Fluid Mechanics - Resistance to fluid motion using Chezy's formula

[Michael V]

Homework Statement

The quantity of water delivered through a buried pipe line that is 2 000 m long and has an inside diameter of 200 mm has over the years steadily decreased to 80 % of its original flow. This is due to encrustation inside the pipe. It was then decided to have the pipe cleaned and lined with cement mortar 5 mm thick. The friction constant of this lining is 80 % better than it is for normal black piping. Calculate the proportion of water that will now be delivered relative to the original quantity, assuming that the total friction loss remains constant and that the static head is negligible.

Homework Equations

Chezy formula: v=C$\sqrt{mi}$
Q=A×v

The Attempt at a Solution

It is assumed that original quantity refers to that delivered when the pipe was clean, thus the information regarding the reduction in diameter to 80% of original is redundant. Written answer in attachment.

 

[nilesh_pat]

The Chezy constant is defined as C=\frac{8fL}{d^2g} where f is the friction coefficient and L is the length of pipe.Assuming that the static head is negligible, the flow rate, Q, can be determined using the Chezy formula, v=C\sqrt{mi}, where m is the slope of the pipe. As the friction coefficient has improved by 80%, the new Chezy constant, C_{new}, can be calculated:C_{new}=\frac{8f_{new}L}{d^2g}=\frac{8\times0.8\times f_{old}L}{d^2g}=\frac{6.4f_{old}L}{d^2g}Therefore, the new flow rate, Q_{new}, can be calculated using the new Chezy constant:Q_{new}=A\sqrt{\frac{6.4f_{old}L}{d^2g}}Relative to the original quantity of water delivered, the proportion of water that will now be delivered is \frac{Q_{new}}{Q_{old}}=\frac{\sqrt{\frac{6.4f_{old}L}{d^2g}}}{\sqrt{\frac{8f_{old}L}{d^2g}}}=\frac{\sqrt{\frac{6.4}{8}}}{1}=\frac{\sqrt{\frac{3}{4}}}{1}=\sqrt{\frac{3}{4}}