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.