MODULE 04 HYDRAULIC CALCULATION.pdf

SECTION 01
HYDRAULICS OF
SPRINKLER SYSTEM
NFPA TRAININGS - Mehboob Shaikh(Mtech. B.Eng., CFPS, CFI,
AMIE)

Purpose of Hydraulic Calculation
• What pipe size will be best?
• Will the water supply provide the flow and pressure necessary to
control/suppress a fire?
AMIE)

How to Conduct Pipe Sizing &
Flow calculation
AMIE)

Occupancy Classification – NFPA 13
• Light Hazard
• Ordinary Hazard Group-1
• Ordinary Hazard Group-2
• Extra Hazard Group -1
• Extra Hazard Group -2
AMIE)

Role of Owners
Certificate
AMIE)

Study of Water
• Hydraulics
• The science which defines the mechanical principles of water at rest or in motion.
• Hydrostatics
• The scientific laws that define the principles of water at rest.
• Hydrokinetics
• The study of water in motion.
AMIE)

Hydraulic Focus
• Pressure • Flow
AMIE)

Pressure Types
• Atmospheric Pressure
• Caused by the weight of air, varies with altitude
• Lower at high altitudes, higher at low altitudes
• 14.7 psi at sea level
• Gage Pressure
• The actual reading on a gage, does not account for atmospheric pressure.
(psig)
• Absolute Pressure
• The sum of atmospheric pressure and gage pressure. (psia)
AMIE)

Pressure Types (continued)
• Static Pressure (Ps)
• The potential energy available within a system when no water is flowing.
• Pressure is created by elevating water above a source, or it can be created
mechanically with pumps or pressure tanks.
AMIE)

Elevation Pressure
• A cubic foot of water results in a static pressure at its
base of 62.4 lbs/ft2
• Converted to square inches a column of water 1-foot
high exerts a pressure of 0.433 lbs/in2
1 ft
62.4 lbs/ft2
1 ft
0.433 lbs/in2
AMIE)

Elevation Pressure (continued)
• Pressure (psi) = 0.433 X Elevation (ft)
5 ft
15 ft
What is the pressure difference?
AMIE)

Elevation Pressure Example (Continued)
• What is the pressure at the hydrant?
Pressure (psi) = 0.433 x Elevation (ft)
psi
91
P
psi
90.93
P
ft
200
x
0.433
P

=
=
P=?
200 ft
6 ft
AMIE)

Elevation Pressure 2nd Example
• How high is the water?
ft
109
ft
108.55
Elevation
0.433
47psi
ft)
Elevation(
0.433
si)
Pressure(p
ft)
Elevation(
ft)
Elevation(
0.433
si)
Pressure(p

=
=
=

=
P=47 psi
? ft
6 ft
AMIE)

• Residual Pressure (PR)
• The pressure at a given point in a conduit or appliance with a specific volume
of water flowing.
AMIE)

• Normal Pressure (PN)
• The pressure created on the walls of pipe
or tanks holding water.
• This is the pressure read by most gages.
• Velocity Pressure (PV)
• The pressure associated with the flow of
water measured in the same direction as
the flow.
PN
PV
AMIE)

Calculating Velocity Pressures
Pn = Pt – Pv
Where:
Pn = normal pressure (psi)
Pt = total pressure (psi)
Pv = velocity pressure (psi)
Velocity pressure can be found as follows:
4
i
2
v
d
0.001123Q
P =
AMIE)

Using Velocity Pressure
• When velocities are high in a closed system the pressure needs to be
accounted for in the calculations
• It can reduce the flows and pressures needed in a system 5-10
percent
• In most sprinkler systems velocities are low and their pressures create
a minor effect, therefore velocity pressures can be ignored.
• It should be used at points where large flows take a 90-degree turn in
the piping.
AMIE)

Flow (Q)
• The quantity (of water) which passes by a given point in a given
period of time
• Generally measured in gallons per minute (gpm) or cubic feet per
second (ft3/sec)
• Uses the term “Q” in most equations
AMIE)

Flow Equation
Q = A x V
• Q = flow in ft3/sec
• A = cross sectional area of pipe in ft2
• V = water velocity in ft/sec
• Q is a constant for any given closed system.
AMIE)

Flow Equation (continued)
Q = A x V = constant flow
When the pipe size changes flow remains
constant:
Q = A1 x V1 = A2 x V2
A1 x V1 = A2 x V2
X gpm
AMIE)

Flow Example 1
• If water is flowing at 5.7 ft/sec in 6-inch pipe, how fast is it flowing when
the pipe size is reduced to 3-inch?
?
6-inch
3-inch
5.7
ft/s
NFPA TRAININGS - Mehboob Shaikh(Mtech. B.Eng., CFPS,
CFI, AMIE)

