(II) Suppose the top surface of the vessel in Fig. 13–58 is su...

Question

(II) Suppose the top surface of the vessel in Fig. 13–58 is subjected to an external pressure P₂ in addition to atmospheric pressure P₀. If P₂ = 0.85 atm and y₂ - y₁ = 2.4 m, determine v₁ for water.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with Bern's equation. So in this problem, we have a container with oil and is subjected to an external pressure PB along with the atmospheric pressure P not. And if the external pressure PB is equal to 0.75 at and the distance between two heights YB and Y A within the fluid at the YB minus Y A is equal to 0.5 m. We need to calculate the velocity va at height Y A at which point the oil is coming out of the container through an opening. We're also told to assume that the atmospheric pressure um one atmosphere is equal to one multiplied by 10 to the five newtons per meter squared. And that the density of oil is 7.5 multiplied by 10 squared kilograms per meter cubed. Below the question, we're given a diagram showing the setup that was described in the problem were also given four choices as are possible answers will come back to those at the end. Now, since we're dealing with um pressures at different and different heights and dealing with fluid flow, we're going to be using Bernel Le's equation. So we call Bern's equation says that P one plus row GH one plus one half row V one squared has to be equal to V two plus row GH two plus one half row B two squared. So here we've taken two points, points, one and two within the fluid. And the pressure plus the quantity of row GH plus the quantity of one half row squared have to be equal to each other. So here um P is the pressure rows, the density of the fluid is acceleration due to gravity. H is the height of the fluid and V is the speed of the fluid. Let's call 0.1. Um Actually at point A in our diagram and 0.2 at the top of the container. So at point A, we have the pressure due to the outside atmosphere acting on the fluid. So that's just gonna be equal to po they won't have plus row G. Our height is gonna be Y A don't have plus one half row D A squared. The speed of 0.1 is going to be va and that's what we're looking for. So this has to equal the pressure at 0.2 and the pressure at 0.2 we, we're told that we have both um PB and the atmospheric pressure. We have PB plus P not for that Mr crusher. Then we'll have plus row G, the very top of the container is located at YB. So I'll have yb multiplying that, then we'll have plus one half row zero squared. Now, the reason why the speed at the very top is zero is if I look at my figure, I notice that the dictator is much, much larger at the top than the tiny little opening at which the fluid is coming out. So since the top of the fluid is much, much larger, that means its motion is going to be much, much less than the speed of the fluid coming out just because it would take a larger volume of fluid to actually move the top surface of the container. So we're gonna set that um to be equal to zero at the top. So the speed at the very top is equal to zero. And that's due to the size of container being much larger than the size of the opening. So we can actually simplify um this expression a little bit. We noticed that the peanuts will cancel out. And what I'm going to do is um move the terms that have the row G Ys all to the right side of the equation. So I wanna be left with the one half row B A squared is equal to PB. And then I'll have plus in both terms, I'm gonna have a row G and then that's gonna get multiplied by the quantity of YB minus Y A. At this point, I need to just um solve for the Y A, they'll have or sorry for the VA. So I'll have VA is equal to, that's gonna be the square root and then I'll have two multiplying the quantity of PB plus row G multiplying the quantity of YB minus Y A and that quantity gets divided by, bro. So at this point, I can actually just plug in values because I have a value for the PB that is the um 0.75 atmospheres. We have the quantity YB minus Y A being equal to 0.5 m. And we know the density of the oil that is 7.5 multiplied by 10 to the 2 kg per meter cubed that we'll have for B A when we plug in our values will be equal to the square root of two multiply with the PB will have um 0.75 atmospheres. But I need to convert that into SI units. And if I look at my question here, um I was given a conversion uh factor atmospheres and newtons per meter squared. So one atmosphere is equal to one multiplied by 10 to the five newtons per meter squared. We'll use that conversion factor that would multiply the 0.75 atmospheres by the fraction of one multiplied by 10 to the five newtons per meter squared divided by one atmosphere. So they'll give me my um pressure in SI unit. Then we'll have plus we'll have the density of the oil, that's the 7.5 multiplied by 10 squared kilograms per meter cubed. That gets multiplied by the acceleration due to gravity, which is the 9.8 m per second squared. And that gets multiplied by the difference in the heights which we were told was 0.5 m. Now, that whole quantity gets divided by the density of the oil. So everything is divided by the 7.5 multiplied by 10 squared kilograms per meter cubed. So everything on the right hand side of that equation is just a number that I can plug into my calculator. And so this turns out have a va equal to 14 meters per second. That's if I keep two significant figures. So that is the answer that I'm I've been looking for. So now let's look at our answer choices that we were given. Now for a, we're given um 3 m per second or choice B we're given 7 m per second or choice C, we're given 14 m per second and choice D we're given 160 m per second. And so if I look at my answer, this corresponds to voice C. So just a quick little recap, we used Benelli um equation to solve for the speed of the fluid that's coming out of the container. But one of the tricks that we ran into is that we had to set the speed of the fluid at the very top of the container equal to zero. And that's because um it's surface area was much, much larger than the area of the hole which the um food was coming out of the container. So keep this in mind if you run into that in future problems. So I hope this has been helpful and I'll see you in the next video.

(II) Suppose the top surface of the vessel in Fig. 13–58 is subjected to an external pressure P₂ in addition to atmospheric pressure P₀. If P₂ = 0.85 atm and y₂ - y₁ = 2.4 m, determine v₁ for water.

Key Concepts

Hydrostatic Pressure

Pascal's Principle

Fluid Density

Watch next

(II) Suppose the top surface of the vessel in Fig. 13–58 is subjected to an external pressure P2 in addition to atmospheric pressure P0. If P2 = 0.85 atm and y2 - y1 = 2.4 m, determine v1 for water.

Key Concepts

Hydrostatic Pressure

Pascal's Principle

Fluid Density

Watch next

(II) Suppose the top surface of the vessel in Fig. 13–58 is subjected to an external pressure P₂ in addition to atmospheric pressure P₀. If P₂ = 0.85 atm and y₂ - y₁ = 2.4 m, determine v₁ for water.