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A tube with radius 0f 1.1 cm and length 0f 5.3m has fluid of viscosity 0.002 Pa-s passing through it: The flow rate ofthe fluid through this tube is 0.0079 m3/s and...

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A tube with radius 0f 1.1 cm and length 0f 5.3m has fluid of viscosity 0.002 Pa-s passing through it: The flow rate ofthe fluid through this tube is 0.0079 m3/s and you may assume the fluid flow aminar:The fluid is inserted into this tube by small pump; andthe fluid exits this tube into air; The tube horizontal, so the points where the fluid enters and exits the tube are atthe same height:What is the absolute pressure of the fluid where it enters the tube? Give your answer in units of Pa

A tube with radius 0f 1.1 cm and length 0f 5.3m has fluid of viscosity 0.002 Pa-s passing through it: The flow rate ofthe fluid through this tube is 0.0079 m3/s and you may assume the fluid flow aminar: The fluid is inserted into this tube by small pump; andthe fluid exits this tube into air; The tube horizontal, so the points where the fluid enters and exits the tube are atthe same height: What is the absolute pressure of the fluid where it enters the tube? Give your answer in units of Pa



Answers

A pipe is horizontal and carries oil that has a viscosity of 0.14 Pa " s. The volume flow rate of the oil is $5.3 \times 10^{-5} \mathrm{m}^{3} / \mathrm{s} .$ The length of the pipe is $37 \mathrm{m},$ and its radius is $0.60 \mathrm{cm} .$ At the output end of the pipe the pressure is atmospheric pressure. What is the absolute pressure at the input end?

So we have to the left a diagram of the system. And so we should first use Bernoulli's equation. Ah, we're going to say that the pressure, the atmospheric pressure plus p g h t o close plus 1/2 row, the Savo squared plus piece of one brother. This equals piece of one plus p g h. Serve wine close 1/2 Roe v suborn squared. We know that the piece of one is gonna equal the atmospheric pressure. Ah, so we can say that he not minus the atmospheric pressure, um would be equal to P g h sub one. Knowing that this term here is gonna be equal to zero s, we can eliminate that term completely. And then this is gonna be a plus 1/2 times row times visa of one squared minus these be knots where the original velocity squared. We're gonna use the equation of continuity. Now the volumetric Florida is going to, of course, stay constant. And given this, we can say that pie are not squared. V not squared equals pi r. Someone swear Davies of one squared pies Cancel out. Of course. Um and we're getting that V not or the original velocity will be our someone square times V's of one divided by are not squared and this is equaling dese of one squared times Visa of one, uh, divided by dean not squared. So at this point, we can say that piece of oh minus piece of a T m would be equal to the density times G h someone plus 1/2 of road The density the sub one squared minus 1/2 times the density of d sub one squared divided by 1 December 2 squared breath Missouri the not squared piece of oh, squared niece of one squared rather Visa One and this entire thing, this entire quantity would be squared. Um, this is also, um, rather weaken further simplify and say that p not equals rather p sub zero minus p a t m equals P G. And then here it would simply be times one minus dese of one to the fourth, divided by decent to to the fourth. So that's how just Well, that's how we would simplify that. And at this point, we can apply Bernoulli's equation. So keep in mind this, and then we have to apply Bernoulli's equation again at levels one and two. So piece of one plus p rather a row. The density G h up one plus 1/2 of Roe v sub one squared would be equal to piece of two, uh, plus P g h rather pg. And then this would be the difference in height. So h someone plus h up to or enter the sum of the heights. Mmm, plus 1/2 of PV two squared. And so we know that Weise of one squared will then be equal to two times G h sub too. Uh, here, The greatest part is that a lot of things cancel out. So this cancels out with this, um, we get that. Ah, BC To is, this is completely eliminated. Um, So again, yeah. And then, uh, each term here has the density term so we can eliminate the density term. So we're getting this from that G eight sub one plus 1/2 V's of one squared equals G eight's of one plus h two. That's essentially how we were to find that the visa one squared equals two times G eight's up to, and now we can use this in this equation. So we can finally say that p not nice. P a t m equals P G eight's of one plus 1/2 times p V one squared, Ah, one minus dese of one to the fourth minus D C to to the fourth. And this is going to equal PG minus other times h of one plus h up to times one minus dese of one divided by someone to the fourth bout of decent to to the fourth. But my apologies. This is actually not two. I'm getting my sub scripts confused. My apologies. Okay. And after that, then we can solve, So p o my p a t m would be equal to the density times the acceleration of gravity. H someone takes up two times one minus uh, the ratio of the diameters to the fourth power. So this would be tend to the third kilograms per cubic meter times 9.8 meters per second, squared times 1.1 meters plus 0.14 meters times one minus 10.5 to the fourth power. And we find that this difference in pressure is gonna eat well, 12,066 past cows. So this would be our final final answer after everything. So we have to relate it. We have to you we have to basically a fly brand new this equation at levels zero and one and then that levels one in to relate them and then find the difference in pressure again. That final answer 12,066 Pascal's That is the end of the solution. Thank you for watching.