Flow Example 1 Solution
2
2
1
1 V
A
V
A =
2
1
1
2
A
V
A
V =
How fast is it flowing when the pipe size is reduced to 2-
inch?
A1 =  r2 =  (3 in)2 = 28.3 in2
A2 =  r2 =  (1.5 in)2 = 7.1 in2
22.7ft/sec
7.1in
c)
)(5.7ft/se
(28.3in
V 2
2
2 =
=
?
6-inch
3-inch
5.7
ft/s
CFI, AMIE)

Flow from an Outlet
• Dependent upon a number of factors
• Size of the orifice
• Construction of the device
• Material used in the device
• Other components near the device (e.g. screens)
• For a sprinkler, that ability is determined experimentally in a
laboratory
AMIE)

Flow from an Outlet (continued)
• Where:
• Q is the flow (gpm)
• di is the diameter of opening (inches)
• Pv is the measured velocity pressure (psi)
• CD is the discharge coefficient of the device
• This is used when testing water supplies to determine the amount of
flow
D
v
2
i C
P
d
29.83
Q 


=
AMIE)

Flow from a Sprinkler
Where:
Q is flow (gpm)
k is k-factor determined in the sprinkler listing
(gpm/psi½)
P is the pressure (psi)
• The diameter of the opening and discharge
coefficient are incorporated into the empirical
determination of k-factor.
P
k
Q 
=
AMIE)

Sprinkler Flow Example
• A sprinkler is being installed with a k-factor of 5.6. If the pressure at
the sprinkler is 20 psi, how much water will exit the sprinkler?
gpm
25.0
Q
psi
20
5.6
Q
P
k
Q
=
=
=
AMIE)

Flow from a Sprinkler (continued)
• The flow equation can be rearranged to solve for pressure or k-factor:
P
k
Q 
=
P
Q
k =
2
k
Q
P 





=
AMIE)

Pressure Calculation Example
• What is the pressure for a sprinkler that has a k-factor of 17.6 and the
expected flow is 83 gpm?
2
17.6
gpm
83
P 





=
( ) psi
22.2
4.716
P
2
=
=
2
k
Q
P 





=
AMIE)

K-factor Calculation Example
• What is the K-factor for an outlet that is flowing 65 gpm at 30 psi?
•
psi
30
gpm
65
k =
11.86
5.48
65
k =
=
P
Q
k =
AMIE)

Friction Loss (PL)
• Occurs when water flows in pipes, hoses, or other system devices
• Caused by water in contact with walls
• Used to account for losses in energy from water making turns or
traveling difficult paths
AMIE)

Formulas for Calculating Friction Loss
• Hazen-Williams formula
• Fire sprinkler systems
• Water-spray systems
• Darcy-Weisbach formula
• Anti-freeze systems
• Water mist systems
• Foam-water systems
• Fanning formula
AMIE)

Hazen-Williams Formula
• Most common for sprinkler calculations
• Assumes water is at room temperature but is still accurate with
temperature variations
• Based on C-factor, flow, and pipe size
• Calculates the amount of friction loss in ONE FOOT of pipe
AMIE)

Hazen-Williams Formula
• Where:
• PL = friction loss (psi/ft)
• Q = flow (gpm)
• C = roughness coefficient (based on pipe material)
• di = interior pipe diameter (inches)
4.87
i
1.85
1.85
L
d
C
4.52Q
P =
AMIE)
You can see the above equation that if Q is raised to the power of 1.85 in the above equation this has the effect that if the flow is
doubled and all other things remain constant the friction loss will increase by almost 4 times if the flow was to triple the friction loss
would almost be 9 times greater. You can also see that the pipe diameter D is raised to the power of 4.87 and that it's in the
denominator on the right-hand side of the equation. Therefore any increase in the pipe size will reduce the friction loss if all other
factors remain the same. If the diameters double, the friction loss will be reduced by almost a factor of 1/32 likewise if the pipe diameter
is tripled The friction loss would be reduced to about 1/243 of its original value.

Roughness Coefficient
Table 22.4.4.7 Hazen-Williams C Values
Pipe or Tube C Value
Unlined cast or ductile iron 100
Black steel (dry systems including pre-action) 100
Black steel (wet systems including deluge) 120
Galvanized (all) 120
Plastic (listed, all) 150
Cement-lined cast or ductile iron 140
Copper tube or stainless steel 150
Asbestos cement 140
Concrete 140
AMIE)

Inside Diameters (di)
List for steel and copper in Table A.6.3.2 and Table A.6.3.5
Nominal
Pipe Size
Schedule
40
Schedule
10
Type K
Copper
CPVC*
1-inch 1.049 1.097 0.995 1.101
1 ¼-inch 1.380 1.442 1.245 1.394
1 ½-inch 1.610 1.682 1.481 1.598
2-inch 2.067 2.157 1.959 2.003
2 ½-inch 2.469 2.635 2.435 2.423
3-inch 3.068 3.260 2.907 2.95
4-inch 4.026 4.260 3.857 N/A
AMIE)

