL = friction loss per 100 feet of hose

Q = flow in gpm

100

This formula is based upon tests and data from many years ago as shown in figure 1. The more modern hoses available today, especially the polyester jacketed lines, tend to have a lower friction loss than would be calculated by traditional formulas.

To account for this, alternative formulas are used in the fire service. Many departments that have hoses in excellent condition use variations of the Underwriters Formula, such as the

*TABLE 1*

**Standard Factors To Determine Friction Loss in Hoses Other than 2 1/2 inches**

Find the friction loss for given flow for 2 1/2-inch hose by departmental methods, then multiply by the factor shown.

Purington Formula (FL = 2Q^{2}) to determine friction loss in 2 1/2-inch lines.

Normally, using either of these formulas, or one of the other available formulas, is equally acceptable due to other unaccounted for uncertainties in friction loss calculations. The nomograph is sealed to indicate both new and old 1½ and 2 1/2-inch hoses to make it compatible with various formulas in use from one department to another.

Regardless of the method used, it is important for the pump operator to be able to determine the friction loss in hose lines in order to determine maximum allowable hose length and proper engine pressures. The engine or pump pressure is determined by adding the required nozzle pressure, the friction loss in the hose, and an allowance to compensate for nozzle height as shown by the formula:

EP = NP + FL ± H where

EP = engine pressure in psi

NP = nozzle pressure in psi

FL = friction loss in psi

+H = head pressure in psi if the nozzle is above the pump (back pressure)

-H = head pressure in psi if the nozzle is below the pump (forward pressure).

The nozzle pressure is considered to be constant depending upon the nozzle in use. Fog nozzles are normally rated for 100 psi, smooth bore nozzles for handlines are usually rated for a maximum of 50 psi, and smooth bore nozzles for master streams (greater than 1 1/4-inch nozzles) for 80 psi. The back pressure (or forward pressure) is determined by multiplying the duration change (in feet) from the pump to the nozzle by 0.434. A working figure of 0.5 pound per foot is generally used by pump operators.

This pressure may be positive and must be overcome by additional pump pressure when stretching to upper elevations in relation to pump location (e.g., upper stories of tall buildings). However, when nozzles are stretched to areas some distance below the pumper, if this forward pressure is unaccounted for at the pumps (negative and subtracted from ideal level pressure), it will add to the nozzle pressure and in some cases make the nozzle unmanageable, cause injuries and, at the very least, an inefficient stream.

For essentially level ground or single-story dwelling evolutions, the engine pressure is determined by adding the nozzle pressure and the friction loss. Since the nozzle pressure is known for each device carried on the apparatus, determining the engine pressure usually reduces the problem of determining the friction loss. The following example shows how friction loss may quickly be determined by use of the fire service nomograph.

Example 2

**A 1 1/2-inch smooth bore master stream tip is connected by an appliance supplied by siamesed hoselines as shown. Since the two discharge lines are of different diameters, the flow through each line will be different and the conventional rule of thumb methods fall short in determining the pressure drop (friction loss) In the hose.**

**The pump operator must set the engine pressure high enough to compensate for the friction loss In the discharge lines and still produce the rated nozzle pressure (80 psi) at the master stream tip. The pump operator should know that a 1 1/2-inch tip will produce a flow of approximately 600 gpm with 80 psi at the nozzle. If in doubt, this flow could also be determined by the nozzle formula or the nomograph as shown in example 1.**

**The pump operator must calculate the friction loss In order to determine the proper pressure at which to pump. The friction loss may be determined mathematically or by the graphical method as shown below:**

*MATHEMATICAL METHOD*

**Friction loss may be determined by the Underwriters Formula (FL = 2Q ^{2} + Q), and the result multiplied by the appropriate factor from Table 1 (one 2 1/2 and one 3-inch siamesed = 0.164).**

**FL = (0.164)(2Q ^{2} + Q)**

**600 Q = 6 100**

**where**

**FL = (0.164)[(2)(6) ^{2} + 6]**

**= (0.164)[(2)(36) + 6]**

**= (0.164)(72 + 6)**

**= (0.164)(78)**

**= 12.8 psi**

**Therefore the pump operator would set the engine pressure at approximately 93 psi (80 + 12.8 = 92.8 psi).**

*GRAPHICAL METHOD*

**Scales 3, 4. and S are used, drawing a line connecting the two known variables (600**

**gpm and mark “L” for a 3 and 2 1/2-inch Siamese).**

**The answer is read directly from the graph where the line drawn crosses the friction loss line (approximately 12.8 psi). Therefore, the pump operator would set the engine pressure at approximately 93 psi (80 + 12.8 = 92.8 psi).**

The nomograph is a quick method of solving conventional hydraulics problems. The nomograph can also solve problems where mathematical solutions would be even more complex, as additional examples will demonstrate.

