# How to calculate flow rate of water through a pipe?

If a water pipe is 15 mm diameter and the water pressure is 3 bar, assuming the pipe is open ended, is it possible to calculate the flow rate or water velocity in the pipe?

Most of the calculations I have found appear to need 2 of these: diameter, flow rate, velocity.

So more specifically can you calculate flow rate or velocity from water pressure and pipe diameter?

Laminar Flow:

If the flow in the pipe is laminar, you can use the Poiseuille Equation to calculate the flow rate:

$$Q=\frac{\pi D^4 \Delta P}{128 \mu \Delta x}$$

Where $$Q$$ is the flow rate, $$D$$ is the pipe diameter, $$\Delta P$$ is the pressure difference between the two ends of the pipe, $$\mu$$ is dynamic viscosity, and $$\Delta x$$ is the length of the pipe.

If your pipe is carrying water at room temperature, the viscosity will be $$8.9\times 10^{-4} \, Pa\cdot s$$. Assuming the pipe is $$5\, m$$ long and that the $$3 \, bar$$ pressure is the gauge pressure, the flow rate is

$$Q = \frac{\pi (0.015)^4(3\times 10^5\,Pa)}{128(8.9\times 10^{-4} \, Pa\cdot s)(5\,m)}=0.0084 \frac{m^3}{s} = 8.4 \frac{l}{s}$$

However, if we calculate the Reynolds number for this flow rate:

$$V = \frac{Q}{A} = \frac{0.0084\frac{m^3}{s}}{\frac{\pi}{4}(0.015m)^2} = 48\frac{m}{s}$$ $$Re = \frac{\rho D V}{\mu} = \frac{(1000\frac{kg}{m^3})(0.015m)(48\frac{m}{s})}{8.9\times 10^{-4}\, Pa\cdot s}= 8\times 10^{5}$$

...we see that this flow is well into the turbulent regime, so unless your pipe is very long, this method is not appropriate.

Turbulent flow:

For turbulent flow, we can use Bernoulli's Equation with a friction term. Assuming the pipe is horizontal:

$$\frac{\Delta P}{\rho}+\frac{V^2}{2}=\mathcal{F}$$

where $$\mathcal{F}$$ accounts for friction heating and is given in terms of an empirical friction factor, $$f$$:

$$\mathcal{F} = 4f\frac{\Delta x}{D}\frac{V^2}{2}$$

The friction factor, $$f$$, is correlated to Reynolds number and pipe surface roughness. If the pipe is smooth, like drawn copper, the friction factor will be about 0.003 in this case. I got that value from "Fluid Mechanics for Chemical Engineers" by de Nevers, table 6.2 and figure 6.10. I also assumed that the Reynolds number will be about $$10^5$$. Substituting the equation for friction heating into Bernoulli's Equation and solving for velocity:

$$V=\sqrt{\frac{2 \Delta P}{\rho \left( 4f\frac{\Delta x}{D}+1 \right)}}$$

If your pipe is some other material with a rougher surface, then this analysis will over-predict the flow rate. I'd suggest looking for tables of friction factors for your particular material if you need higher accuracy.

• In any way I calculate this using the laminar flow calculation, the result is 0,084 m³/s and not 0,0084 m³/s. When I think like a practical guy, 0,084 m³/s seems a lot for such a pipe with this pressure, so I think your result is OK, but what am I missing? Jul 13 '17 at 14:52
• The Poiseuille's Equation given seems to accept dynamic viscosity in terms of Poise. 1 Pa.s = 10 Poise. Thus the 8.9E-04 should actually be 8.9E-03. See hyperphysics.phy-astr.gsu.edu/hbase/ppois.html That should fix things up.
– Tim
Sep 13 '17 at 18:30
• I echo @vasch findings, when I use the Poiseuille Equation provided here, or on many other websites, I get 0.084, not 0.0084. Also another question, what's the unit of measure for f (friction factor) ? May 6 at 1:46

General case

The basic tools for this kind of questions would be Bernoulli's equation, in the case of water, for an incompressible fluid.

$\frac{p}{\rho} + gz + \frac{c^2}{2} = const$

As you stated correctly you would at least need to know the velocity for one point. You can extend Bernoulli with pressure drop terms or combine it with the continuity equation and / or make a momentum balance depending the complexity of the problem. To be clear: I mentioned this tools because they are used for this kind of problem, they will not help you solve yours without you knowing more parameters.

Other possible prerequisites

• you know that the flow is the result from the hydrostatic pressure out of a big enough tank
• you know $\eta$ and $N$ of the pump responsible for the fluid flow

$\eta \equiv \text{efficiency}$

$N \equiv \text{power}$

Basically from what you currently stated, you cannot find out the velocity.

Getting an estimate anyway

You could assume that the pressure at the entry is constant and no flow occurs there. Neglecting friction losses and height differences you would get

$\frac{p_{in}}{\rho} + gz + \frac{c_{in}^2}{2} = \frac{p_{out}}{\rho} + gz + \frac{c_{out}^2}{2}$

$\frac{p_{in}}{\rho} = \frac{p_{out}}{\rho} + \frac{c_{out}^2}{2}$

$\sqrt{\frac{2(p_{in}-p_{out})}{\rho}} = c_{out} = 20 \frac{\mathrm{m}}{\mathrm{s}}$

$\dot{V} = cA = 10.60 \frac{\mathrm{L}}{\mathrm{min}}$

$\rho \equiv 1000\frac{\mathrm{kg}}{\mathrm{m}^3}$

$p_{out} \equiv 1 \mathrm{bar}$

$A \equiv \text{cross-sectional area of the pipe}$

This would do for a ballpark estimate. Alternatively you could get a bucket and measure how much water you can collect in a minute.

• In my set up, I know the water pressure at the start of the pipe. (it's mains water pressure so no pump or head of water, but there is a gauge on the pipe.) Mar 17 '16 at 12:26
• Is this an existing setup? How accurate do you need the result to be? Why can't you just measure the flow rate? Mar 17 '16 at 12:44
• Yes I can measure the flow rate at the end of the pipe, actually the end of the pipe is a small hole acting as a flow restrictor. I was just curious to know if the math behind the measured result is complex. Mar 17 '16 at 15:39
• Not really, since you are only interested in the flow rate. For a stationary flow the flow rate is constant or in general you have mass conservation. Everything the flows through the pipe has to flow out of the pipe eventually. Velocity can be calculated by $c A = \dot{V} = const$ Mar 17 '16 at 16:59

In order to calculate the average velocity or the flow rate, you need to determine, whether the flow is laminar or turbulent, and also calculate the friction coefficient.

The pressure drop in a circular pipe is given by :

$$\Delta P = \lambda \cdot \frac{L \cdot\rho}{2 \cdot D}\bar{u}^2$$

where:

• $$\lambda$$ is the pipe friction coefficient (See below(
• $$L$$ is the length of the pipe (for L 1200[m])
• $$\rho$$ is the density of the medium (for water 1000[kg/m])
• $$D$$: the diameter of the pipe (This is a design parameter)
• $$\bar{u}^2$$: the mean flow velocity in [m/s].

Obviously the parameters that are missing in this case are:

• $$\lambda$$ is the pipe friction coefficient
• $$\bar{u}$$ the mean flow velocity in [m/s].

Given the above we can calculate the average flow velocity as :

$$\bar{u}^2 \lambda= \Delta P \cdot \frac{2 \cdot D}{L \cdot\rho} \tag{eq.1}$$

For lambda again the determination of the mean flow velocity is crucial, and the calculation of the Reynolds number. Reynolds will be calculated in this case as:

$$Re = \frac{\bar{u} D}{v}$$

where:

• $$v$$ is the kinematic viscosity of the liquid ($$= 1.787[m^2/s]$$)

If Reynolds is:

• less than 2300 (Re<2300) then the flow is laminar and lambda is calculated as $$\lambda = \frac{64}{Re}$$

• more than 2300 (Re>2300) then the flow is turbulent and lambda is calculated by the colebrook equation (or the Darcy friction factor) as

$$\frac{1}{\sqrt{\lambda}} =-2 \ln\left(\frac{2.51}{Re \sqrt{\lambda}} + 0.269\cdot \frac{k}{D} \right)$$

where:

• $$k$$ is the absolute roughness in mm

Notice that this equation can be solved iteratively. So you need to assume a value of lambda, e.g. $$\sqrt{\lambda_{assumption}}=1$$, then do the calculations on the right hand and arrive at a value for $$\sqrt{\lambda_{calculated}}$$, then make the calculated value your new assumption and repeat until the difference between the $$\sqrt{\lambda_{assumption}}$$ and $$\sqrt{\lambda_{calculated}}$$ is very small.

At this point you should have everything you need to calculate the pressure drop due to friction.

## how to solve this

Steps to solve this:

1. Assume an initial value for $$Re$$. E.g. $$Re_0 =2000$$.

2. For the value you assumed, check whether the flow is laminar or turbulent. For the example $$Re_i =2000$$, the flow is laminar. If the flow is laminar then:

$$\lambda = \frac{64}{Re_i}$$

if the value is greater than 2300 then use the other equation.

1. Calculate an estimate value for $$\lambda$$ (eg $$\lambda_i$$).

2. Then you can go back to eq.1 and calculate an estimate for :

$$\bar{u}_i = \sqrt{\frac{\Delta P}{\lambda_i} \cdot \frac{2 \cdot D}{L \cdot\rho}}$$

1. Calculate an estimate for Reynolds based on that velocity

$$Re_{i+i} = \frac{\bar{u_i} D}{v}$$

1. Check whether $$Re_{i+i}$$ is approximately equal $$Re_{i}$$. If they are equal then stop. If they are not equal then return to step 2 and repeat the process.

Therefore the average velocity should be:

$$\bar{u} <2300\frac{v}{ D}$$

substituting that into