Hazen-Williams Example
If a pressure gage is reading 40 psi at one end of a 32-foot section of 2-
inch schedule 40 pipe (C = 120) flowing at 110 gpm, what will a gage at
the other end read?
40 ?
2-inch schedule 40 pipe
32 ft
4.87
1.85
1.85
4.87
i
1.85
1.85
L
(2.067in)
(120)
m)
4.52(110gp
d
C
4.52Q
P =
=
PL= 0.112 psi/ft
AMIE)

Hazen-Williams Example (continued)
• What will a gage at the other end read?
• PL = 0.112 psi/ft
• Friction Loss = 0.112 psi/ft x 32 ft = 3.6 psi
• Gage Pressure = 40 psi – 3.6 psi  36 psi
40 ?
2-inch schedule 40 pipe
32 ft
AMIE)

Fittings
• Energy losses through fittings are caused by turbulence in the water
• To determine losses through fittings “equivalent length” is used
• NFPA has a table to provide equivalent pipe lengths
• Table is based on schedule 40 steel in a wet pipe system with C Values
of 120.
AMIE)

Equivalent Length Chart
Fittings & Valves Fittings & Valves Expressed in Equivalent Feet of Pipe
¾ in 1 in
1 ¼
in
1 ½ in 2 in 2 ½ in 3 in 3 ½ in 4 in 5 in 6 in 8 in 10 in 12 in
45° Elbow 1 1 1 2 2 3 3 3 4 5 7 9 11 12
90° Standard
Elbow
2 2 3 4 5 6 7 8 10 12 14 18 22 27
90° Long Turn
Elbow
1 2 2 2 3 4 5 5 6 8 9 13 16 18
Tee/Cross 3 5 6 8 10 12 15 17 20 25 30 25 50 60
Butterfly Valve - - - - 6 7 10 - 12 9 10 12 19 21
Gate Valve - - - - 1 1 1 1 2 2 3 4 5 6
Swing Check - 5 7 9 11 14 16 19 22 27 32 45 55 65
AMIE)

Adjusting Equivalent Lengths
• NFPA 13 table is based on schedule 40 steel pipe for a wet system
• All others need to be adjusted for:
• Change in pipe material
• C-factor other than 120
• Change in interior diameter
• Other than those for schedule 40 steel
AMIE)

Adjusting for C-Factor
Table 22.4.3.2.1 C Value Multiplier
Value of C 100 130 140 150
Multiplying
Factor
0.713 1.16 1.33 1.51
• Begin with the equivalent length value from the table
• Multiply the length by the factor above for the
appropriate C-factor
AMIE)

Adjusting for Inside Diameter
• Begin with the equivalent length value from the table
• Multiply the length by the factor above calculated for
the inside diameter of the pipe being used
4.87
diameter
inside
Pipe
Steel
40
Schedule
diameter
inside
Actual
Factor 







=
AMIE)

Fittings (continued)
• All fittings must be accounted for in the calculations
• Including tees, elbows, valves, etc.
• Some may have pressure loss or equivalent length values from manufacturer’s
listing information
• Special provisions:
• Fittings connected directly to sprinklers
• Fittings where water flows straight through without changing direction
AMIE)

Equivalent Length Exercise
• What is the equivalent pipe length of Type K copper tube which used
for a 3-inch standard turn 90-degree elbow?
AMIE)

Equivalent Length Exercise Solution
• NFPA 13 Table 23.4.3.1.1 :
• 3-inch 90-degree elbow = 7 ft of pipe
• Adjustments are needed for:
• Type K Copper
• Interior diameter
AMIE)

(continued)
• Adjustment for material (C-factor)
• Copper has a C-Factor of 150
• Per Table 22.4.3.2.1: Multiplier = 1.51
• Adjustment for inside diameter
• 3-inch copper has an inside diameter of 2.907-inch
77
.
0
068
.
3
907
.
2
87
.
4
=






=








=
4.87
i.d.
Pipe
Steel
40
Schedule
i.d.
Actual
Factor
AMIE)

(continued)
• Apply the factors:
• Equivalent pipe length per Table 22.4.3.1.1 = 7 ft
• Adjustment for C-factor = 1.51
• Adjustment for diameter = 0.77
• The equivalent length for a 3-inch Type K Copper standard turn elbow
is:
7 ft x 1.51 x 0.77 = 8.14 ft
AMIE)

Concept of Storage Sprinklers
• A precise design approach is necessary to meet the challenges of
these occupancies, an effort that includes using storage fire
sprinklers.
CMDA CMSA ESFR RACK
49
NFPA 13 Training- Mehboob Shaikh(M Tech. | B.Eng. | AMIE |
CFPS | CFI)

50
CFPS | CFI)

CMDA: control-mode density area sprinklers
• The “control-mode” in a control mode density area (CMDA) sprinkler
refers to the fact that these heads provide wetting and cooling to
control a fire until first responders can arrive. “Density-area” refers to
how CMDA systems are designed.
• A set of density-area curves specifies the amount of water flow
required for a given area. The system and its water source are
engineered based on them to provide the necessary flow and
pressure.
51
CFPS | CFI)

