Channeling fluid flow through pipes is an everyday occurrence.  We see it in plumbing, industrial plants, and fluid power (hydraulic) systems.  A proper understanding of the basics of fluid flow in pipes–including measurement and losses–is essential for proper configuration of these systems.  In this monograph we’ll concentrate on incompressible fluids; compressibility introduces some complications.

Classical Flow Metering: Theory

Euler’s inviscid equation states that

X_i-\frac{1}{\rho}\frac{\partial p}{\partial x_i} = \frac{\partial v_i}{\partial t} + \left( v\cdot grad \right)v_i

The fact that it is inviscid is important; we will see why shortly.

Slater and Frank (1947) show that this can be transformed to Bernoulli’s Equation, thus

\rho V + p + \frac{ 1}{ 2} \rho v^2 = constant

or, for passage from State 1 to State 2,

\rho_1 V_1 + p_1 + \frac{ 1}{ 2} \rho_1 u_1^2 = \rho_2 V_2 + p_2 + \frac{ 1}{ 2} \rho_2 u_2^2 .

Bernoulli’s Equation is basically an expression of conservation of energy.  There are three terms on each side:

  1. The first term is the energy from an external force field.  For practical applications, this means gravity.
  2. The second is the pressure energy.
  3. The third is the velocity energy.  We use u for velocity as a borrowing from CFD, u being the velocity in the x-direction and v being the velocity in the y-direction.  We’ll stick with one-dimensional flow in this case.

For our purposes we will ignore the external force field/gravity term.  Leaving out the gravity term, Bernoulli’s Equation can be written thus:

p_1 + \frac{ 1}{ 2} \rho_1 u_1^2 = p_2 + \frac{ 1}{ 2} \rho_2 u_2^2

where p is the pressure and \rho is the density.

Let us now add the assumption that the fluid is incompressible, thus

\rho_1 = \rho_2 = \rho

Dividing through by the density yields

\frac{p_1}{\rho} + \frac{ 1}{ 2} u_1^2 = \frac{p_2}{\rho} + \frac{ 1}{ 2} u_2^2

At this point the business about being inviscid comes into play; both Euler’s and Bernoulli’s Equations assume no energy losses due to viscosity.  We could try something really fancy (like starting with the Navier-Stokes equations, which were actually first formulated by Saint-Venant) or we could do something simplistic like add a head (energy per unit mass) loss, like this:

\frac{p_1}{\gamma} + \frac{ 1}{ 2g_c} u_1^2 - h_L= \frac{p_2}{\gamma} + \frac{ 1}{ 2g_c} u_2^2

Note that we have also divided through by the acceleration due to gravity g_c ; this is so that the units of head loss will be in unit length, which will match them with a manometer.  \gamma is the unit weight of the fluid.

Rearranging terms yields

\frac{p_1 - p_2}{\gamma} - h_L= \frac{ 1}{ 2g_c} \left( u_2^2 - u_1^2 \right)

Now let us turn to continuity of mass flow, which (unless a leak is sprung somewhere) is not subject to loss, or

\rho_1 A_1 u_1 = \rho_2 A_2 u_2

Since the fluid is incompressible, the densities cancel out.  At this point we also make another simplifying assumption: the velocity of the fluid across the cross-section of the pipe or restriction is uniform, or at least that the variation across the cross-section is not significant.  Solving for u_1 ,

u_1 = \frac{A_2}{A_1} u_2

Substituting this into the last form of Bernoulli’s Equation yields

\frac{p_1 - p_2}{\gamma} - h_L= \frac{ u_2^2}{ 2g_c} \left( 1 - \left( \frac{A_2}{A_1} \right)^2 \right)

If we are reading a U-tube manometer, the difference in column heights will be

\Delta H = \frac{p_1 - p_2}{\gamma}

Substituting this bring us to

\Delta H - h_L= \frac{ u_2^2}{ 2g_c} \left( 1 - \left( \frac{A_2}{A_1} \right)^2 \right)

Assuming the uniform flow across the cross-section, the flow rate is

Q = u_1 A_1 = u_2 A_2

Substituting and solving for the flow rate, we have at last

Q = A_2 \sqrt{\frac{2g_c\left( \Delta H - h_L \right)}{\left( 1 - \left( \frac{A_2}{A_1} \right)^2 \right)}}

Up to now, it probably seems that we’ve pulled several “rabbits out of a hat.”  These are simplifying assumptions that are based on experimental experience, the physical reality of the experiment, or both.  But we have saved the biggest rabbit for last: we will dispense with the head loss term h_L (Mott (1994)) and restate the equation above as

Q = C_d A_2 \sqrt{\frac{2g_c\Delta H}{\left( 1 - \left( \frac{A_2}{A_1} \right)^2 \right)}}

C_d is referred to as the coefficient of discharge; 0 < C_d < 1 .  We can rearrange this equation to determine C_d experimentally as follows:

C_d = \frac{Q}{A_2 \sqrt{\frac{2g_c\Delta H}{\left( 1 - \left( \frac{A_2}{A_1} \right)^2 \right)}}}

Classical Flow Metering: Implementation

Now that we have all of this theory, we ask ourselves, “What’s it good for?”  There are two ways we can take this; the first thing we’ll consider is flow measurement.  In short, we can set up a place in the flow where we change the flow area and, by measuring the pressure difference, we can determine the flow.  This is the way flow has been measured in pipes for many years.

Let’s consider a couple of examples.  The first is a sharp-edged orifice, simply a plate placed in the flow stream as shown below.

Sharp-Edged Orifice. From Lewitt, E.H. (1923) Hydraulic and Fluid Mechanics. London: Sir Isaac Pitman & Sons, Ltd.

Here we see a U-tube manometer which measures pressure on the upstream and downstream side of the orifice.  The fluid velocity varies as we have shown.  The placement of the orifices is an important problem in using this type of device.

The major problem with this type of orifice is that it will measure fluid flow, but generates significant losses in use.  A more efficient measuring device from that standpoint is the venturi meter, shown below.

Venturi Meter. From O’Brien, M.P., and Hickox, G.H. (1937). Applied Fluid Mechanics. New York: McGraw-Hill Book Company. George Hickox worked for the Tennessee Valley Authority; the building of the original TVA system had many technological spin-offs.

The setup is similar except for the (God forbid) mercury manometer, which worked but, because of its hazards, is mercifully rare.  The practical drawback to a venturi flowmeter is the long pipe length it occupies, which can be difficult to include in a busy piping system.

The efficiency variation can be seen in the coefficients of discharge, which vary from around 0.6 for a sharp-edged orifice to just under unity for a venturi meter.  There are also flow meters which occupy a middle ground between the two, such as the rounded edge manometer.

One might calibrate a flow meter by measuring its discharge coefficient C_d against the Reynolds Number R_e .  The meaning of R_e is discussed in the monograph Wind Tunnel Testing.  To compute R_e  the diameter is customarily the large diameter and the velocity the average velocity (uniform velocity across the pipe) as described earlier.  Here are some pointers on getting the Reynolds number right.  If water is being used in the test, typical properties are given in the monograph Variation in Viscosity.

As an example, consider a venturi where the inlet pipe diameter is 1.48″ and the throat diameter is 0.9″.  A water flow of 4 gallons/minute (GPM) flow through the venturi at room temperature.  The water manometer shows a head differential of 0.8″.  What is the discharge coefficient?

First, we do some preliminary conversions.  The cross-sectional area of the venturi inlet pipe is 1.72 and of the throat 0.64  The flow in the inlet pipe is

Q = \frac{4 \frac{gal}{min} \times 231 \frac{in^3}{gal}}{60 \frac{sec}{min}} = 15.4 \frac {in^3}{sec}

Note that we’re using inches instead of feet.  For problems of this scale, it’s really more convenient, even though it’s not really a “proper” way to do it.  This means that g_c = 386.4 \frac{in}{sec^2} .  Since we now have a completely consistent set of units, we can substitute and compute the discharge coefficient:

C_d = \frac{Q}{A_2 \sqrt{\frac{2g_c\Delta H}{\left( 1 - \left( \frac{A_2}{A_1} \right)^2 \right)}}} = \frac{15.4}{0.64 \sqrt{\frac{2 \times 386.4 \times 0.8}{\left( 1 - \left( \frac{0.64}{1.72} \right)^2 \right)}}} = 0.904

