Trapezoidal to rectangular transition with bed lowering

Given data

Trapezoidal section: $b=6$ m, side slope 2H:1V ($z=2$), $y_1=2.5$ m, $Q=60$ m³/s.

$$A_1=(b+zy)y=(6+2\times2.5)(2.5)=27.50\ \text{m}^2,\quad V_1=\frac{60}{27.50}=2.1818\ \text{m/s}$$ $$E_1=y_1+\frac{V_1^2}{2g}=2.5+\frac{2.1818^2}{19.62}=2.7426\ \text{m}$$

Rectangular section: $b_2=6$ m, so $q_2=\dfrac{60}{6}=10$ m²/s, $y_{c2}=\sqrt[3]{\dfrac{100}{9.81}}=2.168\ \text{m}$.

(i) Bed lowered by 0.6 m

Energy above the new bed: $E_2=E_1+0.6=3.3426$ m.

$$y_2+\frac{q_2^2}{2gy_2^2}=y_2+\frac{5.0968}{y_2^2}=3.3426$$

Take the subcritical root, since the upstream flow is subcritical ($T_1=6+4\times2.5=16$ m, $D_1=A_1/T_1=1.719$ m, $Fr_1=2.1818/\sqrt{9.81\times1.719}=0.53$):

$$y_2=2.5724\ \text{m}$$

Water surface level in the rectangular section (datum: bed of the trapezoidal channel) $=y_2-0.6=1.9724$ m, compared with 2.500 m upstream.

Change in water surface level $=1.9724-2.5=-0.5276$ m, i.e. a fall of 0.528 m.

(ii) Water surface to fall by only 0.30 m

Let the bed be lowered by $x$. Then the downstream water level is $2.5-0.3=2.2$ m, so $y_2=2.2+x$. Energy balance, $E_1+x=E_2$, gives

$$(2.2+x)+\frac{5.0968}{(2.2+x)^2}=E_1+x\;\Rightarrow\;2.2+\frac{5.0968}{(2.2+x)^2}=2.7426$$ $$(2.2+x)^2=\frac{5.0968}{2.7426-2.2}=9.3929\;\Rightarrow\;2.2+x=3.0648$$ $$x=0.8648\ \text{m},\quad y_2=3.065\ \text{m}$$

Answer: (i) depth in the rectangular section $=2.57$ m; the water surface falls by 0.53 m; (ii) the bed must be lowered by $\approx0.86$ m (depth $3.06$ m).