Fire 11

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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

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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)

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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 wont 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.

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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 DischargeDrafting 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).

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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

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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 = 8Hb. 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

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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 = 30d2PWork Problem: How many GPMs are flowing through a 2 hoselinewith a

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

GPM = 30d2P

GPM = 30 (1.125)250 (nozzle size converted into adecimal)

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 PWork Problem: How many GPMs are flowing through a 2 hoseline

without a nozzle attached to it at 50 psi?

GPM (open butt) = 27d2P

GPM = 27 (2.5)250

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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 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

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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 surgesbackwards. 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

fromthe 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

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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 to 3

We see how 1 X 2 = 3

Inverting the 2 we get

Now we take the and multiply it to the fifth power

5 = X X X X = 1/32

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

the

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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 100note: This formula is only for 2 diameter hoselines.

The friction loss figure represents friction loss per 100 ft of 2

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 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 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

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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

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

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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 bothfunctions

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.

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Fire Hydrants

1) Dry Barrel Hydrants (see figure 3.1)

a. Dry barrel hydrants do not have water stored in the fire hydrantitself. 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.

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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.

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1. One pumper suction hose outlet (usually 4)

2. One regular hose outlet (usually 2 )

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 of1000 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 inchesare 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 hydrants

static pressure.

ii. Open a discharge gate for one hoseline. Again look atyour suction gauge. The pressure will be lower because

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some of the static pressure will have been used for the first

firefighting line. This is the hydrants 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

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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 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



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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

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



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

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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 collapseviolently 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 Pumps1) 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 vanepriming pump unit is used to prime centrifugal pumps.

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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 stagepump.

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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 theseries 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.

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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 pumps

capacity. The transfer valve should be switched to the

parallel position when evolutions or circumstances require a

pump to deliver more than 70% of its 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.

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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 pistons 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 1600s. In the 1900s they were adapted towork 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 Hoselines

As discussed earlier, the formula for finding friction loss in a 2 fire hose

is:

FL= 2Q2 + Q; where Q= GPM 100.This formula only applies to:

a. 2 diameter fire hose per 100 ft length

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b. 2 diameter fire hose flowing 100 gpm or greater

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

100

gpm is:

FL = 2Q2 + Q; where Q = GPM 100Work Problems:

Find the friction loss in a 500 ft length of 2 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 fire hose flowing 250gpm.

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 fire hose flowing 80 gpm.

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

FL = 2 (.8)2 + (.8) Q = 80 100

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

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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 diameters.

1) Finding Equivalent Lengths

Before we can use the friction loss formula on hoses other than 2 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 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

hoselines. We will need to convert hoses of all other diameters to a 2

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 13

1 7.76

3 .4

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3 .17

4 .09

Work Problem: Using the rule of thumb factor, convert the

following to an equivalent 2 length:

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

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

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

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 .074

1 .129

3 2.5

3 5.8

4 11.0

Work Problem: Using the conversion factor, convert the

following to an equivalent 2 length:

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

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

100 ft of 4 hose = 100

11 or 9 ft of 2

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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 .

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

200 x 13 = 2600 ft or 200 .074 = 2700 ft (forthis 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 + 1)

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

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Step 4: Find Friction loss

FL = 2Q2 + Q Q = GPM 100

FL = 2 (.4)2 + (.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

hose.

a. Converting Siamesed Hoselines to Equivalent 2 Lengths

To convert siamesed lines to an equivalent 2 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.

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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 ruleof 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 hoses 3.75

2 2 hoses .28

3 2 hoses .13

4 2 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


FL = 2Q2 + Q Q = GPM 100

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

FL = 2 (25) + 5 Q = 5


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FL = 55 x 1.68 (168 ft 100)

FL = 92.4 psi


below.


600 ft + 600 ft = 1200 ft

1200 ft 2 = 600 ft



600 ft x .28 = 168 ft

Step 3: Find total equivalent length of evolution

(equivalent length of siamesed hose + single 2 hose)

168 ft + 300 ft = 468 ft


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 x 4.68 (468 ft 100)

FL = 70.2 psi

Work Problem: The fire engine below is supplying three 2

hoselines to a portable deluge flowing 600 gpm.

