Contraction to 1.25 m: is flow possible?

Given data

$b_1=4$ m, $y_1=1.8$ m, $V_1=1.4$ m/s, contracted width $b_2=1.25$ m (no hump).

$$Q=b_1y_1V_1=4\times1.8\times1.4=10.08\ \text{m}^3/\text{s},\qquad q_1=\frac{10.08}{4}=2.52\ \text{m}^2/\text{s}$$ $$E_1=1.8+\frac{1.4^2}{19.62}=1.8999\ \text{m}$$

(a) Is the flow possible?

At the contraction, $q_2=\dfrac{10.08}{1.25}=8.064$ m²/s.

$$y_{c2}=\sqrt[3]{\frac{q_2^2}{g}}=1.878\ \text{m},\qquad E_{min2}=1.5y_{c2}=2.818\ \text{m}$$

Since the available energy $E_1=1.900$ m $<E_{min2}=2.818$ m, the flow is not possible with this specific energy (the flow would be choked and the upstream level would rise).

(b) Discharge possible with the given specific energy

The maximum flow through the contraction occurs with critical flow at $E_{min2}=E_1$:

$$y_c=\frac23E_1=1.2666\ \text{m},\qquad q_{max}=\sqrt{gy_c^3}=4.465\ \text{m}^2/\text{s}$$ $$Q_{max}=q_{max}b_2=4.465\times1.25=5.581\ \text{m}^3/\text{s}$$

So the discharge should not exceed about 5.58 m³/s.

(c) Depths

• At the contracted section: critical, $y=y_c=1.267$ m.
• Upstream: with $Q=5.581$ m³/s, $q_1=1.3952$ m²/s and $E_1=1.8999$ m (subcritical root):

$$y_1+\frac{1.3952^2}{19.62y_1^2}=1.8999\;\Rightarrow\;y_1=1.872\ \text{m}$$

Answer: not possible for 10.08 m³/s; the discharge must be reduced to $Q\approx5.58$ m³/s; depth at the contraction $=1.27$ m (critical), depth upstream $\approx1.87$ m.