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 aircooled 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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FLUID POWER FORMULAS
Basic Formulas: Pressure (psi) = Force (pounds) / Area (in^{2})Force (pounds) = Area (in^{2}) x Pressure (psi)Area (in^{2}) = Force (pounds) / Pressure (psi) Fluid Power Horsepower (hp) = Pressure (psi) x pump flow (gpm) / 1,714 Torque (ft.lbs.) = Horsepower (hp) x 5,252 / Speed (rpm)Horsepower (hp) = Torque (ft.lbs.) x Speed (rpm) / 5,252Speed (rpm) = Horsepower (hp) x 5,252 / Torques (ft.lbs.)Cylinder Formulas:Piston cylinder area (in^{2}) = Diameter squared x .7854
= 3.1416 x Radius squared
Rodend cylinder area (in^{2}) (Annulus end area)
= Cylinder area (in^{2}) – Rod area (in^{2})
Cylinder force (pounds)
= Pressure (psi) x Area (in^{2})
Cylinder speed feet/minute) (divide by 60 for feet/second)=19.25 x Flow rate (gpm) /Area (in^{2}) Cylinder time (seconds)
= Area (in^{2}) x Cylinder stroke (in.) x .26 /Flow rate (gpm)
Cylinder flow rate (gpm) = 12 x 60 x Cylinder speed (ft./sec.) x Area (in^{2}) / 231Cylinder volume capacity (gallons)= 3.1416 x Radius squared (in.) x Cylinder stroke / 231Hydraulic Motor Formulas:Fluid motor torque(in.lbs.) = Pressure (psi) x Fluid motor displacement / 6.28 = Horsepower (hp) x 63,025 / Speed (rpm) = Flow rate (gpm) x Pressure (psi) x 36.77 / Speed (rpm)Fluid motor speed (revs./min.)= 231 x Flow rate (gpm) / Fluid motor displacement (in^{3}/rev.)
Fluid motor power (hp output)
= Torque output (in.lbs.) x Speed (rpm) / 63,025
Fluid motor torque / 100 psi (in.lbs.)= Fluid motor displacement (in^{3}/rev.) / .0628 Fluid motor flow rate (gpm)= Motor speed (rpm) x Motor displacement(in^{3}/rev.) / 231
Pump Formulas:
Pump outlet flow (gpm)
= Speed (rpm) x Pump displacement (in^{3}/rev.) / 231 Pump speed (rpm)
= 231 x Pump flow rate (gpm) / Pump displacement (in^{3}/rev.)Pump input horsepower (hp)
= Flow rate output (gpm) x Pressure (psi) / 1,714 x Efficiency factor (overall %)
Pump Efficiency (Overall in %)(Output horsepower / Input horsepower) x 100 Pump displacement (in^{3}/rev.) = Flow rate (gpm) x 231 / Pump speed (rpm) Pump torque (in.lbs.) = Horsepower (hp) x 63,025 / Speed (rpm) = Pressure (psi) x Pump displacement (in^{3}/rev.) /6.28 Miscellaneous Formulas, Data and "Rules of Thumb":Reservoir capacity (gallons) = Length (ins.) x width (ins.) x height (ins.) / 231 Reservoir cooling capacity(BTU/hr.)2 x Temperature difference between reservoir walls and air (degrees F) x area of reservoir (ft^{2}) (Based on adequate air circulation)Heat radiating capacity of a steel reservoir expressed in horsepower (hp)Horsepower (hp)= 0.001 x reservoir surface area (ft^{2}) x temperature Difference (degrees F) between oil and surrounding air Heat equivalent of fluid power(BTU/hr.) = Pressure (psi) x Flow (gpm) x 1.5 (Note: One horsepower = 2,545 BTU/hour)One British Thermal Unit (BTU) is the amount of heat required to raise the temperature of one pound of water one degree Fahrenheit.Heating hydraulic fluid = 1 watt will raise the temperature of 1 gallon of oil 1 degree Fahrenheit per hour Compressibility of hydraulic oil = Volume reduction is approximately 1/2% for every 1,000 psi of fluid pressure Compressibility of water = Volume reduction is about 1/3% for every 1,000 psi of fluid pressure Estimating pump drive horsepower (hp) = Approximately 1 hp of drive for every 1 gpm (flow) at 1,500 psi (pressure) Pump idling horsepower (hp) = Approximately 5% of a pumps full rated horsepower will be required when a pump is running unloaded"Guidelines" for flow velocity in hydraulic lines: Pump suction lines 2 – 4 ft./sec. Pressure lines to 500 psi