Find the friction loss of the hoses.

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400 ft + 350 ft + 450 ft = 1200 ft

1200 ft 3 = 400 ft



400 ft x .13 = 52 ft


FL = 2Q2 + Q Q = GPM 100

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

FL = 2 (36) + 6 Q = 6


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 .

a. Converting Wyed Hoselines to Equivalent 2 Lengths

To convert wyed hoselines to an equivalent 2 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.

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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 hoses 3.75

2 2 hoses .28

3 2 hoses .13

4 2 hoses .08


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

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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 hose #1 = 100 GPM 1 hose #2 = 100

GPM

100 GPM + 100 GPM = 200 GPM

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


FL = 2Q2 + Q Q = GPM 100

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

FL = 2 (4) + 2 Q = 2


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 isdischarging water.


below. A fire truck is pumping through 400 ft of 2 hose that is wyed to

two 2 hoselines each flowing 200 gpm.

Step 1: Find the average length of the wyed hoses

200 ft + 300 ft = 500 ft

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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



70 ft + 200 ft = 270 ft

Step 4: Find total GPM flowing from all hoses

Total GPM is calculated by adding the GPMs

flowing from each hoseline or discharge.

2 hose #1 = 200 GPM 2 hose #2 = 200

GPM

200 GPM + 200 GPM = 400 GPM

GPM = 400 (this figure will be used to find Q)


FL = 2Q2

+ Q Q = GPM 100

FL = 2 (4)2 + 4 Q = 400 100

FL = 2 (16) + 4 Q = 4


FL = 36 x 2.7 (270 ft 100)

FL = 97.2 psi

Chapter 6: Fire Ground (Field)

Calculations

General Information

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In the previous chapter we discussed how engine pressure was found using

various formulas and conversion factors. If youre sitting in a classroom

taking an exam, these methods for finding the proper engine pressure is

adequate. But, as you all know, finding the proper engine pressure is most

critical and valuable at the fire scene. Pump operators do not have the luxury

of booting up their laptop computers or pulling out a calculator during anincident. Operators need to have some sort of pre-determined method to

quickly deliver water to the hoselines with the proper pressure. This method is

termed Fire Ground Calculations or Field Calculations. These

calculations are not designed to be exact, but rather to be quick and close.

Field Calculations also require the operator to memorize certain constants

such as nozzle pressures, appliance losses, and even memorizing the friction

loss in commonly used hoselines. Below is an outline of Field Calculation

constants that will need to be memorized.

1) Basic Formula

The basic formula for Engine Pressure (EP) is:

EP= Nozzle Pressure + Friction Loss (hose) + Back Pressure + Appliance

Loss

EP = NP + FL + BP + APP

The components (NP, FL, BP, and APP) of this formula will be explainedbelow.

2) Appliance Loss Figures (APP)

Appliance Friction Loss Figure

Deluge (Turret or Deluge) 25 psi

Dry Standpipe connection 25 psi + BP

Wye or Siamese connection 5 psi

Ladderpipe 80 psi

Sprinkler System (no fire) 100 psi + BP

Sprinkler System (with fire) 150 psi + BP

3) Nozzle Pressures (NP)

Handlines Master Streams

Barrel Tip (solid stream) 50 psi 80 psi

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Fog Nozzle (fog or straight stream) 80 psi 100 psi

These nozzle pressure figures are to be used for work problems where

nozzle type, and not pressure, is given.

4) Back Pressure

The field calculation for back pressure is 5 psi per 10 of grade or5 psi

per story above the first floor.

5) Friction Loss in Hoselines

You will be required to memorize the boldface friction loss figures in the

chart below. The calculated figure is next to the field calculation figure to

show how the field calculation figure is obtained.

