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Pumping and Aerial Apparatus Driver/Operator Handbook 3rd Edition > Ch 7 Theoretical Pressure Calculations > Flashcards

Ch 7 Theoretical Pressure Calculations Flashcards

1
Q

Primary function of the driver

A

provide water to crews

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2
Q

Total Pressure Loss (TPL)

A

TPL = Friction loss + elevation

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3
Q

Two methods to determining Friction loss

A
  • Actual test and calculations
  • Measure at both ends of hose line subtract difference
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4
Q

Friction Loss Formula

A

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)

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5
Q

Coefficients

not required to be memorized

A

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

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6
Q

Testing for friction loss

A
  • Test one diameter of hose at a time
  • Smooth bore nozzle with pitot tube
  • Any nozzle when using flowmeter
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7
Q

Appliance pressure loss (APL)

A
  • 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)
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8
Q

Determining Elevation Pressure

A
  • Water exerts pressure at .434 psi per foot elevation
  • Difference between nozzle and pump
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9
Q

Elevation Pressure Loss Formula for Hills

A

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

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10
Q

Elevation Pressure Loss Formula for Buildings

A

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

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11
Q

Hose Layout Applications

A
  • 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
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12
Q

Determining Pump Discharge Pressure

A
  • 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
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13
Q

Determining Net Pump Discharge

A

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

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14
Q

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

A

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

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15
Q

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

A

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

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16
Q

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

A

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

17
Q

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

A

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

17
Q

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

A

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

17
Q

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

A

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)

18
Q

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

A

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

19
Q

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

A

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

20
Q

Pump Discharge Pressure of
Wyed Lines

A
  • 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

Pumping and Aerial Apparatus Driver/Operator Handbook 3rd Edition (20 decks)

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