Two reservoirs which differ in surface elevation

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The old pressure is the same as a new pressure. Tell A P prime and the new floor. It, though, is twice the old florid. So that means we need to change the radius. So now we're just gonna change the radius to New Radius.

Okay, So if we said this year the poor souls for Mullah would be Delta p prime equals Delta P is going to imply that build up your prime is eat by pi times. Peter is the same. We're keeping the length of the pipe. Same radius, which is I'm going to call it our prime power for new radius times to prime.

So I'm going to write down instead of Q prime. I'm gonna write twice the Q Okay, Twice que you close eight Pi pi. This is going to be the same over our power for que. Now, let's see here this to cancel out this to cancel out and Q and Q cancel out. So we only have the terms to buy our prime power, for it calls one by our power for right.

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Electrical Engineering. Mechanical Engineering. Advanced Math. Advanced Physics. Anatomy and Physiology. Earth Science. Social Science. Political Science. Literature Guides. Popular Textbooks. Two reservoirs, which differ in surface elevation by 40 m,are connected by m of new pipe of diameter 8 cm. Want to see the step-by-step answer?

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Tagged in. A: Click to see the answer. The cross sectional area of a pipe is simply going to be the area of the circle Pi r squared and a square. A subtitle would be also pi r squared. So here we're going to use these values and we're going to divide it by two because this is, of course, the diameter.

So we have pie dese of one over two squared eat. And then for the second cross sectional area, we have pie diesel attitude about by two squared. And this right here is going to Teo give us all of our Constance needed in order to use this equation of continuity. So solving for this now we have we're going to solve for the V two. Of course, we're trying to isolate that.

So the sub two would equal the cross sectional area of the first section Times of velocity of the first section divided by the cross sectional area of the second section Plugging in for a Constance from here and here we have the cross sectional area pie d sub one over two squared times.

Piece of one divided by pi d sub two squared over two pies. Cancel out and we can simply so we can simply put plug in for are constants. December one is, of course, that point two nine eight four divided by two squared times six six centimeters per second and then we're going to divide this bye point three nine three seven two by two squared and solving. For this, we find that thief velocity in the second section is going to be thirty seven point five centimeters per second.

So again, as you can see the cross sectional area, if the cross sectional area decreases, the velocity increases in that particular section. So one can say that the velocity is inversely proportional to the cross sectional area in that specific section. And again, that final answer is going to be thirty seven point five centimeters per second in the section and the second in the second section of this pipe. And I see a final answer. Thank you for watching.