shape given in the figure below The main channel of the stre

shape given in the figure below. The main channel of the stream can be taken as rectangular with a width of 1.5 m. Once the water level in the channel excceds a depth of 1.0 m, the channel width increases to 2.5 m 0.020 The channel bed slope is 0.001, and the channel has a Manning\'s roughness of n The current water depth (normal depth) is 0.50 m as indicated in the figure. An engineer wishes to measure the flow rate by installing a broad-crested weir. 2.5 m 0.5 m 0.5 m 1.0 m 0.5 m 1.5 m Water level before the weir is installed a) What is the minimum weir height that will force critical flow over the weir, assuming frictionless flow in the upstream vicinity of the weir? Answer: P0.175 m b) If a weir of height 1.0 m is installed in the channel, what will be the depth of water over the weir? You can assume that the discharge in the channel is the same as in part a), Answer: yc- 0.166 m

Solution

(a)

Let the weir height be H m.

Using Manning\'s equation,

Discharge (Q) = (1/n) x A x (A/P)^(2/3) x S^0.5

Before weir is installed,

A = 1.5 x 0.5 = 0.75 m²

P = 0.5 + 1.5 + 0.5 = 2.5 m

S = 0.001

n = 0.02

So, Q = (1/0.02) x 0.75 x (0.75/2.5)^(2/3) x 0.001^0.5 = 0.531 m³/s

V_initial = Q / A = 0.531 / 0.75 = 0.709 m/s

At the location of weir, critical flow will occur. So, Froude number = 1

V_weir / (g y_c)^0.5 = 1

Area of flow over weir, A_weir = 1.5 y_c

So, V_weir = Q / A_weir = 0.531 / 1.5 y_c = (0.354 / y_c) m/s

Now, (0.354 / y_c) / (g y_c)^0.5 = 1

=> (0.354 / y_c)² = g y_c

=> y_c³ = 0.354² / g = 0.354² / 9.81

=> y_c = (0.354² / 9.81)^(1/3) = 0.234 m

Initial total energy head (E_i) = 0.5 + V_initial² / 2g = 0.526 m

Total energy head over weir (E_w) = H + y_c + V_weir² / 2g = H + 0.234 + (0.354/0.234)² / (2 x 9.81) = H + 0.351

Since frictionless flow is considered, minimum weir height can be obtained by equating E_i and E_w

So, 0.526 = H + 0.351

=> H = 0.175 m

(b)

Let water depth be y m.

The water will try to flow at critical depth so as to achieve minimum specific energy.

Q = 0.531 m³/s

Weir height (H_w) = 1 m

Since above 1 m , width of the channel increases, so, B = 2.5 m

Area of flow over weir (A_w) = 2.5 y m²

Flow velocity over weir (V_w) = Q/A_w= 0.531 / (2.5 y) = (0.213 / y) m/s

For critical flow, Froude number = 1

=> V_w / (g y)^0.5 = 1

=> V_w² = g y

=> (0.213 / y)² = g y

=> y³ = 0.213² / 9.81

=> y = (0.213² / 9.81)^(1/3) = 0.166 m