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2 Water is flowing out of a hose at a fluid velocity of 10 m/s pointed directly up in the air. Since the water leaving the hose is in contact with the air, it is al...

Question

2 Water is flowing out of a hose at a fluid velocity of 10 m/s pointed directly up in the air. Since the water leaving the hose is in contact with the air, it is always at P=] atm _ Use Bernoulli to determine how high this fountain of water will go. (Hint: assume that the water at the top of the fountain is not moving, i.e. v = 0.)

2 Water is flowing out of a hose at a fluid velocity of 10 m/s pointed directly up in the air. Since the water leaving the hose is in contact with the air, it is always at P=] atm _ Use Bernoulli to determine how high this fountain of water will go. (Hint: assume that the water at the top of the fountain is not moving, i.e. v = 0.)



Answers

Find the horizontal and vertical forces to hold stationary the nozzle shown in Fig. P5.37. The fluid flowing through it is $10^{\circ} \mathrm{C}$ liquid water; $A_{1}=1.0 \mathrm{m}^{2}, A_{2}=0.25 \mathrm{m}^{2}, V_{1}=20 \mathrm{m} / \mathrm{s}, p_{2}=p_{\text {urm }}$ and $p_{1}=p_{\text {atm }}+30 \mathrm{kPa}$. Neglect gravity.

These questions, we are given this situation whereby it has pied and at the park coming up? Yeah, there's another pint coming out in this manner and out in constant that Amita. Okay, so we have p one because two hemispheres, uh, you want it was too one meters per second. Then we are given that p two because 21 and monsieur. And he asked you find me too. In this case, there's a drop off 10 meters, and then we have given that Petri, this one embassy, us and then Vetri. It's also something we need to determine. Okay, so in this question are we need to determine b two and b tree. So to solve this problem will be using but an easy question. Yes. So, for me to as long change in height. So this is, uh this is the question that we will be using, uh, have been released. Equation. We doubt the height. Okay. All right. So p one is to m Osias. He too is one and says V one is one meters per second. You can rearrange you want to find me too. And so we have people lined by this p to, uh, last off road. You want square? Okay, then we to you just be too. Don't happy you have my room. Uh, plus we want square. It's where Okay, So putting in the numbers S O P one minus p two would just be one m Osias. So, Pascal, if we want 101 tree time said to the fine the density of water said 1000 and last meal elsewhere. Take a square root and you should be getting 14 country you a second. So this is the answer for you too. Hey, then next we want to find Hold me tree. Okay, so we'll be using binary separation again this time. We need to use the equation with height. So you look like this. You want a glass half role? You want square plaster over tree each one. Because to me too class have Rome tree. Sorry, V P Tree square less Rocchi each tree. Okay, so, uh, we have the numbers. P one is to 80. N Petri is 1 80 m. He won this. Want me? Just a second. Maybe have each one minus a street would be 10 meters. Okay, so So we way We want to find me tree. So we put the tongue retreat at the left Insight that we have P one by this Petri A plus half room me one squared class role g each one minus each treat. Okay, so the tree would be to that toppy by my room. Plus, he wants where? Less to, uh, g their package. And we need to take this skirt. Okay. We think either numbers two times one part of the year. One tree tends tend to the five. Work by those in what's once way. Last two times. That won't be one. It's 10. Take this. Gertrude's. We should be getting 20 meters per second. Okay, so this is the answer. Well, Mitri.

Good day in this question we will be solving for the escape velocity already. We do. So you think the Bernoulli's equation where Bernoulli's equation is he won Does one have We'll be one squared was H1 Rajaji Is equal to P two blush one half. Garavito squared has age two. Rogie. We could we could we can cancel this group here since it is a large tongue. Therefore our every one is addressed Or zero m/s. So we are left with B one plus H one Taraji. It was pito plus one half Robidoux squared plus H two Raji. So our The 1 -22 is equal to 500 kill Abascal or five. I am standard a power of five pascal. So rearranging this we will get B one. My speed too plus each one minus age. To urology we will be grouping the height one and high too Is equal to 1/2 a widow square. So we have rearranging which will this will give us since the and open water down colleagues Using the Figure 14. The ash too the height one, It's just equal to H two. Therefore, weekend cancel this figure here and we are left with Be 1 -22 equals one half roe V two squared To find the video. We have B2 equals two times be 1 -22 divided by raw and square root. So we have the density is the density of water which is 1000 kg per cubic meter three. So substituting we have squared of two times wife Time stand to the power of five pascal Divided by 1000 kg per cubic meter. Therefore, our escape velocity is equal to 31 when six m/s. Thank you.

