Chapter 1: Water at Rest and In Motion

Six Principles of Fluid Pressure

1) Fluid pressure is perpendicular to any surface on which it acts.

2) Beneath the surface of a liquid at rest, the pressure is the same in all directions (upward, sideward, downward).

3) Pressure applied to a confined fluid from without is transmitted equally in all directions.

4) The pressure of a liquid in an open vessel is proportional to its depth.

5) The pressure of a liquid in an open vessel is proportional to the density of the liquid.

6) Liquid pressure at the bottom of a vessel is unaffected by the size and shape of the vessel as long as the height of water remains the same.

Pressure – Height – Density Relationship

Formulas

a. Pressure (P) = .434 X Height or P = .434H

b. Height / Head (H) = 2.31 X Pressure or H = 2.31P

Work Problems: Using the formulas above, solve the following:

a. Find the pressure at the bottom of a standpipe filled with water 100 feet high.

P = .434H

P = .434 (100)

P = 43.4 psi

* The pressure in this formula is often referred to as back pressure (BP) in pumping operations. This back-pressure may be encountered during high-rise operations, while using dry standpipes, or pumping up or down hills. BP = .434H

b. The static pressure in a fire hose connected to a standpipe is 150 psi. How high will that static pressure raise the water in the standpipe?

H = 2.31P

H = 2.31 (150)

H = 346.5 ft

Back Pressure

1) Multi-Story Buildings

The average height per story is 10-12 feet

BP = .434H

Therefore, BP per story is .434 (12) or 5.2 psi per story

As a rule of thumb, 5 psi per story above the first floor is used for calculating BP in high-rise buildings.

Work Problems: Using the rule of thumb, find the BP for the following:

a. Fire on the 10th floor level of a 20-story office building

BP = 5 X 9

BP = 45 psi

note: the fire is only 9 floors above ground level (this can be tricky)

b. Fire on the roof top of a 20-story office building

BP = 5 X 20

BP = 100 psi

Note: in this case, the fire is actually 20 floors above the ground floor

Because it is on the roof and not the floor level. (Tricky too)

2) Uphill vs. Downhill

A. Uphill:

When pumping on a grade, either uphill or downhill, pump operators must take into consideration the pressure loss or gain caused by BP.

When pumping uphill, the pump has to work harder to get the water to the desired location because gravity is acting on the water and holding it back. The pump pressure must be increased to overcome the back pressure.

Example: A fire engine is pumping water uphill through a hoseline that is

80ft above the firetruck.

BP = .434H

.434 (80)

34.7 psi

The pump operator will have to increase the pump pressure by 34.7 psi to make up the difference in back pressure.

B. Downhill

When pumping downhill, the pump does not have to work as hard because gravity is acting on the water, helping to move it through the fire hose. This means that the BP gained will be in addition to the pump pressure reading. You won’t see the pressure increase on your pump gauge, but the hoseman will feel it on the hoseline.

Example: A fire engine is pumping water downhill through a hoseline that

is 60ft below the firetruck.

BP = .434H

.434 (60)

26.04 psi

The pump operator will have to decrease the pump pressure by 26.04 psi

To negate the pressure increase caused by back pressure.

3) Types of Pressure

a. Static Pressure - Pressure of water at rest

b. Flow Pressure - Pressure of water flowing from nozzle

c. Residual Pressure – Pressure remaining in water main or inlet side of

Fire pump after water is flowing

Chapter 2: Velocity and Discharge

Drafting Operations

1) Theory

Drafting is a way in which a fire pump uses atmospheric pressure to draw water into the fire pump from a static water source (see figure 2.1). Atmospheric pressure, at sea level, is 14.7 psi. Fire pumps are capable of expelling its air through use of a priming pump. If all the air within a pump is displaced and a good seal is maintained (no leaks or loose fittings), the fire pump can create a vacuum-like atmosphere within the fire pump. Once this vacuum-like atmosphere is obtained, the atmospheric pressure outside the pump will be greater than the pressure within the fire pump, and water will be forced into the pump through the drafting hose. This means that using the formula H = 2.31P, we can see how atmospheric pressure will force water into the fire pump up to 33.9 feet high (H = 2.31 X 14.7 psi).

Figure 2.1 – Drafting Operations

The theoretical lift of 33.9 feet is nearly impossible to obtain. Since the fire pump cannot produce an absolute vacuum (0 psi), and there is friction loss in the drafting hose, a practical lift of 22-25 feet is more realistic.

2) Factors Affecting Priming Operations (trying to obtain a vacuum)

a. Loose hose connections, loose covers, open gates, open valves, too long or too small suction hose

b. Defective priming pump

c. Depth of water source

note: 1250 gpm pumpers should only take 30 seconds to prime

1500 gpm pumpers should only take 45 seconds to prime

3) Factors Affecting Lift

a. Altitude

b. Weather

c. Water temperature

d. Too long or too small suction hose

Velocity

1) Formulas

For velocity, you will be given either the height of the water or the pressure. Therefore, two formulas for velocity will be discussed.

a. Velocity = 8 X Height of water or V = 8H

b. Velocity = 12.1 X Pressure of nozzle or V = 12.1P

Work Problems:

What is the velocity of water flow is the nozzle pressure is 60 psi?