So for a horizontal pipe, the flow rate is given by the equation. Pi times are to the other four times p two minus P one over eight times Ika. So we're given that the flow rate is 5.3 times 10 to the negative. Five meters per second equals high. The radius is 0.6 centimeters. So we need to change that leader. So times 10 to the negative, too. And then to the power for and then Pete, who is what we're trying to find out and p one is atmospheric pressure. So one points you won the times 10 to the power five divided by eight and the viscosity is given as 0.14 and the length have to pipe is 37 meters. So solving this equation for P two gives P two is equal to 6.4 times 10 to the power five best gal

Lola. So we have, um, a problem where we're thinking about a drinking fun kind like I've drawn here and like they show in the textbook. And I just start solving problems by missing out all the information they've given in terms of some just writing. Some variable names, too, were told that the diameter of the nozzle, which I called the end, is 0.6 serious centimeters and the diameter off pump eyes 1.2 centimetres. That's no label DP, and the water from the nozzle shoots to maximum height, which I called H of 12 centimeters. And before you get to the novel going from the ground where the pump is up to the nozzle, we're told that that's 1.1 meters, which I labeled wide, so up to you what you call these variables and we want to find, um, the gauge pressure at the pump so called P G a two pump. And, um, who was it told to ignore viscosity? And that this will be an underestimate, since realistically, there would be viscosity in this system so we can solve this problem using Bernoulli's equation, and that's basically like an energy conservation equation So if we raise and hers of Gage pressures we have at the pump side, we have a pressure. Plus, um, whatever we want to call this height at the bottom, we can set this to you zero like a potential energy term and, ah, kinetic term. And it's row here. This is the density ofwater that's just flowing through this system and hasn't input velocity. And she'll be the gray JJ pressure at the nozzle end with rosy. Why are wise really this height here, up to the nozzle, That's 1/2 ro. And this will be some output velocity square, and we want to find the gate for sure the pump here so we can rearrange us equation. This is what we're looking for. Mantis. Factoring out this 1/2 road term and something else we can think about is that they get the pressure, a tte, the nozzle and it's just gonna be atmospheric pressure. So ah, the actual pressure is to find as the gauge pressure plus the pressure of the atmosphere. So when the actual pressure is equal to the atmosphere, which is the case of the nozzle, that means that the gauge pressure is zero seeing your end of this term so I can rewrite this again. Um, let's see. And I can factor out the density ro. So we know that the density of water, that's why they want to look up in a table. And in Chapter 10 why is the height that were given? We don't know these input or output velocities. We could think about what's happening and basically do a cinematics problem for the output velocity. Seriously, the output velocity of the house is sort of like you can think of the input for the initial velocity of this projectile. So the waters, um so watching in the air, you could just think of this as some kind of projectile motion, and it's going to read a maximum height H. And when we know that when it's at that maximum height, it's last day, that moment is zero. And we can just basically set a new sort of first making a person as a new um, problem here. So we can set this to be, um, because zero potential energy point here with star, and we can then write down that energy will be conserves that the candidate energy at the input. It will be the equal to the potential energy at the math. Um, height. Oh, h So that means that 1/2 em the and I won't label it here. This is really the output velocity, right? This is yeah. So we can rearrange this equation than canceling out the masses so that Thea Pomposity is to G H. And we want to write a gin terms of meters still 12 times 10 to the minus two years since we were told I was 12 centimeters and this is in our final answer. So you want to keep his money, come disses you can your calculations of your approximating at the end. We should get a number close to this. So that's just considering, you know, a whole separate problem for basically a project out sort of motion for the water coming out of the nacelle bone. We can find the impulse velocity using the continuity creation. So that means that you know, this is an inn compressible fluid flowing to the system to know that the cross sectional area of this pipe, at the input times input velocity would know we were looking for will be the same a cz the output cross sectional area times the output velocity which we just calculated. This is just like a conservation of the well. I am flow rate here and we're looking for being And were we can solve this equation because we aren't told what thes cross sectional area czar we are told. The diameters we can calculate that is the area would be pi R square, which is really, um, and the output. This is the nozzle That's the jammer over, too. And the input is at the pump. So that's while they will dp over too. So you can solve this factor of pie and the factor of 1/2 squared the numerator and denominator. So then we have, um, an equation terms off what we just calculated the out and the diameters that were given. And since we're just going to divide these, you don't have to convert to meters. We can just write them both in centime years, since that unit's going to cancel and rewriting the velocity. And this carries money Did just as he can't hear calculation, you should get for that output velocity. We're sorry for the improv lossy, which is a bossy at the pump. Um, and remember where you want to keep his bandages as you can. That way we're approximating at the end for the gauge pressure of the pump. So that was what we needed. Need to calculate those velocities to plug this back into burn. Oi separation. The density of water times. Do you? Why? Minus 1/2 me out squared minus fiends. Where? So everything. Now we have to plug into this equation. He confined the density of water and texts of 1,000 kilograms per meter cubed, and we're told that wise 1.1 meters, So that's maybe a little nasty looking there. But just make sure carryings many digits as you can through scoring each of these terms. Um, this is well deployed here and she an answer of about proximity to choose to giving it. It is 1.2 times 10 to the four units per meter square


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