• The “control-mode” in a control mode density area (CMDA) sprinkler
refers to the fact that these heads provide wetting and cooling to
control a fire until first responders can arrive. “Density-area” refers to
how CMDA systems are designed.
• A set of density-area curves specifies the amount of water flow
required for a given area. The system and its water source are
engineered based on them to provide the necessary flow and
pressure.
52
CFPS | CFI)

• Does that mean CMDA sprinklers are almost identical to average fire
sprinklers ?
• Yes in Both Shape and Functions
• The two things separating a CMDA sprinkler from a regular sprinkler
are larger K-factors and higher temperature ratings.
53
CFPS | CFI)

CMSA: Control-mode special application
sprinklers
• Like CMDA sprinklers, control-mode special application (CMSA)
sprinklers are designed for “control” functions: wetting and cooling to
prevent fire spread. And like most storage sprinklers, CMSA sprinklers
usually have large K-factors. However, they differ from CMDA
sprinklers in two ways.
54
CFPS | CFI)

CMSA: Control-mode special application
sprinklers
1.CMSA sprinklers have unique sprinkler deflectors that produce
different water droplet sizes and spray patterns. This makes CMSA
sprinklers suited to “special applications;” in other words, high-
challenge storage occupancies. The unique deflector also requires the
sprinklers to carry a different listing.
2.CMSA sprinklers don’t use the design concepts employed by CMDA
sprinklers. No density/area curves are involved. Instead, different
variables are analyzed to calculate the necessary flow and pressure
for a given system
55
CFPS | CFI)

ESFR: Early-suppression fast response
sprinklers
• Unique among fire sprinklers, early-suppression fast-response (ESFR)
sprinklers provide fire suppression instead of fire control. The goal of
ESFR sprinklers isn’t preventing fire spread until firefighters can fully
extinguish it. Rather, they are meant to activate quickly and attack a
fire directly. The idea is that early fire suppression requires less total
water, allows less fire spread, and ultimately makes sprinkler systems
less expensive
56
CFPS | CFI)

ESFR: Early-suppression fast response
sprinklers
• ESFR sprinklers have large K-factors and feature uniquely designed
deflectors meant to produce large, high-momentum droplets that won’t
evaporate before penetrating a fire plume. Another major distinguishing
characteristic of ESFR sprinklers is fast-response elements designed to
operate sooner than standard sprinkler elements.
57
CFPS | CFI)

In-rack fire sprinklers
• In-rack sprinklers work just like their name suggests: sprinkler risers
and branch pipes are installed with storage racks to put sprinklers as
close to potential fires as possible. Of course, doing so creates
unique infrastructure challenges for warehouses—storage racks
become more permanent when plumbing is involved. But it solves the
issues posed by height and obstruction.
• Various ESFR and Control Mode ceiling level sprinklers introduced
since 1980 have led to ceiling-only protection dominating the storage
sprinkler market. But as storage buildings have evolved, the
limitations of ceiling-only sprinkler protection have become apparent
58
CFPS | CFI)

SECTION 02
HYDRAULIC CALCULATION
USING NFPA 13
AMIE)

Design Elements
AMIE)

Design Elements- System Layout
AMIE)

Tree System
Riser
Cross Main
Branch lines and sprinklers are
fed from only one direction
AMIE)

Grid System
Allows smaller cross mains and
branch lines since each sprinkler
is fed by at least two paths.
AMIE)

Design Elements- System Layout –
Loop System
Allows smaller cross mains
because each branch line is fed
from two directions.
AMIE)

Hydraulic Calculation Method
1. Define the Hazard
2. Analyze the Structure
3. Analyze the Water Supply
4. Select the Type of System
5. Select the Sprinkler Type(s) and Locate Them
6. Arrange the Piping
7. Arrange Hangers and Bracing (where needed)
8. Include System Attachments
9. Hydraulic Calculations
10. Notes and Details for Plans
11. As-Built Drawings
AMIE)
The Layout Process

Density/Area Method
• Density is the flow of water that lands in a single square foot under
the sprinkler
• Measured in flow divided by unit area
• English units: gpm/ft2
• Flow required from a sprinkler is calculated by multiplying selected
density by the coverage area
AMIE)

Density/Area Curves
AMIE)

Density/Area Example 1
• A sprinkler system has been installed with standard spray sprinklers
spaced 10 feet by 11 feet 6 inches apart. If this is an Ordinary Hazard
Group 2 occupancy and the discharge density is 0.2 gpm/ft2, what is
the minimum required flow from a sprinkler?
• Coverage Area:
A = 10 ft x 11.5 ft = 115 ft2
• Density times area equals flow:
0.2 gpm/ft2 x 115 ft2 = 23 gpm
AMIE)