10 – 15 ft./sec.
Pressure lines of 500 to 3,000 psi 15 – 20 ft./sec. Pressure lines over 3,000 psi 25 ft./sec. Fluid velocity of oil flow in a pipe (ft./sec.)
= Flow rate (gpm) x 0.3208 / Inside area of pipe (in^{2})
Common Fluid Power Equivalents:One U.S. gallon = 231 in^{3} 4 quarts or 8 pints 128 liquid ounces 128 liquid ounces 133.37 ounces in weight 8.3356 pounds 3.785 liters One imperial gallon =1.2 U.S. gallons One liter = 0.2642 U.S. gallons One cubic foot = 7.48 gallons 1728 cubic inches 62.4 pounds of water One Bar at sea level = 14.504 psi 0.98692 atmosphere 33.6 foot water column 41 foot oil column
Approximately ½ psi decrease for each 1,000 feet of elevation change
One inch mercury (hg.) = 0.490 psi 1.131 ft. water
One horsepower =
33,000 ft.lbs./min. 550 ft.lbs./sec. 42.4 BTU/min. 2,545 BTU/hour 746 watts or 0.746 kilowatts (kw)One psi = 2.0416 inches of mercury (hg.) 27.71 inches of water 0.0689 bar
One atmosphere =
1.013 bar 29.921 inches of mercury (hg.) 14.696 psi 760 mm of mercury (hg.)One foot water column = 0.432 psi One foot oil column = 0.354 psi
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LUBRICATION FORMULAS
Calculating Bearing Requirements for Oil Lubricants:
V = A x T
V = Volume in terms of lubeoil replacement rate in cubic inches per hour (in^{3}/hr)A = Bearing surface area in square inches (in^{2}) (Sized differently based on bearing type)
T = Film thickness…generally .001 inch… but it may vary based on oil type and application
Calculating Bearing Requirements for Grease Lubricants: V = A x TV = Volume in terms of lubegrease replacement as cubic inches per four hour (in^{3}/4 hrs) A = Bearing surface area in square inches (in^{2}) (Sized differently based on bearing type)
T = Film thickness…generally .002 inches…but it may vary based on grease type and application
Note: Quite often requirements are expressed in metric terms. To convert to metric, calculate volume requirements as noted above and simply multiply by 16.39 to convert to cubic centimeters per hour...cc^{3}/hr... (oil) or cubic centimeters per four hours...cc^{3}/4 hrs... (grease).
Common Bearing Types:(Necessary to know for calculating areas.)Plain Bearings: Area (in^{2}) = 3.14 x Shaft diameter (ins.) x Length of bearing (ins.)Slides, Gibs and Ways: Area (in^{2}) = Area of largest contact surface AntiFriction Bearings: Area (in^{2}) = Shaft diameter squared x number of rows
Gears: Area (in^{2}) = 3.14 x Pitch diameter of gear (ins.) x width of gear (ins.)
Sizing Example:Plain bearing with 6 inch shaft and 6 inch long bearing surface using oil. Area (in^{2}) = 3.14 x 6(ins.) x 6 (ins.) = 113.04 (in^{2}) Volume (in^{3}/hr.) = 113.04 (in^{2}) x .001 (in.) = .113 (in^{3}/hr.) lube oil replacement rate
Should this need to be converted to metric, the requirement for this single bearing application would be 1.85 cubic centimeters per hour. .113 (in^{3}/hr.) x 16.39 = 1.85 (cc^{3}/hr.)
Each and every bearing or lube point on a machine would be calculated in this fashion and when done, the replacement rates for all points would be added together to determine the total system lubrication requirement.
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ELECTRICAL FORMULAS
Volts: (E)

