Ch 7 Theoretical Pressure Calculations

Question 1

Primary function of the driver

Answer 1

provide water to crews

Question 2

Total Pressure Loss (TPL)

Answer 2

TPL = Friction loss + elevation

Question 3

Two methods to determining Friction loss

Answer 3

• Actual test and calculations
• Measure at both ends of hose line subtract difference

Question 4

Friction Loss Formula

Answer 4

FL= CQ²L
C= friction loss coefficient
Q= flow rate in hundreds of GPM (flow/100)²
L= hose length in hundreds of feet (length/100)

Question 5

Coefficients

not required to be memorized

Answer 5

1 = 150
1 ¾ with1 ½ inch coupling= 15.5
2 ½ inch= 2
3 inches with 2 ½ inch coupling= 0.8
3 inches with 3-inch coupling= 0.677
4 inches= 0.2
5 inches= 0.08

Question 6

Testing for friction loss

Answer 6

• Test one diameter of hose at a time
• Smooth bore nozzle with pitot tube
• Any nozzle when using flowmeter

Question 7

Appliance pressure loss (APL)

Answer 7

• Friction loss is insignificant where flow through appliance is less than 350 GPM
• Flowing greater than 350 GPM a 10 psi per appliance
• 25 psi friction loss in all master stream appliances regardless of flow
  -master stream
  -aerial waterway systems (including nozzle)
  -standpipe(including FDC, outlet, and piping)

Question 8

Determining Elevation Pressure

Answer 8

• Water exerts pressure at .434 psi per foot elevation
• Difference between nozzle and pump

Question 9

Elevation Pressure Loss Formula for Hills

Answer 9

EP= (0.5)H
EP= elevation pressure in psi
0.5= constant
H= Height

Question 10

Elevation Pressure Loss Formula for Buildings

Answer 10

EP= 5 psi x (number of stories -1)

Question 11

Hose Layout Applications

Answer 11

• Two types
  1. Simple hose lay outs
  2. Complex hose layouts
• Simple hose layouts
  -Includes single hose lines, equal lengths of multiple hose lines, equal wyed lines, and equal Siamese lines
• Multiple Hose Lines (equal length)
  -When determining equal length lines whose diameter is the same it is only necessary to calculate for one line
  -When diameters vary must calculate for each line and set for higher pressure
• Wyed Hose lines (Equal Length)
  -Include operation of 2 ½-, 3-, or 4-inch hose line
  -keep attack lines at same length and diameter to avoid 2 different nozzle pressures
  -10 psi if over 350 gpm
• Siamese Hose lines (Equal Length)
  -Use hose same diameter and length to supply
  -Friction loss is 25 percent less than single hose line
• Standpipe
  -Predetermined pressure
• Multiple Hose Lines (unequal length)
  -Total pressure loss in the system is based on the highest loss of the two lines
• Wyed Hose Lines (unequal)
  -Gate down line required less pressure at the wye
• Master Streams/Elevated Waterway
  -Sometimes supplied by multiple hose lines with siamese
  -Calculate 25 psi for friction loss

Question 12

Determining Pump Discharge Pressure

Answer 12

• PDP= NP+TPL
  -PDP= pump discharge pressure
  -NP= nozzle pressure
  -TPL= total pressure loss in psi (appliances, friction, elevation)
• Solid stream nozzle
  -handline 50 psi
  -master stream 80 psi
• Standard fog nozzle= 100 psi
• Low pressure fog= 50-75 psi
• Net pump discharge pressure (NPDP) if a pumper is required to discharge 150 psi and has an intake pressure of 50 psi, it needs to add an additional 100 psi

Question 13

Determining Net Pump Discharge

Answer 13

NPDP= PDP- Intake reading
from draft NPDP=PDP + incoming pressure

Question 14

You’re flowing three 2 ½-inch hoses 400 feet long operated with 1-inch tips what is the total pressure loss?

mult. hose, equal hose

Answer 14

GPM=29.7xD²x√NP
GPM= (29.7)(1)²(50)=210.01
FL= CQ²L
C= 2
Q= gpm/100 = 210/100 = 2.1
L= feet/100= 400/100 = 4
FL= (2)(2.1)²(4) = 35.28 psi per hose line

Question 15

What is the friction loss in 400 feet of 2 ½-inch hose wyed into two 1 ½-inch hose lines 100 feet flowing 95gpm
1 ½-inch hose

wyed, equal hose

Answer 15

FL= CQ²L
C= 24
Q= 95/100= .95
L= 100/100= 1
FL= (24)(0.95)²(1) = 21.66
2 ½-inch hose FL= CQ²L
C= 2
Q=gpm hose line 1+(gpm hose line 2)100
Q= 190/100= 1.9
L= 400/100= 4
FL= (2)(1.9)²(4) = 28.88 psi
Total Pressure Loss (TPL)= 21.66 + 28.88 = 50.54 psi

