A hydraulic jump is formed in a 5.0 m wide rectangular channel with sequent depths of 0.2 m and 0.8 m. The discharge in the channel, inm3/s, is:a)2.43b)3.45c)4.43d)5.00Correct answer is option 'C'. Can you explain this answer?

Question

Answer

Given data:

Width of the rectangular channel (B) = 5.0 m

Initial depth of the channel (h1) = 0.2 m

Final depth of the channel (h2) = 0.8 m

Approach:

To find the discharge in the channel, we can use the energy equation for a hydraulic jump. The energy equation is given by:

(V1^2)/(2g) + h1 = (V2^2)/(2g) + h2

where V1 and V2 are the velocities of the flow before and after the jump, and g is the acceleration due to gravity.

Since the width of the channel is constant, the discharge (Q) can be obtained by multiplying the velocity with the cross-sectional area of flow.

Q = V * A

where A is the cross-sectional area of flow.

Solution:

From the energy equation, we can write:

(V1^2)/(2g) + h1 = (V2^2)/(2g) + h2

Substituting the given values:

(V1^2)/(2g) + 0.2 = (V2^2)/(2g) + 0.8

V1^2 - V2^2 = 2g(h2 - h1)

Since the flow is incompressible, the continuity equation can be used:

Q1 = Q2

V1 * A1 = V2 * A2

Since the width of the channel is constant, A1 = A2 = B * h

V1 * B * h1 = V2 * B * h2

Dividing both sides by B:

V1 * h1 = V2 * h2

From the above two equations, we can write:

V1^2 - V2^2 = 2g(h2 - h1)

V1 * h1 = V2 * h2

Substituting the values:

(V2 * h2)^2 - V2^2 = 2g(h2 - h1)

V2^2 * (h2^2 - 1) = 2gh(h2 - h1)

V2^2 = (2gh(h2 - h1))/(h2^2 - 1)

Q = V2 * B * h2

Substituting the values:

Q = ((2gh(h2 - h1))/(h2^2 - 1)) * B * h2

Q = (2 * 9.81 * (0.8 - 0.2) * 0.8) / (0.8^2 - 1) * 5.0

Q = 4.43 m^

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