# What are the Units of flow coefficient Cv? And relationship between flow coefficient Cv formulas of incompressible and compressible fluids

I tried to understand flow coefficent Cv of a valve. There are many similar discussions on the Internet, e.g., https://www.pneumatictips.com/what-is-the-flow-coefficient-cv/

There are two formulas for Cv, one for incompressible fluids: $$C_v = Q \sqrt{\frac{SG}{\Delta P}} \tag{1}$$

and another for compressible fluids: $$C_v = \frac{Q}{960\sqrt{\frac{\Delta P(P_1+P_2)}{SG * T}}} = \frac{Q}{960}\sqrt{\frac{SG}{\Delta P}\frac{T}{(P_1+P_2)}} \tag{2}$$

Obviously, the units of Cv given by the above two formulas are completely different, while (1) gives the units of the Cv: $$\rm \frac{Gallon}{Min}$$ [Notes: As David Bailey pointed out my mistake, above units should be $$\rm \frac{Gallon}{Min} \frac{1} {\sqrt{PSI}} = \frac{GPM}{\sqrt{PSI}}$$ ]

(2) gives the units of the Cv: $$\rm \frac{SCFH.\sqrt{^\circ R}}{PSI} = \frac{Ft^3.\sqrt{^\circ R}}{H.PSI} \sim \frac{Gallon}{Min} \frac{\sqrt{^\circ R}}{PSI} \qquad {\color{red} \neq} \qquad \frac{Gallon}{Min} \frac{1} {\sqrt{PSI}}$$

WHY different? Can someone give me some explanation?

Furthermore, I never figured out how formula (2) was derived, any suggestion？

Any reply would be greatly appreciated.

The compressible fluid flow coefficient formula is just the incompressible formula with different units and with the ideal gas law applied under a particular convention.

I can find references to the compressible fluid flow formula going back to Brockett in 1952, but I cannot find a published derivation online. After reverse engineering the formula, however, it turns out to be relatively simple, but with confusing units and one less-than-obvious convention. I have since found a paper copy of an article that is consistent with my derivation: "Development of a universal gas sizing equation for control valves" (Buresh, James F., and Charles B. Schuder. ISA Transactions 1964 Vol. 3 Iss. 4, Pages 322-328).

Incompressible Flow

Let's first review the inspiration for the incompressible flow coefficient formula. The volume flow rate through a valve of area $$A$$ is just $$Q_v=A v_o$$ where $$v_o$$ and $$A$$ are the flow velocity and area of valve outlet. For a horizontal valve, from Bernoulli's principle we expect $$\frac{1}{2}\rho v_i^2+P_i = \frac{1}{2}\rho v_o^2+P_o \tag{Eq. A.1}$$ For zero inlet velocity, $$v_i$$, we have $$v_o^2 = 2\frac{P_i-P_o}{\rho} \tag{Eq. A.3}$$ so the volumetric flow through a valve is $$Q_v=A\sqrt{\frac{2 \Delta P}{\rho}} = C_v \sqrt{\frac{\Delta P}{S_l}} \tag{Eq. A.3}$$ where $$\Delta P \equiv P_i-P_o$$, $$C_v=A\sqrt{2/\rho_L}$$ is the naive volumetric flow coefficient, and $$S_l=\rho/\rho_L \tag{Eq. A.4}$$ is the specific gravity relative to a reference liquid density $$\rho_L$$.

Compressible Flow

There is much discussion in the literature about the best way to parameterize $$C_v$$. For example, Turnquist reported that the experimental volume flow in 32 different valves was better described with $$Q_v =C_v \left(1399 - 636 \frac{\Delta P}{P} \right) \sqrt{\frac{P\Delta P}{S_g T}} \tag{Eq. A.5}$$ For large pressure differences, the flow rate for a compressible fluid has a more complex relationship to pressure that depends on factors such as the heat capacity ratio $$\gamma$$, e.g. $$Q_v =C A \sqrt{2\frac{P_i}{\rho_i}\left(\frac{\gamma}{\gamma-1}\right)\left[(P_o/P_i)^{2/\gamma}-(P_o/P_i)^{(\gamma+1)/\gamma}\right]} \tag{Eq. A.6}$$

For small pressure differences, however, the gas can be treated as an incompressible fluid, but by convention with different units and and with the flow coefficient defined for a standard density.