Question 16

What is the friction loss in two 3-inch hose lines with 2 ½-inch couplings 1,000 feet long supplying siamese with 300 feet 2 ½-inch hose with 1 ¼-inch tip at 50 psi

Siamese

Answer 16

2 ½-inch hose GPM= 29.7xD²x√NP
D= 1.25
NP= 50
GPM= (29.7)(1.25)²(50) = 328.14
FL= CQ2L
C= 2
Q= 328100= 3.28
L= 300100= 3
FL= (2)(3.282)²(3) = 64.55psi
3-inch hose with 2 ½-inch coupling FL=CQ2L
C= 0.2
Q= 328100= 3.28
L= 1000100= 10
FL= (0.2)(3.282)²(10)= 21.52 psi
TPL= 64.55 + 21.52= 86.07 psi

Question 17

Fire on fifth floor Two 2 ½-inch hose lines 150 feet connect to standpipe. 200 feet of 1 ¾-inch hose flowing 125 gpm from fog nozzle. What is total pressure loss?

Standpipe

Answer 17

Standpipe based on 10 feet per floor 40 feet
FL=CQ²L
C= 0.374
Q=125/100= 1.25
L=40/100= 0.4
FL= (0.374)(1.252)²(0.4)=0.23 psi
Elevation Pressure EP= (5 psi) x (number of stories -1)
EP= (5) (5-1) = 20 psi
2 ½-inch hose connected to FDC FL=CQ²L
C= 0.5
Q= 125/100= 1.25
L= 150/100= 1.5
FL= (0.5)(1.252)²(1.5)= 1.17 psi
TPL= 48.44 + 0.23 + 20 + 1.17= 69.84 psi

Question 17

Pumper supplying 800 feet of 5-inch to master stream at 50 feet with 2-inch tip at 100 psi

Answer 17

GPM= 29.7xD²x√NP
D= 2
NP= 100
GPM = (29.7)(2)²(100)= 1188
FL=CQ2L
C= 0.08
Q= 1188/100= 11.88
L= 800/100= 8
FL= (0.08)(11.882)²(8)= 90.33 psi
EP= 0.5H
EP= 0.5 x 50 = 25 psi
TPL= 90.33 + 25 + 25= 140.33 psi

Question 17

700 feet of 5-inch supplying three hose lines attached to manifold. Two hose lines are 150 feet of 3-inch with 2 ½-inch couplings supplying portable master stream at 80 psi with 1½-inch tip. Third hose line 150 feet 2 ½-inch hose flowing 275 gpm

Master Stream

Answer 17

3-inch hose with 2 ½-inch coupling 29.7xD²x√NP
D= 1.5
NP=80
GPM= (29.7)(1.52)²(80)= 597.7
FL=CQ²L
C= 0.2
Q= 598/100= 5.98
L= 150/100= 1.5
FL= (0.2)(5.98)²(1.5) = 10.73 psi
TPL= 10.73 + 25= 35.73 psi (25 psi added for master stream device)
2 ½-inch hose line FL=CQ²L
C= 2
Q= 275/100= 2.75
L= 150/100= 1.5
FL= (2)(2.75)²(1.5) = 22.69 psi
5-inch hose FL=CQ²L
C= 0.08
Q=598+275/100 =873100= 8.73
L= 700/100= 7
FL= (0.08)(8.732)²(7)= 42.68 psi
TPL= 35.73 + 42.68 + 10= 88.4 psi (10 psi for flow greater than 350 gpm)

Question 18

Determine pump discharge pressure required to supply two 2 ½-inch hose lines one is 300 feet long and other is 500 feet long with 250 gpm fog nozzle

NP= Nozzle Pressure in psi
TPL= Tota

PDP= Pump Discharge Pressure in psi

Answer 18

Line 1 FL=CQ²L
C= 2
Q= 250/100= 2.5
L= 300/100= 3
FL= (2)(2.52)²(3)= 37.5 psi
Line 2 FL=CQ²L
C= 2
Q= 250/100= 2.5
L= 500/100= 5
FL= (2)(2.52)²(5)= 62.5 psi
PDP is set for highest line which is line 2
PDP= NP + TPL
PDP= 100 + 62.5= 162.5 psi

Question 19

Pumper being supplied by hydrant with 20 psi on the intake gauge is discharging water at 170 psi

Net Pump Discharge Pressure
NPDP= PDP – Intake Pressure
NDNP= Net pump d

Answer 19

NPDP = PDP – Intake Pressure
NPDP = 170 – 20
= 150 psi

Question 20

Pump Discharge Pressure of
Wyed Lines

Answer 20

• First find TPL of wyed lines
  TPL
  =
  FL of line w/ highest FL
  +
  FL of supply (in the FL flormula, gpm(Q) = sum of all nozzle gpms)
  +
  appliance
• PDP of wyed lines = NP + TPL wyed lines