Calculating Minimum Pipe Pressure Requirement for an Irrigation System

Question

I am setting up an irrigation system for my ranch and I have a hard time calculating the minimum pipe pressure requirements to reduce cost. I need to know the maximum pressure the pipe has to cope with. The pipe would transfer water from a high elevated pond (with enough water to feed the line constantly) to a barn down below. There is a 166 meter height difference between the inlet and outlet. I can easily calculate the upper bound pressure as 16.6 bars, however as I get it, the friction loss plays an important role too. The pipe's length is 1200 meters and made of polyethylene (Absolute roughness of 0.007). We need to open and close the faucet on the exit pint on demand (important) so the water head may reach the full height of 166 meters when the outlet is closed. The question is: do I have to order the 16.6 bar pipe or a lower grade pipe would do?

Answer 1

One way to calculate the pressure drop consist of the following steps:

• Determine if the flow is laminar or turbulent
• Calculate the friction coefficient
• Put them all together and calculate the pressure drop $\Delta P$.

I'll start from the end. The pressure drop in a circular pipe is:

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

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 there are two parameters that are missing in this case:

• the diameter of the pipe $D$
• the volume flow rate of material $\dot{V} [m^3/s]$

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

$$\bar{u}= \frac{\dot{V}}{A}= \frac{\dot{V}}{\pi \frac {D^2} 4} $$ $$\bar{u}= \frac{ 4 \dot{V}}{\pi D^2} $$

Therefore the initial equation can be rewritten so that it encapsulates the main two design parameters as :

$$\Delta P = \lambda \cdot \frac{L \cdot\rho}{2 \cdot D}\left(\frac{ 4 \dot{V}}{\pi D^2}\right)^2$$ $$\Delta P = \lambda \cdot \frac{ 8 L \cdot\rho}{\pi^2}\cdot\frac{ \dot{V}^2}{ D^5}$$

Now only pareameter that is not known is $\lambda$.

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:

$$ Re = \frac{\bar{u} D}{v} = \frac{\frac{ 4 \dot{V}}{\pi D^2} D}{v} $$

$$ Re = \frac{4 }{\pi \cdot v}\frac{\dot{V}}{D} $$

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

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

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.

Pressure drop due to height

The pressure drop due to height is simply given by:

$$\delta P_h = \rho\cdot g \cdot \Delta H$$

Total pressure

The total pressure will be given by:

$$\Delta P_{total} =\Delta P_h + \Delta P_{friction} + \Delta P_{elements} $$

where:

• $\Delta P_{elements}$ is the pressure drop on valves, elbows and any other elements apart from the straight pipe. I will assume this is zero for this application, but you need to consider it.

Final thought

Although the above is the rough calculation for the pressure drop, IMHO you should go for at least two pumps (one at the base and one at the middle). That will allow you to use more commonly used pipes and avoid expensive and difficult to find components. Also that might lower overall the cost of the pump, and make it easier to procure and maintain. You will probably incur though some initial setup costs.

This is relevant to question

Answer 2

If the lower grade pipe is not rated for that pressure, it'll bust. The solution is to transfer the water to a water reservoir/tank in a lower elevation, which acts to relieve the excessive pressure by reducing the head. But you might have to adjust the pipe size to maintain the flow rate.