Glossary of Leak Terms

Compressors
The type of compressor most likely to be used for an industrial compressed air system depends largely on size, cost, and reliability requirements:

• Rotary screw compressors in sizes up to 500-600 hp are very popular because of their high reliability and low maintenance requirements.
• Centrifugal compressors are often used in sizes ranging from about 150 hp up to over 10,000 hp. The larger size models are relatively low in cost and small in physical size compared to reciprocating compressors.
• Reciprocating compressors are one of the oldest air compressor technologies, but are commonly used today only in sizes up to 25 hp or so. These compressors are often used for light-duty applications or in startup industrial enterprises because they are reliable and low cost.
• Rotary vane compressors are not commonly used as they tend to be energy inefficient and require higher maintenance than other compressor designs.

Compressed Air System Components

Figure 1: Compressed Air System Diagram (Reprinted with permission from Compressor Engineering Data Book. Copyright 1974, 1996 by Scales Air Compressor Corp. All rights reserved)

A typical compressed air system consists of compression, cooling, storage, and distribution equipment, as shown in Figure 1.

• Intake Filtering (1): Incoming air must be filtered to remove dust and other contaminants.
• Compression (2): The filtered air is compressed (typically to 80 to 110 psi) with motor-driven screw, centrifugal, or reciprocating compressors.
• Cooling (3): Compressing air raises its temperature dramatically, so cooling is required. Much of the energy “lost” in making compressed air is in the form of removed heat. Cooling is also important in the process of drying air. Much of the water vapor condenses as the air is cooled, making it easy to drain away.
• Air Storage (4): A tank called an air receiver typically is placed downstream of the cooler to provide surge capacity for the system. Some systems provide additional receiver tanks in the process area to accommodate widely variable demand.
• Drying (5): Cooled, pressurized air still carries a significant amount of moisture and lubricants from the compression process, virtually all of which must be removed before the air can be used. Drying compressed air can be very energy intensive.
• Distribution (6): A system of distribution pipes and regulators convey compressed air from the central compressor plant to process areas. This system includes various isolation valves, fluid traps, intermediate storage vessels, and even heat trace on pipes to prevent condensation or freezing in lines exposed to the outdoors. Pressure losses in distribution typically are compensated for by higher pressure at the compressor discharge.
• Point of Use (7): At the intended point of use, a feeder pipe with a final isolation valve, filter, and regulator carries the compressed air to hoses that supply processes or pneumatic tools.

Convert all input parameters to SI units for calculation purposes and convert the results back to local units.

The Equations

The pressure loss in the pipe is due to friction within the fluid. Friction losses can only be quantified if the flow regime is known. The flow regime may be determined from the Reynolds number, calculated from the following formula:

rho x V x d
Re = —————-
mu

where
Re is the Reynolds number (dimensionless)
rho is the density in kg per cubic meter
V is the flow velocity in m/s
d is the inside diameter of the tube in meters
mu is the absolute viscosity expressed in Pa.s

If the Reynolds number is below 2,000 the flow regime may be assumed to be laminar. If Re is above 4,000 it can be assumed to be turbulent. Between these figures the flow is in some intermediate regime that at one end may be laminar and at the other end may be fully turbulent.

Absolute viscosity (also known as dynamic viscosity) is related to kinematic viscosity by the following relation:

mu = rho x neta

where neta is the kinematic viscosity expressed in square meters per second.

For completeness here is the relation between flow (Q in cubic meters per second) and velocity in a circular conduit:

4 x Q
V = ————–
pi x d x d

where pi = 3.14159… (extend to taste)

In a laminar flow regime the Hagen-Poiseuille equation should be used:

128 x mu x L x Q
deltaP = ———————-
pi x d x d x d x d

where deltaP is the differential pressure in pascal, i.e. the difference between the pressure at the input end of the pipe and the discharge end of the pipe, and L is the length of tube in meters.

For turbulent flow the more general Darcy-Weisbach equation for flow must be used:

f x rho x L x V x V
deltaP = ——————————-
2 x d

where f is a dimensionless quantity known as the Darcy-Weisbach skin friction factor.