V = 12.1 P

V = 12.1 60

V = 12.1 (7.75)

V = 93.78 fps

What is the velocity of water flow from a water tank 50 ft high?

V = 8 H

V = 8 50

V = 8 (7.07)

V = 56.56 fps

Flow Velocities

The velocity of water varies inversely with the cross section of the hoseline and nozzle tip. What does this mean??

With the same nozzle pressure:

Changing to a smaller nozzle tip will increase nozzle velocity / pressure

Changing to a larger nozzle tip will decrease nozzle velocity / pressure

The inverse relationship between velocity and nozzle size simply means that when one increases, the other decreases and vice versa.

Nozzle Discharge – Gallons per Minute (GPM)

1) Discharge Formulas

a. With Nozzle

Discharge (GPM) = 30 X Diameter of Nozzle2 X Nozzle Pressure

GPM = 30d2P

Work Problem: How many GPMs are flowing through a 2 ½” hoseline with a

1 1/8” nozzle tip with a nozzle pressure of 50 psi?

GPM = 30d2P

GPM = 30 (1.125)2 50 (nozzle size converted into a decimal)

GPM = 30 (1.27) (7.07)

GPM = 269.37

b. Without Nozzle (open butt)

To find the discharge pressure of a hoseline without a nozzle, simply use 90% of the original discharge formula. The hose diameter will substitute as the nozzle size in this case.

GPM (open butt) = 90% X 30 X Hose Diameter2 X Pressure

GPM (open butt) = 27d2 P

Work Problem: How many GPMs are flowing through a 2 ½” hoseline

without a nozzle attached to it at 50 psi?

GPM (open butt) = 27d2 P

GPM = 27 (2.5)2 50

GPM = 27 (6.25) (7.07)

GPM = 1193.06

Nozzle Reaction

1) Formula

Nozzle Reaction (NR) = 1.57 X Nozzle Diameter2 X Nozzle Pressure

NR = 1.57d2 P

note: In theory, the nozzle reaction will always be greater than the actual

nozzle reaction felt by the firefighter because:

a. The hoseline is in contact with the ground, and this absorbs

some of the nozzle reaction.

b. Bends in the hoseline as it is laid out will help to absorb some

of the nozzle reaction.

Work Problem: What is the nozzle reaction of a 2 ½” hoseline with a

1 1/8” nozzle tip flowing 50 psi?

NR = 1.57 d2 P

NR = 1.57 (1.125)2 (50) (nozzle size converted to a decimal)

NR = 1.57 (1.27) (50)

NR = 99.70 lbs

2) Safety Factors to Consider

a. Handling Hose Lines

i. Bends near the nozzle tend to straighten out. The hoseline should be straight at least 10 feet back of the nozzle

ii. Nozzle reaction from fog streams is less than straight streams

iii. Open and close nozzle slowly because:

1. Initial nozzle reaction is greater than the nozzle reaction when water is flowing

2. Sudden closing of nozzle sends pressure surges backwards. This is called a “water hammer”. A water hammer can break the hoseline, fire pump, and/or water main.

iv. When using handlines on ladders, the nozzle reaction could cause the ladder to lift-off and fall away from the building. To help avoid this dangerous situation from occurring firefighters should:

1. Fasten the ladder to the window sill

2. Set the base of the ladder further away from the building

b. Ladder Truck Operations

i. If the hose should burst, the ladder / boom will whip violently

ii. Always try and shoot the fire stream in-line with the ladder. Never turn the nozzle more than 15 degrees from the center of the ladder.

Work Problem: What is the nozzle reaction of a ladder pipe

operation flowing 80 psi from a 2” nozzle tip?

NR = 1.57 d2 P

NR = 1.57 (2)2 (80)

NR = 1.57 (4) (80)

NR = 502.4 lbs (force)

Friction Loss

1) Effect of Flow Pattern

a. Laminar Flow – Low Flow Velocities

b. Turbulent Flow – High Flow Velocities

i. Friction loss in hose affected by:

1. Inner lining of hose

2. Age of hose

3. Thickness of hose lining

4. Type of hose jacked weave (will it expand or not)

2) Friction Loss in Hoses

a. Friction loss varies with quality of hose

b. Friction loss varies directly with length of hose line (the longer the hoseline, the greater the friction loss).

c. Friction loss varies approximately as the square of the velocity of flow (the faster the flow velocity, the greater the friction loss)

Example:

If the flow velocity is doubled – Friction loss is 4 times

greater (2)2 = 4

If the flow velocity is tripled – Friction loss is 9 times

greater (3)2 = 9

d. For a given velocity, friction loss varies inversely as the fifth power of the hose diameter (the bigger the hose diameter, the less the friction loss)

Example:

If the hose size was doubled from 1 ½” to 3”

We see how 1 ½” X 2 = 3”

Inverting the 2 we get ½

Now we take the ½ and multiply it to the fifth power

½ 5 = ½ X ½ X ½ X ½ X ½ = 1/32

We can now conclude that if the hose size is doubled, the

New friction loss is only 1/32 as much as the original figure.

e. For a given velocity of flow, friction loss is nearly independent of pressure. In other words, he velocity of flow, and not the pressure, is the determining factor in friction loss.

