Consider the arrangement of a collector tank, pump and reservoir tank shown in Figure 17.

Taking the pump pressure requirement equation from Section 8 iv:

And assuming the kinetic energy term is negligible gives us:

(36)

This is the general equation for calculating the pressure requirement for a pump.

Combined with two other Equation (20) which gives the* *power that needs to be supplied to the pump (W_{in}):

and Equation (21) which gives the electrical power required (W_{e}):

it is possible to derive the pump specification for a series of different diameter delivery pipes, assuming a given volumetric flow rate.

**Numerical Example**

A reservoir tank is 100m uphill from a water source, the difference in height between the two is 20m. It is proposed to use a pump to push the water up to the reservoir tank at a flow rate of 0.5LPS. Two pipe diameters of ½” and 1″ are available to link the two. What size pumps are required (in Watts) for each pipe diameter, assuming a pump efficiency of 50% and what electrical current is required assuming the smaller pump is chosen and we are stealing the electricity from a 110V pylon ?

**Answer **

We need to find the required pump pressure for each diameter of delivery pipe. We know the difference in height between the source and the reservoir tank (h_{2 }– h_{1}) is 20m. For a flow rate of 0.5LPS the frictional head loss chart gives the following results for ½” and 1″ dia. pipes:

Flow Rate (LPS) | Frictional Head Loss (m/100m) ½” dia. | Frictional Head Loss (m/100m) 1″ dia. |
---|---|---|

0.5 | 49.14 | 3.86 |

As we have 100m of pipe in each case these values can be inserted directly into Equation 36 to give the pump pressure requirement:

For ½” dia. pipe:

For 1″ dia. pipe:

Taking the density of water as ρ= 1000 Kg/m^{3}_{ }and the acceleration due to gravity as g = 9.81m/s/s.

These two pump pressures can now be inserted into Equation (20) to give the pump power requirement, taking the pump efficiency (γ ) as 0.5 and converting the volumetric flow rate (Q) from LPS to m^{3}/s:

For ½” dia. pipe:

For 1″ dia. pipe:

Note that the power requirement for the 1″ dia. delivery pipe is less than half of that for the ½” dia. pipe.

Taking the smaller pump (for the 1″ dia. delivery pipe) and using Equation (21) we get a electrical current requirement (I) of:

Assuming no electrical losses W_{e} = W_{in}, so: