Question 6.SP.1: The water passage in Fig. S6.l is 10 ft (3 m) wide normal to......

(BG units)
In free-surface flow such as this where the streamlines are parallel, the water surface is coincident with the hydraulic grade line. Writing an energy equation from the upstream section to the downstream section, where V = V_x,

6+{\frac{V_{1}^{2}}{2g}}=3+{\frac{V_{2}^{2}}{2g}}      (1)

From continuity,      {\mathfrak{s}}(10)V_{1}=3(10)V_{2}     (2)

Substituting Eq. (2) into Eq. (1) yields

\begin{array}{l l}{{V_{1}=8.02\,\mathrm{f}\mathrm{ps}}}&{{V_{2}=1605\,\mathrm{fps}}}\\\\ {{Q=A_{1}V_{1}=A_{2}V_{2}=481\,\mathrm{cfs}}}\end{array}

Next take a free-body diagram of the control volume (CV) of water shown in the figure and apply the momentum equation (6.7a),

F_{1}-F_{2}-F_{x}=\rho Q(V_{2}-V_{1})

where F, represents the force of the structure on the water (CV) in the horizontal direction, and the F,s and V,s are understood to have no y components.

where F_x represents the force of the structure on the water (CV) in the horizontal direction, and the F^,s and V^,s are understood to have no y components.

From Eq. (3.16), we have F₁ = -γh_c1A ₁ and F₂ = -γh_c2A ₂• Hence

62.4(3)(10\times6)-62.4(1.5)(10\times3)-F_{x}\,=\,1.94(481)(16.05-8.02)

and             F_{x}\,=\,+936\ 1{\mathrm{b}}\,=\,936\ 1{\mathrm{b}}\leftarrow

The positive sign means that the assumed direction is correct. The force of the water on the structure is equal and opposite, namely,

(F_{W/S})_{x}\:=\:936\ 10\to

Note that the momentum principle will not permit us to obtain the vertical
component of the force of the water on the shaded structure, because the
pressure distribution along the bottom of the· channel is unknown. We can
estimate the pressure distribution along the boundary of the structure and along
the bottom of the channel by sketching a flow net and applying Bernoulli’s
principle. Then we can find the horizontal and vertical components of the force
by computing the integrated effect of the pressure-distribution diagram.

(SI units)

Energy:         2+\frac{V_{1}^{2}}{2(9.81)}\,=\,1+\frac{V_{2}^{2}}{2(9.81)}                     (3)

Continuity:     2(3)V_{1}\,=\,1(3)V_{2}      (4)

Substituting Eq. (4) into Eq. (3) yields

\begin{array}{l l}{{V_{1}\,=\,2.56\,{\mathrm{m/s}},\,\,\,\,\,V_{2}\,=\,5.11\,{\mathrm{m/s}}}}\\\\ {{Q\,=\,A_{1}V_{1}\,=\,A_{2}V_{2}\,=\,15.34\,{\mathrm{m}}^{2}/{\mathrm{s}}}}\end{array}

Applying momentum equation (6.7a) to the free-body diagram,

\begin{array}{c}{{F_{1}-F_{2}-F_{x}=\ \rho Q(V_{2}-V_{1})}}\\\\ {{9.81(1)(3)-9.81(0.5)(1)(3)-F_{x}=\ 1.0(15.34)(5.11-2.56)}}\\\\ {{F_{x}=\ +.91\ \mathrm{kN}\ -\ 4.91\ \mathrm{kN}\ =\ 4.91\ \mathrm{kN}\ -\ 2.56)}}\end{array}

So   (F_{W/S})_{x}\;=\;4.91\;\mathrm{kN}\rightarrow

\sum F_{x}\,=\,\imath\dot{n}(\Delta V_{x})\;=\;\rho G(\Delta V_{x})\;=\;\rho G(V_{2x}-\ V_{1x})                    (6.7a)

F=\gamma h_c A                 (3.16)