But of course, part of the question. Usually applying the banal situation between our faces of the tank. At 50 I bought the races. So when all this equation is the one upon Rocchi, Plus, we want square upon twice cheap. Plus Z one is equal to P two upon the road, cheap plus B two square upon twice cheap. Plus, they do. Um, here we have, um P one is 50 into 10 to the power tree of one. So is how 1002 9.8. Plus, this will be whole zero. The one is two. This is zero plus zero plus Z two is at. So from this we'll get h two b 11.968 Therefore, the high. Yeah, this four. Um, therefore height. Uh huh. Of the water rises is 7.0 968 m. Um then, for a second one again, apply Bonilla's equation between surface of the tank and the pipe is exit. That is, um, P one upon road G plus, we even square upon twice cheap. Plus, the one is equal to uh, p e upon Rajhi plus z squared upon Y c G plus said Z E Um after putting the values, we will get the V A 7.794 meter per second. Therefore, the velocity of the pipe is 7.794 Then for the third one again applied the Vinales equation between surface of the tank and at any section in the horizontal pipe, That is, um, be one upon, uh oh gee plus V one square upon twice G plus z One is a girl to be H upon row G plus V square upon twice. See U S Z Edge. So we have the one is 15 to 10 to the power three upon 1002 9.81 plus zero plus two is equal to the age of 100,000. In 29.81 plus 7.7 9/4 were upon twice 9.81 90 That is a B H will be 39.247 kg baskets. Therefore, the pressure in the horizontal part of the pipe is 39.247 kg. Paschal

So here we're going to let um the area at the top of the tank Equal 10 times the area of the valve at the bottom of the tank. We can say that. Then the area at the top multiplied by the velocity at the top would be equal and then the area of the ball valves multiplied by the velocity at the bottom, We're at the valve. The velocity at the bottom would be equaling to the velocity at the top, multiplied by the area at the top, divided by the area of the valve. And so this would simply be equal to 10 times the velocity at the top. We can apply Bernoulli's equation and say that the pressure at the top yeah plus one half times the row the density times the velocity at the top squared plus row G. H. Equalling the Pressure at the bottom plus 1/2 row velocity at the bottom squared. And so we can see that as the pressure at the top of the tank is equal to the pressure at the valve. These to cancel out and we can actually solve for the velocity at the bottom. Velocity at the bottom squared would then be equal to the velocity at the top squared plus to G. H. And so, but we know what the velocity at the top is, 1/10 of the velocity at the bottom. So we can actually say that the velocity at the bottom squared would be equaling to the velocity at the bottom Squared divided by 100 Plus two G. H. And so solving for the velocity at the bottom, this would be equal and then the square root of 200 G. H. Divided by 99. We can use the kinetic relationship and say that then final velocity squared minus initial velocity squared equaling to G. H. And so velocity V, given that this is zero is equal in the square root of two G. H. And at this point you can solve for the velocity of the water at the valve or at the bottom. This would be equal to the square root of 200 multiplied by 9.8 m per second squared, Multiplied by 1.0 m, extend the square root. This would be divided by 99 And this is giving us then four .45 m/s. And we can say that the speed of the water at the valve when it is falling from rest at the top of from the top of the tank V. So this would be falling from rest from the top. This would simply be equal to two, multiplied by 9.8 m per second squared, Multiplied by again 1.0 m. And this is giving us then 4.4, m/s. So we can see very close that is the end of the solution. Thank you for watching


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