Standard hydraulics formulas and rule of thumb methods fall short when considering the operating characteristics of adjustable flow fog nozzles operated at pressures above or below the standard 100 psi. Since nozzles such as these have several flow settings that can be changed by the nozzleman, the pump operator may not know which flow setting is selected and therefore what the friction loss and engine pressure should be. In a practical sense, this is not normally a concern on the fireground as long as the pressure selected by the pump operator is within a fairly broad acceptable range.

However, there still may be situations in which it is required to determine the flow characteristics of adjustable flow nozzles. The easiest way would probably be to obtain the manufacturer’s data for the nozzle; however, this information may not always be available. Once again, the two methods for solving this problem are compared.

**Example 3a**

**An Elkhart Model SFL 1 1/2-inch adjustable flow nozzle has settings for 40, 60, 95, and 125 gpm (all flows rated at 100 psi nozzle pressure). Determine the flow through the nozzle if the pressure is 175 psi and the flow selected is 95 gpm.**

*MATHEMATICAL METHOD*

**The nozzle flow formula (Q = 29.7 d ^{2} √p) may be used if an equivalent nozzle diameter for the fog nozzle is determined. This could be done using the rated flow conditions as follows:**

**Q = 29.7 d ^{2} √p**

**where**

**O = flow (95 gpm)**

**d = equivalent nozzle diameter**

**p = nozzle pressure (100 psi)**

**Q = 29.7 d ^{2} √p**

**95 = 29.7 d ^{2} √100**

**95 = (29.7)d ^{2}(10)**

**95 = (297)d ^{2}**

**95 / = d ^{2} 297**

**0.319 = d ^{2}**

**√0.319 = d 0.565 = d**

**This equivalent diameter (0.565) and new nozzle pressure (175 psi) may then be substituted into the formula to determine the new flow:**

**Q = 29.7 d ^{2} √P**

**Q = (29.7)(0.565) ^{2} √l75**

**Q = (29.7)(0.319)(13.2)**

**Q = 125 gpm**

*GRAPHICAL METHOD*

**Nozzles not already shown on the nomograph may be added by drawing a line from the rated flow (95 gpm) to the rated nozzle pressure (100 psi). The nozzle mark may then be added to the nomograph at the point where the line crosses the nozzle scale.**

**The flow at 175 psi nozzle pressure would then be determined connecting the new nozzle mark and the 175 psi nozzle pressure, and extending the line to the flow scale.**

**The answer is read directly off the flow scale as 125 gpm.**

**Example 3b**

**The 1 1/2-inch Elkhart Model SFL nozzle described in example 3a is connected to a new I 72*>mi;ri uid^iiafyc ime, OJO ICCI m ICIIIJIM. The pumper is equipped with a flow indicator that the pump operator uses to set the pump speed. In this hypothetical situation, assume the nozzleman has set the nozzle for 95 gpm, but the pump operator assumes that maximum flow (125 gpm) has been set at the nozzle.**

**If the pump operator attempted to actually flow 125 gpm through the nozzle, example 3a showed that a nozzle pressure of approximately 175 psi would be required. The pump pressure would therefore be equal to 175 psi plus the friction loss in the hose. The friction loss would be determined by the method developed in example 2.**

*MATHEMATICAL METHOD*

**Friction loss in 1 1/2-inch hose (new) could be determined by figuring the friction loss for 2 1/2-inch hose and multiplying by the appropriate correction factor listed in Table 1.**

*GRAPHICAL METHOD*

**Friction loss in new 1 1/2-inch hose is determined by placing a line across the 175 gpm mark through the 1 1/2-inch hose (new) mark and extend it to intersect the friction loss line.**

