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 معلومات عن تصميم نظام مكافحة الحريق2

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تاريخ التسجيل : 30/07/2008
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مُساهمةموضوع: معلومات عن تصميم نظام مكافحة الحريق2   السبت نوفمبر 07, 2009 1:57 am

Ms flowing from each hoseline or discharge.
2 ½ hose #1 = 200 GPM 2 ½” hose #2 = 200 GPM
200 GPM + 200 GPM = 400 GPM
GPM = 400 (this figure will be used to find Q)
Step 5: Find Friction loss
FL = 2Q2 + Q Q = GPM  100
FL = 2 (4)2 + 4 Q = 400  100
FL = 2 (16) + 4 Q = 4
FL = 36 (per 100 ft of hose)
FL = 36 x 2.7 (270 ft  100)
FL = 97.2 psi


Chapter 6: Fire Ground (Field) Calculations
General Information
In the previous chapter we discussed how engine pressure was found using various formulas and conversion factors. If you’re 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 an incident. 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 explained below.
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
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 or 5 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
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 100’ length 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 ½” handline with a fog tip nozzle flowing 100 gpm
EP = NP + FL + BP + APP
EP = 80 + (35 X 2) + 0 + 0
EP = 80 + 70
EP = 150 psi
b) 400’ of 2 ½” handline with a barrel tip nozzle flowing 250 gpm on a
hill 50’ above the fire truck
EP = NP + FL + BP + APP
EP = 50 + (15 X 4) + (5 X 5) + 0
EP = 50 + 60 + 25 + 0
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 the average 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 ½” 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.

EP = NP + FL + BP + APP

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 ½” 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 = NP + FL + BP + APP
EP = 100 + 40 + 0 + 25
EP = 165 psi
7) Wyed Hoselines
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 all friction 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.

EP = NP + FL + BP + APP

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 = 35 psi for every 100’ of 1 ½” hose flowing 100 GPM
FL = 10 X 2
FL = 70 psi

Step 3: Find FL in the 2 ½”hoseline before the wye
Find the total GPM in the 2 ½” 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 ½” hose is 200’
FL = 10 psi for every 100’ of 2 ½” hose flowing 200 GPM
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 ½” 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
Step 5: Plug all the figures into the formula
EP = NP + FL + BP + APP
EP = 80 + 90 + 0 + 5
EP = 175 psi
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, this formula is only for every 100’ of 2 ½” hose. Hoses of all other diameters need to be converted to an equivalent length.
Cool 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 ½” hoselines is 150 psi
The EP for the 2 ½” 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 ½” hoselines the proper pressure. As for the 2 ½” 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 ½” 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 ½” 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 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.
EP = NP + FL + BP + APP
Step 1: Find NP
The NP for a master stream using a fog nozzle is 100 psi (given)
NP = 100 psi
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 ½” 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
Step 5: Plug all the figures into the formula
EP = NP + FL + BP + APP
EP = 100 + 42 + 0 + 80
EP = 222 psi
10) Relay Pumping
Relay pumping is used when the distance from the water supply (fire hydrant) to the incident is longer than the supply lines carried by a single 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 ½”
FL for 2 ½” 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 ½”
FL for 2 ½” flowing 250 gpm = 15 per 100’
1000  100 = 10
FL = 15 X 10
FL = 150
Add 20 psi residual pressure
EP = 150 + 20
EP = 170 psi
Engine #3: EP = NP + FL + BP + APP
NP = 80 psi
FL = 60 psi (400’ 2 ½” 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
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
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 be used
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 ½ of the hose diameter. This means that a 1” hoseline should not use a nozzle with a diameter of greater than ½”.
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 45 will produce the greatest horizontal reach. However for firefighting purposes, the maximum effective horizontal range of a fire stream can be obtained from a fire stream angled between 30-34


The formula for finding the horizontal reach of a fire stream is:
Horizontal Reach = ½ x Nozzle Pressure + 26 feet
HR = ½ NP + 26
This formula is based on a nozzle size of ¾”. For every 1/8” over
¾”, 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 = ½ NP + 26 (+5 for every 1/8” over ¾”)
Step 1: Find difference in eighths between 1” and ¾”
1” (8/Cool minus ¾” (6/Cool = 2/8
1” is 2 eighths over ¾”
Step 2: Multiply 5 by the total number of eighths over ¾
5 x 2 = 10
Step 3: Add the figure in step two into the formula
HR = ½ NP + 26 + 10
Step 4: Solve problem (using adjusted formula in step 3)
HR = ½ NP + 26 + 10
HR = ½ (50) + 26 + 10
HR = 25 + 26 + 10
HR = 61 feet
Work Problem: What is the horizontal reach of a fire stream flowing 60
psi through a 1 1/8” tip (nozzle)?
HR = ½ NP + 26 (+5 for every 1/8” over ¾”)
Step 1: Find difference in eighths between 1 1/8” and ¾”
1 1/8” (9/Cool minus ¾” (6/Cool = 3/8
1 1/8” is 3 eighths over ¾”
Step 2: Multiply 5 by the total number of eighths over ¾
5 x 3 = 15
Step 3: Add the figure in step two into the formula
HR = ½ NP + 26 + 15
Step 4: Solve problem (using adjusted formula in step 3)
HR = ½ NP + 26 + 15
HR = ½ (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 90 will 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

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 is because
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 ¾”. For every 1/8” over
¾”, 5 feet must be added to the 26.
Work Problem: What is the vertical reach of a fire stream flowing 40
psi through a 1” tip (nozzle)?
VR = 5/8 NP + 26 (+5 for every 1/8” over ¾”)
Step 1: Find difference in eighths between 1” and ¾”
1” (8/Cool minus ¾” (6/Cool = 2/8
1” is 2 eighths over ¾”
Step 2: Multiply 5 by the total number of eighths over ¾
5 x 2 = 10
Step 3: Add the figure in step two into the formula
VR = 5/8 NP + 26 + 10
Step 4: Solve problem (using adjusted formula in step 3)
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 ¾”)
Step 1: Find difference in eighths between 1 1/8” and ¾”
1 1/8” (9/Cool minus ¾” (6/Cool = 3/8
1 1/8” is 3 eighths over ¾”
Step 2: Multiply 5 by the total number of eighths over ¾
5 x 3 = 15
Step 3: Add the figure in step two into the formula
VR = 5/8 NP + 26 + 15
Step 4: Solve problem (using adjusted formula in step 3)
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 ½” 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 to a fire hydrant (intake side of pump) and supply lines to the siamese 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 ½” 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 ½” 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 ½” hose outlets for each riser
i. 1 per floor level (optional for 1st floor)
ii. Roof outlet requires a two-way 2 ½” 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 the valve allowing water to flow through the hose. With this having been said, 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 ½”
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 – Factory Mutual)
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 Sprinkler System)
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.

EP= FL ( 2 ½” hoses to siamese) 2 2-1/2” lines flowing 200 gpm 3 psi
FL (2 ½” hose on fire floor) 1 2-1/2” line flowing 200 gpm 10 psi
FL (1 ½” 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
Find the average length of the wyed hoses
100 ft + 100 ft = 200 ft
200 ft  2 = 100 ft
1 ½” flowing 100 gpm = 35 psi / 100’
FL = 35 psi
Step 2: Find friction loss in 2 ½” hose
Total flow = 200 gpm (both 1 ½” hoses)
2 ½” hose flowing 200 gpm = 10 / 100’
FL = 10 psi
Step 3: Find NP and Appliance loss
NP = 80 psi
Appliance Loss = 5 psi (2 ½” to 1 ½” 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 ½” hose
200 gpm (both 1 ½” hoses)
divided by 2 (number of siamesed hoses)
Each hose is flowing 100 gpm
2 ½” 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 Systems
General Information
1) History
Sprinkler systems were developed around the 1850’s. 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 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
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