Consider a collector or spring tank at an elevation h_{1}, supplying a reservoir tank at an elevation h_{2}.

Taking the equation from Section 8 i:

And assuming the kinetic energy term is negligible gives us:

Simplifying gives us:

(31)

This means that for the natural flow condition the actual head (h_{1} – h_{2}) must equal the frictional head (f_{h}) “burned off” by the pipe. This means scanning the frictional head loss chart (interpolating where necessary) to find the frictional head loss that exactly equals the actual head for the given pipe diameter and length. The following numerical example will demonstrate this process.

**Numerical Example**

A spring tank is 500m from a reservoir tank, the difference in height between the two is 5m. It is proposed to use 1″ diameter pipe to link the two, what will be the volumetric flow rate of the water assuming a natural flow situation ?

**Answer **

Knowing h_{1} – h_{2} = 5, we can write Equation (31) as:

Now there are 500m of pipe and we are looking for a frictional head loss in the chart per 100m of pipe. So divide the actual head loss by 5 (as we have 500m of pipe). So we are looking for the frictional head loss per 100m of 1m on the chart for 1″ dia. pipe.

Scanning the chart gives us the following values for 1″ dia. pipe:

Flow Rate (LPS) | Frictional Head Loss (m/100m) |
---|---|

0.19 | 0.62 |

0.25 | 1.07 |

Using the interpolation method (see Appendix 5) we can find the flow rate to give us exactly 1m of frictional head loss per 100m of 1″ dia. pipe.

The variables for the general interpolation Equation 30 are given below:

X_{1} = 0.62, X_{2} = 1.07, Y_{1} = 0.19, Y_{2} = 0.25 and X_{3} = 1.

Substitute these into the general interpolation Equation 30 to give:

And so, the approximation is Y_{3} = 0.2406 which can be rounded up to **0.24 LPS**.

This is the flow that will produce a frictional head loss of 5m over 500m of 1″ dia. pipe and is therefore the natural flow rate of the system.