Because mass, but not gas volume, is conserved, I found it conceptually easier to start with the mass flow per unit time $$Q_m =\rho Q_v = \rho C_v \sqrt{\frac{\Delta P}{S_l}} = C_v \sqrt{\rho \rho_L \Delta P} \tag{Eq. A.7}$$

The simplest assumption about how to modify the incompressible formula for compressible fluids is to assume the ideal gas law, $$PV=nRT$$, so the fluid density will be $$\rho =n/V = P/RT = \rho_0 \frac{T_0}{P_0}\frac{P}{T} \tag{Eq. A.8}$$ where the reference gas density ($$\rho_0$$) is defined at some reference temperature ($$T_0$$) and pressure ($$P_0$$). The mass flow (Eq. A.7) can now be rewritten as $$Q_m = C_v \sqrt{\rho_L \rho_0 \frac{T_0}{P_0}\frac{P\Delta P}{T} } \tag{Eq. A.9}$$ By a standard convention (see note below), the volumetric flow coefficient is defined at the reference density: $$Q_v = \frac{Q_m}{\rho_0} = C_v \sqrt{\frac{\rho_L}{\rho_0} \frac{T_0}{P_0} \frac{P\Delta P}{T} } \tag{Eq. A.10}$$ and the flow coefficient is : $$C_v \equiv Q_v \sqrt{\frac{\rho_0}{\rho_L} \frac{P_0}{T_0} \frac{T}{P\Delta P} } \tag{Eq. A.11}$$ This is a general form, but Equations 1 & 2 in the question are defined under the following conventions:

• Flow rates are given in SCFH (Standard Cubic Feet per Hour) for gases, but in GPM (US Gallons Per Minute) for liquids. The conversion factor is 1 GPM = 8.02 SCFH.
• The standard reference gas temperature and pressure is 60°F (519.7°R) and 1 atmosphere (14.696 PSI).
• The specific gravity for compressible gas flow is defined relative to air, not water as is the case for incompressible liquid flow. Since at the standard pressure and temperature, the water density is $$\rho_{water} = 999.0\,\mathrm{kg/m^3}$$ and the air density is $$\rho_{air} = 1.222\,\mathrm{kg/m^3}$$, the specific gravity for gases is $$S_g=817.5S_l$$, where $$S_g\equiv \rho_0/\rho_{air}$$ and $$S_l\equiv \rho_0/\rho_{water}$$.
• All temperatures in (°R) and and pressures in (PSI).
• The average pressure is $$P=(P_1+P_2)/2$$.

So Equation A.11 can be rewritten as $$C_v = \frac{Q_v(\mathrm{GPM})}{Q_v(\mathrm{SCFH})} \sqrt{ \frac{\rho_{air}}{\rho_{water}}} \sqrt{ \frac{P_0}{T_0}} \,Q_v(\mathrm{SCFH}) \sqrt{\frac{\rho_0}{\rho_{air}} \frac{T}{P\Delta P}} \tag{Eq. A.12}$$ and inserting the numerical values gives $$C_v = \frac{1}{8.02} \sqrt{\frac{1}{817.5}} \sqrt{\frac{14.696\,\mathrm{PSI}}{519.7\mathrm{°R}}} \,Q_v(\mathrm{SCFH}) \sqrt{ \frac{S_gT}{P\Delta P} } \tag{Eq. A.13}$$ This reduces to $$C_v = \frac{Q_v(\mathrm{SCFH})}{1364\,\mathrm{PSI}^{-1/2}\mathrm{°R^{1/2}}} \sqrt{ \frac{S_g T}{P\Delta P} } \tag{Eq. A.14}$$ or if using $$P_1+P_2$$ instead of $$P=(P_1+P_2)/2$$ $$C_v = \frac{Q_v(\mathrm{SCFH})}{964\,\mathrm{PSI}^{-1/2}\mathrm{°R^{1/2}}} \sqrt{ \frac{S_g T}{(P_1+P_2)\Delta P} } \tag{Eq. A.15}$$

If you round "964" to "960", this is just a rearranged version of the question's Equation 2, with it clear that "960" is not a dimensionless constant but has units.

Notes:

• There are various versions of the formula with slightly different coefficients, e.g. this 1961 article on "Comparing Gas flow Formulas for Gas Valve Sizing" has "1360", "1364", and "1390". Aside from possibly using slightly different values for water and air densities, the coefficient also depends on whether the input, output, or average pressure is used in the definition of $$C_v$$.

• According to Buresh & Schuder, the above formula (Eq. A.15)is accurate for $$\Delta P /P < 0.02$$, but looking at some of the figures in that paper and Brockett, it seems to often work reasonably well up to $$\Delta P /P \sim 0.3$$.

• Getting the correct $$T$$ and $$P$$ dependence is confusing. It helped me to remember $$C_v$$ is the volume of fluid that flows through the valve per unit time for a given pressure. According to Bernoulli's principle we expect this to be inversely proportional to $$\sqrt{S}$$, i.e. the denser the fluid the lower the flow. Since a fixed value for the specific gravity is used in the flow formula, but actual gas density is proportional to $$P/T$$, we must have $$C_v \propto \sqrt{T/P}$$ to compensate.

• Using this fixed value is consistent with the ISA-75.01.01 Standard on "Flow Equations for Sizing Control Valves" that states that "Volumetric flow rates for compressible fluids … refer to normal or standard conditions". Accordingly, equation A.14/15 is a definition of $$C_v$$ that corrects the observed volume flow $$Q_v$$ at some temperature and pressure to give the flow coefficient that would be observed under standard conditions (1013.25 mbar / 14.7 PSI and 288.6 K / 60°F).

• The original question had a mistake in the the units for the incompressible flow coefficient, which are actually $$\frac{GPM}{\sqrt{PSI}}$$.

• Give me SI units any day.

