How to calculate acceleration in a funnel

[James O'Neill]

I have seen some tidal turbines that have funnel ducts around them to improve efficiency.
So how would one calculate water acceleration in a funnel? If you want figures:
water velocity - 3 m/s
funnel opening A (largest) - 8 meters
funnel opening B (narrowest) - 5 meters
Distance between A and B - 4 meters

 

[Geofleur]

Do you actually want the acceleration of an element of water ($\frac{d^2\textbf{r}}{dt^2}$) as it moves through the funnel, or just the difference in velocity between an element entering the funnel and one that is leaving ($\textbf{v}_{out} - \textbf{v}_{in}$)?

 

[James O'Neill]

I want to calculate the water velocity as it exists the smallest part of the funnel. So probably

the difference in velocity between an element entering the funnel and one that is leaving

 

[Geofleur]

The fluid flux can be written $\rho u$, where $\rho$ is the density and $u$ the velocity. If you multiply $\rho u$ times the area $A_{big}$ of the funnel at the big end and times a time interval $\Delta t$, that will give you the amount of fluid entering in time $\Delta t$. This quantity should be equal to $\rho u$ times the area $A_{small}$ at the small end, times $\Delta t$, as long as the fluid is incompressible (which is a good approximation here). You can then solve the resulting equation for the velocity at the small end. Finally, you can take the difference between the two velocities to get what you want.

 

[James O'Neill]

So if you say density, do you mean the density of sea water?
If so, then:
ρ = 1024 kg/m³
u = 4 m/s
A_big = 54 m²
Δt = 5 mins
A_small = 27 m²

ρu x A_big = 4096 x 54 = 221 184
221 184 x Δt = 221 184 x 5 = 1 105 920
ρu x A_small = 4096 x 27 = 110 592 x 5 = 552 960

Is this right so far? (It doesn't seem right)

 

[Geofleur]

It's not quite right, but it's headed in the right direction! The velocity on the little side is unknown, so your second $\rho u$ should not also be 4096. Also, it's good to put the units in when you write a number, so that would be 4096 kg/m$^2$ s. You will have $\rho u_{big} A_{big} = \rho u_{small} A_{small}$, with $u_{big}$ known, and you'll want to solve for $u_{small}$.

 

[Merlin3189]

Since volume out must equal volume in and density is constant, you can simplify this to
Ain x Vin = Aout x Vout or Vout = Vin x Ain / Aout = 3 x 64 /25 = 7.68 m/sec
There's a link here.

I can't follow your calculation because an 8m input and 5m output doesn't seem to tie in with 54 m² and 27 m², unless the pipe changes shape as well as size. (In which case, there's no point in saying 8m and 5m alone!)
In the calculation above, I just assumed the input and output are similar and the linear measurements are corresponding lengths.

But what is worrying me here is the pressure difference that this represents. It seems very big. So how do you know the water is going to flow at 3 m/sec into this funnel?

(PS. I regret I will be out of touch for the next 10 days, so won't be able to respond to any further posts till then.)

 

[James O'Neill]

A_big = 8 m x 6.75 m = 54 m²
A_small = 4 m x 6.75 m = 27 m²
So ρu_bigA_big = ρu_smallA_small
4096 kg/m²s x 54 m² / 27 m² = 8192 kg/m²s
Thus ρu_small = 8192 kg/m²s ?? Is this right ??
What would the velocity be then?? Perhaps 8192 / 1024 kg/m³ (ρu_small / water density) = 8 m/s
That gives me 8 m/s which is 0.32 m/s higher than Merlin3198's answer.
If I did my calculations right, both methods prove to be right within half a m/s of each other.

 

[Nidum]

You are looking at basically Bernoulli flow but with a very uncertain set of end conditions and complicated by the fact that there is a great big turbine in the hole .

 

[Geofleur]

Abig = 8 m x 6.75 m = 54 m2
Asmall = 4 m x 6.75 m = 27 m2

Does the funnel have a rectangular, as opposed to a circular, cross section? I still don't understand where the 6.75 m is coming from. At any rate, if I take the areas you calculated, I get

$\rho u_{big} A_{big} = \rho u_{small} A_{small} \rightarrow u_{small} = \frac{A_{big}}{A_{small}} u_{big} = \frac{54}{27} 3 = 2\cdot 3 = 6$ m/s.

We would get the same answer as Merlin if we used the same areas he did. Notice that if the areas on each end were the same, we would get $u_{small} = u_{big}$, which makes sense. However, also note that, if we let $A_{small}$ get really small, we can make $u_{small}$ as big as we wish! But in reality, changing the area at the small end will affect the rate at which fluid enters the funnel.

So I agree with the others' concerns about how one would know that the water is indeed entering the funnel at 3 m/s.

 

[James O'Neill]

The funnel has a rectangular cross section.
If the funnel is circular, does one use the circumferences?
Also, if the tide is 3 m/s, won't the water enter the funnel at 3 m/s?

 

[Merlin3189]

The funnel has a rectangular cross section.
If the funnel is circular, does one use the circumferences?
Also, if the tide is 3 m/s, won't the water enter the funnel at 3 m/s?

Area is what matters, so
Rectangle is fine, so long as you say length & width.
For a circle we can calculate the area from any specified measurement. And since we only need the ratio of the areas, we can use the ratio of the lengths² whether they be radii, diameters or circumferences.

If the tide is 3m/s, it will not enter the funnel at that speed.

The water must slow down in order to create some pressure to force the water through the funnel. (Pse excuse intuitive rather than Physics language. I'm sure others can do that better than me.)
The water going through the funnel must speed up. Volume in = volume out, so if the area gets smaller, the water must go faster.
But the same mass of water is going out as coming in, so if it speeds up there is an increase in KE.
Energy is never free, so where does it come from? The pressure of the water going in is higher than the pressure of the water coming out.
Pressure times area is force. Force times velocity is power.
So the water being pushed into the funnel is having work done on it and it is this work which provides the energy to increase the KE.

Where does this pressure come from? From the water slowing down as it approaches the obstruction. In fact a lot more water than goes through the funnel is also slowed down and the total loss of KE of all that water is what provides the energy to speed up the water through the funnel.