Fluid
Power Formulas
The
following formulas are readily available
in many engineering textbooks, fluid power
design guides, and hydraulic handbooks.
Every effort has been made to insure the
accuracy of the formulas and the examples
shown. However, it is possible that a typographical
error or two has slipped in. Please double
check any results that don't seem right.
Hydraulic
Pump Calculations
Hydraulic
Cylinder Calculations
Hydraulic
Motor Calculations
Fluid
& Piping Calculations
Heat
Calculations
Pneumatic
Valve Sizing
Conversions
Hydraulic
Pump Calculations
Horsepower
Required to Drive Pump
GPM
X PSI X .0007 (this is a 'ruleofthumb'
calculation)
How
many horsepower are needed to drive
a 10 gpm pump at 1750 psi?

GPM
= 10
PSI = 1750
GPM X PSI X .0007 = 10 X 1750
X .0007 = 12.25 horsepower


Pump
Output Flow (in Gallons Per Minute)
RPM
X Pump Displacement / 231
How
much oil will be produced by a 2.21
cubic inch pump operating at 1120
rpm?

RPM
= 1120
Pump Displacement = 2.21 cubic
inches
RPM X Pump Displacement / 231
= 1120 X 2.21 / 231 = 10.72 gpm


Pump
Displacement Needed for GPM of Output Flow
231
X GPM / RPM
What
displacement is needed to produce
7 gpm at 1740 rpm?

GPM
= 7
RPM = 1740
231 X GPM / RPM = 231 X 7 / 1740
= 0.93 cubic inches per revolution


Hydraulic
Cylinder Calculations
Cylinder
Blind End Area (in square inches)
PI
X (Cylinder Radius) ^2
What
is the area of a 6" diameter
cylinder?

Diameter
= 6"
Radius is 1/2 of diameter = 3"
Radius ^2 = 3" X 3"
= 9"
PI X (Cylinder Radius )^2 = 3.14
X (3)^2 = 3.14 X 9 = 28.26 square
inches


Cylinder
Rod End Area (in square inches)
Blind
End Area  Rod Area
What
is the rod end area of a 6" diameter
cylinder which has a 3" diameter
rod?

Cylinder
Blind End Area = 28.26 square
inches
Rod Diameter = 3"
Radius is 1/2 of rod diameter
= 1.5"
Radius ^2 = 1.5" X 1.5"
= 2.25"
PI X Radius ^2 = 3.14 X 2.25 =
7.07 square inches

Blind
End Area  Rod Area = 28.26 
7.07 = 21.19 square inches


Cylinder
Output Force (in Pounds)
Pressure
(in PSI) X Cylinder Area
What
is the push force of a 6" diameter
cylinder operating at 2,500 PSI?

Cylinder
Blind End Area = 28.26 square
inches
Pressure = 2,500 psi
Pressure X Cylinder Area = 2,500
X 28.26 = 70,650 pounds
What
is the pull force of a 6" diameter
cylinder with a 3" diameter rod
operating at 2,500 PSI?
Cylinder
Rod End Area = 21.19 square inches
Pressure = 2,500 psi
Pressure X Cylinder Area = 2,500
X 21.19 = 52,975 pounds

Fluid
Pressure in PSI Required to Lift Load (in
PSI)
Pounds
of Force Needed / Cylinder Area
What
pressure is needed to develop 50,000
pounds of push force from a 6"
diameter cylinder?

Pounds
of Force = 50,000 pounds
Cylinder Blind End Area = 28.26
square inches
Pounds of Force Needed / Cylinder
Area = 50,000 / 28.26 = 1,769.29
PSI
What
pressure is needed to develop 50,000
pounds of pull force from a 6"
diameter cylinder which has a 3: diameter
rod?

Pounds
of Force = 50,000 pounds
Cylinder Rod End Area = 21.19
square inches
Pounds of Force Needed / Cylinder
Area = 50,000 / 21.19 = 2,359.60
PSI


Cylinder
Speed (in inches per second)
(231
X GPM) / (60 X Net Cylinder Area)
How
fast will a 6" diameter cylinder
with a 3" diameter rod extend
with 15 gpm input?

