Problem

Referring to Figure 1, determine the rate at which water is being withdrawn from the syste...

Referring to Figure 1, determine the rate at which water is being withdrawn from the system at point A when the pump discharge pressure is 350 kPa and the pressure at point A is 250 kPa. The storage tank floats on the line with water at a height of 20 m, as shown. Sketch the HGL for the system. Determine at what rate the storage tank is being filled (or emptied) under these conditions.

FIGURE 1

Illustration for Example.

Example

A peak hourly flow rate of 100 L/s is required at point A, as illustrated in Figure 1. Pressure at that point is not to drop below 150 kPa. Determine (a) the required pressure at the pump station if the demand is satisfied without any distribution storage tank, and (b) the required pressure head at the pumps if the storage tank is used to help to meet peak demand by floating on the line at elevation 20 m, as shown.

Solution

(a) Compute the drop in pressure head from the pump station to the point of withdrawal, as follows:

For Q = 100 L/s and D = 250 mm, read S = 0.024 on the Hazen–Williams nomograph (Figure 2). Compute hL = S × L = 0.024 × 2000 = 48 m.

At point A, the pressure is not to drop below 150 kPa. Using Equation 1, one sees that 150 kPa is equivalent to 0.1 × 150 or 15 m of pressure head. Since flow occurs in the direction of sloping HGL, the 48 m of head loss must be added to the 15-m minimum pressure head requirement at point A. The HGL is shown in Figure 1. The pressure head is then 48 + 15 = 63 m, and using Equation 2, we find that the required pressure at the pump station is P = 9.8 × 63 ≈ 620 kPa.

(b) In this part of the problem, the storage tank is contributing flow to the withdrawal point. The head loss in the 500 m of pipeline is 20 – 15 = 5 m. The slope of the HGL is then 5/500 = 0.01, and from the Hazen–Williams nomograph, with D = 200 mm and S = 0.01, read Q = 32 L/s.

Because the total withdrawal is still 100 L/s, the flow from the pump station must be 100 – 32 = 68 L/s. From the Hazen–Williams nomograph, with Q = 68 L/s and D = 250 mm, read S = 0.012. The head loss between the pump station and point A is then hL = 0.012 × 2000 = 24 m, and the required pressure head at the pump station is then 15 + 24 = 39 m; the required pressure is P = 9.8 × 39 ≈ 380 kPa.

It can be seen in this problem that the required pump capacity decreases from 100 L/s at a discharge pressure of 620 kPa to a capacity of 68 L/s at a pressure of 380 kPa when peak hourly demand is partially satisfied by water from a distribution storage tank.

FIGURE 2

A nomograph that provides a graphical solution to the Hazen–Williams equation for water flowing in circular pipes under pressure, with C = 100.