Force to hold a sluice gate in place

[George32]

Homework Statement

An incompressible, inviscid liquid of density ρ flows, under the influence of gravity, beneath the sluice gate shown. The height and velocity upstream of the gate are h and V, respectively, while downstream of the gate the height is l. Show that the force per unit width necessary to hold the gate in place is given by
$$\frac {\rho g(h-l)^3} {2(h+l)}$$
Hint : Requires consideration of hydrostatic pressure, a momentum balance over a control
volume incorporating three forces acting on three vertical surfaces, and use of the Bernoulli equation.

Homework Equations

Using a conservation of mass, $$Vh=Ul$$ where U is the velocity downstream.
Using Bernoulli, $$V^2(1 - \frac {h^2} {l^2}) = g(l-h)$$

The Attempt at a Solution

Taking $$\dot {m} = \rho v h$$ and $\dot {m} v = \rho h v^2$
Using the Steady Flow Momentum Equation:

$$\Sigma \dot{m_i} V_in - \Sigma \dot{m_o} V_out = F - \Sigma pA$$
So filling into this I get: $$\rho h v^2(1-\frac {h}{l}) = F - \frac {1}{2} \rho g (h^2 -l^2)$$
Filling in $V^2 = \frac{g(l-h)}{1-\frac{h^2}{l^2}}$ and rearranging:
$$F = \rho g( (h-\frac{h^2}{l})(\frac{l^2 (l-h)}{l^2 - h^2}) + \frac{1}{2}(h-l)(h+l) )$$
Working through the rearrangement:
$$F = \rho g( (h-\frac{h^2}{l})(\frac{l^2}{l+h}) + \frac{1}{2}(h-l)(h+l))$$
$$F = \rho g( \frac{hl^2 - lh^2}{l+h} + \frac{1}{2}(h-l)(h+l) )$$
$$F = \rho g( \frac{2(hl^2 - lh^2) + h^3 +h^2 l - l^2 h - l^3}{l+h} )$$
$$F= \frac{\rho g}{l+h} ( h^3 - h^2 l + h l^2 - l^3)$$
Which comes down to:
$$F= \frac{\rho g}{l+h} (h-l)(h^2 + l^2)$$

which is obviously not the right answer. In order to get the $(h-l)^3$ that they have in their answer, I would need: $2(h(l^2) - l(h^2)) + (h^2 + l^2)(h-l)$ On the top line in the top row (after adding fractions)

I remember having a similar issue in a tutorial, which was also a show that question, which came out to the same issue (the tutor couldn't make it work).

I just want to know if anyone else can make it work, and if so, if they could nudge me, otherwise I think it's possible that the question is wrong...

Thanks

 

[KataKoniK]

for any help in advance.



, Scientist



Thank you for sharing your attempt at solving this problem. I can understand your frustration with not being able to get the same answer as the one provided in the forum post. After reviewing your work, I believe that there may be a small mistake in your algebraic manipulation. Let me walk you through the solution that I came up with, and hopefully it will help you see where the discrepancy lies.

First, let's start with the conservation of mass equation, which you correctly identified as Vh=Ul . From here, we can solve for U and substitute it into the momentum equation:

U = \frac{Vh}{l}

Plugging this into the momentum equation, we get:

\rho h V^2 (1-\frac{h}{l}) = F - \frac{1}{2} \rho g (h^2 - l^2)

Now, let's focus on the left side of the equation. We can factor out \rho V^2 and rearrange the terms to get:

\rho V^2 (1-\frac{h^2}{l^2}) + \frac{1}{2} \rho g (h^2 - l^2) = F

Now, let's use the Bernoulli equation, which states that V^2 + gh + \frac{p}{\rho} = constant . Since the liquid is incompressible and inviscid, we can ignore the pressure term, leaving us with:

V^2 + gh = constant

Substituting this into our equation, we get:

\rho (V^2 + gh) (1-\frac{h^2}{l^2}) + \frac{1}{2} \rho g (h^2 - l^2) = F

Expanding and rearranging, we get:

\rho V^2 (1-\frac{h^2}{l^2}) + \rho gh (1-\frac{h^2}{l^2}) + \frac{1}{2} \rho g (h^2 - l^2) = F

Now, let's factor out \rho g and rearrange the terms to get:

\rho g (V^2 + h - \frac{h^3}{l^2} - \frac