GPM
= 6
Net Cylinder Area = 28.26 square
inches
(231 X GPM) / (60 X Net Cylinder
Area) = (231 X 15) / (60 x 28.26)
= 2.04 inches per second
How
fast will it retract?

Net
Cylinder Area = 21.19 square inches
(231 X GPM) / (60 X Net Cylinder
Area) = (231 X 15) / (60 x 21.19)
= 2.73 inches per second


GPM
of Flow Needed for Cylinder Speed
Cylinder
Area X Stroke Length in Inches / 231
X 60 / Time in seconds for one stroke
How
many GPM are needed to extend a 6"
diameter cylinder 8 inches in 10 seconds?

Cylinder
Area = 28.26 square inches
Stroke Length = 8 inches
Time for 1 stroke = 10 seconds
Area X Length / 231 X 60 / Time
= 28.26 X 8 / 231 X 60 / 10 =
5.88 gpm

If
the cylinder has a 3" diameter
rod, how many gpm is needed to retract
8 inches in 10 seconds?

Cylinder
Area = 21.19 square inches
Stroke Length = 8 inches
Time for 1 stroke = 10 seconds
Area X Length / 231 X 60 / Time
= 21.19 X 8 / 231 X 60 / 10 =
4.40 gpm


Cylinder
Blind End Output (GPM)
Blind
End Area / Rod End Area X GPM In
How
many GPM come out the blind end of
a 6" diameter cylinder with a
3" diameter rod when there is
15 gallons per minute put in the rod
end?

Cylinder
Blind End Area =28.26 square inches
Cylinder Rod End Area = 21.19
square inches
GPM Input = 15 gpm
Blind End Area / Rod End Area
X GPM In = 28.26 / 21.19 * 15
= 20 gpm


Hydraulic
Motor Calculations
GPM
of Flow Needed for Fluid Motor Speed
Motor
Displacement X Motor RPM / 231
How
many GPM are needed to drive a 2.51
cubic inch motor at 1200 rpm?

Motor
Displacement = 2.51 cubic inches
per revolution
Motor RPM = 1200
Motor Displacement X Motor RPM
/ 231 = 2.51 X 1200 / 231 = 13.04
gpm


Fluid
Motor Speed from GPM Input
231
X GPM / Fluid Motor Displacement
How
fast will a 0.95 cubic inch motor
turn with 8 gpm input?

GPM
= 8
Motor Displacement = 0.95 cubic
inches per revolution
231 X GPM / Fluid Motor Displacement
= 231 X 8 / 0.95 = 1,945 rpm


Fluid
Motor Torque from Pressure and Displacement
PSI
X Motor Displacement / (2 X PI)
How
much torque does a 2.25 cubic inch
motor develop at 2,200 psi?

Pressure
= 2,200 psi
Displacement = 2.25 cubic inches
per revolution
PSI X Motor Displacement / (2
x PI) = 2,200 X 2.25 / 6.28 =
788.22 inch pounds


Fluid
Motor Torque from Horsepower and RPM
Horsepower
X 63025 / RPM
How
much torque is developed by a motor
at 15 horsepower and 1500 rpm?
Horsepower
= 15
RPM = 1500
Horsepower X 63025 / RPM = 15 X
63025 / 1500 = 630.25 inch pound

Fluid
Motor Torque from GPM, PSI and RPM
GPM
X PSI X 36.77 / RPM
How
much torque does a motor develop at
1,250 psi, 1750 rpm, with 9 gpm input?

GPM = 9
PSI = 1,250
RPM = 1750
GPM X PSI X 36.7 / RPM = 9 X 1,250
X 36.7 / 1750 = 235.93 inch pounds
second

Fluid
& Piping Calculations
Velocity
of Fluid through Piping
0.3208
X GPM / Internal Area
What
is the velocity of 10 gpm going through
a 1/2" diameter schedule 40 pipe?

