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The application of Appropriate Technology

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

Figure 17

Figure 17

Taking the pump pressure requirement equation from Section 8 iv:

P_p=\frac{1}{2}\rho.v_2^2+\rho.g.h_2-\rho.g.h_1+f_h.\rho.g

And assuming the kinetic energy term is negligible gives us:

P_p=\rho.g.(h_2-h_1+f_h) 

(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 (Win):

W_{in} = P_p .Q/\gamma

and Equation (21) which gives the electrical power required (We):

W_e = I.V

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 (h2 – h1) 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:

P_p=1000\times 9.81\left (20 + 49.14\right )=678263.4~Pa

For 1″ dia. pipe:

P_p=1000\times 9.81\left (20 + 3.86\right )=234066.6~Pa

Taking the density of water as ρ= 1000 Kg/m3 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 m3/s:

For ½” dia. pipe:

W_{in}=678263.4\times 0.0005/0.5=\bf 678.3~{Watts}

For 1″ dia. pipe:

W_{in}=234066.6\times 0.0005/0.5=\bf{234.1~Watts}

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:

I =W_e/V

Assuming no electrical losses We = Win, so:

I=234.1/110=\bf{2.13~Amps}