Volts

= Square root of (Watts x Ohms)


= Watts / Amperes


= Amperes x Ohms

Ohms: (R)

Ohms

= Volts / Amperes


= Volts squared / Watts


= Watts / Amperes squared

Watts: (W)

Watts

= Volts squared / Ohms


= Amperes squared x Ohms


= Volts x Amperes

Amperes: (I)

Amps

= Volts / Ohms


= Watts / Volts


= Square root of (Watts / Ohms)

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PNEUMATIC FORMULAS
Air valves are sized for flow capacity (Cv) based on given cylinder piston size, stroke and travel time requirements. Cv is actually a flow coefficient that measures the amount of air a device can pass. The following formula can be used for air valve sizing:
Cv =Area (in^{2}) x Length (ins.) x Compression factor Pressure drop factor x Time (secs) x 29
Area = Effective cylinder piston area in square inches
(A = 3.1416 x radius^{2 }– or – diameter^{2 }x .7854)
Note: For the rod end (annulus end) of the cylinder, the same area formulas apply, but to calculate accurately, one must take the cylinder area (in^{2}) minus the rod area (in^{2}) in using this valve sizing formula for determining return stroke Cv rating.
Length = Simply the total cylinder stroke length in inches
Compression factor = Taken from the table based on supply pressure rating.
Pressure drop factor = Taken from the table....10 or 15 psi drop is a good guideline for using in this formula
Time = Required cylinder stroke time in seconds

Pressure Drop Factor PD for Various Pressure Drops

Supply Pressure

Compression Factor

2

5

10

15

20

PSI

CF

PSID

40

3.7

9.9

15.3

20.5

23.6

N/A

50

4.4

10.8

16.7

22.6

26.4

29

60

5.1

11.7

18.1

24.6

29

32

70

5.8

12.5

19.3

26.5

31.3

34.8

80

6.4

13.2

20.5

28.2

33.5

37.4

90

7.1

13.9

21.6

29.8

35.5

39.9

100

7.8

14.5

22.7

31.3

37.4

42.1

110

8.5

15.2

23.7

32.8

39.3

44.3

120

9.2

15.8

24.7

34.2

41.0

46.4

130

9.8

16.4

25.6

35.5

42.7

48.4

140

10.5

16.9

26.5

36.8

44.3

50.3

150

11.2

17.5

27.4

38.1

45.9

52.1

Sizing Example:
6 inch bore cylinder with 2 inch rod and 15 inch stroke.... 2 second travel time....100 psi supply pressure....and 15 psi pressure drop factor will be used:
1. Calculate piston area in square inches
(A = 6 ins. x 6 ins. x .7854 = 28.2 in^{2})
Note that this is the cylinder extend area, to calculate the cylinder return area, the rod area must be subtracted from
(A = 2 ins. x 2 ins. x .7854 = 3.14 in^{2})
Cylinder return area is then 28.2 in^{2} – 3.14 in^{2} = 25.06 in^{2}
2. Simply apply application variables to the Cv sizing formula:
Cv = 28.2 in^{2} x 15 ins. x 7.8 = 3,299 = 1.52 Cv 37.4 x 2 secs x 29 2,169
3. Select a valve that meets this 1.52 Cv rating.
Many fluid power engineering and data resources have flow charts to simplify this sizing process, but in absence of those charts, this information should help to size pneumatic valve requirements.
Air Flow Rates
Standard Cubic Feet Per Minute (SCFM) One cubic foot of gas (air) per minute at standard conditions of 68 degrees F, 14.69 psi and a relative humidity of 36%.
Cubic Feet Per Minute (CFM): One cubic foot of gas (air) per minute at actual conditions...ie: at actual temperature and compressed or expanded pressure.
Free Air Flow: The volume of air at normal atmospheric conditions which enters a vacuum system due to the lower pressure caused by the pump or vacuum in a tank.
Expanded Air Flow: Air flow inside a vacuum system, same as CFM.
SCFM and Compressor Horsepower Requirements: To calculate pneumatic cylinder air consumption in SCFM and convert it to required air compressor horsepower, please request an RHM Fluid Power Data Book which includes quick reference charts for these purposes.
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Note: This information is provided as a quick reference resource and is not intended to serve as a substitute for qualified engineering assistance. While every effort has been made to ensure the accuracy of this information, errors can occur. As such, neither OCSI or its employees will assume any liability for damage, injury or misapplication tied to the use of this information.