Nozzle Size Hose Size GPM Friction Loss per 100

feet

Desktop Field

Calculation

Calculation

1/2 - 5/8 1 30 30 30

40 47.3 45

50 68.2 70

3/4" 1 1/2 100 39 35

3/4" 1 3/4 125 34 35

150 46.6 45

200 77 75

1 2 1/2" 200 10 10

1 1/8 2 1/2" 250 15 15

1 1/4" 2 1/2" 300 21 21

1 3/8 4 500 4.95 5

1 1/2" 600 7 7

1 3/4" 800 12.2 10

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2 1000 18.9 20

Notes: 4 hoseline used as a supply line for master streams (ladder

pipe, deluge, and snorkel) and relays. Friction loss figures are for each 100length of hose Items in bold to be memorized for final

Work Problems:

Using Fire Ground calculations, find the proper EP for the following:

a) 200 of 1 hose flowing 30 gpm at 60 psi

EP= NP + FL + BP + APP

EP = 60 + (30 X 2) + 0 + 0

EP = 60 + 60

EP = 120 psi

Note: FL figures must be multiplied by the number of 100 of

hose

When there are no figures for BP or APP, use 0

b) 200 of 1 handline with a fog tip nozzle flowing 100 gpm


EP = 80 + (35 X 2) + 0 + 0

EP = 80 + 70

EP = 150 psi

b) 400 of 2 handline with a barrel tip nozzle flowing 250 gpm on

a

hill 50 above the fire truck


EP = 50 + (15 X 4) + (5 X 5) + 0

EP = 50 + 60 + 25 + 0

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EP = 135 psi

6) Siamesed Hoselines

When two or more hoselines are used to supply water to a desired point or

appliance, calculations are simplified by calculating the friction loss in theaverage length of the siamesed hoselines. Each hoseline will deliver its

equal share of water because the pressure applied by the fire pump will

equalize in the hoselines. The discharge rate (GPM) will be divided by the

number of siamesed hoselines when determining gpm for each hoseline.

The average length of the siamesed hoses is 600

total length number of hoses

(600 + 600) 2 = 600

The average flow of the hoses is 250 gpm

Total gpm number of hoses

500 gpm 2 = 250

Using fire ground calculations, we know that each 100 length

of 2 hose flowing 250 gpm has a friction loss of 15 psi.

600 100 = 6

6 X 15 = 90psi

The total friction loss in the siamesed hoses is 90 psi.

Work Problem: Using Fire Ground Calculations, find the EP of the

following evolution.


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Step 1: Find NP

The NP for a master stream using a fog nozzle is 100

NP = 100 psi

Step 2: Find FL

The average length of siamesed hoses is 400:

(400 + 350 + 450) 3

1200 3 = 400

The average flow of the siamesed hoses is 200 GPM

600 GPM 3 = 200

FL = 10 psi for every 100 of 2 hose flowing 200 GPM

FL = 10 X 4

FL = 40 psi

Step 3: Find BP

There is no BP for this problem

BP = 0

Step 4: Find APP

The appliance loss figure for a Deluge is 25 psi

APP = 25 psi

Step 5: Plug all the figures into the formula


EP = 100 + 40 + 0 + 25

EP = 165 psi

7) Wyed Hoselines

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For wyed lines of equal diameter with nozzles of the same size, the friction

loss for the average length of wyed lines will be considered. Find the

average length and treat as one line. This means that the nozzle pressure

of only one hose will be added to the NP portion of the EP formula. The

hoseline supplying the wyed lines (before the wye) must provide the total

amount of GPM to all the wyed lines. The total GPM will be used for allfriction loss calculations behind (pump side) the wye. For all calculations

in front of the wye (nozzle side of wye), use the discharge of only one

hoseline. The following example should make this a little clearer.

Work Problem: Using Fire Ground Calculations, find the EP of the

following evolution.