Density/Area Method (continued)
• Fire Rectangle: “…the design
area shall be a rectangular
area having a dimension
parallel to the branch lines at
least 1.2 times the square
root of the area of sprinkler
operation used…”
• Different remote area
geometry may be required
by other authorities.
Area
Design_
2
.
1
CFI, AMIE)

Density/Area Curves
• Total Number of Sprinklers to Calculate
• Design Area ÷ Area Per Sprinkler
• Number of Sprinklers per Branch Line
S
Area
Design
1.2
Where:
S is the distance between sprinklers on the branch line
AMIE)

Density/Area Example 2
• The sprinkler system in an OH2 occupancy has a discharge density of 0.2
gpm/ft2 over 1500 ft2 (selected from Figure 11.2.3.1.1), each sprinkler covers 115 ft2,
how many sprinklers will be in the design area?
1500 ft2 ÷ 115 ft2 = 13.04
14 sprinklers
• If sprinklers along the branch line are 10 ft apart, how many sprinklers/line
are calculated?
line
sprinkler/
5
4.65
10ft
1500ft
1.2 2

=
AMIE)

Density/Area Example 2 (continued)
Which sprinklers on the 3rd line should be added?
1
7
6
5
4
3
2
10
9
8
A
E
D
C
B
AMIE)

Area Adjustments
• Dry-Pipe Systems
• Increase area 30% (Section 11.2.3.2.5)
• Double Interlock Pre-action Systems
• Increase Area 30% (Section 11.2.3.2.5)
• Extra Hazard Occupancy with High Temperature Sprinklers
• Decrease Area 25%, but minimum of 2000 ft2 (Section 11.2.3.2.6)
AMIE)

Area Adjustments (continued)
• Quick Response Sprinklers (11.2.3.2.3)
• Area of operation can be reduced 25 to 40% depending on ceiling height
when:
• Wet pipe system only
• Light or ordinary hazards
• 20 ft maximum ceiling height
• No unprotected ceiling pockets
• No less than 5 sprinklers in design area
• Area may be less than 1500 ft2
AMIE)

Quick Response Area Adjustment
0
5
10
15
20
25
30
35
40
0 5 10 15 20
Ceiling Height (ft) v. %
Reduction
• Ceiling Height <10 ft
• Reduction is 40%
• Between 10 and 20 ft
• Y = (-3x/2)+55
• Ceiling Height is 20 ft
• Reduction is 25%
• Over 20 ft Ceiling Height
• No reduction allowed
AMIE)

Area Adjustments (continued)
• Sloped Ceilings
• Area of operation is increased by 30% if pitch exceeds 2 in 12 (rise in run). This is
an angle of 9.46°
rise
run
CFI, AMIE)

Multiple Adjustments Example 1
• Compound adjustments based on original area of operation selected
from Figure 11.2.3.1.1.
• Dry-pipe system installed under slope of 4 in 12
• 30% increase for dry system
• 30% increase for slope
• Using 1500 ft2 as the selected operation area
• 1500 ft2 x 1.3 x 1.3 = 2535 ft2 design area
• There is no change in the density.
AMIE)

Multiple Adjustments Example 2
• Compound adjustments based on original area of operation selected
from Figure 11.2.3.1.1.
• QR sprinklers under 3 in 12 slope, ceiling height is 20 ft
• 25% decrease for ceiling height
• 30% increase for slope
• Using 1500 ft2 as the operation area
• 1500 ft2 x 0.75 x 1.3 = 1463 ft2 design area
• There is no change in density.
AMIE)

Why examine only the most remote
area?
The logic of examining only the set number of heads at the most
remote portion of the building is as follows:
Keeping pipe diameters and minimum water flow requirements the same
throughout the building, because of physics and hydraulics, if the minimum
required water pressure and gpm are met at the most remote section of the
building, as you move closer to the riser, water pressure and gpm will
automatically be greater
AMIE)

City Water Mains
• Information from the local water authority
• Flow testing near the site
• Need the following information:
• Static Pressure
• Residual Pressure
• Residual Flow
AMIE)

Water Supply Summary
• If the system demand is NOT within the capacity of the water supply,
alterations are need to the supply or to the system
• If the supply is too low on flow:
• arrange a secondary water source (e.g. tank, lake, pond, etc.)
• If the supply is too low on pressure:
• install a fire pump
• use larger pipe to reduce friction loss
• maintain higher water level in an elevated tank
• install tank at higher elevation
AMIE)

Step-by-Step Calculations
1. Identify hazard category
2. Determine sprinkler spacing
3. Determine piping arrangement
4. Calculate amount of water needed per sprinkler
5. Calculate number and location of open sprinklers in the
hydraulically most demanding area
6. Start at most remote sprinkler and work towards the water supply
calculating flows and pressures
7. Compare demand with supply
AMIE)

Example
• Ordinary Hazard Group 2 occupancy
• 12 ft ceiling height
• Quick Response standard spray sprinklers with a k-factor of 5.6
• Wet pipe sprinkler system
• Sprinklers on 10 ft x 12.5 ft spacing
AMIE)

