Conversion Tables and Formulas - by Nelsen Corporation

Volume 
1 U.S. Gallon231.0 in³
0.137 ft³
3.785 liters
.00379 meters³
0.83 Imp gal
0.238 42-gal barrel
1 Imperial Gallon1.2 U.S. gal
1 Cubic Foot7.48 U.S. gal
0.0283 meter³
1 Liter0.2642 U.S. gal
1 Cubic Meter35.315 ft³
264.2 U.S gal
1 Acre Foot43,560 ft³
325,829 U.S. gal
1 Acre Inch3,630 ft³
27,100 U.S. gal

Length 
1 Inch2.54 centimeters
1 Meter3.28 feet
39.37 inches
1 Rod16.5 feet
1 Mile5280 Feet (1.61 Kilometers)

Capacity 
1 Cubic Foot Per Second
(2nd foot) (C.F.S)		449 gpm
1 Acre Foot Per Day227 gpm
1 Acre Inch Per Hour454 gpm
1 Cubic Meter Per Minute264 gpm
1,000,000 Gal. Per Day595 gpm

Head 
1 Pound Per Square Inch (p.s.i)2.31 ft. head of water
2.04 in. mercury
0.07 kg/cm2
1 Foot of Water

0.433 lb/in2
.855 inc. mercury
1 Inch of Mercury (or vacuum)

1.132 ft of water
1 Kilogram Per Square Cm14.22 lb/in2

1 Atmosphere (at sea level)14.7 lb/in2
34.0 ft of water
1 Acre Foot10.35 meters of water
1 Meter of Water3.28 feet of water

To Find Capacity of a Tank or Cistern    
Diameter
of Tank
in Feet
Squared		x .7854 xHeight
of Tank
in Feet 		x 7.48Capacity
U.S. Gallons

Horsepower 
1 H.P. Equals.746 kilowatts of 746 watts
33,000 ft lbs per minute
550 ft lbs per second
H.P. Input EqualsHorsepower input to motor
1.34 x kilowatts input to motor
Water H.P. EqualsHorsepower required to lift water at
G.P.M. x total head (in ft.)
3960
Brake H.P. EqualsH.P. delivered by motor
H.P. required by pump
H.P. input x motor efficiency
1.34 x KW input x motor efficiency
Water horsepower
Pump efficiency
G.P.M x total head (ft.)
3960 x pump efficiency
G.P.M x total head (lbs/in2)
103,000 x pump efficiency
Electric Power 
ACAlternating Current Power
DCDirect Current
EVolts
IAmperes
WWatts
KWKilowatts
Apparent PowerVolts x amperes = Volt amperes
Apparent PowerE I
Useful Power WE I x P.F.
Power Factorratio of useful power to
apparent power
Power FactorW
e i = PF
KW Hr.Kilowatt Hour
Single PhaseE x I x PF
Power W
3 Phase Power W1.73 x E x I x PF
Where EAverage voltage between phases
IAverage current in each phase
Electric Power 
Efficiency EqualsPower Output
Power Input
Motor Efficiency EqualsH.P. Output
K.W. input x 1.34
Pump Efficiency EqualsG.P.M. x total head (ft.)
103,000 x B.H.P
Efficiency 
Efficiency EqualsPower Output
Power Input
Motor Efficiency EqualsH.P. Output
K.W. input x 1.34
Pump Efficiency EqualsG.P.M. x total head (ft.)
103,000 x B.H.P

Pressure/Feet of Head   
Lbs. Per Sq. In.Feet HeadFeet HeadLbs. Per Sq. In.
12.3110.43
24.6220.87
36.9331.3
49.2441.73
511.5452.17
613.8562.6
716.1673.03
818.4783.46
920.7893.9
1023.09104.33
1534.63158.66
2046.182012.99
2557.722517.32
3069.273021.65
4092.364025.99
50115.455030.32
60138.546034.65
70161.637038.98
80184.728043.31
90207.819047.65
100230.910051.97
110253.9811056.3
120277.0712060.63
125288.6212564.96
130300.1613069.29
140323.2514073.63
150346.3415077.96
160369.4316082.29
170392.5217086.62
180415.6118097.45
190438.9190108.27
200461.78200119.1
225519.51225129.93
250577.24250140.75
275643.03275151.58
300692.69300173.24
325750.41325216.55
350808.13350259.85
375865.89375303.16
400922.58400346.47
5001154.48500389.78
100023091000433.09

A column of water 1 inch square by 2.31 feet high weighs 1 pound. Therefore, one pound of pressure per square inch (PSI) is equal to 2.31 feet of head.

A pressure of .433 pounds per square inch will support a column of water 1 inch square by 1 foot high. Therefore, one foot of head is equal to .433 PSI.

To convert pressure in pounds per square inch (PSI) to head in feet: multiply by 2.31. To convert head in feet to pressure in pounds per square inch (PSI): multiply by .433.


Storage Capacity of Well Casing and Vertical Tanks		   
DiameterGal. Per Ft.DiameterGal. Per Ft.
2 in.0.168.5 ft.424.5
3 in.0.379 ft.475.9
4 in.0.659.5 ft.530.2
5 in.110 ft.587.5
6 in.1.511 ft.711
8 in.2.612 ft.846
10 in.4.113 ft.993
12 in.5.914 ft.1151
16 in.10.415 ft.1322
18 in.13.216 ft.1504
20 in.16.317 ft.1698
24 in.23.518 ft.1904
30 in.36.719 ft.2121
36 in.52.920 ft.2352
42 in.7221 ft.2591
48 in.9422 ft.2845
54 in.11923 ft.3109
60 in.14724 ft.3384
66 in.177.825 ft.3672
72 in.211.526 ft.3971
78 in.248.227 ft.4283
84 in.287.928 ft.4606
90 in.330.529 ft.4841
96 in.37630 ft.5288

Flow of Water/Gravity or Tank Pressure

The approximate flow of water in GPM through a length of pipe due to the force of gravity can be easily determined by the formula:D x 100 devided by L

Determine the vertical distance in feet (D) between the pipe inlet and the pipe outlet. Multiply this distance by 100 and divide that amount by the total length of the pipe in feet (L).

Refer to the appropriate friction loss table for the size and type of pipe. Read down the appropriate column to the  number of feet as determined by the formula above. Read across to the left to determine the approximate flow rate  through the pipe. The flow at the lower end of the pipe will be at zero pressure.

Example

A 300' length of 1" plastic pipe runs from an inlet point to a point to discharge 40 feet lower. The approximate flow rate would be 40 x 100 divided by 300 = 13.3. Referring to the friction loss tables gives a flow rate between 14 and 16, or about 15 GPM.

For pressure tanks the formula is the same, except that the pressure in the tank should be converted to vertical feet of head and added to the vertical distance if any. Again the flow at the end of the pipe will be at zero pressure.

Contact:
George Hogg
WTC

Email:
info@waterbygeorge.ca

Cell:
780 871 9283

Postal Address:
PO Box 12699
Lloydminster, AB T9V 0Y4

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