Step 1: Find NP

Find the nozzle pressure of only one nozzle

The NP for a handline using a fog nozzle is 80 psi

NP = 80 psi

Step 2: Find FL in the wyed hoselines

The average length of wyed hoses is 200:

(150 + 250) 2

400 2 = 200

The flow of one of the wyed hoses is 100 GPM


FL = 10 X 2

FL = 70 psi

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Step 3: Find FL in the 2 hoseline before the wye

Find the total GPM in the 2 hoseline (total of all discharge)

The total flow of all the wyed hoses is 200 GPM

100 GPM + 100 GPM = 200 GPM

The length of the 2 hose is 200


FL = 10 X 2

FL = 20 psi

Step 4: Find the total Friction loss by all hoses

Total the friction loss in the wyed lines and the 2 line

Add the results from steps 2 and 3

70 + 20 = 90

FL = 90 psi

Step 5: Find BP

There is no BP for this problem

BP = 0

Step 4: Find APP

The appliance loss figure for a Wye is 5 psi

APP = 5 psi



EP = 80 + 90 + 0 + 5

EP = 175 psi

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Do problems on worksheet provided for extra practice. Contact the

instructor if you need a worksheet or have any questions.

Remember: If you encounter a work problem that has a GPM value

not covered in the Field Calculation Chart, or if you forget a FL of a

hose, you can always find it by using 2Q2 + Q. But remember, thisformula is only for every 100 of 2 hose. Hoses of all other

diameters need to be converted to an equivalent length.

8) Supplying Multiple Hoselines

Some incidents require that hoselines of different lengths and diameters be

used simultaneously from the same fire truck. The pump operator must be

able to quickly determine the proper pump pressure for each of the

different lines. Below is an example of a common hose evolution

involving different hose diameters and lengths

The EP for each of the 1 hoselines is 150 psi

The EP for the 2 hoseline is 95 psi

What engine pressure should the operator pump?

Modern fire trucks have multiple discharge outlets, each equipped with

individual gates and pressure gauges. The pump operator would have

to set the pump speed at the pressure of the highest discharge pressure.

In the above example, the pump pressure would have to be set at 150

psi. This would give the 1 hoselines the proper pressure. As forthe 2 hoseline, the operator would have to choke down or only

partially open the gate valve to obtain the desired pressure of 95 psi. If

the 2 gate valve was fully opened, the pressure would be too high

for the hoseline, and if the pump speed was lowered to 95 psi, the

pressure would be insufficient for the 1 hoselines.

Some older fire trucks have multiple discharges, but only one pressure

gauge. This makes it very difficult when pumping multiple hoselines

requiring different pressures. One method of pumping these types of

evolutions is to take the average pressure of all hoselines and set the

pump pressure to that average. This only works if the different

pressures are moderately close. The old timers used to set the pump to

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the highest hose pressure and choke down on the other lines that

require lower pressures. They would check the pressure by stepping

on the hose and feeling for the perfect hardness.

9) Aerial Streams (Ladderpipe and Platform Operations)

Aerial streams are master stream nozzles that are elevated to heights up to

100 through use of an aerial ladder or an aerial platform. These ladders or

platforms are mounted directly onto an apparatus called a Ladder Truck or

Snorkel. Ladder trucks or Snorkel trucks are not required to have their

own pumps, although some models do. Usually, in an aerial stream

operation, the truck providing the aerial stream will set up operations in a

way that best utilizes their aerial stream. Once the aerial is in place, a fire

truck with a pump will provide water to the aerial truck. It is important for

the pumper truck to deliver the proper pressure so that an effective fire

stream is delivered.

As noted earlier in this chapter, the APP loss for a ladderpipe operation is

80 psi. This figure is only for the friction loss of components after the

supply lines and before the nozzle (siamese, hose, BP of elevation). This

means that operators will have to find the friction loss in the supply lines

(method for finding FL in siamesed lines) and find the appropriate FL for

the nozzle used. These figures will be added to the constant APP loss of

80 psi to get the engine pressure.

Work Problem: Find the EP of the above ladderpipe operation.


