How to Calculate the SCFM Required To Actuate a Pneumatic Cylinder

When designing with pneumatic cylinders, it is often desirable to calculate the standard cubic feet per minute (SCFM) of air required to operate the cylinder. This article will explain how to do that calculation.


Let's assume, for example, that you are using a double acting (reciprocating) air cylinder with a 2-inch bore and a 12-inch stroke. Let's further assume that it reciprocates at 10 cycles per minute using 90-P.S.I.G. of air. If you want to know what the required scfm flow-rate of the air to the cylinder is, use the following equation:

Q2 = P2 x Q1 / P1

Where;
Q2 equals the quantity of compressed air required to operate the cylinder based on the volume of the cylinder bore for a given stroke, forward and return.

P1 equals the atmospheric pressure, which is 14.7 pounds per square inch.

P2 equals the gauge pressure of the compressed air being used to activate the cylinder.

Q1 = the piston area * (2 * stroke) * the number of cycles per minute / 1728.

Q2 equals the volume of uncompressed air required.

The area for a 2-inch bore = Pi/4 * D^2 = 3.14-in^2
12-inch stroke, Pressure, P2 = 90-Psig
Cycles = 10 cycles /min
Dividing by 1728 converts cubic inches to cubic feet.

Q1 = Piston area * (2 * stroke) * cycles/minute / 1728

Q1 = 3.14-in2 x (2 x 12) x 10 / 1728 = 753.98 / 1728 = .4363-ft3/ min (cubic feet per minute)

Q2 = P2 x Q1 / P1 = 104.7 x .4363 / 14.7 = 3.11 scfm (standard cubic feet per minute)

So, your compressor and valve must be able to supply 3.11 standard cubic feet per minute of air to operate your cylinder for this design.

When doing these calculations, disregard the piston rod cross-sectional area, and assume that the temperature remains constant. The value "P2" equals the gauge pressure plus the atmospheric pressure, which is 14.7 p.s.i.

The photo is courtesy of Parker Automation Products

How to Calculate The Pressure Drop Through An Orifice

Sometimes we engineers and designers spend too much time and money designing around large water fittings because we think that mold performance will be significantly hampered by the use of small fittings. It’s not necessarily true, and I want to show you why, and what IS important.

Darcy’s equation is used to calculate pressure losses in fluid system pipes and drilled passages.

Darcy's equation is: (f * (L/D) * ((v^2) / 2g))

To calculate pressure loss, two additional factors are introduced as follows:

Pressure Losses = (f * (L/D) * ((v^2) / 2g)) * (.433 * Sg )

where,
f = a friction factor,
L = the length of the pipe or passage,
D = the diameter of the pipe, or passage, in inches,
v = the fluid velocity in feet per second
g = gravitational acceleration constant (32.2 ft per second / per second)

Sg = the specific gravity of the fluid, water is approximately 1, for other fluids consult your literature or fluid supplier.

The .433 * Sg factor converts what would be called "head loss", in units of feet, into pounds per square inch. Since L/D is a ratio, the units are unimportant as long as the same units are used for both length and diameter.

The factor “f” (friction factor) can be found on charts available from hose and fitting suppliers, and in most Fluid Dynamics texts. These charts usually require you to be able to define the surface roughness in the passage. The one that is probably the most familiar to engineers is Moody's diagram.

There are two equations that I use to calculate the friction factor, that have been found to be appropriate for molds. The first, used for calculations involving turbulent flow, is called Blasius’ Law, and is as follows:

f = .3164 / ((Reynolds Number)^.25)

The second, used for Laminar flow, is:

f = 64 / (Reynolds Number)

One more calculation is required. Darcy’s equation requires the fluid velocity, in feet per second, as input. Since we rarely know this, we must convert the quantity that we usually know, which is usually the flow rate in gallons per minute, into feet per second.

The conversion from gallons per minute to feet per second is as follows:

v = (GPM * 231 in^2 / 60 seconds per minute / Area in inches^2 / 12)

To calculate the area of our hole diameter (Area in inches input), we use the familiar A = ((D ^ 2) * Pi) / 4

GPM is gallons per minute; and there are 231 cubic inches in a gallon. We divide by 60 to convert from minutes to seconds; the A is area of the hole diameter, and the division by 12 converts inches to feet.

Finally, we have the tools required to be able to calculate the pressure drop through our fittings. We’ll do two sample calculations, one each for a .375 and .718” diameter holes, to compare the difference.