3) Formulas (more details in Chapter 5)

a. Friction Loss = 2 X Q2 + Q

FL = 2Q2 + Q Q = GPM 100

note: This formula is only for 2 ½” diameter hoselines.

The friction loss figure represents friction loss per 100 ft of 2 ½” hose.

i.e. If the hoseline is 600 ft, the friction loss figure must be

Multiplied by 6 to get the total friction loss in that hoseline.

Work Problem: Find the friction loss in a 2 ½” hoseline 100 ft in length

Flowing 300 gpm.

FL = 2Q2 + Q Q = GPM 100

FL = 2 (3)2 + 3 Q = 300 100

FL = 2 (9) + 3 Q = 3

FL = 21 # per 100 ft length

FL = 21# (since there is only 100 ft of hoseline)

c. Engine Pressure = Friction Loss + Nozzle Pressure

EP = FL + NP

Work Problem: An engine is pumping through 600 ft of 2 ½” hoseline to a nozzle that is flowing 200 gpm at 100 psi nozzle pressure. Find the engine pressure.

EP = FL + NP FL = 2Q2 + Q Q = GPM 100

EP = FL + 100 FL = 2(2)2 + 2 Q = 200 100

EP = 60 + 100 FL = 2(4) +2 Q = 2

EP = 160 psi FL = 10 per 100ft hoseline

FL = 10 X 6 (# of hundreds of feet of hose)

FL = 60

d. Engine Pressure = Friction Loss + Nozzle Pressure + Back Pressure

EP = FL + NP + BP

Work Problem: An engine is pumping through 300 ft of 2 ½” hoseline

to a nozzle that is flowing 200 gpm at 100 psi nozzle

Pressure. The nozzle is on a hill that is 60 ft higher

Than the fire pump. Find the engine pressure.

EP = FL + NP + BP

FL = 2Q2 + Q Q = GPM 100 BP = .434H

FL = 2 (2)2 + 2 Q = 200 100 BP = .434 (60)

FL = 2 (4) + 2 Q = 2 BP = 26.04

FL = 10 per 100 ft hoseline

FL = 10 X 3 ( # of hundreds of feet of hose)

FL = 30

EP = FL + NP + BP

EP = 30 + 100 + 26

EP = 156 psi

Note: If the nozzle was below the fire pump (downhill or in a Basement), the BP would have to be subtracted from the EP.

Remember that the gravity would cause the pressure to increase, thus giving the firefighter on the hoseline too much pressure.

Chapter 3: Water Distribution System

General Remarks

1) Public Water Systems are designed to perform two functions.

a. Provide water for domestic, commercial and industrial use

b. Provide water for fire protection

i. Combination systems use same water source for both functions

ii. Separate systems use “potable” water for domestic use and use brackish, salt or treated sewage water for firefighting.

2) Fire hydrants must be installed and maintained in accordance to standards set forth by the American Water Works Association.

3) The term “Fire Plug” originated from the old days when pipes were made from hollowed out wood and buried underground. These pipes were gravity fed from water sources located in the area. If there were a need for water, the fire department would expose the pipe by digging up the ground. Once exposed, a hole was drilled into the wooden pipe. Water would then flow out of the hole by means of gravity. When the fire was out, a tapered wooden plug was inserted into the hole and the pipe would be re-buried.

Fire Hydrants

1) Dry Barrel Hydrants (see figure 3.1)

a. Dry barrel hydrants do not have water stored in the fire hydrant itself. Instead, the water is stored in the piping below the hydrant. When the operating stem is opened, water will begin to fill the hydrant. This type of hydrant is used in areas where freezing can occur. If the water is stored in the hydrant, it could freeze and that hydrant would be useless in a fire situation. Because the water in the piping below the hydrant is constantly moving, it usually does not freeze.

b. Main Parts:

i. Dry Barrel

ii. Footpiece

iii. Bonnet

iv. Operating Stem

v. Main Valve

vi. Drain

2) Wet Barrel Hydrants (see figure 3.2)

a. Unlike the dry barrel hydrant, the wet barrel hydrant has water in the hydrant right up to the discharge outlet. These hydrants should not be used in areas where freezing could occur. These are the types of hydrants commonly used in Hawai`i. However, there are some dry barrel hydrants in use today in Hawai`i.

b. Wet barrel hydrants have fewer parts than dry barrel hydrants.

c. Each outlet has an independent valve on a threaded stem with

operating nut on opposite side of barrel.

Figure 3.1 – Dry Barrel Hydrant

Figure 3.2 – Wet Barrel Hydrant

3) Other Information

d. Hydrant Outlets

i. Every hydrant must have at least two outlets.

1. One pumper suction hose outlet (usually 4”)

2. One regular hose outlet (usually 2 ½”)

ii. Outlets must not be less than 18 inches from the ground level.

iii. Outlets must have a cap and chain.

e. Hydrant Spacing

i. Hydrants should not be spaced more than 250 ft apart in commercial / industrial areas and should have a minimum flow of 1000 gpm.

ii. Hydrants should not be spaced more than 350 ft apart in residential areas and should have a minimum flow of 1000 gpm.

iii. Hydrants should not be spaced more than 700 ft apart in rural areas and should have a minimum flow of 1000 gpm.

f. Branch Connection

i. The minimum size water main supplying fire hydrants in Honolulu is 8 inches. Main sizes smaller than 6 inches are not suitable for providing fire protection.

ii. Each wet barrel hydrant has its own gate valve. This gate valve is located somewhere near the hydrant, and its location is indicated on the fire hydrant. In case the hydrant should be damaged (if a car knocks one over) the gate valve can be used to stop the flow of water to that one hydrant without interrupting the flow to other hydrants on that same water main.

