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The application of Appropriate Technology

Consider a similar situation to Worked Example 1, except that a control valve has been inserted into the line, immediately before the discharge point (a tap).

Figure 16

Figure 16

Taking the equation from Section 8 ii:

P_2-P_1 = \rho.g.h_1- \rho.g.h_2-frac{1}{2}.\rho.v_2^2 -f_h.\rho.g

Assuming the kinetic energy term to be negligible and rewriting P2 – P1 as the residual pressure at the valve or ΔP we get:

\Delta P = \rho.g.(h_1 - h_2) - f_h.\rho.g

If we divide by ρ.g, then we can convert this equation from pressure (Pa) to head (m):

\Delta H=(h_1-h_2)-f_h

(34)

Where ΔH is the residual head at the valve.

Numerical Example

A reservoir tank is 100m from a valve and tap arrangement, the difference in height between the two is 5m. It is proposed to use 100m of 1″ dia. pipe to link the two. Assuming the volumetric flow rate of the water is set by adjusting the valve to 0.25LPS, what is the residual head at the valve ?

Answer

Knowing h1 – h2 is 5m we can substitute this into Equation (34) to give us:

\delta H=5-f_h

Now we know the volumetric flow rate is set at 0.25LPS, so consulting the friction chart for 1″ diameter pipe we find that the frictional head loss is 1.07m/100m for this flow rate. We have 100m of pipe, so therefore the head loss (fh) is 1.07m.

This means that the residual head at the valve is:

\Delta H =5-1.07 = \bf{3.93m}

See Appendix 4 Part 6 for the acceptable residual heads at valves and taps.