For the .375 diameter hole, and using 6 gallons per minute of 60-degree water, we’ll calculate the Reynolds Number:

Re = (3160*Q) / [(Diameter in inches)) * (viscosity in centistokes)]

Re = (3160*6) / (.375 * 1.12) = 37920 / .42 = 45,142

Pressure Losses = (f * (L/D) * ((v^2) / 2g)) * (.433 * Sg )

f = .3164 / ((Reynolds Number)^.25) = .0217

A= .375^2 * Pi/4 = .11045

v = (6 * 236 / 60 / .11045 / 12 ) = 17.43

Press. Loss = .0217 * (.375 / 1) * [(17.43^2) / (2 * 32.2)] * (.433 *1) = .139 p.s.i.

For the .718 diameter hole, and using 6 gallons per minute of 60-degree water, we’ll calculate the Reynolds Number:

Re = (3160*Q) / [(Diameter in inches)) * (viscosity in centistokes)]

Re = (3160*6) / (.718 * 1.12) = 37920 / .80 = 23,577

Pressure Losses = (f * (L/D) * ((v^2) / 2g)) * (.433 * Sg )

f = .3164 / ((Reynolds Number)^.25) = .0255

A= .718^2 * Pi/4 = .40489

v = (6 * 236 / 60 / .40489 / 12) = 4.75

Press. Loss = .0255 * (.718 / 1) * [(4.75^2) / (2 * 32.2)] * (.433 *1) = .0054 p.s.i.

Now, on a percentage basis, there is a huge difference, but on an absolute basis, you are not likely to notice any difference in mold performance by changing between these two fittings, in this situation.

A quick look at the equation shows us that the only non-linear relationship is the velocity, which gets squared. Therefore, proportional increases in velocity will have a greater effect on pressure drop that any other factor. This is why it is so important to actually engineer your mold design.

Let’s see what would happen if you used a fitting with a .125 diameter hole.

For the .125 diameter hole, and using 6 gallons per minute of 60-degree water, we’ll do the calculations:

Re = (3160*Q) / [(Diameter in inches)) * (viscosity in centistokes)]

Re = (3160*6) / (.125 * 1.12) = 18960 / .14 = 135,428

Pressure Losses = (f * (L/D) * ((v^2) / 2g)) * (.433 * Sg )

f = .3164 / ((Reynolds Number)^.25) = .0165

A= .125^2 * Pi/4 = .01227

v = (6 * 236 / 60 / .01227 / 12) = 156.86

Press. Loss = .0165 * (.125 / 1) * [(156.86^2) / (2 * 32.2)] * (.433 *1) = 21.829 p.s.i.

Keep in mind that 6 gallons per minute will provide a Reynolds Number of 8200 in six .344 diameter water lines, assuming 60-degree water with no glycol added, so it’s not likely that you would ever end up in a situation where you would try to push 6 gallons per minute through a .125 diameter hole.

The water fittings used on a mold are most often dictated by shop standards. If the molding department only has hoses with “353” series fittings, then that is what you will usually have to design around. If you are designing for a molder that has standardized on “251” series fittings, this may or may not be causing you problems. The point of the article is this: it is not always immediately obvious how much affect your choice of pipe tap and water-fitting sizes are having on mold performance. If calculations can be done in minutes, or seconds, does it make any sense to skip over this critical step? You can make better decisions if you have more information. As always, over design can be expensive, and under design can be disastrous.

This is a segment of an article that I originally wrote for Moldmaking Technology Magazine. The link to the original article is here:

http://www.moldmakingtechnology.com/articles/090604.html

One very real problem with small fittings is that they are often used with small inside diameter hoses. Do these calculations for 20 feet of hose with small inside diameter. You may find that your largest pressure drops are outside the mold. If you need to use small fittings, be sure to reduce down at the end of the hose, and use as large a hose diameter as possible.
It can be a little time consuming to run through all of these calculations manually, so it is best to either create a spreadsheet, or purchase a program, that will allow you to do “what if” scenarios quickly. Once this is programmed, the calculations can be done in a matter of seconds, or minutes.

How to Do Basic Hydraulic Cylinder Calculations

Designing with hydraulic cylinders is very common in everything from injection molds to log splitters to Backhoes and more. Here are some of the most basic calculations used in these applications.