4) Estimating Available Flow From Fire Hydrants

a. In order to estimate the amount of flow we have available in a hydrant, we must first find the percentage of drop in pressure between static and residual pressure.

i. Open hydrant with suction hose connected to fire truck. At this time take note as to what your intake (suction) pressure gauge is reading. This is the hydrant’s static pressure.

ii. Open a discharge gate for one hoseline. Again look at your suction gauge. The pressure will be lower because some of the static pressure will have been used for the first firefighting line. This is the hydrant’s residual pressure.

iii. Subtract the residual pressure from the static pressure and convert that number into a percentage. This is the percentage of drop in pressure between static and residual.

Work Problem: A hydrant is connected to your fire truck with no

Firefighting lines flowing. Your suction gauge reads 60

psi. After opening one firefighting line, your suction

Gauge now reads 55 psi. What is the percentage of drop in pressure?

Static Pressure = 60 psi

Residual Pressure = 55 psi

Pressure Difference = 5 psi

To find the drop in a %, simply divide the difference in

pressure by the static pressure.

5 psi 60 psi = .083 or 8.3%

b. Applying percentage of drop in pressure to practical situations.

i. Once the percentage of drop between static and residual pressure is found, that number can be used to help estimate the number of additional hoselines the hydrant can supply. The estimates are based on hoselines of the same diameter utilizing nozzles of the same diameter also. The following chart shows the general rule of thumb regarding additional hoselines:

10% or less 3 more hoselines

11-15% 2 more hoselines

16-25% 1 more hoseline

more than 25% no more hoselines

Chapter 4: Fire Service Pumps

Introduction

1) Three Basic Types of Fire Pumps

a. Piston Type Fire Pump

b. Rotary Type Fire Pump

c. Centrifugal Type Fire Pump

2) Pump Mounting on Apparatus

a. Mid-ship (middle) – 2 ways

i. Between road transmission and rear axle in line with drive shaft (most common)

ii. Ahead of clutch and transmission with flywheel and power take off. This type allows for direct engine power to pump transmission connection. It allows for driving and pumping simultaneously.

b. Front Mounting

i. From front of engine crankshaft connected to pump transmission. This type also allows for driving and pumping simultaneously.

3) Pump Ratings

a. Standard pumper capacity ratings start from 500 gpm and increase in 250 gpm increments (NFPA 19 Specification) up to 2000 gpm.

i. 500 gpm, 750 gpm, 1000 gpm, 1250 gpm, 1500 gpm, 1750 gpm, 2000 gpm

ii. Pumpers must have one 2 ½” gated outlet per 250 gpm rated capacity.

b. Fire pumps are designed to perform as follows:

100% of rated capacity at 150 psi net pumps pressure

70% of rated capacity at 200 psi net pumps pressure

50% of rated capacity at 250 psi net pumps pressure

Net pump pressure is found by subtracting suction or inlet pressure from the discharge pressure.

Work Problem: A fire pump is discharging 200 psi through a 2 ½”

Hoseline and is receiving 50 psi from a fire

Hydrant. What is the net pump pressure?

Net Pump Pressure = Discharge Pressure – Intake Pressure

Net Pump Pressure = 200 psi – 50 psi

Net Pump Pressure = 150 psi

Work Problem: A pumper has a capacity rating of 1000 gpm.

Using net pump pressure finds the efficiency of the pump:

At 150 psi net pump pressure _______gpm

At 200 psi net pump pressure _______gpm

At 250 psi net pump pressure ¬¬¬¬¬¬_______gpm

The answers are 1000 gpm (1000 gpm X 100%), 700 gpm

(1000 gpm X 70%) and 500 gpm (1000 gpm X 50%).

note: Fire pumps are most efficient at 150 psi or less.

4) Cavitation

a. Causes of cavitation

i. Lift too high for volume and pressure discharged

ii. Suction hose too small for volume and pressure discharged

iii. Suction strainer or hose clogged

iv. Partial collapse of hose lining

v. Temperature of water too high

b. Signs of cavitation

i. Pump vibrations with loud pinging noises. This is caused by air bubbles that form in the pump that collapse violently when they enter the impeller.

ii. “Pump running away” This happens when the pump is pumping air or steam instead of water. The pump speed will increase with no increase in discharge pressure or volume. The pump operator will hear the engine revving and running away. If this should occur, shut down pumping operations immediately.

Centrifugal Fire Pumps

1) General Information

a. Water enters the centrifugal pump through the eye and is delivered to the impeller through the vanes. The impeller increases the speed of the water and discharges it through the volute. (see figure 4.1)

figure 4.1 – Centrifugal Fire Pump

c. Centrifugal pumps may be connected in stages to each other. Centrifugal pumps can have 1 – 4 stages. Each impeller and volute is one stage in a multi-stage centrifugal pump.

