In a rectangular channel, there occurs a jump corresponding to Fr₁ = 2.5. Then the critical depth expressed in terms of initial depth y₁ is:

Explanation:

Fraud number is given by

\(F_{r1} = {V_1 \over {√{gy_{1}}}}\)  ........ (1)

Discharge per unit width (q) is given by

q = V1 × y1

Substituting V1 from the above equation, we get

V1 = Fr1√gy1

\(q = F_{r1} \sqrt {gy_1} × y_1 = F_{r1} \sqrt {g} × y _{1}^{3/2}\)   ........ (2)

Now, we know that, critical depth for rectangular channel is given by

\(y_{c}=\left(\frac{q^{2}}{g}\right)^{1/3}\)

Putting q from equation 2 in the above equation, we get yc as

\(y_{c}=\left(\frac{\left(F_{r1}\sqrt{g}y_{1}^{3/2}\right)^{2}}{g}\right)^{1/3}\)

\(y_{c}=\left(F_{r1}^{2}× g× y_{1}^{3}\over g\right)^{1/3} =\left(F_{r1}^{2}× y_{1}^{3}\right)^{1/3}\ \\ y_c =F_{r1}^{2/3}× y_{1}\)

Putting Fr1 as 2.5 in the above equation, we get

yc = (2.5)2/3 × y1 = 1.84 × y1

$y_1$