Bernoulli's Equation: Example

by Jared Rovny

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    00:01 Let’s go ahead and apply an example of Bernoulli’s principle and see if we can solve this example problem which asks, if you have a tank of water that’s 2 meters and has an open top with an opening of 0.5 meters away from the base, what is the velocity of the fluid that will be escaping at 0.5 meters away from the base of this tank. So, just using Bernoulli’s principle where you get to conserve the quantity that we just wrote out, see if you can solve this by comparing Bernoulli’s principle at the top of the 2 meters of water and Bernoulli’s principle 0.5 meters away from the base of this bucket. We’re going to do that now. If you solve this problem, hopefully it looks something like this. You have a tank. There’s 2 meters of water in this tank. Then it has a hole and this is where the water is escaping. This distance is half a meter from the bottom of the tank.

    00:55 What we’re going to do in order to see how fast the water would be leaving this hole is compare Bernoulli’s principle here at the top of the water and then here when it’s leaving the tank.

    01:05 So here position at 1, we can say we have pressure P1 plus ½ ρv1 squared plus ρgh1 from the height of the water. At position 2, we have P2 plus ½ the density times the velocity squared.

    01:28 This is the velocity we’re looking for plus the density of the water times g times h2 which is much lower.

    01:34 It’s only at 0.5 meters. But Bernoulli’s principle, the idea is that these things will be the same.

    01:39 That’s all we have to do. So, let’s do that. P1, let’s actually write the whole thing, plus ½ ρv1 squared plus ρgh1 equals and then this will be equal to the second quantity. Now, what we’re going to do after we’ve written this entire expression out is try to simplify by looking at these different variables and seeing if any of them could be simplified. So first of all, what are these pressures, these different pressures? These are just the opposites. These are the absolute pressures.

    02:10 So, the pressure pushing down on this fluid for example and the pressure pushing on this fluid from the atmosphere are the same. So, these are just the atmospheric pressures.

    02:19 So, they’re no different from each other because neither one of this is closed off.

    02:23 The problem, there is a special point made that this was an open top container. This is important because this means the atmosphere is able to push down on the water. If we had a closed container like this with the water coming at the bottom, we would have a different scenario because as this water tried to move down, there would be a vacuum created at the top of this container.

    02:42 But that’s not our situation. We have an open top container. So, we have atmospheric pressure pushing in this open nozzle and in the top of the open tank. Because this atmospheric pressure term is the same on both sides, we can get rid of that. The second thing we can do is look at this expression here, ½ ρv1 squared. This is referring to the velocity of the top of your tank of water.

    03:04 Unless you have a very, very thin tank, so for example if you have this case with an opening here, this is 0.5 meters from the base. You had water flowing this way. Then if water is flowing out here, you'd have a very quick velocity of this water flowing downwards. But we don’t have this case.

    03:19 We have a big tank of water. So, we’re going to consider the velocity of this water moving downwards as very slow especially relative to the velocity of the water escaping from the small spout at the bottom.

    03:29 Now, we’ve simplified this greatly. We just need to find this velocity term. So, let’s first of all rearrange this so that we have that a little more easily accessible. Subtract this term from both sides.

    03:41 So, we have ρgh1 minus ρgh2. Let’s divide the density of water from both sides, so that now we have v2 squared. Multiply both sides by 2. It’s 2 times g, which we can factor out, times the difference in heights here. This, as you see, is h1 minus h2, where h1 is the tall height of 2 and h2 is the low height of 0.5 meters. This is just equal to 2 times g times 2 minus 0.5 or 1.5.

    04:15 This is equal to the square of the velocity at the bottom of the tank. Plugging in some numbers, we can say that this is approximately equal to 2 times 10 times 1.5. Let’s put this in parenthesis so we don’t get confused with any decimal places. This is equal to 30. Taking the square root, we simply have that the velocity at the bottom of the tank is equal to the square root of 30 meters per second. This quantity we know must be somewhere between 5 and 6 because 5 squared is 25 and 6 squared is 36. One is too low and one is too high.

    04:52 So, we’re going to be somewhere in between and in fact, if you calculated this, you would get an answer close to 5.5 meters per second as the velocity of the water leaving the bottom of the tank.

    05:04 This is a basic example of the application of Bernoulli’s principle which basically says that you can conserve the energy or the somewhat modified energy equation which has pressures in it from one point to another.

    About the Lecture

    The lecture Bernoulli's Equation: Example by Jared Rovny is from the course Fluids.

    Included Quiz Questions

    1. √60 m/s
    2. √30 m/s
    3. √20 m/s
    4. 0 m/s
    5. Area of the hole has to be given in order to solve this problem

    Author of lecture Bernoulli's Equation: Example

     Jared Rovny

    Jared Rovny

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