Step 1: Find NP

The NP for a master stream using a fog nozzle is 100 psi

(given)

NP = 100 psi

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Step 2: Find FL in the supply hoselines

The average length of wyed hoses is 200:

(200 + 200) 2

400 2 = 200

The average flow of the hoses is 300 gpm

Total gpm number of hoses

600 gpm 2 = 300

Using fire ground calculations, we know that each 100 length

of 2 hose flowing 300 gpm has a friction loss of 21 psi.

200 100 = 2

2 X 21 = 42 psi

FL = 42 psi

Step 3: Find BP

The BP for ladderpipe operations is included in the APP loss

BP = 0

Step 4: Find APP

The appliance loss figure for a ladderpipe operation is 80 psi

APP = 80 psi



EP = 100 + 42 + 0 + 80

EP = 222 psi

10) Relay Pumping

Relay pumping is used when the distance from the water supply (firehydrant) to the incident is longer than the supply lines carried by a single

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fire truck. A relay operation consists of two or more fire trucks, in

concession, providing water to the next fire truck.

Each fire truck, except for the truck pumping the firefighting lines, should

provide 20 psi residual pressure to the next fire truck. To accomplish this,

pump operators must pump 20 psi above the friction loss of the relay

hose. If the 20 psi is not added, the receiving truck will have 0 psi coming

in and will not be able to deliver any water to the next fire truck. The gpm

used for finding the friction loss is determined by the amount of water

flowing through the firefighting lines.

Engine #1: EP = FL + 20 psi

Pumping 250 gpm through 1500 of 2

FL for 2 flowing 250 gpm = 15 per 100

1500 100 = 15

FL = 15 X 15

FL = 225

Add 20 psi residual pressure

EP = 225 + 20

EP = 245 psi

Engine #2: EP = FL + 20 psi

Pumping 250 gpm through 1000 of 2

FL for 2 flowing 250 gpm = 15 per 100

1000 100 = 10

FL = 15 X 10

FL = 150

Add 20 psi residual pressure

EP = 150 + 20

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EP = 170 psi

Engine #3: EP = NP + FL + BP + APP

NP = 80 psi

FL = 60 psi (400 2 flowing 250 = 4 X 15)

BP = 0

APP = 0

EP = 80 + 60 + 0 + 0

EP = 140 psi

The quickest method in setting up a relay operation is to pump all relaying fire

trucks at the same pressure (except truck pumping firefighting lines). The pressure

used should be that of the truck with the longest hose lay. This will ensure

adequate pressure to all trucks in the relay operation. Once the relay operation is

set up, adjustments can be made to the pressure. In a relay operation it is difficult

to tell exactly how much hose is initially laid out, especially if the fire truck did not

lay out its entire compliment of hose. After water is flowing, there will be time to

fine- tune the operation.

Chapter 7: Fire Streams

1) General Information

A good fire stream will extinguish a fire in the shortest period of time with a

minimum amount of water. A good fire stream must have a sufficient amount

of volume and reach to get to the seat of the fire and cool burning materials

below their ignition temperature. Some characteristics of a good fire stream

are as follows:

a. Does not break up before reaching the fire

b. Compact enough to reach the height or distance needed

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c. Compact enough so that:

I. 90% of its volume fits within a 15-inch circle

OR

ii. 75% of its volume fits within a 10-inch circle

b. With no wind, a good fire stream should be able to enter a room

through a window and strike the ceiling with enough force to

splatter well enough to extinguish a fire (indirect attack)

There are several factors that affect a fire stream:

a. Air resistance (friction of fire stream traveling through air)

b. Gravity

c. Wind Conditions

1. Moderate tail winds will increase the horizontal reach but

will decrease the vertical reach

2. Head winds will raise the vertical reach but will shorten

horizontal reach

d. Condition of nozzle

2) Nozzle Size and Pressure

Each nozzle tip has an optimal nozzle pressure. If the pressure is

substantially lower or higher than the rated (recommended) nozzle

pressure, the fire stream produced by that nozzle will be inefficient or

ineffective. In other words, if a pump operator does not provide the proper

nozzle pressure, the fire stream produced will break up before reaching the

fire. As a general rule of thumb, nozzle pressure recommendations are as

follows:

Handlines Master Streams

Barrel Tip (solid stream) 50 psi 80 psi

Fog Nozzle (fog or straight stream) 80 psi 100 psi

note: For safety reasons handlines and master stream devices (deluge,

turret, ladder pipe) should not be pumped from the same pumper

at

the same time. If a master stream device should suddenly shut

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down, the pressure being used for the master stream device

could be

absorbed by the handlines causing them to burst or injuring the

firefighters on those handlines. Separate fire trucks should beused

for incidents requiring the simultaneous use of handlines and

master stream devices

The nozzle size must also be suited for the diameter of hoseline that is

being used. As a general rule, the diameter of the nozzle should not

exceed of the hose diameter. This means that a 1 hoseline should not

use a nozzle with a diameter of greater than .

3) Horizontal Reach

Firefighters may encounter situations requiring the use of long-range fire

streams. Some examples are:

a. Fires producing extreme heat

b. Unusual structural conditions

c. Dangerous fires

i. Flammable tanks

ii. Gas tanks

iii. Reactive materials

d. Limited access

i. Junk yards

ii. Lumber yards

iii. Brush fires

In theory, a fire stream angled at 45will produce the greatest horizontalreach. However for firefighting purposes, the maximum effective horizontal range of

a fire stream can be obtained from a fire stream angled between 30-34

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The formula for finding the horizontal reach of a fire stream is:

Horizontal Reach = x Nozzle Pressure + 26 feet

HR = NP + 26

This formula is based on a nozzle size of . For every 1/8

over

, 5 feet must be added to the 26.

Work Problem: What is the horizontal reach of a fire stream flowing

50

psi through a 1 tip (nozzle)?

HR = NP + 26 (+5 for every 1/8 over )

Step 1: Find difference in eighths between 1 and

1 (8/8) minus (6/8) = 2/8

1 is 2 eighths over

Step 2: Multiply 5 by the total number of eighths over

5 x 2 = 10

Step 3: Add the figure in step two into the formula

HR = NP + 26 + 10

Step 4: Solve problem (using adjusted formula in step 3)

HR = NP + 26 + 10

HR = (50) + 26 + 10

HR = 25 + 26 + 10

HR = 61 feet

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Work Problem: What is the horizontal reach of a fire stream flowing

60

psi through a 1 1/8 tip (nozzle)?

HR = NP + 26 (+5 for every 1/8 over )

Step 1: Find difference in eighths between 1 1/8 and

1 1/8 (9/8) minus (6/8) = 3/8

1 1/8 is 3 eighths over


5 x 3 = 15


HR = NP + 26 + 15


HR = NP + 26 + 15

HR = (60) + 26 + 15

HR = 30 + 26 + 15

HR = 71 feet

4) Vertical Reach

Firefighters may encounter situations requiring the use of long vertical fire

streams. Some examples are:

a. Multi-storied buildings

b. Hillside fires

c. Use of fire streams to disperse contaminants or smoke

In theory, a fire stream angled at 90will produce the greatest vertical

reach. However for firefighting purposes, the maximum effective vertical

range of a fire stream can be obtained from a fire stream angled between

60-75

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When using a vertical fire stream on a multi-story building, firefighters

should consider the following factors:

a. The third floor is considered the highest story that a fire stream may

be

applied effectively from the street level.

a. The fire stream should not be angled greater than 50. This isbecause

an angle is needed so that the stream may enter the building and

deflect off the ceiling towards the fire. If the angle is too steep, the

stream will not reach the fire within the structure, but rather hit the

ceiling and fall straight down.

b. If using a deluge or deck gun (turret), park the apparatus on the

opposite side of street from the fire to help achieve the effective

angle of discharge (50).

The formula for finding the vertical reach of a fire stream is:

Vertical Reach = 5/8 x Nozzle Pressure + 26 feet

VR = 5/8 NP + 26

This formula is based on a nozzle size of . For every 1/8

over

, 5 feet must be added to the 26.