*MATHEMATICAL METHOD*

**An alternate method which is much quicker and produces results consistent with friction losses in newer 1 1/2-inch hoses is to use the formula FL = 25Q ^{2}.**

**FL = 25Q ^{2}**

**FL = 25(125/100)2**

**FL = 25(1.25) ^{2}**

**FL = 25(1.56)**

**FL = 39.1 psi per 100 feet**

**Since there is 350 feet of hose in this example, the friction loss would be 3.5 times greater than the loss in only 100 feet of hose, or**

**FL = 350/100 X 39.1**

**= 137 psi**

*GRAPHICAL METHOD*

**The friction loss is read from the graph as being approximately 39 psi/100. Since there is 350 feet of hose In this example, the friction loss would be 3.5 times greater than the loss in only 100 feet of hose or**

**= 136.5 psi**

**FL = 350/100 X 39**

In this situation, the pump operator, relying only on flow indication, would be forcing the engine to develop excessive pressure at the pump discharge. The pump discharge pressure in this situation would be approximately 312 psi (175 psi + 137 psi), which is above the hose test pressure used by many departments.

Therefore, without realizing it, the pump operator can risk damaging the hose and injuring the hose handlers. The example demonstrates the importance of good communications on the fireground and why pressure gauges and hydraulics calculations are important.

These examples should also demonstrate that on those occasions when hydraulic calculations are required, the fire service nomograph can be relied upon to provide accurate answers in almost no time at all.

**Example 4**

**Determine the flow through an adjustable 50 to 300 gpm automatic flow nozzle If the engine pressure Is 125 psi and the discharge line is 2 inches in diameter, rubber Jacketed, and 200 feet long. These relatively new “automatic” nozzles are self adjusting to vary the nozzle opening and thereby the flow as the pressure is varied, while maintaining a fairly constant nozzle pressure of approximately 100 psi. As determined In the previous example, a mathematical solution is prohibitively complex and the nomograph Is the best tool to solve the problem.**

**Since the automatic nozzle adjusts the flow to maintain a nozzle pressure of 100 psi, the friction loss in this example would be the difference between the engine pressure (125 psi) and the nozzle pressure (100 psi), or 25 psi total. The friction loss per 100 feet is, therefore, one-half this value, since the hose layout has 200 feet of hose, or 12½ psi per 100 feet. By drawing a line on the nomograph (as shown in figure 2) from the 12 1/2 psi point through the 2-inch hose mark, and extending it to the flow line, the flow is readily determined to be 125 gpm.**

A modified nomograph, which automatically determines friction loss for the actual hose length used, is shown in figure 3. It is used just like the first nomograph, except a pivot line on the far right side is used in conjunction with the flow and hoze size scales. To tion loss, a line is extended to the pivot line from the known flow and hose size. From the pivot line, a line is then drawn to the hose length line. The total friction loss is then read directly from the total friction loss per 100 feet by the number of 100-foot sections in the hose layout. One final example is given to show how this modified nomograph would be used to solve a typical problem.

**Example 5**

**For the layout shown below, determine the proper engine (pump) pressure**

**The nomograph (figure 3) shows how the pump pressure is determined. Line 1 shows that the flow through the 1 1/4-inch tip would be slightly more than 400 gpm. Line 2, drawn from the flow mark through the 3-inch hose (with 2 1/2-inch couplings) mark is extended to the pivot line. From this point, a line (line 3) Is drawn back to the 500-foot mark for hose length. The total friction loss is read directly from the graph where line 3 intersects the friction loss line. The friction loss indicated (80 psi) plus the master stream nozzle pressure (also 80 psi) would be added to determine the proper pump pressure (160 psi).**

**Notice that the pivot line on the far right side of figure 3 is shown as a broken line at the top. This serves as a reminder that flow conditions that result in the pivot point in this area are generally considered excessive, and a larger hose size of parallel lines should be used to reduce the friction losses.**

Additional modifications could also bo made to the nomographs to customize it to show only those nozzles and hose sizes commonly used by a department. Nozzle types and hose sizes or combinations can be added by plotting known conditions on the graph and adding new marks on the graph as shown in example 2. Once modified, each department member should have no problem in being able to solve any typical hydraulics problem.