Example: Plan View
80
ft
123 ft
38
ft
10 ft
12.5 ft
5 ft
5 ft
5
ft
5.5
ft
AMIE)

Example (continued)
1. Select hazard category: OH2
2. Determine sprinkler spacing: 10 ft x 12.5 ft
3. Determine piping arrangement: Done
4. Calculate amount of water per sprinkler
a) Select Density/Area Method
b) Pick point from density/area curve: 0.2 gpm/ft2 over 1500 ft2
c) 0.2 gpm/ft2 x 125 ft2 = 25 gpm/sprinkler
AMIE)

Example (continued)
1. Select hazard category: OH2
2. Determine sprinkler spacing: 10 ft x 12.5 ft
3. Determine piping arrangement: Done
4. Calculate amount of water per sprinkler: 25 gpm
5. Calculate number & location of open sprinklers
a) Area Adjustment(s):
QR Reduction: % = (-3x/2) + 55 = [-3(12)/2] + 55 = 37%
1500 ft2 x 0.63 = 945 ft2
b) 945 ft2 ÷ 125 ft2 per sprinkler = 7.56 = 8 sprinklers
c) 1.2(945)0.5/10 = 3.7 = 4 sprinklers per branch line
AMIE)

Example: Hydraulic Remote Area
80
ft
123 ft
38
ft
10 ft
12.5 ft
5 ft
5 ft
5
ft
5.5
ft
AMIE)

Determine the Starting Pressure
• Most remote sprinkler needs 25 gpm
• Sprinkler k = 5.6
• Starting information for the first sprinkler:
• 25 gpm at 19.9 psi
• Next work back to the water supply adding pressure
losses and flows throughout the system
19.9psi
5.6
25
k
Q
P
2
2
=






=






=
AMIE)

Information Needed for Calculations
• Select initial pipe sizes
• Locate nodes on all places where:
a. Flow (Q) takes place,
b. Type of pipe or system changes (C), and
c. Diameter (di) changes.
• Layout calculation paths starting with primary path then attachment
paths
• Fill in hydraulic calculation sheets
AMIE)

Hydraulic Calculation Paths
38
ft
10 ft
12.5 ft
5 ft
5 ft
5
ft
5.5
ft
Locate the system nodes:
1
2
3
4
5
6
7
8
BL1 BL2
TOR
AMIE)

Full System Hydraulic Calculation
• An electronics factory is being built.
• Water supply tests were done near the site and produced the
following information:
• Static pressure = 90 psi
• Residual pressure = 60 psi
• Flow at 60 psi = 1000 gpm
AMIE)

The Layout Process
AMIE)

Identify the Hazard
• In accordance with NFPA 13 hazard classifications, an electronics
factory is classified as an Ordinary Hazard Group I occupancy.
AMIE)

The Layout Process
completed
AMIE)

Flow (gpm)
Available Water Supply
Water Supply
Pressure
(psi)
100
Flow Test
Summary Sheet
0
60 psi residual
pressure at 1000 gpm
90 psi static
pressure
20
80
120
110
100
90
70
60
50
40
30
10
200 300 400 500 600 700 800 900 1000 1100
AMIE)

System Details
• Type of System:
• Wet pipe system
• Type of Sprinkler: TY3121
• Standard spray quick response upright sprinkler with a K-factor of 5.6
• Typical Sprinkler Spacing:
• Sprinklers are 10 ft apart on the branch lines, and 12.5 ft between branch
lines
AMIE)

200 ft
Electronics Factory Plan View
100
ft
53
ft
5 ft from north wall and 6 ft from west wall
12.5
ft
10
ft
Branch lines are
Schedule 40
Mains are
Schedule 10
N
5 ft from south wall and
6.5 ft from east wall
AMIE)

All branch lines
are on a 1 ft
riser nipple
Electronics Factory Elevation View
N
15
ft
5
ft
Gate Valve
Alarm Check Valve –
Viking J-1
Riser is 1 ft away from
the east wall.
Long Turn
Elbow
12.5 ft between
branch lines
18
ft
4-inch PVC
(ID – 4.240 inches)
42 ft
7
ft
3-inch
Schedule 10
AMIE)

Electronics Factory Isometric View
N
15
ft
1 ft riser nipple
1 ½-inch
1 ½-inch
1-inch
1 ¼-inch
1 ¼-inch
1 ½-inch
main and riser
3-inch
Underground
4-inch PVC
AMIE)

Select a Design Approach
• Use the density/area method
• A point from the density/area curves need to be selected
AMIE)

Check for Area Adjustments
• Quick response sprinklers (in light hazard or ordinary hazard with wet
pipe system, reduce design area based on maximum ceiling height,
where it is less than 20 ft)
• Original design area, from the area/density curve, is 1500 ft2.
• Wet pipe system, ordinary hazard, and a ceiling height of 18 ft
AMIE)