This article shows you how to do the three most basic, and common, calculations for designing with hydraulic cylinders. These three would be the forward force of the cylinder, the return force, and flow rate required for the forward portion of the cycle. The return flow rate and force are lower because the volume and area are reduced by the presence of the piston rod, and therefore is not included in these calculations. The working portion of the hydraulic cylinder cycle should be the piston forward movement, for the above-mentioned reasons.

The first choice to be made is the units you wish to work with, inch or metric.

To calculate the forces generated by the piston, you must enter the appropriate diameter or diameters and the pressure supplied to the cylinder, in the appropriate units. In order to calculate the required volumetric flow rate, you must also use the length of stroke and the time is seconds.
The familiar relationships for area are used to calculate these answers:
Piston forward area = (Piston diameter)^2 x Pi / 4

Piston return area = ((Piston diameter)^2 - (Rod diameter)^2) x Pi / 4

Force = Pressure x Area

GPM = (((Pi * (D^2)) / 4) * S) * (60 / 231) / T

where,
GPM = gallons per minute
D = Piston diameter in inches
S = Stroke in inches
T = time in seconds
the 60 factor converts seconds to minutes
If you have a hydraulic cylinder with a 2.000 diameter piston, a 1.000 diameter rod, a 12-inch stroke, and want to extend the rod in .5 seconds, and the system pressure is 1000 p.s.i., the calculations for force and gallons per minute are as follows.
Piston forward area = 2^2 x Pi/4 = 4 x Pi / 4 = Pi = 3.142 inches^2 (rounded)
Piston return area = ((2)^2 - (1)^2) x Pi / 4 = 2.356 inches^2

Forward Force = 1000 x 3.142 = 3142 pounds
Return Force = 1000 x 2.356 = 2356 pounds
Since the time to extend the piston rod is only affected by the diameter of the piston, we can ignore the piston's rod size.

GPM = (((Pi * (D^2)) / 4) * S) * (60 / 231) / T
GPM = ((3.142 x 4)/4) * S) x (60/231) / .5 = 19.58 gallons per minute.
Of course, if you are going to a have a continuously cycling hydraulic cylinder, you must do the flow rate calculations for both the piston side, and the rod side.
Do not exceed the manufacturers recommendation for pressure in your hydraulic cylinder. Make sure that there are pressure and temperature relief valves for safety, and if in doubt, get professional advice.
Picture courtesy Parker Hannafin.

How to Calculate Hoop Stress, Internal Pressure Only

It is often desirable to calculate how much a cavity, hydraulic cylinder, or some other cylindrically shaped tube will expand due to internal pressure. This is applicable when designing molds for thin walled parts, with cavities that are mounted on the face of a plate. If the cavity expands more than the wall thickness of the plastic part will shrink, when the injections pressure stops, the strained cavity walls will contract, transmitting the force through the plastic to the core. This has been seen to be so severe, at times, that a mold could not be opened by the molding machine.

There are other reasons to minimize cavity expansion, including controlling the size of the molding and preventing flash in certain mold designs.

I am supplying 4 choices of calculations:

1. Stress at the outside of the hoop

2. Stress at the inside of the hoop

3. Increase of the inner radius of the hoop

4. Increase of the outer radius of the hoop

All calculations assume zero external pressure. The stress calculated is the tangential stress. The stress must be kept below the fatigue stress endurance limit for the material chosen to get the maximum number of cycles.

One of the material properties that you need to know is the Modulus of Elasticity.

The Machinery Handbook defines Modulus of Elasticity as follows: "Modulus of Elasticity, E, (also called Young' s modulus) is the ratio of unit stress to unit strain within the proportional limit of a material in tension or compression". In general, the Modulus of Elasticity, for steel is 30,000,000 p.s.i. Some steels are as low as 28,000,000 and some as high as 34,000,000 p.s.i.

The shear Modulus of Elasticity, G, also known as the Modulus of Rigidity, is also the ratio of stress to strain within the proportional limit of a material, but in shear. It is usually expressed as a percentage of the Modulus of Elasticity, E, in tension. The Modulus of Elasticity in shear, G, is usually about 38% of the Modulus of Elasticity in tension, E.

Some values for Modulus of Elasticity for some common metals are as follows:

Aluminum, Alcoa QC-7: 10,300,000 p.s.i.

Brush-Wellman's Moldmax XL Beryllium Copper: 17,000,000 p.s.i.

Brush-Wellman's Protherm Beryllium Copper: 20,000,000 p.s.i.

Ampco MoldMATE 90: 22,000,000 p.s.i.

Steel: 30,000,000 p.s.i.