GPM
= 10
Internal Area = .304 (see note
below)
0.3208 X GPM / Internal Area =
.3208 X 10 X .304 = 10.55 feet
per second
Note:
The outside diameter of pipe remains
the same regardless of the thickness
of the pipe. A heavy duty pipe has
a thicker wall than a standard duty
pipe, so the internal diameter of
the heavy duty pipe is smaller than
the internal diameter of a standard
duty pipe. The wall thickness and
internal diameter of pipes can be
found on readily available charts.
Hydraulic
steel tubing also maintains the same
outside diameter regardless of wall
thickness.
Hose
sizes indicate the inside diameter
of the plumbing. A 1/2" diameter
hose has an internal diameter of 0.50
inches, regardless of the hose pressure
rating.

Suggested
Piping Sizes

Pump
suction lines should be sized
so the fluid velocity is between
2 and 4 feet per second.

Oil
return lines should be sized so
the fluid velocity is between
10 and 15 feet per second.

Medium
pressure supply lines should be
sized so the fluid velocity is
between 15 and 20 feet per second.

High
pressure supply lines should be
sized so the fluid velocity is
below 30 feet per second.

Heat
Calculations
Heat
Dissipation Capacity of Steel Reservoirs
0.001
X Surface Area X Difference between
oil and air temperature
If
the oil temperature is 140 degrees,
and the air temperature is 75 degrees,
how much heat will a reservoir with
20 square feet of surface area dissipate?

Surface
Area = 20 square feet
Temperature Difference = 140 degrees
 75 degrees = 65 degrees
0.001 X Surface Area X Temperature
Difference = 0.001 X 20 X 65 =
1.3 horsepower
Note:
1 HP = 2,544 BTU per Hour

Heating
Hydraulic Fluid
1
watt will raise the temperature of
1 gallon by 1 degree F per hour
and
Horsepower
X 745.7 = watts
and
Watts
/ 1000 = kilowatts

Pneumatic
Valve Sizing
Notes:
Valve
Sizing for Cylinder Actuation
SCFM
= 0.0273 x Cylinder Diameter x Cylinder
Diameter x Cylinder Stroke / Stroke
Time x ((PressurePressure Drop)+14.7)
/ 14.7
Cv
Required = 1.024 x SCFM / (Square
Root of (Pressure Drop x (PressurePressure
Drop+14.7)))
Pressure
2 (PSIG) = PressurePressure Drop

Air
Flow Q (in SCFM) if Cv is Known
Valve
Cv x (Square Root of (Pressure Drop
x ((PSIG  Pressure Drop) + 14.7)))
/ 1.024

Cv
if Air Flow Q (in SCFM) is Known
1.024
x Air Flow / (Square Root of (Pressure
Drop x ((PSIGPressure Drop) + 14.7)))

Air
Flow Q (in SCFM) to Atmosphere
SCFM
to Atmosphere = Valve Cv x (Square
Root of (((Primary Pressure x 0.46)
+ 14.7) x (Primary Pressure x 0.54)))
/ 1.024
Pressure
Drop Max (PSIG) = Primary Pressure
x 0.54

Flow
Coefficient for Smooth Wall Tubing
Cv
of Tubing =(42.3 x Tube I.D. x Tube
I.D. x 0.7854 x (Square Root (Tube
I.D. / 0.02 x Length of Tube x 12)

Conversions
To Convert 
Into 
Multiply By 
Bar 
PSI 
14.5 
cc 
Cu. In. 
0.06102 
°C 
°F 
(°C x 1.8) + 32 
Kg 
lbs. 
2.205 
KW 
HP 
1.341 
Liters 
Gallons 
0.2642 
mm 
Inches 
0.03937 
Nm 
lb.ft 
0.7375 
Cu. In. 
cc 
16.39 
°F 
°C 
(°F  32) / 1.8 
Gallons 
Liters 
3.785 
HP 
KW 
0.7457 
Inch 
mm 
25.4 
lbs. 
Kg 
0.4535 
lb.ft. 
Nm 
1.356 
PSI 
Bar 
0.06896 
In. of HG 
PSI 
0.4912 
In. of H20 
PSI 
0.03613 
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