Work Problem: What is the vertical reach of a fire stream flowing 40

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psi through a 1 tip (nozzle)?

VR = 5/8 NP + 26 (+5 for every 1/8 over )

Step 1: Find difference in eighths between 1 and

1 (8/8) minus (6/8) = 2/8

1 is 2 eighths over


5 x 2 = 10


VR = 5/8 NP + 26 + 10


VR = 5/8 NP + 26 + 10

HR = 5/8 (40) + 26 + 10 (convert 5/8 to decimal on

calc.)

HR = 25 + 26 + 10

HR = 61 feet

Work Problem: What is the vertical reach of a fire stream flowing 80

psi through a 1 1/8 tip (nozzle)?

VR = 5/8 NP + 26 (+5 for every 1/8 over )

Step 1: Find difference in eighths between 1 1/8 and

1 1/8 (9/8) minus (6/8) = 3/8

1 1/8 is 3 eighths over


5 x 3 = 15


VR = 5/8 NP + 26 + 15


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VR = 5/8 NP + 26 + 15

HR = 5/8 (80) + 26 + 15 (convert 5/8 to decimal on

calc.)

HR = 50 + 26 + 15

HR = 91 feet

Note: to increase horizontal or vertical range, increase pressure

by 1 psi for every foot needed. (i.e. if 10 more feet of reach is needed, increasing the

nozzle pressure by 10 psi would get the approximate distance needed)

Chapter 8: Standpipe Systems

General Remarks

1) Where required

a. Tall Buildings

b. Large Buildings

c. Special Occupancies

2) Enforcement in Hawai`i

a. Building Department Uniform Building Code

b. Fire Department Uniform Fire Code

Types of Standpipe Systems in Hawai`i

1) Dry Standpipe System (designed for Fire Department

use)

Dry standpipe systems are required in certain types of occupancies and are

installed to help provide a water supply throughout the occupancy. A dry

standpipe system provides 2 hose outlets to each floor of a building.

These outlets are connected to a pipe, called a riser, which is connected to

a siamese connection located on the street level at the front of the

building. When firefighters need water, a fire truck will have to connect toa fire hydrant (intake side of pump) and supply lines to the siamese

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connection (discharge side of pump). Once the fire truck begins

discharging water to the siamese connection, water will fill the risers and

will be distributed to all the 2 outlets. Each outlet has an individual

shut-off, and firefighters can connect their firefighting lines to the desired

outlet. Once the hoselines are connected and in place, firefighters can then

open the 2 outlets to allow water to flow through their hoselines.

The following are guidelines for dry standpipe systems:

a. Required in buildings 4 or more stories

b. Riser Size (pipe) 4 6 inches (found within stairwells)

c. Fire Department Siamese Connection

i. Located on street front of building

ii. 2 or 4 way connection for Fire Department use

d. 2 hose outlets for each riser

i. 1 per floor level (optional for 1st floor)

ii. Roof outlet requires a two-way 2 connection

2) Wet Standpipe System (designed for occupant / tenant

use)

Wet standpipe systems are similar to the dry standpipe system, but there

are a few differences. Wet standpipes have hose cabinets on each floor.

These hose cabinets contain 1 fire hose with a nozzle. In case of a fire,

tenants can open the hose cabinet, pull out the hose and then open thevalve allowing water to flow through the hose. With this having been said,

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this type of system requires that water be provided and pressurized up to

each hose cabinet at all times. Buildings can either use county water

pressure, or have some type of pressure booster, such as a pump.

