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What is the Relationship Between the Flow Coefficient (Cv) and Resistance Coefficient (K)?

by Jeff Sines, Senior Product Engineer at Engineered Software, Inc.

Accuracy of hydraulic calculations is critical for the proper design, operation, and determination of cost for many types of piping systems in residential, commercial, and industrial applications. It is crucial that the engineer understand and apply the correct formula to prevent costly mistakes in the sizing and selection of equipment, operating within safety limits, and avoiding unnecessary modifications later in the process.  One aspect that leads to mistakes is the misuse of coefficients that characterize the hydraulic performance of devices that have a fluid flowing through them.

For example,  the hydraulic performance of a control valve is characterized by its Flow Coefficient over its range of travel (Cv in US units, Kv in SI units), whereas the hydraulic performance of other piping systems devices (such as isolation valves, check valves, tees, and other fittings) is typically characterized by a dimensionless Resistance Coefficient (K). The two are inversely related in that the flow coefficient represents how much flow capacity an obstruction allows, whereas the resistance coefficient represents how much resistance to flow the obstruction presents.

Both concepts express the Conservation of Energy for a flowing fluid through an obstruction.

The Resistance Coefficient

The Resistance Coefficient, K, is a dimensionless value given by the equation:

(1) \[ K=f\frac{L}{D}\]

Where

f = Darcy Friction Factor, dimensionless

L / D = Equivalent Length Ratio of the obstruction, dimensionless

L = Length of pipe in feet

D = Inside diameter of the pipe in feet

 

The Resistance Coefficient is used in a form of the Darcy Equation below:

(2) \[ h_L=K\frac{v^2}{2g}\]

Where

hL = hydraulic energy lost (head loss) from the fluid due to friction, in ft

v = average fluid velocity, in ft/sec

g = gravitational acceleration, 32.2 ft/sec^2

 

The head loss, in imperial units, is related to pressure drop, in psi, by the fluid density in lb/ft3:

(3) \[ h_L=\frac{144dP}{\rho}\]

Re-arranging the Equation 2 to solve for K and substituting into Equation 3 yields:

(4) \[ K=\bigg(\frac{2g}{v^2}\bigg)h_L=(2g)(144)\bigg(\frac{dP}{\rho v^2}\bigg)\]

Conceptually, the Resistance Coefficient represents:

\[ K\rightarrow (Constant)\bigg(\frac{\text{Change in Hydraulic Energy of the Fluid (dP term)}}{\text{Kinetic Energy of the Fluid } (\rho v^2\ term)}\bigg)\]

The Flow Coefficient

In its simplest form, the Flow Coefficient is given by the equation:

(5) \[ C_v=\frac{Q}{\sqrt{\frac{dP}{\bigg(\dfrac{\rho_{fluid}}{\rho_{water}}\bigg)}}}\]

 

Where:

Q = volumetric flow rate (in gpm)

dP = pressure drop across the obstruction (in psi)

ρ  fluid = density of fluid (lb/ft3)

ρ water = density of water at standard conditions, 62.37 lb/ft3

 

This makes gpm/sqrt(psi) the units of the Flow Coefficient, which makes it difficult to see what the Cv represents conceptually. The equation can be manipulated to bring the concept into focus.

The volumetric flow rate, Q in gpm, is equal to a unit conversion constant times the flow area, times the average fluid velocity:

(6) \[ Q\frac{gal}{min}=\bigg(\frac{7.4805\ gal}{ft^3}\bigg)\bigg(\frac{60\ sec}{min}\bigg)\bigg(A\frac{ft^2}{}\bigg)\bigg(v\frac{ft}{sec}\bigg)=448.8\ Av\frac{gal}{min}\]

Where:

A = cross-sectional area of the flow path, in ft2

 

Squaring both sides of Equation 5 and substituting in Equation 6 for Q gives:

(7) \[ C_v^2=\Bigg[\frac{Q}{\sqrt{\frac{dP}{\bigg(\frac{\rho_{fluid}}{\rho_{water}}\bigg)}}} \Bigg]^2=\bigg[\frac{(448.8Av)^2}{dP}\bigg]\bigg(\frac{\rho_{fluid}}{\rho_{water}}\bigg)=\bigg[\frac{(448.8A)^2}{\rho_{water}}\bigg]\bigg(\frac{\rho v^2}{dP} \bigg)\]

 

Since the area and reference density of water is constant, conceptually the square of the Cv represents:

\[ C_v^2\rightarrow(Constant)\bigg(\frac{\text{Kinetic Energy of the Fluid}}{\text{Change in Hydraulic Energy of the Fluid}}\bigg)\]

Relationship Between Cv and K

To determine the relationship between Cv and K, Equation 4 and Equation 7 can be re-arranged to solve for dP and equated to each other:

Solve Equation 4 for dP:

(8) \[ dP=\frac{K\rho v^2}{(2g)(144)}\]

 

Solving Equation 7 for dP:

(9) \[ dP=\bigg[\frac{(448.8A)^2}{\rho_{water}}\bigg] \bigg(\frac{\rho v^2}{C_v^2} \bigg)\]

 

Equating Equations 8 and 9 yields:

(10) \[ \frac{K \rho v^2}{(2g)(144)}=\frac{\rho (448.8Av)^2}{\rho_{water} C_v^2}\]

 

Simplifying:

(11) \[ K=\frac{(2g)(144)(448.8)^2 A^2}{(62.37)C_v^2}\]

 

For circular flow passage with diameter, d in inches:

(12) \[ A=\frac{\pi}{4}D^2=\frac{\pi}{4}\bigg(\frac{d}{12}\bigg)^2\]

Where: 

A = pipe flow area in ft2

D = pipe diameter in feet

d = pipe diameter in inches

 

Finally, substituting Equation 12 into Equation 11:

(13) \[ K=\frac{(2)(32.2)(144)(448.8)^2\bigg[\frac{\pi}{4}\bigg(\frac{d}{12}\bigg)^2\bigg]^2}{(62.37)C_v^2}\]

Simplifying the unit conversions into one constant leads to the familiar equation showing the relationship between Cv and K:

 

(14) \[ K=890.9\frac{d^4}{C_v^2}\]

The following example shows the application of the above relationship in PIPE-FLO® Professional. 

 

The above image specifies a fixed K value of 1.6 on a pipeline. This causes a 3.558 psi pressure drop across the pipe.

Using equation 14 and an inner pipe diameter of 3.548 in, the equivalent Cv value is calculated to be 297. This also causes a 3.557 psi pressure drop. 

 

Summary

Another example of confusion found with the application of coefficients is seen in spray nozzle manufacturing which uses a "Nozzle Discharge Coefficient” shown as a K value. However, a review of their literature shows that they define K = Q/sqrt(dP). This makes the nozzle discharge coefficient equivalent to the flow coefficient used for control valves. Using the nomenclature K for the nozzle discharge coefficient confuses the issue even more.

Because engineers view the hydraulic performance of devices differently, mistakes can be made if the proper equations are not applied. As this paper points out, these coefficients essentially represent the Conservation of Energy by defining how much hydraulic energy changes as the fluid flows through an obstruction.

Correctly understanding the concepts and applying methodologies will prevent costly mistakes in sizing and selection of equipment.  Also the correct methods will ensure operating safety limits are properly established to avoid unnecessary future modifications. 

Take at look at this article for the Relationship between the Flow Coefficient (Cv) and the Discharge Coefficient (Cd), commonly used for safety relief valves and flow meters.