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 upstream, (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
- Derive a relation expressing conservation of mass.
- 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- 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.
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this has been bugging me quite a bit, i'll come back to this.
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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.
Best Answer
The formula for the Froude number has a misprint. It should be $Fr = \frac{u_1}{\sqrt{gh_1}}$.
Part (2): Bernoulli's equation along the bottom streamline is
$$\frac{1}{2} \rho u_1^2 + p_a + \rho g h_1 = \frac{1}{2} \rho u_2^2 + p_a + \rho g h_2$$
Subtracting the ambient pressure $p_a$ from both sides and dividing both sides by $(\rho g h_1)/2$ we get
$$\tag{1}\frac{u_1^2}{g h_1} + 2 = \frac{u_2^2}{g h_1} + 2\frac{h_2}{h_1}$$
From mass conservation we have $u_2 = \frac{h_1}{h_2} u_1$. Substituting for $u_2$ and using $Fr = \frac{u_1}{\sqrt{g h_1}}$, (1) reduces to
$$Fr^2 + 2 = Fr^2 \left(\frac{h_2}{h_1}\right)^{-2} + 2 \frac{h_2}{h_1}$$
Multiplying both sides by $\left(\frac{h_2}{h_1} \right)^2$ and rearranging we get
$$\tag{2} 2 \left(\frac{h_2}{h_1}\right)^3 - (Fr^2 +2) \left(\frac{h_2}{h_1}\right)^2 + Fr^2 = 0 $$
We see immediately that one root of this cubic equation is $x := \frac{h_2}{h_1} = 1$, and so (2) must factor as
$$\tag{3} (x-1)(\alpha x^2 + \beta x + \gamma) = 0$$
Solving for $\alpha$, $\beta$ and $\gamma$ to equate the LHS of (2) and (3) we get
$$\alpha = 2, \quad \beta = -Fr^2, \quad \gamma = - Fr^2$$
Thus, the other solution for $x = \frac{h_2}{h_1}$ must be the positive root of the quadratic equation
$$2x^2 - Fr^2 x - Fr^2 = 0,$$
which is
$$\frac{h_2}{h_1} = x = \frac{-(-Fr^2) + \sqrt{(-Fr^2)^2 - 4(2)(-Fr^2)}}{2(2)} = \frac{Fr^2 + Fr\sqrt{Fr^2 + 8}}{4}$$
Part (3):
To solve for the force of the gate $F_{net}$ acting on the fluid use a control volume that includes all of the fluid between station (1) and (2). The pressure force at station (1) is
$$W\int_0^{h_1} \rho g z \, dz = \frac{1}{2}W\rho gh_1^2,$$
and, similarly, the back pressure at station (2) is $-\frac{1}{2}W\rho gh_2^2$.
The total force from pressure and the gate must balance the net momentum change in the control volume according to
$$-F_{net} + \frac{1}{2}W\rho gh_1^2 - \frac{1}{2}W\rho gh_2^2 = \rho W h_1u_1(u_2 - u_1) $$
You should be able to complete the solution from here.