The following are guidelines for dry standpipe systems:

a. Required in buildings 4 stories or more

Note: Not required in buildings equipped with an automatic

sprinkler system.

b. Riser size (pipe) 2 2

c. Outlet (Fire hose cabinet) on each floor level

d. Building fire pumps may be needed to meet flow and pressure

requirements (UL Underwriters Laboratory FM FactoryMutual)

3) Combination / Combined Systems

It is not uncommon to find occupancies having a combination of systems

for fire protection. Examples of combination systems are:

a. Combination System (Wet standpipe and Dry standpipe)

b. Combined System (Dry standpipe and Automatic SprinklerSystem)

Pumping Operations:

The following is an example of a typical dry standpipe operation. Using fire

ground calculations, figure out the engine pressure of a fire truck pumping this

evolution.

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EP= FL ( 2 hoses to siamese) 2 2-1/2 lines flowing 200 gpm 3 psi

FL (2 hose on fire floor) 1 2-1/2 line flowing 200 gpm

10 psi

FL (1 firefighting lines) 1 1-1/2 line flowing 100

gpm 35 psi

FL (Appliance for siamese)

25 psi

FL (Wye on the fire floor) 5

psi

BP (Back pressure 11 floors)

50 psi

NP (Nozzle pressure) 80

psi

Engine Pressure =

208 psi

note: The way I like to figure this one out is to break up the evolution into 3

parts.

First, figure the friction loss for the evolution on the fire floor.

Step 1: Find the friction loss in the wyed hoses

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Find the average length of the wyed hoses

100 ft + 100 ft = 200 ft

200 ft 2 = 100 ft

1 flowing 100 gpm = 35 psi / 100

FL = 35 psi

Step 2: Find friction loss in 2 hose

Total flow = 200 gpm (both 1 hoses)

2 hose flowing 200 gpm = 10 / 100

FL = 10 psi

Step 3: Find NP and Appliance loss

NP = 80 psi

Appliance Loss = 5 psi (2 to 1 wye)

Total other losses = 85 psi

Step 4: Combine figures for steps 1 - 3

35 + 10 + 85

130 psi

Second, find friction loss for hoses supplying the siamese.

Step 1: Find the friction loss in the siamesed hoses

Find the average length of the wyed hoses

100 ft + 100 ft = 200 ft

200 ft 2 = 100 ft

Find the flow for each 2 hose

200 gpm (both 1 hoses)

divided by 2 (number of siamesed hoses)

Each hose is flowing 100 gpm

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2 flowing 100 gpm = 3 psi / 100

FL = 3 psi

Step 2: Find Appliance loss

Appliance Loss = 25 psi (siamese connection)

Step 3: Combine figures for steps 1 - 2

3 + 25

28 psi

Third, find back pressure and add this figure to the totals of the above

steps.

BP = 10 x 5 (fire on 11th floor = 10 floors above ground)

BP = 50

Now we can add all the figures from the 3 parts.

Part 1 = 130

Part 2 = 28

Part 3 = 50

Total = 208 psi

Chapter 9: Automatic Sprinkler

SystemsGeneral Information

1) History

Sprinkler systems were developed around the 1850s. These early systems

were made up of perforated piping and were not automatic. In 1878, USA

saw its first automatic sprinkler system and shortly afterward, Federick

Grinnel began rapid commercial development of these automatic sprinkler

systems. Nowadays, law requires automatic sprinkler systems in certain

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occupancies. These requirements vary from local jurisdictions. Hawai`i

law requires automatic sprinkler systems in the following occupancies:

a. High Rise Buildings (required since 1972)

b. Hotels

c. Large Retail Stores

d. Hospitals

e. Large Assembly Buildings

f. Basements

g. Hazardous Storage Areas

These modern sprinkler systems are very efficient, extinguishing

approximately 96% of all fires before firefighters arrive at the scene.

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 theseoccupancies 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 160and above.

Examples of these occupancies are:

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a. Warehouses

b. Laundries

c. Manufacturing Occupancies

Extra (high) hazard occupancies require sprinkler heads to open when

exposed to temperatures of 325and above. Examples of theseoccupancies 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).

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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 willflow 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 systems 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 )

Most sprinkler heads have a 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 = Pressure + 15

Sprinkler Discharge = 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 = P + 15

Discharge = (40) + 15

Discharge = 20 + 15

Discharge = 35 (per head)

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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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