# Fisherman in a boat...

#### hoot

10 miles upstream from a dam. A 50 sq. ft. gate is opened 100 feet above the low water level side. The lake is 300 feet across by 30 feet deep. How long does it take before the boat starts to move with the current? I have used V1= sq rt 2gh and a1v1=a2v2 and t=(10x5280)/v2, but I don't think it works this way. I am thinking about doing a simple test in a bathtub and see what happens. This is what happens when the fish are not biting....

Thank you...

#### ChipB

PHF Helper
10 miles upstream from a dam. A 50 sq. ft. gate is opened 100 feet above the low water level side. The lake is 300 feet across by 30 feet deep.
I assume what you mean is the boat is on the surface of a lake contained by a dam, 10 miles from the dam. I don't understand the gate - if the lakes is only 30 feet deep how can the gate be 100 ft above the bottom of the lake? What's important here is how far below the surface of the lake is the gate, and also what is the gate's height? The deeper the gate in the water, the faster water will flow through it.

How long does it take before the boat starts to move with the current?
The current starts to pick up as soon as the pressure change from the gate being opened reaches the boat. The speed of sound in water is about 0.92 miles/second, so the current will start to move at t= 10.9 seconds. It will of course be a barely perceptible movement at first.

As for how fast the boat moves as the water level declines - that's a very difficult thing to model. You are correct that velocity of water fow through the gate is sqrt(2gh), but what we don't know is whether the water that flows out the gate is all from that same depth of the lake, or does some water get "sucked down" from higher layers? And as water at the bottom of the lake starts to flow toward the gate the viscosity of the fluid would cause water at adjacent depths to be dragged forward as well, which complicates things quite a lot. It would require some pretty sophisticated simulations to figure this out.

Last edited:

#### hoot

Thanks ChipB

Chip, I would let the gate be 10 ft hi X 5 ft wide and be located at the bottom of the high side, so the center of the gate would be at 25 ft below the water. I would also say that the on the high side, there is enough water so that the level doesn't change. The 100 ft high spillway does not need to be considered, but to clarify, the gate is 100 feet above the top level of the low side.

Do you have a reference as to the pressure solution method you used?

#### ChipB

PHF Helper
I would also say that the on the high side, there is enough water so that the level doesn't change.
In this case I suggest that you can use A1V1 = A2 V2, together with the previous formula. It's not a question of "how soon" will the boat feel the current, but rather what is the rate of current flow?

The velocity of flow through the gate is (approximately) sqrt(2gh) = sqrt(2 x 32.2 ft/s^2 x 25 ft) = 40 ft/s

For conservation of mass flow, A1V1=A2V2:

50 ft^2 x 40 ft/s = 2000 ft^3/s = 300 ft x 30 ft x V2

V2 = 0.22 ft/s

So if we assume that the water fow is constant from surface to bottom and side to side of the lake, and that there is a river feeding the lake at 2000ft^3/s of water to keep the water level constant, the boat would move on average at 0.22 ft/s.

The 100 ft high spillway does not need to be considered, but to clarify, the gate is 100 feet above the top level of the low side.
So the dam is built at the top of what used to be a 100 ft waterfall - interesting. Either that or the lake used to be 130 feet deep but silt has filled in the bottom so its now only 30 feet deep, right?

Do you have a reference as to the pressure solution method you used?
I haven't used a pressure solution. If you mean V=sqrt(2gh) , that's a result of Bernoulli's principal:

v^2/2 + gz + p/rho = constant (see: http://en.wikipedia.org/wiki/Bernoulli's_principle )

where p = pressure, rho=density of the fluid, and z = elevation above some reference plane. In this case we choose the surface of the water as the reference plane, where v=0, p=0, and z=0, and hence the value of the constant is 0. Therefore at depth h (z=-h), which is where the open gate causes p= 0 we have:

v^2/2 -gh = 0,
v = sqrt(2gh)

Hope this helps.

Last edited:

#### hoot

still following and looking....

ChipB and others, I am looking for the time that it takes for the boat to start moving due to the opening of the gate. If I use the .22 ft/sec and 52800 ft away and solve for t, I get 66 hours, which isn't correct.

From Chip: The current starts to pick up as soon as the pressure change from the gate being opened reaches the boat. The speed of sound in water is about 0.92 miles/second, so the current will start to move at t= 10.9 seconds. It will of course be a barely perceptible movement at first.

Is this the correct answer? This is the pressure reference I was referring to.

#### ChipB

PHF Helper
ChipB and others, I am looking for the time that it takes for the boat to start moving due to the opening of the gate. If I use the .22 ft/sec and 52800 ft away and solve for t, I get 66 hours, which isn't correct.
Of course that's not correct. 0.22 ft/s is the average speed of water flow in the lake, and has nothing to do with "how soon" the boat starts to move. What you calculated is how long it would take the boat to reach the dam, assuming that the lake is maintained at constant depth by incoming water flow.

From Chip: The current starts to pick up as soon as the pressure change from the gate being opened reaches the boat. The speed of sound in water is about 0.92 miles/second, so the current will start to move at t= 10.9 seconds. It will of course be a barely perceptible movement at first.

Is this the correct answer? This is the pressure reference I was referring to.
Yes, it's correct. You can google "speed of sound in water" for reference.