The stress, due to internal pressure only, calculations require only the outer and inner radii, and the internal pressure. The increase in outer wall requires that you also use the Modulus of Elasticity quantity, which is described in step 2.

The calculation for increase of inner radius requires an additional piece of information, Poisson's ratio. For steel this is approximately .30, for copper it is .34. These can be looked up in the Machinery's Handbook, Mark's Standard Handbook for Mechanical Engineer's, or a variety of other places, including web searches.

The calculation for increase of inner radius requires an additional piece of information, Poisson's ratio. For steel this is approximately .30, for copper it is .34. These can be looked up in the Machinery Handbook, Mark's Standard Handbook for Mechanical Engineer's, or a variety of other places, including web searches.

The nomenclature for the following equations is as follows:

P = Pressure inside hoop
R = Outer radius
r = Inner Radius
E = Modulus of Elasticity
Po = Poisson's Ratio

Stress at Inside of Hoop = (P * ((R ^ 2 + r^ 2) / (R ^ 2 - r^ 2)))

Stress at Outside of Hoop = ((P * 2) * ((r ^ 2) / (R ^ 2 - r ^ 2)))

Increase of Inner Radius = (P * (r / E)) * (((R ^ 2 + r ^ 2) / (R ^ 2 - r ^ 2)) - Po)

Increase of Outer Radius = (P * (R / E)) * ((2 * (r ^ 2)) / (R ^ 2 - r ^ 2))

Simply substitute the numbers from your problem, or application, into the appropriate equation, and do the calculations. Obviously, it's much quicker and easier to do this with software, or a spreadsheet.

It is advisable to start any engineering calculation with a sketch, or diagram, of the problem that you want to solve. It helps you to visualize the relationships involved.

How to Calculate Thermal Expansion

Calculating linear thermal expansion and contraction, or change of size due to change in temperature, is a common engineering calculation. Materials that will be welded, glued, or rigidly fastened together should have similar coefficients of thermal expansion, or breakage and cracking may occur. Knowing the expansion due to temperature change is extremely importnat in mold design, and can cause problems with running fits, location misalignment, and more.

Fortunately, the calculation for size change due to temperature change is fairly simple:

Size Change = initial size x temperature change x coefficient of expansion.

The important thing to remember, as with all calculations, is to keep all of the terms of the equation in the same units.

In order to use this equation, you need to know the coefficient of thermal expansion for the material in question. Here is a list of some common coefficients of thermal expansion, for different materials, in inches per inch per degree Fahrenheit:

Aluminum - .000013 or 12.8 x 10^-6 in scientific notation
Beryllium Copper - .0000093 or 9.3 x 10^-6
Brass - .000014 or 10.4 x 10^-6
Brick - .0000032 or 3.2 x 10^-6
Cement - .000006 or 6.0 x 10^-6
Concrete - .000008 or 8.0 x 10^-6
Glass, hard - .0000033 or 3.3 x 10^-6
Glass, Plate - .000005 or 5.0 x 10^-6
Glass, Pyrex - .0000022 or 2.2 x 10^-6
Gold - .0000084 or 8.4 x 10^-6
Graphite - .0000044 or 4.4 x 10^-6
Lead - .0000151 or 15.1 x 10^-6
Marble - .0000065 or 6.5 x 10^-6
Porcelain - .0000017 or 1.7 x 10^-6
Rubber - .0000428 or 42.8 x 10^-6
Silver - .0000107 or 10.7 x 10^-6
Steel - .0000064 or 6.4 x 10^-6
Tin - .000013 or 13.0 x 10^-6
Titanium - .0000048 or 4.8 x 10^-6
Wood, oak parallel to grain - .0000027 or 2.7 x 10^-6
Wood, oak across the grain - .000003 or 3.0 x 10^-6
Zinc - .0000165 or 16.5 x 10^-6

You can use the internet, the Machinists Handbook, Marks Standard Handbook for Mechanical Engineers, and many other resources to find the coefficient of thermal expansion for materials not listed here.

OK, suppose you have a piece of steel that is 12.0000 inches long, and has a 100 degree increase in temperature. What will the new length be?

The change in length = 12.0000 x 100 x .0000064 = .00768

The new length will be 12.0000 = .00768 = 12.00768

As shown in the list, .0000064 inch/inch/degree F is the coefficient of thermal expansion for steel. This can also be represented as 6.4 x 10^-6 in scientific notation.

If you want software to automate this calculation, go to www.dzynsource.com