A number of equations exist to determine the correct friction factor for a given flow. For laminar flow:

64
f = ——-
Re

For turbulent flow, the Colebrook-White equation may be used:

1 epsilon 2.51
——- = -2 x log10(————– + ——————-)
sqrt(f) 3.7 x d Re x sqrt(f)

where epsilon is the absolute roughness of the inside diameter of the tube and is expressed in meters.

Since the friction factor appears on both sides of the equation a numerical solution must be obtained.

Where a Reynolds number between 2,000 and 4,500 is obtained the friction factor is more difficult to predict. Interpolation routines may be developed based on the Moody diagram but these are generally specific to a limited range of flow conditions. A more general relation has been recently derived by Chue that covers all flow regimes including the intermediate region:

1
——- = -2 x log10(G)
sqrt(f)

2.512 epsilon
where G = ((1 – gamma) x antilog10(-sqrt(Re)/16) + (gamma x (—————– + ————–))
Re x sqrt(f) 3.7 x d

1
and where gamma = ————————————————-
Re – 3057.2516
1 + exp(———————-)
227.52765

Chue’s equation must be solved numerically in order to obtain the friction factor. Note also that at low flow rates the friction factor given by Chue may be severely inaccurate.

Using the Equations
In using the friction loss equations alone, so called “minor” losses are neglected. Minor losses are energy losses that occur in fittings, bends, couplings and valves. In short lengths and in high flow rates minor losses can become significant and will need to be included. In simple systems using the pipe friction loss equations alone will provide a reasonably accurate indication of the performance of the system.

How the pipe pressure loss equations are to be used will depend on whether the starting point is a required flow rate or if the flow capacity of a line needs to be determined from the available pressure loss.

Specified Flow Rate
Where the required flow rate is known the process of determining the associated pressure loss for a given tube diameter is fairly straightforward. A possible routine for determining the pressure loss is given below:

1. Calculate Reynolds number.
2. If Reynolds number is less than 2,000 go to step 3 else go to step 4.
3. Calculate deltaP from Hagen-Poiseuille.
4. If Reynolds number is greater than 4,000 go to step 5 else go to step 6.
5. Use Colebrook-White to determine friction factor, go to step 7.
6. Use Chue to determine friction factor.
7. Use Darcy-Weisbach to determine pressure loss.

Specified Pressure Loss
Knowing the difference between the pressure at the inlet to the pipe and the discharge from the pipe, the flow rate can be obtained, but this approach requires more iteration. The starting point is to take a guess at the friction factor and iterate until it’s value doesn’t change significantly from one iteration to the next. One routine might look like this:

1. Assume a friction factor.
2. Calculate flow rate from Darcy-Weisbach.
3. Calculate Reynolds number from the flow rate.
4. Put the new Reynolds number in to Chue to obtain a revised friction factor.
5. If the revised friction factor has not changed much from that used in step 2 then go to step 7.
6. Go to step 2 with the new friction factor.
7. The friction factor is stable from one iteration to the next so the flow rate is now correct for the pressure loss.

If the flow regime is expected to be turbulent then the Colebrook-White equation may be substituted for Chue in the above routine.

References
1. Fluid Mechanics, 1st SI Metric Edition, Streeter & Wylie, McGraw-Hill
2. Proceedings of the Institution of Civil Engineers, Part 2 Research and Theory, March 1984, Volume 77, pages 43-48, Technical Note 399, A pipe skin friction factor of universal applicability, S H Chue BE BSc(Spec) PhD.

Unit Conversions
To calculate psi from bar multiply by 14.5, although for a more accurate result divide by 0.06894757.
To calculate bar from pascals divide by 100,000.
To calculate centipoise from Pa.s multiply by 1,000. Note that 1 cP is equal to 1 mPa.s.
To calculate centistokes from square meters per second multiply by 1,000,000. Note that 1 cS is equal to 1 square millimeter per second.
1 US gallon is equal to 3.78 liters, there are 1,000 liters in 1 cubic meter.
1 foot is 0.3048 meters.
To obtain pounds per cubic foot from kg per cubic meter, divide by 16.01846.