Final answer:
Using the Chezy formula, we can show that in a circular pipe of diameter d, the velocity of flow will be at its maximum when the depth of flow y at the center is 0.81d. This is because the hydraulic radius R reaches its maximum value when y is around 0.81d, given R is derived from the flow area and wetted perimeter, which are dependent on y.
Step-by-step explanation:
To show that the velocity of flow in a circular pipe of diameter d will be maximum when the depth of flow y at the center is 0.81d, we can use the Chezy formula. The Chezy formula is given by V = C (R S)^0.5, where V is the velocity of flow, C is the Chezy coefficient, R is the hydraulic radius, and S is the slope of the energy grade line. For a circular pipe flowing partially full, the hydraulic radius R is equal to the area of flow divided by the wetted perimeter. Mathematically, when the pipe is flowing full (y = d), the hydraulic radius is R = d/4.
However, when the pipe is flowing partially full, we need to determine the relationship between the depth of the flow and the hydraulic radius. It can be shown that the hydraulic radius will be a maximum when the depth of flow is approximately 0.81d, meaning that it's at this depth where the velocity is at its peak. This occurs because the hydraulic radius describes the efficiency of the flow channel, with a greater hydraulic radius leading to a higher velocity, assuming slope and roughness are constant.
To achieve this, we would need to calculate the area of the flow and the wetted perimeter for different values of y, ranging from 0 to d, and find the point where dA/dy (the derivative of the area with respect to y) equals the wetted perimeter. This will occur at the optimum y value, which empirical observations suggest to be around 0.81d for a semicircular section. Using calculus, we can find this local maximum to illustrate that the flow velocity is maximized at this depth.