2) Capabilities and Limitations

a. Volume (gpm) varies directly as the pump speed. The faster the pump speed, the greater the volume.

b. Pressure (psi) varies as the square of the pump speed. If you double the pump speed, the pressure increases 4 times (22 = 4)

c. Centrifugal pumps are not self-priming. A separate rotary vane priming pump unit is used to prime centrifugal pumps.

3) Single-Stage Centrifugal Fire Pumps

a. Basic Designs

i. Single suction impeller

1. Limited up to 750 gpm pumpers

ii. Double suction impeller

1. 1000 – 1500 gpm pumpers

iii. Double Volute design

1. Limited to single stage pumps

2. High efficiency at rated capacity

b. In general, single stage pumps have a high efficiency rating (about 70%), which is generally slightly higher than multi-stage pumps.

4) Two-Stage Centrifugal Fire Pumps

a. Basic design

i. Two impellers with separate volute chambers for each impeller

b. Limited use in fire service

i. Pumps may be front mounted for smaller trucks

1. Provides vehicle movement while pumping

2. Able to provide high-pressure for operations (300-400 psi).

5) Parallel – Series Two-Stage Centrifugal Pumps

a. Basic design

i. Two impellers with separate volutes

ii. Addition of transfer or changeover valve for parallel or series pumping operations

iii. The efficiency of parallel – series two-stage pumps is around 65-70%. This is slightly less than a single stage pump.

b. Parallel (Volume) Operations (see figure 4.2)

i. When a pump is placed in the parallel position, water enters both stages of the pump simultaneously from the suction side. This means that the pump will be able to deliver twice the volume at half of the pressure.

ii. An example of this is: A 1000 gpm rated capacity pumper operating at 150 psi net pump pressure. In the parallel position, each impeller will deliver 500 gpm at 150 psi. Note that the pump will double the volume of water, but at half the speed.

figure 4.2 – Parallel – Series pump in parallel position

c. Series (Pressure) Operations (see figure 4.3)

i. When a pump is placed in the series position, water enters one stage of the pump from the suction side. The water is then delivered to the second stage via the first stage of the pump. This means that the pump will be able to deliver half the volume at twice the pressure.

ii. An example of this is: A 1000 gpm rated capacity pumper operating at 150 psi net pump pressure. In the series position, the first impeller will deliver 500 gpm at 150 psi to the second stage. The second impeller will discharge the 500 gpm, but it will double the pressure. In the series position, this pump will deliver 500 gpm at 300 psi. Note that the pump will double the pressure, but at half the volume.

figure 4.3 – Parallel – Series pump in series position

d. Transfer Valve

i. A transfer valve is the device that is used to change a pump from series to parallel or vice versa. The pump operator must decide which position would best meet the needs of a given situation.

ii. Transfer valves may be either powered or manual

iii. Transfer valves may be disk type or cylinder type

iv. Transfer valves are normally left in the series position for normal day-to-day operations.

v. As a general rule, the transfer valve should be kept in the series position when pumping up to 70% of the pump’s capacity. The transfer valve should be switched to the parallel position when evolutions or circumstances require a pump to deliver more than 70% of it’s rated capacity.

vi. When switching the transfer valve from one position to another, the pump pressure should be lowered to below 60 psi. This is especially crucial when switching from parallel to series because the pressure will immediately double. This sudden increase in pressure could damage the pump, hoses, or seriously injure the firefighters on the hoselines.

6) Piston Pumps

a. General Remarks

i. Piston pumps were the first pumps developed for firefighting.

ii. Piston pumps are preferred by fire departments that get their water supply mostly by drafting operations.

iii. Piston pumps have a high efficiency (around 75-80%)

b. Basic Design

i. Use of piston to displace water

1. Single action (water gets moved on piston’s up stroke)

2. Double action (water gets moved on up stroke and down stroke)

ii. Piston pumps are positive displacement pumps. This means that they are self-priming

iii. Different gear ratios between engine and pump are needed to obtain higher pressure.

7) Rotary Fire Pumps

a. General Remarks

i. Rotary fire pumps were first used by the fire service around the 1600’s. In the 1900’s they were adapted to work with the steam engines.

ii. Rotary pumps are preferred where all pumping operations involve drafting, high lift conditions, or long suction hoses.

Chapter 5: Friction Loss Calculations

Basic 2 ½” Hoselines

As discussed earlier, the formula for finding friction loss in a 2 ½” fire hose is:

FL= 2Q2 + Q; where Q= GPM 100.

This formula only applies to:

a. 2 ½” diameter fire hose per 100 ft length

b. 2 ½” diameter fire hose flowing 100 gpm or greater

The formula for finding friction loss in a 2 ½” fire hose flowing less than 100

gpm is:

FL = 2Q2 + ½ Q; where Q = GPM 100

Work Problems:

Find the friction loss in a 500 ft length of 2 ½” fire hose flowing 200 gpm.

FL = 2Q2 + Q Q = GPM 100

FL = 2 (2)2 + 2 Q = 200 100

FL = 2 (4) + 2 Q = 2

FL = 10 per 100’ length

FL = 10 x 5 (500 ft 100)

FL = 50 psi

Find the friction loss in a 800 ft length of 2 ½” fire hose flowing 250 gpm.

