Question 7.11: What is the horizontal force required to hold the thin sluic...

This question asks for the horizontal component of the external force acting on the sluice gate in the open and closed positions. We know that this force component must balance the horizontal force of the air and water on the gate in each case. In the necessary sketches for this problem (Figure 7.20A, 7.20B), the horizontal forces of air and water on the gate are shown in the expected directions. The horizontal external force is shown pointing in the negative y direction. Our horizontal force balance on the gate in each case is

F_water − F_air − F_E = 0                                     (A)

In the closed position the pressure distribution in the water is hydrostatic and it is constant in the air. There is no need for a momentum balance, since we can write the net hydrostatic force applied by both fluids on the gate by considering the pressure distribution on each side as shown in Figure 7.20B. The net horizontal force of the air and water is found to be:

F_water − F_air = \left(\frac{\rho gD_1}{2}\right)D_1L-\left(\frac{\rho gD_2}{2}\right)D_2L=\left[\frac{1}{2}\rho gD^2_1L-\frac{1}{2}\rho gD^2_2L\right]

Using (A) the horizontal external force needed to keep the gate stationary in the closed position is found to be

F_E=\frac{1}{2}\rho gD^2_1L-\frac{1}{2}\rho gD^2_2L                                    (B)

This positive value confirms that in the closed position the external force points to the left as assumed in Figure 7.20B.

In the open position we choose a mixed CV as shown in Figure 7.20C and use Eq. 7.20b,

\int_{CS}^{}{(ρv)(u • n)dS= F_{B_Y}+ F_{S_Y}}

noting that the body force is zero in the y direction. On the inlet we have u= V₁j, n = − j, and u • n = −V₁. On the exit, u= V₂j, n = j, and u • n = V₂. The momentum transport terms are found to be (ρV^2 _2 A_2 −ρV^2 _1 A_1), where A₁ = D₁L and  A₂ = D₂L. We next examine each section of the control surface shown in Figure 7.20C, dropping those on which the surface forces do not act in the y direction. The remaining surfaces are the inlet, the decal surfaces on the left and right sides of the gate exposed to air (on which the surface forces cancel each other), the vertical surfaces at the exit exposed to air and water, respectively, and the decal surface adjacent to the bottom of the channel where we have shown a reaction shear force R_shear acting to the left due to friction. We must also include the external horizontal force F_E. Since the gate is completely inside and thus part of the mixed CV, this force is transmitted to the gate across some section of the control surface. The horizontal momentum balance becomes

We are told that pressure distribution on the inlet and exit surface is hydrostatic. When we sketch the pressure distributions on the inlet, exit and air–exit surface as shown in Figure 7.20D, we see that the sum of the three integrals containing these pressures yields

\int_{inlet}^{}{p_1(z)dS}+\int_{air-exit}^{}{-p_AdS}+\int_{exit}^{}{-p_2(z)dS}

=\left[p_A+\frac{\rho gD_1}{2}\right]D_1L − p_A(D_1 − D_2)L-\left[p_A+\frac{\rho gD_2}{2}\right]D_2L

=\frac{1}{2}\rho gD_1^2L-\frac{1}{2}\rho gD_2^2L

Thus the horizontal component of the external force needed to hold the gate open is

F_E=\left(\frac{1}{2}\rho gD_1^2L-\frac{1}{2}\rho gD_2^2L\right)-\left(\rho V^2_2A_2-\rho V^2_1A_1\right)-R_{shear}

In most cases the reaction force due to shear on the bottom is neglected because it is unknown and presumed to be small. Note that it is also possible to solve this problem with the CV containing fluid only shown in Figure 7.20E.