Area Reduction For QR Sprinklers
0
5
10
15
20
25
30
35
40
0 5 10 15 20
Ceiling Height (ft) v. %
Reduction
• y = % reduction in area
• x = ceiling height
28%
55
2
54
y
55
2
3(18)
y
55
2
3x
-
y
=
+
−
=
+
−
=
+
=
AMIE)

Design Area (continued)
• Starting with 1500 ft2 design area
• Applying the 28% reduction in area:
100% - 28% = 72%
1500 ft2 * 0.72 = 1080 ft2
• New design area is 1080 ft2
• Density remains at 0.15 gpm/ft2
AMIE)

Design Area (continued)
• Design area is 1080 ft2
• Each sprinkler, spaced 10 ft x 12.5 ft, is covering 125 ft2
• How many sprinklers are in the design area?
• 1080 / 125 = 8.64 sprinklers = 9 sprinklers
AMIE)

Forming the Design Area
• Continue to add branch lines until 9 sprinklers are
included
lines
branch
the
to
parallel
Area
Design
1.2
sprinklers
4
3.94
ft
10
ft
1080
1.2 2

=
AMIE)

200 ft
Remote Area
100
ft
53
ft
12.5
ft
10
ft
Branch lines are
Schedule 40
Mains are
Schedule 10
N
1
2
3
4
5
6
7
8 9
AMIE)

1. Select initial pipe sizes
2. Locate nodes on all places where:
a) Flow (Q) takes place,
b) Type of pipe or system changes (C), and
c) Diameter (di) changes.
3. Layout calculation paths starting with primary path then
attachment paths
4. Fill in hydraulic calculation sheets
completed
AMIE)

N
15
ft
1 ft riser nipple
1 ½-inch
1 ½-inch
1-inch
1 ¼-inch
1 ¼-inch
1 ½-inch
main and riser
3-inch
Underground
4-inch PVC
1
2
3
4
5
6
7
8 9
TOR
FF
Node Locations - Isometric
AMIE)

• Select initial pipe sizes
• Locate nodes on all places where:
• Flow (Q) takes place,
• Type of pipe or system changes (C), and
• Diameter (di) changes.
• Layout calculation paths starting with primary path then attachment
paths
• Fill in hydraulic calculation sheets
AMIE)

Balancing Flows
AMIE)

SECTION 03
USING ROOM DESIGN
METHOD
AMIE)

Room Design Method
The concept of the Room Design Method is that all
sprinklers within a room will operate, along with some
sprinklers in adjacent rooms of light hazard occupancy,
if the openings aren’t protected
Cl. 11.2.3.3

Room Design Method
Cl. 11.2.3.3.1 through 11.2.3.3.7
• Room design method allows for
calculation of the sprinklers in the Largest
room, so long as the calculation produces
the greatest Hydraulic demand among
selection of rooms and communicating
spaces.
115
Cl. 11.2.3.3
Communicates through unprotected openings with
three rooms.
Not necessarily this room is the largest in hydraulic
demand

Room Design Method
• Corridors are rooms and should be considered as
such.
• Walls can terminate at a substantial suspended
ceiling and need not be extended to a rated floor
slab above for this section to be applied.
• When opting for this approach, the designer must
use the room that is also the most hydraulically
demanding in terms of water supply and pressure
• The enclosing walls of such a room must provide a
fire resistance rating equal to the water supply
duration requirements
116
Cl. 11.2.3.3
Table 11.2.3.1.2
Light Hazard 30 minutes
Ordinary Hazard 60 – 90 minutes
Extra Hazard 90 – 120 minutes

Room Design Method (continued)
• Light Hazard
• Doors must have automatic or self closers, or
• Calculations must include two sprinklers from each adjoining space for no opening
protection
• A minimum lintel of depth of 8 in. (200 mm) is required for openings and the
opening shall not exceed 8 ft. (2.4 m) in width
• It shall be permitted to have a single opening of 36 in. (900 mm) or less without a
lintel, provided there are no other openings to adjoining spaces.
• Ordinary and Extra Hazard
• Doors must have automatic or self closers with appropriate fire resistance ratings.
• Corridors/Narrow Rooms
• When protected with a single-row of sprinklers, calculate maximum of 5 sprinklers
or 75 feet
117
The selection of the room and communicating space sprinklers
to be calculated shall be that which produces the greatest
hydraulic demand

Room Design Method Example
Which room is the most demanding? Light Hazard, no door closers
118
1
9
8
7
6
5
4
3
2
10
11
12 14
13
15
TO APPLY ROOM DESIGN
METHOD
All communicating rooms must
have fire rating equal to water
supply duration as per table
11.2.3.1.2 including between
the corridor and the adjacent
apartment suites/dwelling units
and INTERIOR WALL WITHIN the
unit

Sprinkler
Calculated
Remarks
R1 R2 R3 R4 R5 R6 R7 R8 R9 R10 R11 R12 R13 R14 R15
1 6 2 2 2 10
2 2 2 2 2 6
3 8 3 2 2 12
4 4 2 2 1 7
5 4 4 2 2 1 2 11
6 7
Corridor 5
In Accordance with clause
11.2.3.3.6
7 1 2 2 2 5
8 1 1 1 2
9 1 2 2 1 2 6
10 3 4 2 2 1 2 10
11 3 6 2 2 1 1 1 2 12
12 1 1 2 3
13 1 1 2 3
14 1 1 2 3
15 6 3 2 1 2 11
Nos. of Sprinkler from adjoining Communicating Spaces
Room
No.
Sprinklers
in Room
No. of Unprotected
openings
Corridor

Limitations of Room Design method
• All Rooms shall be enclosed with walls having a fire resistance equal
to water supply duration as per the given table
• The Annex states that “walls may terminate at a substantial
suspended ceiling”, leaving it up to the Authority Having Jurisdiction
to decide what “substantial” means
• This method is really intended for the Light Hazard occupancy where
a high degree of compartmentation is more common
• While requirements are set forth for walls and horizontal openings,
there is no mention of vertical protection if you have a multi-story
building
CFPS Exam Preparation- Mehboob Shaikh(CFPS, CFI, AMIE)

Limitations of Room Design method
• If there were some kind of unprotected vertical opening in the room,
perhaps an open dumb waiter, Do not use the room design method.
• With the room design method, a careful review of the original
sprinkler design is needed when Walls are moved or eliminated.
CFPS Exam Preparation- Mehboob Shaikh(CFPS, CFI, AMIE)

SECTION 04
USING NFPA 13D
AMIE)

NFPA 13D DESIGN APPROACHES
▪ Up to (2) head design Listed Flows and Pressures
▪ 10-minute water supply
NFPA 13D Training- Mehboob Shaikh(M Tech. | B.Eng. | AMIE |
CFPS | CFI)
123

EXAMPLE PROJECT : SOLUTION
CFPS | CFI)
124
Garage Dining/Living Room
Bedroom-1
Bedroom-3 Bedroom-2
Bath
Kitchen
C
C
C
C
30’
50’
20’
18’ x 12.5’
12’ x 12.5’
C
2.5’ x 7.5’
8’ x 12.5’
9.5’
3’
12 x 12.5’
13’ x 13.5’
9.5’
4’
Fireplace
Range Tub/shower
False 4" x 10" Beams
(Dropped Fluorescent Ceiling)
2.5’
2.5’

STEP 01 : Determine if NFPA 13D is applicable ?
Is this a one and two-family dwelling?
Since it is a single-family home
Yes
CFPS | CFI)
125

STEP 02 : Selection of Sprinkler
CFPS | CFI)
126

STEP 03 : Which Area Require Protection?
CFPS | CFI)
127
Will this closet require a sprinkler? Section
8.3.3
• Sprinkler not required in clothes closets
– cloth closet
• Area does not exceed 24 SF – 23.75 SF
• Non or limited combustible finish –
Drywall & Ceiling

CFPS | CFI)
128
How many Sprinkler bedroom-1 will require
Bedroom 1 requires one sprinkler, since its
area is less than 14 ft x 14 ft. The same
applies to Bedrooms 2 and 3.
No. of Sprinklers = 01 Nos.

CFPS | CFI)
129
Will this bathroom require a sprinkler? Section 8.3.2
Sprinkler not required in bathrooms 5.1 sq. meters (55 sq. ft.) or
less – Bathroom area is 100 SF

CFPS | CFI)
130
Will this Kitchen require a sprinkler? Yes
Kitchen requires two sprinkler, since its area is
more than 14 ft x 14 ft.
8’min.
3’

CFPS | CFI)
131
Will this Garage require a sprinkler? Section
8.3.4
Sprinklers not required in garages, open
attached porches, carports, and similar
structures.

CFPS | CFI)
132
Will this Dining/LR require a sprinkler?
10”
4”
2”max.
1” Min
2” Max.
8’
Min

STEP 04 : Final Protection Scheme
CFPS | CFI)
133
Garage Dining/Living Room
Bedroom-1
Bedroom-3 Bedroom-2
Bath
Kitchen
C
C
C
C
30’
50’
20’
18’ x 12.5’
12’ x 12.5’
C
2.5’ x 7.5’
8’ x 12.5’
9.5’
3’
12 x 12.5’
13’ x 13.5’
9.5’
4’
Fireplace
Range Tub/shower
False 4" x 10" Beams
(Dropped Fluorescent Ceiling)
2.5’
2.5’
8’
Min

STEP 05 : Hydraulic Calculation and pipe sizing
Refer to Hydraulic Calculation PDF
AMIE)

Practice Project
CFPS | CFI)
135

SECTION 06
STORAGE SPRINKLERS
AMIE)

Refer to NFPA 13 plans and Calculation PDF
AMIE)

Thank You
AMIE)

MODULE 04 HYDRAULIC CALCULATION.pdf

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