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PNEUMATICS - HYDRAULICS - VACUUM - ELECTRONICS - AUTOMATION - VALUE ADD

Hydraulic Pump Calculations

Horsepower Required to Drive Pump:

GPM x PSI x .0007 (this is a 'rule-of-thumb' calculation)

Example: How many horsepower are needed to drive a 5 gpm pump at 1500 psi?

GPM = 5

PSI = 1500

GPM x PSI x
.0007 = 5 x 1500 x .0007 = 5.25 horsepower

Pump Output Flow (in Gallons Per Minute):

RPM x Pump Displacement ÷ 231

Example: How much oil will be produced by a 2.5 cubic inch pump operating at 1200 rpm?

RPM = 1200

Pump Displacement = 2.5 cubic inches

RPM x Pump Displacement ÷
231 = 1200 x 2.5 ÷ 231 = 12.99 gpm

Pump Displacement Needed for GPM of Output Flow:

231 x GPM ÷ RPM

Example: What displacement is needed to produce 5 gpm at 1500 rpm?

GPM = 5

RPM = 1500

231 x GPM ÷ RPM
= 231 x 5 ÷ 1500 = 0.77 cubic inches per revolution

Cylinder Calculations

Cylinder Blind End Area (in square inches):

PI x (Cylinder Radius)2

Example: What is the area of a 6" diameter cylinder?

Diameter = 6"

Radius is 1/2 of diameter = 3"

Radius2 = 3" x 3" =
9"

π 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

Example: 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"

Radius2 = 1.5" x 1.5" = 2.25"

π x Radius2 = 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

Example: 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

Example: 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)

Example: 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

Example: 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

Example: 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 x 15 = 20 gpm

Hydraulic Motor Calculations

GPM of Flow Needed for Fluid Motor Speed:

Motor Displacement x Motor RPM ÷ 231

Example: How many GPM are needed to drive a 3.75 cubic inch motor at 1500 rpm?

Motor Displacement = 3.75 cubic inches per revolution

Motor RPM = 1500

Motor Displacement x Motor RPM ÷ 231 = 3.75 x 1500 ÷ 231 = 24.35 gpm

Fluid Motor Speed from GPM Input:

231 x GPM ÷ Fluid Motor Displacement

Example: How fast will a 0.75 cubic inch motor turn with 6 gpm input?

GPM = 6

Motor Displacement = 0.75 cubic inches per revolution

231 x GPM ÷ Fluid Motor Displacement = 231 x 6 ÷ 0.75 = 1,848 rpm

Fluid Motor Torque from Pressure and Displacement:

PSI x Motor Displacement ÷ (2 x π)

Example: How much torque does a 2.5 cubic inch motor develop at 2,000 psi?

Pressure = 2,000 psi

Motor Displacement = 2.5 cubic inches per revolution

PSI x Motor Displacement ÷ (2 x π) = 2,000 x 2.5 ÷ 6.28 = 796.19 inch pounds

Fluid Motor Torque from Horsepower and RPM:

Horsepower x 63025 ÷ RPM

Example: How much torque is developed by a motor at 12 horsepower and 1750 rpm?

Horsepower = 12

RPM = 1750

Horsepower x 63025 ÷ RPM = 12 x 63025 ÷ 1750 = 432.17 inch pound

Fluid Motor Torque from GPM, PSI and RPM:

GPM x PSI x 36.77 ÷ RPM

Example: How much torque does a motor develop at 1,200 psi, 1500 rpm, with 10 gpm input?

GPM = 10

PSI = 1,500

RPM = 1200

GPM x PSI x 36.7 ÷ RPM = 10 x 1,500 x 36.7 ÷ 1200 = 458.75 inch pounds second

Fluid and 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 ÷ .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

Example: If the oil temperature is 140 degrees fahrenheit, and the air temperature is 75 degrees fahrenheit, how much heat will a reservoir with 20 square feet of surface area dissipate?

Surface Area = 20 square feet

Temperature Difference = 140ºF - 75ºF = 65ºF

0.001 x Surface Area x Temperature Difference = 0.001 x 20 x 65 = 1.3 horsepower

Heating Hydraulic Fluid Notes:

• 1 HP = 2,544 BTU per Hour

• 1 watt will raise the temperature of 1 gallon by 1ºF per hour

• Horsepower x 745.7 = watts

• Watts ÷ 1000 = kilowatts

COMPRESSED AIR FORMULAS

Electrical cost = HP x .746 x hours x Kw cost / motor efficiency

Example: 50 hp air compressor that runs 8 hours a day 5 days a week for a year with a $.06 Kw electric rate and a 90% efficient electric motor.

50 hp x .746 x 2080 hours x $.06 / .90 = $5,172.27 per year

Compressor RPM = motor pulley diameter x motor rpm / compressor pulley diameter.

Motor pulley diameter = compressor pulley diameter x compressor RPM / motor RPM

Compressor pulley diameter = motor pulley diameter x motor RPM / compressor RPM

Motor RPM = compressor pulley diameter x compressor RPM / motor pulley diameter

Gallons= cubic feet / .134

Cubic Feet = gallons x .134

Pump up time (minutes) =

V (tank size) x (final pressure – initial pressure)

7.48 x atmospheric pressure x pump delivery (cfm)

Example: 7.5 hp compressor rated at 24 cfm with an 80 gallon tank – unit starts at 100 psi and turns off at 150 psi.

80 gallons x (150 psi – 100 psi)

7.48 x 14.7 psi x 24 cfm

4,000 = 1.51 minutes

2,638

Pressure drop and horsepower: Every 1 psi of pressure drop equals 0.5% in horsepower

Heat and horsepower : Rejected heat from an air-cooled compressor is equal to total machine horsepower x 2,545 BTU per hour

Example: 50 hp compressor with 3 hp fan motor will produce…

53 hp x 2,545 = 134,885 BTU per hour

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