Question 6.11: As in Examples 6.9 and 6.10, water flows through a pipeline ......
As in Examples 6.9 and 6.10, water flows through a pipeline of 1.2 m diameter that has a hydraulic gradient of 1 in 250 and a roughness k of 0.60 mm. Find the velocity of flow and the discharge using the Hydraulic Research chart (Fig. 6.15), assuming the pipeline is flowing full.
Step 1: Find the chart that relates to k = 0.60 mm.
Step 2: Express the gradient in terms of m per 100 m. Thus 1/250 is 0.4 m per 100 m.
Step 3: Find the 0.4 m/100 m hydraulic gradient line in the right-hand margin of the chart. Note that these lines slope from top right towards bottom left at the same angle as the numbers are printed in the margin (For future reference, hydraulic gradients > 0.6 m/100 m have the number printed within the chart about 10 mm from the top.)
Step 4: Find the line representing D = 1.2 m in the bottom margin of the chart. The numbers are printed vertically to correspond with the vertical lines representing diameter.
Step 5: Find the point of intersection between the 0.4 m/100 m line and D = 1.2 m line.
Step 6: From this intersection point move horizontally across the chart and read the velocity printed in the left-hand margin (note, horizontal lines, horizontal numbers in the margin). Thus V = 2.35 m/s, about the same as in Example 6.9.
Step 7: From the intersection point move up the line sloping from bottom right towards top left, and read the discharge from the sloping numbers: Q = 2750 litres/s (2.750 m³/s). Again, this is more or less as in the previous examples.