FL = 2Q2 + Q Q = GPM 100

FL = 2 (2.5)2 + 2.5 Q = 250 100

FL = 2 (6.25) + 2.5 Q = 2.5

FL = 15 per 100’ length

FL = 15 x 8 (800 ft 100)

FL = 120 psi

Find the friction loss in 400 ft of 2 ½” fire hose flowing 80 gpm.

FL = 2Q2 + 1/2 Q Q = GPM 100

FL = 2 (.8)2 + ½ (.

Q = 80 100

FL = 2 (.64) + .32 Q = .8

FL = 1.6 per 100’ length

FL = 1.6 x 4 (400 ft 100)

FL = 6.4 psi

Finding friction loss for hoses other than 2 ½” diameters.

1) Finding Equivalent Lengths

Before we can use the friction loss formula on hoses other than 2 ½” in diameter, we must first find the equivalent length. In other words, we will be converting the hose to reflect the length in a 2 ½” diameter.

This is like converting feet to inches. Suppose we wanted to convert 5 feet into inches; it is understood that we must multiply the number of feet by 12 (the number of inches in a foot). 5 feet converted to inches would be: 5 x 12 = 60 inches.

Keeping that principle in mind, we only have a friction loss formula for 2 ½” hoselines. We will need to convert hoses of all other diameters to a 2 ½” length. This is called finding the “equivalent length”. It is easier to find the equivalent length, than it is to learn a new friction loss formula for each possible hose diameter.

There are two methods in which we can find an equivalent length.

a. Using Rule of Thumb

To find the equivalent length of a fire hose using the rule of thumb,

simply multiply the total length of hose by the rule of thumb factor. These factors are listed below and will be provided for you on the exam. You are not required to memorize the factors.

Diameter of Hose Rule of Thumb Factor

1” 91

1 ½” 13

1 ¾” 7.76

3” .4

3 ½” .17

4” .09

Work Problem: Using the rule of thumb factor, convert the following to an equivalent 2 ½” length:

100 ft of 1” hose = 100 x 91 or 9100 ft of 2 ½”

100 ft of 1 ½” hose = 100 x 13 or 1300 ft of 2 ½”

100 ft of 4” hose = 100 x .09 or 9 ft of 2 ½”

b. Using Conversion Factor

To find the equivalent length of a fire hose using conversion factor,

divide the total length of hose by the conversion factor. These factors are listed below and will be provided for you on the exam. You are not required to memorize the factors.

Diameter of Hose Conversion Factor

1” .011

1 ½” .074

1 ¾” .129

3” 2.5

3 ½” 5.8

4” 11.0

Work Problem: Using the conversion factor, convert the following to an equivalent 2 ½” length:

100 ft of 1” hose = 100 .011 or 9090 ft of 2 ½”

100 ft of 1 ½” hose = 100 .074 or 1351 ft of 2 ½”

100 ft of 4” hose = 100 11 or 9 ft of 2 ½”

2) Finding friction loss using equivalent length

Using either equivalent length method (rule of thumb or conversion factor), we can now find the friction loss for hoses other than 2 ½”.

Work Problem: Find the friction loss in a 1 ½” diameter hoseline 200 ft in

Length flowing 100 gpm.

Step 1: Find equivalent length

200 x 13 = 2600 ft or 200 .074 = 2700 ft

(for this example we will use 2600 ft)

Step 2: Find friction loss

FL = 2Q2 + Q Q = GPM 100

FL = 2 (1)2 + 1 Q = 100 100

FL = 2 (1) + 1 Q = 1

FL = 3 (per 100 ft of hose)

FL = 3 x 26 (2600 ft 100)

FL = 78

Work Problem: Find the total friction loss in a 1 ½” diameter hoseline 200 ft in length, connected to a 1” diameter hoseline 200 ft. in length flowing 40 gpm.

Step 1: Find equivalent length of 1 ½” hose

200 x 13 = 2600 ft or 200 .074 = 2700 ft (for this example we will use 2600 ft)

Step 2: Find equivalent length of 1” hose

200 x 91 = 18,200 ft or 200 .011 = 18,182 ft

(for this example we will use 18,200 ft.)

Step 3: Find total equivalent length of evolution (1 ½” + 1”)

2600 ft + 18,200 ft = 20, 800 ft

Step 4: Find Friction loss

FL = 2Q2 + ½ Q Q = GPM 100

FL = 2 (.4)2 + ½ (.4) Q = 40 100

FL = 2 (.16) + .2 Q = .4

FL = .52 (per 100 ft of hose)

FL = .52 x 208 (20800 ft 100)

FL = 108.16

3) Finding Friction Loss in Siamese Hose Lays

Some hose evolutions involve the use of siamesed hoselines. This means that two or more hoses are running parallel to each other and are supplying water to the same discharge. This discharge can be either a deluge or another hoseline. In order to find friction loss in this type of evolution, we must first convert all siamesed lines into an equivalent length of 2 ½” hose.

a. Converting Siamesed Hoselines to Equivalent 2 ½” Lengths

To convert siamesed lines to an equivalent 2 ½” length, we must first find the average length of the siamesed hoses. This is done by taking the total length of all siamesed hoses and dividing this figure by the number of hoses being siamesed.

The example above shows a typical siamese operation. To find the average length of siamesed hose:

Step 1: Find total length of siamesed hose

600 ft + 600 ft = 1200 ft

Step 2: Divide the total length of siamesed hose by the number of

siamesed hoses.

1200 ft 2 = 600 ft

b. Using rule of thumb factors to find equivalent length

To find the equivalent length of siamesed hoses, we will need to multiply the average length of the siamesed hoses by the rule of thumb factor. These factors are listed below and will be provided for you on the exam. You are not required to memorize the factors.

Number of Siamesed Hoses Rule of Thumb Factor

2 – 1 ½” hoses 3.75

2 – 2 ½” hoses .28

3 – 2 ½” hoses .13

4 – 2 ½” hoses .08

Work Problem: Find the friction loss for the evolution shown

below.

Step 1: Find the average length of siamesed hoses

600 ft + 600 ft = 1200 ft

1200 ft 2 = 600 ft

Step 2: Find equivalent length of siamesed hoses (this is the average length multiplied by the factor)

600 ft x .28 = 168 ft

Step 3: Find friction loss

FL = 2Q2 + Q Q = GPM 100

FL = 2 (5)2 + 5 Q = 500 100

FL = 2 (25) + 5 Q = 5

FL = 55 (per 100 ft of hose)

FL = 55 x 1.68 (168 ft 100)

FL = 92.4 psi

Work Problem: Find the friction loss for the evolution shown

below.

Step 1: Find the average length of siamesed hoses

600 ft + 600 ft = 1200 ft

1200 ft 2 = 600 ft

Step 2: Find equivalent length of siamesed hoses (this is the average length multiplied by the factor)

600 ft x .28 = 168 ft

Step 3: Find total equivalent length of evolution (equivalent length of siamesed hose + single 2 ½” hose)

168 ft + 300 ft = 468 ft

Step 4: Find Friction loss

FL = 2Q2 + Q Q = GPM 100

FL = 2 (2.5)2 + 2.5 Q = 250 100

FL = 2 (6.25) + 2.5 Q = 2.5

FL = 15 (per 100 ft of hose)

FL = 15 x 4.68 (468 ft 100)

FL = 70.2 psi

Work Problem: The fire engine below is supplying three 2 ½” hoselines to a portable deluge flowing 600 gpm.

Find the friction loss of the hoses.

Step 1: Find the average length of siamesed hoses

400 ft + 350 ft + 450 ft = 1200 ft

1200 ft 3 = 400 ft

Step 2: Find equivalent length of siamesed hoses (this is the average length multiplied by the factor)

400 ft x .13 = 52 ft

Step 3: Find friction loss

FL = 2Q2 + Q Q = GPM 100

FL = 2 (6)2 + 6 Q = 600 100

FL = 2 (36) + 6 Q = 6

FL = 78 (per 100 ft of hose)

FL = 78 x .52 (52 ft 100)

FL = 40.56 psi

4) Finding Friction Loss in Wyed Hose Lays

Some hose evolutions involve the use of wyed hose lays. This means that one hose is split into two or more hoselines by use of a wye or water thief appliance. In order to find friction loss in this type of evolution, we must first convert all wyed hoselines into an equivalent length of 2 ½”.

a. Converting Wyed Hoselines to Equivalent 2 ½” Lengths

To convert wyed hoselines to an equivalent 2 ½” length, we must first find the average length of the wyed hoses. This is done by taking the total length of all wyed hoselines and dividing this figure by the number of hoses being wyed.

The example above shows a typical wyed operation. To find the average length of wyed hose:

Step 1: Find total length of wyed hose

200 ft + 200 ft = 400 ft

Step 2: Divide the total length of siamesed hose by the number of

siamesed hoses.

400 ft 2 = 200 ft

b. Using rule of thumb factors to find equivalent length

To find the equivalent length of wyed hoses, we will need to multiply the average length of the wyed hoses by the rule of thumb factor. These factors are listed below and will be provided for you on the exam. You are not required to memorize the factors.

Number of Wyed Hoses Rule of Thumb Factor

2 – 1 ½” hoses 3.75

2 – 2 ½” hoses .28

3 – 2 ½” hoses .13

4 – 2 ½” hoses .08

Work Problem: Find the friction loss for the evolution shown below.

Step 1: Find the average length of the wyed hoses

200 ft + 200 ft = 400 ft

400 ft 2 = 200 ft

Step 2: Find equivalent length of wyed hoses (this is the average length multiplied by the factor)

200 ft x 3.75 = 750 ft

Step 3: Find total equivalent length of evolution (equivalent length of siamesed hose + single 2 ½” hose)

750 ft + 400 ft = 1150 ft

Step 4: Find total GPM flowing from all hoses

Total GPM is calculated by adding the GPMs flowing from each hoseline or discharge.

1 ½ hose #1 = 100 GPM 1 ½” hose #2 = 100 GPM

100 GPM + 100 GPM = 200 GPM

GPM = 200 (this figure will be used to find Q)

Step 5: Find Friction loss

FL = 2Q2 + Q Q = GPM 100

FL = 2 (2)2 + 2 Q = 200 100

FL = 2 (4) + 2 Q = 2

FL = 10 (per 100 ft of hose)

FL = 10 x 11.5 (1150 ft 100)

FL = 115 psi

Note: Finding friction loss for siamesed and wyed lines are very similar. The rule of thumb figures are the same when finding equivalent length

The only difference is when finding friction loss in wyed lines, the total gpm must be found by adding the gpm of each individual line that is discharging water.

Work Problem: Find the friction loss for the evolution shown below. A fire truck is pumping through 400 ft of 2 ½” hose that is wyed to two 2 ½” hoselines each flowing 200 gpm.

Step 1: Find the average length of the wyed hoses

200 ft + 300 ft = 500 ft

500 ft 2 = 250 ft

Step 2: Find equivalent length of wyed hoses (this is the average length multiplied by the factor)

250 ft x .28 = 70 ft

Step 3: Find total equivalent length of evolution (equivalent length of siamesed hose + single 2 ½” hose)

70 ft + 200 ft = 270 ft

Step 4: Find total GPM flowing from all hoses

Total GPM is calculated by adding the GP

2) Design & Installation per Occupancy Classification

Sprinkler systems are usually activated when a sprinkler head is exposed to extreme heat. Sprinkler heads have links or fuses that are designed to open at a pre-determined temperature. Once this temperature is reached, the link or fuse will break and water will begin to flow. Each sprinkler head has its own fuse or link, and only those exposed to the pre-determined temperature will flow water. In other words, if a fire starts in a corner of the room, only the sprinkler heads affected by the fire will activate. The heads in the opposite corner may not be exposed to enough heat to activate them.

Light hazard occupancies (low combustibility) require sprinkler heads to open when exposed to temperatures around 135 - 150. Examples of these occupancies are:

a. Churches

b. Schools

c. Office Buildings

Ordinary hazard occupancies (moderate quantity of combustibles) require sprinkler heads to open when exposed to temperatures of 160 and above.

Examples of these occupancies are:

a. Warehouses

b. Laundries

c. Manufacturing Occupancies

Extra (high) hazard occupancies require sprinkler heads to open when exposed to temperatures of 325 and above. Examples of these occupancies are:

a. Airplane Hangars

b. Occupancies dealing with explosives

c. Occupancies dealing with flammable liquids or gases

Basic Types of Automatic Sprinkler Systems

1) Wet-pipe System

Wet-pipe sprinkler systems are the most common in use today. These systems contain water under pressure to each individual sprinkler head. When a head is exposed to a pre-determined temperature, the fuse or link will break and water will begin to flow.

2) Dry-pipe System

Dry-pipe systems are usually installed in occupancies where there is a chance of the water freezing in the lines. Dry-pipe systems have air or nitrogen under pressure to each sprinkler head. The pressure in these lines is slightly above the water pressure, and this pressure difference is what keeps the water out of the sprinkler lines. When a sprinkler head is activated, the air will begin to expel, and the air pressure will drop. As the air pressure drops, water will begin to advance throughout the lines and flow through the activated heads.

3) Preaction System

Pre-action systems are usually installed in areas or occupancies that are concerned about water damage from broken or faulty sprinkler lines or heads. Water is stopped at the feeders (in the walls before the pipes supplying the sprinkler heads) by a valve. This valve is electronically activated by a heat-detecting device within the area. Once the heat-detecting device detects heat, a signal is sent to the valve and the valve opens. Water will then flow to all heads, but will only discharge through the activated heads. If a forklift or some other type of equipment breaks a sprinkler line, water will not immediately discharge because the valve is holding back the water flow and not the sprinkler heads (unlike the wet-pipe or dry-pipe systems).

4) Deluge System

Deluge systems are generally installed in hazardous areas requiring the immediate application of water. This system is very similar to the preaction system, except all sprinkler heads are open (no activating device). Once the heat-detecting device activates the valve, water will flow from all heads within the area.

Pumping Operations

Sprinkler systems are installed in accordance with building and fire codes, and therefore usually designed with adequate pressure to supply water. However, there may be circumstances when a fire pumper is needed to supplement the system. Examples are:

a. City water main broken or out of order

b. Building fire pump not working

When pumping into sprinkler systems, fire pumps may attach hoses to the sprinkler system’s siamese connection (similar to a standpipe connection). It is recommended that:

a. Initial water pressure be 100 psi

b. Minimum of 2 supply lines (2 ½”)

Most sprinkler heads have a ½” discharge opening. Each head can cover approximately 100 (10 x 10) square feet. The discharge for sprinkler heads can be found using the following formula:

Sprinkler Discharge = ½ Pressure + 15

Sprinkler Discharge = ½ P + 15

note: this formula is to find discharge per head (must multiply # of heads flowing)

Work Problem: A sprinkler system has 8 sprinkler heads activated at 40 psi. Find the total gpm (discharge)

Discharge = ½ P + 15

Discharge = ½ (40) + 15

Discharge = 20 + 15

Discharge = 35 (per head)

Total Discharge = 35 x 8 (# of heads flowing)

Total Discharge = 280 gpm

The pressure of a sprinkler head can be found using the following formula:

Sprinkler Pressure = 2 x (Sprinkler Discharge – 15)

P = 2 (Dis – 15)

Work Problem: An activated sprinkler head is flowing 35 gpm. What is the pressure of that sprinkler head?

Pressure = 2 ( Dis – 15)

Pressure = 2 (35 – 15)

Pressure = 2 (20)

Pressure = 40 psi

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