# Deriving the ideal equation for sluice gate problems - incompressible inviscid fluid

I'm trying to figure this problem out, i know the correct answer, and i know the method, but for some reason i'm missing something and i'm just not getting there, any help would be greatly appreciated.

Consider a Sluice Gate (lock gate) as pictured,

A brief Description is.

A large reservour if fluid of depth $$h_1$$ is held stationary in $$x < 0$$ to the left of a vectical gate at the origin of width W. The gate is raised and fluid streams steadily through the opening between the bottom of the gate and the base of the reservoir.

Far upstream (the left of the gate) the fluid moves with constant velocity $$u_1$$ in the x horizontal direction and uniformly through the depth $$h_1$$. far downstream, (to the right of the gate) the fluid assumes a depth $$h_2$$ and moves with constant velocity $$u_2$$ in the horizontal x direction.

The job is to

1. Derive a relation expressing conservation of mass.
2. Use Bernoulli's equation along a suitable streamline in order to show two >possible choices of $$\frac{h_2}{h_1}$$ are $$\frac{h_2}{h_1} = 1$$ and >$$\frac{h_2}{h_1} =\frac{Fr^2+Fr\sqrt{Fr^2+8}}{4}$$ where $$Fr$$ is the Froude number $$Fr = \frac{u_1}{\sqrt{gh}}$$ Then finally
3. Find the net force on the raised gate

So,

1. Deriving a relationship expressing conservation of mass is just understanding that as the flow empties from one side, it must rush into the other at the same rate, and so we have $$Q = u_1 A_1 = u_2 A_2 \implies Q = u_1 h_1W = u_2 h_2 W$$ as our relation.

2. this has been bugging me quite a bit, i'll come back to this.

3. with This question, i believe the correct method is to pick a control volume at about the sluice, then combine the above conservation of mass equation with bernoulli's equation for a steady flow. so here i go...

We express Bernoulli's equation as $$\frac{1}{2} \rho u^{2}_1+\rho g h_1 - \frac{1}{2}\rho u^{2}_{2}+\rho g h_2= p_{constant_1} - p_{constant_2} = P_{c}$$ where we have h is the elevation, $$\rho$$ the density and u is the flow velocity. we have that the pressure also cancels out. and the above equation happens because bernoulli's is a constant along any streamline in the flow.

now using this line of logic, i find myself coming up with $$F = \rho g \frac{\left(1-\frac{h_2}{h_1}\right)^3}{2\left(1+\frac{h_2}{h_1}\right)}$$ which means im missing a step as the correct answer is actually

$$F_{net} = h_1^2 W\rho g \frac{\left(1-\frac{h_2}{h_1}\right)^3}{2\left(1+\frac{h_2}{h_1}\right)}$$

This has been really bothering me for the last few, so any help would be greatly appreciated. Thank you for taking the time to read this.

I think the missing point is that the force you want is not in the unit N/m^3, as the term (1−h2/h1)^3/2*(1+h2/h1) = K is unit less so in order to get the correct answer you should find the force as point load by (pgh* K * [Area = h W ]) as pgh is the pressure at the bottom pgh* K * h W = pgW h^2

You need to re-work the flow rate $$Q$$ as indicated below:

The Bernoulli energy equation may be applied in the cases where there is a negligible loss of total head from one section to another, or where the magnitude of the head loss is already known.

$$H_1 = H_2$$, so

$$y_1 + \dfrac {V_1^2}{2g} = y_2 + \dfrac {V_2^2}{2g}$$

Expressing the velocities in terms of Q, the above equation becomes

$$y_1 + \dfrac {Q^2}{2gW^2y_1^2} = y_2 + \dfrac {Q^2}{2g W^2 y_2^2}$$

Simplifying and re-arranging the terms, one obtains

$$Q = Wy_1\sqrt {\dfrac {2gy_1}{(y_1/y_2 )+1}}$$

See this paper for the derivation of the force, $$F_g$$ on the sluice gate. https://www.kuet.ac.bd/webportal/ppmv2/uploads/1582306005Expt%201%20Flow%20beneath%20sluice%20gate.pdf