• Super nice answer! Thank you! David.
– John
Apr 22 at 3:17
• Relation about specific gravity S_g and S_l is not very clear, and the number 813 appearing in the formula is also not so clear. Although they should be right.
– John
Apr 22 at 11:58
• @John Thanks for the feedback. I tried to clarify things, but realize I have confused things even more. I am working on it. Apr 22 at 15:08
• I am reading. Thank you very much, because I know you have spent a lot of time on my question.
– John
Apr 22 at 15:19
• @John I am still not happy with the clarity of my explanation, but added a note at the end explaining why we must have a $C_v\propto\sqrt{T/P}$. I will think about this more later, and suggestions are welcome. Apr 22 at 16:30

Flow coefficient is a way of expressing the relationship between flow rate and pressure drop. So we could say $$Q$$ is a function of $$\Delta P$$:

$$Q = f(\Delta P)$$

For an incompressible fluid through a valve it might be possible to derive a good theoretical model for $$C_v$$, but any fluid, you could determine the relationships experimentally for instance by applying a number of different pressure differences and measuring the resulting flow rate; then you would find an equation that fitted the resulting data points. The link you provided acknowledges that (2) is not universally used by all engineers, it is a model you can use as a design guideline in certain situations. For an experimental (empirical) relationship, units don't have to be strict, if the valve was subject to some other variable, e.g. age, you could just arbitrarily add a unit of 'years in service' to the equation and suddenly $$C_v$$ has different units again.

To attempt to explain the difference: if you heat up a liquid (let's assume you are not near its boiling point!) and repeat your pressure drop/flow rate experiment, then this will likely have the same result because none of the fluid's properties have changed much.

However, if you repeat the experiment for a gas with different temperatures, you are likely to see a difference. If it is an ideal gas for instance we know that:

$$pv = nRT$$

Because the (specific) volume of the fluid changes with temperature and pressure, you need to account for that in determining the relationship of flow rate and pressure drop: temperature will have an effect, but also the total pressure of the system will have an effect and in the case of (2) it seems that $$P_{average} = 1/2(P_1 + P_2)$$ has been used as an approximation of how different the pressure is.

$$Q = f(\Delta P, P_{average}, T)$$

If $$\Delta P$$ is really big or small or you are dealing with a gas with unusual physical properties, the whole equation might break down! So always consider the operating conditions too.

• Thank you very much for your nice reply. Compressible fluid is much more complicated, formula (2) might be only an empirical relationship, but is there any deep explaniation? I've searched the web without success.
– John
Apr 21 at 3:39
• I can't come up with a mathematical derivation of the relationship. It would be nice to know who first came up with that equation wouldn't it! This product info sheet from 'Johnson controls' uses 963 instead of 960, it's a mystery! Apr 21 at 19:48

Under the hint of David Baily's answer of my question, we may have a simple relation between formula (2) and (1).

Let's derive from (2) to (1).

We rewrite (2)

\begin{align} C_v &= \frac{Q{(\rm SCFH)}}{964} \sqrt{\frac{SG_{\color{red}c}}{\Delta P}\frac{T}{(P_1+P_2)}} \tag{2} \\ &= \frac{Q{(\rm SCFH)}}{1364} \sqrt{\frac{1}{\Delta P} \frac{\rho}{\rho_{air}}\frac{T}{P}} \end{align}

According to above accepted answer by David Baily, $$\rho_{air} = \frac{\rho_{water}}{817.5},$$ and according to https://www.pneumatictips.com/what-is-the-flow-coefficient-cv/ : $$P = \frac{P_1+P_2}{2} =14.7 ~{\rm PSI}, \qquad T=520 ~{\rm ^\circ R}\sim 60 {\rm ^\circ F}.$$ Then \begin{align} C_v &= \frac{Q{(\rm SCFH)}}{1364} \sqrt{\frac{1}{\Delta P} \frac{817.5\rho}{\rho_{water}}\frac {520~ {\rm ^\circ R}}{14.7 ~ {\rm PSI}}} \\ &= \frac{Q{(\rm SCFH)}}{8.02} \sqrt{\frac{1}{\Delta P} \frac{\rho}{\rho_{water}}\frac { {\rm ^\circ R}}{{\rm PSI}}} \\ &= \frac{Q{(\rm SCFH)}}{8.02} \sqrt{\frac{SG_{\color{red}i}}{\Delta P} }\sqrt{\frac {\rm ^\circ R}{\rm PSI}} \\ &= Q{(\rm GPM)} \sqrt{\frac{SG_i}{\Delta P} }\sqrt{\frac {\rm ^\circ R}{\rm PSI}} \\ % &= \frac{Q{(\rm GPM)}}{8.02\times8.00} \sqrt{\frac{SG_i}{\Delta P} }\sqrt{\frac {\rm ^\circ R}{\rm PSI}} \\ % &{\color{red}\neq}~ Q{(\rm GPM)} \sqrt{\frac{SG_i}{\Delta P} }\sqrt{\frac {\rm ^\circ R}{\rm PSI}} \tag{1} \end{align}