We said we’d like to compute the Reynolds number for this discharge coefficient.  Reynolds number is

Re = \frac {VD_1}{\nu}

The diameter for this case is the inlet diameter, as noted in Getting the Reynolds Number Right.  We need first to compute the average velocity, which is

V_1 = \frac {Q}{A_1} = \frac {15.4}{1.72} = 8.95\,\frac{in}{sec}

Using the value of kinematic viscosity given in Variation in Viscosity and inserting the appropriate unit conversions yields

Re = \frac {8.95 \frac{in}{sec} \times 1.48 {in}}{10.877 \times 10^{-6} \frac {ft^2}{sec} \times 144 \frac {in^2}{ft^2}} =  8,459

Other Flow Meters

As flow meters go, classical flow meters are not as common as they used to be to measure flow.  We think of other methods as “new” but the Rotameter was first patented in 1908.  At the right is a variation of the concept dating from 1914.  Rotameters are still very commonly used, although their accuracy can be highly variable.  Today we have devices such as paddle wheel and magnetic flowmeters to more accurately measure the flow and to report the results to computer controlled systems.

“Minor” Losses in Systems

Another important application of this theory is the prediction of “minor” losses in such restrictions as valves, elbows, enlargements or contractions in the piping and other flow constrictions in the system.  The use of the term “minor” is misleading; careless design can turn these into major losses very quickly, and the designer must be diligent in avoiding them if he or she wants a successful design.  (Unless, of course, you’re trying to restrict fluid flow…)

Picking up where we left off in the theory, dividing both sides by the small area yields

u_2 = C_d \sqrt{\frac{2g_c\Delta H}{\left( 1 - \left( \frac{A_2}{A_1} \right)^2 \right)}}

Eliminating the square root,

u_2^2 = C_d^2 \frac{2g_c\Delta H}{\left( 1 - \left( \frac{A_2}{A_1} \right)^2 \right)}

Solving for the head differential,

\Delta H = \frac{u_2^2}{2g_c} \frac{\left( 1 - \left( \frac{A_2}{A_1} \right)^2 \right)}{C_d^2}

Since we normally prefer to consider the velocity in the incoming pipe u_1 , we substitute to

\Delta H = \frac{u_1^2}{2g_c} \frac{\left( \frac{A_1}{A_2} \right)^2\left( 1 - \left( \frac{A_2}{A_1} \right)^2 \right)}{C_d^2}

and then

\Delta H = \frac{u_1^2}{2g_c} \frac{\left( \left( \frac{A_1}{A_2} \right)^2 - 1 \right)}{C_d^2}

We normally prefer to determine the second fraction on the right hand side experimentally as a loss or resistance coefficient K , so at last

\Delta H = K \frac{u_1^2}{2g_c}

The derivation indicates that, unlike the discharge coefficient, the loss or resistance coefficient can and frequently does exceed unity.  It is obviously possible to use this equation to determine K experimentally by solving for it.  This is done by rearranging the previous equation as follows:

K = \Delta H \frac{2g_c}{u_1^2}

As an example, consider a gate valve with an inlet diameter of 1.48″.  Water is flowing through this and the manometer is likewise a water manometer.  For a measured through flow of 15.9 gallons per minute, the manometer reads a differential height of 1.3″.  What is the K value of the gate valve at this position and flow?

First, we compute the cross-sectional area of the inlet pipe (the customary “reference area” in this case.)  That of course is

A = \frac {\pi D^2}{4} = \frac{\pi \times 1.48^2}{4} = 1.72\,in^2 .

The flow must be converted as follows


The average velocity is thus the flow divided by the area, or


Since g_c = 386.4\frac{in}{sec^2} , the loss coefficient is thus

K = \Delta H\frac{2g_{c}}{u_{1}^{2}}=1.3\,in.\frac{2\times386.4\frac{in}{sec^{2}}}{\left(35.58\frac{in}{sec}\right)^{2}}=0.79

Flow in Pipes

It may seem like the long way around, but with the topic of flow metering and “minor” losses out of the way, we now consider the issue of losses in pipes.  For straight pipes, the “classic” implementation of this is the Moody Chart, first published around World War II.  One of these is shown below (from this site):


Note that the roughness values k are in millimetres, not feet, as is customary in U.S. textbooks.

Although the chart can be used “as is” correlations have been developed, which are certainly more convenient in a spreadsheet than the chart.  Some of those are shown on the chart above; another one (good for turbulent flow in the range 5000 < R_e < 1 \times 10^8 ) can be found in Mott (1994):

f = \frac{1}{4\left[ log\left( \frac{\epsilon}{3.7 D}+\frac{5.74}{R_e^{0.9}} \right) \right]^2}

where R_e is the Reynolds Number and the fraction and \frac{\epsilon}{D} is the relative roughness of the pipe, or the ratio of the height of the asperities along the wall of the pipe to the diameter.  Keep in mind that the logarithm in this equation is a common, base 10 logarithm, not a natural logarithm.

As an example, consider a 12 gpm flow of water at room temperature through a 12′ straight stretch of 1/2″ Type L copper tubing.  What is the head loss we would expect?

The first thing we need to note is that Type L copper tubing has an I.D. of 0.545″ = 0.0452′.  Again Mott (1994) gives typical values of roughness for this type of tubing as 5 \times 10^{-6} ft.  This means that the relative roughness of the pipe is \frac{5 \times 10^{-6}}{0.0452'} = 0.00011   (Some textbooks invert this ratio; to be consistent with the Moody chart above, we’ll stick with the arrangement we have.)  So now we have the properties of the copper tubing we need.

The head loss (the ordinate on the Moody Chart) we would expect is given as follows:

{\it h_f}=1/2\,{\frac {fL{u}^{2}}{D g_c}}

The Reynolds number (the abscissa on the Moody Chart) is given earlier, and the reader is exhorted to review Getting the Reynolds Number Right.

One thing we can do to make things a little simpler (this from Streeter (1966)) is to rewrite both formulas to allow direct input of the flow.  Doing this yields

{\it h_f}=8\,{\frac {fL{Q}^{2}}{{D}^{5}{\pi }^{2}g_c}}


R_e=4\,{\frac {Q}{\pi \,D\nu}}

It is important that the units of flow are consistent with the rest of the units.  Since we’re going to be doing this in feet, 12\,gpm=0.0267\,\frac{ft^3}{sec} .

Knowing the Reynolds Number and the relative roughness, we can proceed.  The Reynolds number (with properties of water from Variation in Viscosity) is

R_e=4\,{\frac {0.0267}{\pi \,0.0452\,10.877 \times 10^{-6)}}} = 68,910

This is in the turbulent range; we can use the formula above.  The friction factor is thus

f = \frac{1}{4\left[ log\left( \frac{0.000005}{3.7\,0.0452}+\frac{5.74}{68910^{0.9}} \right) \right]^2} = 0.0199

Comparison with the Moody Chart should show that this is very close.

The head loss is thus

{\it h_f}=8\,{\frac {0.0199\,12\,{0.0267}^{2}}{{0.0452}^{5}{\pi }^{2}\,32.2}} = 22.19' = 266.4"

The inches conversion is done because most manometers–physical and digital–measure water head differences in inches.

Equivalent Length

We can combine our work on both minor losses and pipe losses by considering the concept of equivalent length of pipe.  Thus, we can say that, for a given diameter of pipe, a restriction has the same effect on the pressure drop of a system as a certain length of pipe L_e would have.  That length of pipe is given by the equation

L_e = D \frac{K}{f}

The friction factor f is determined from the Moody chart (discussed earlier) in the turbulent region only.

Equivalent length is a good practical way of looking at the losses in a section of pipe with all of its restrictions without getting into “apples and oranges” comparisons.  Since they are in series, the equivalent and actual lengths of the pipe are simply added.


Losses in fluid elements such as valves and pipes can be considered in two ways.  Orifices such as the sharp-edged orifice and the venturi can be used to measure the flow in the system.  The theory can be flipped to enable us to estimate losses in the system and thus avoid unwanted energy, pressure and flow degradation of the fluid system.