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A fire hose has an inside diameter of 6.40 cm. Suppose such a hose carries a flow of 40.0 L/s starting at a gauge pressure of $1.62 \times 10^{6} \mathrm{N} / \mathrm{m}^{2}$ . The hose goes 10.0 $\mathrm{m}$ up a ladder to a nozzle having an inside diameter of 3.00 $\mathrm{cm} .$ Calculate the Reynolds numbers for flow in the fire hose and nozzle to show that the flow in each must be turbulent.

Case-I: $N_{\mathrm{R}}=791250$ Here $N_{\mathrm{R}}=791250>3000$ Therefore the flow of oilis turbulent.Case-II: $N_{\mathrm{R}}=1688000$ Here $N_{\mathrm{R}}=1688000>3000$ Therefore the flow ofoil is turbulent.

Physics 101 Mechanics

Chapter 12

Fluid Dynamics and Its Biological and Medical Applications

Fluid Mechanics

University of Washington

Simon Fraser University

University of Sheffield

Lectures

04:16

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A horizontal pipe $10.0 \m…

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Unreasonable ResultsA …

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The Venturi tube shown in …

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The horizontal pipe shown …

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Another fire hose The pump…

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you can say that the velocity at the start of the hose can be found by using the relationship thieve volumetric Flory equaling the cross sectional area at the start of the host times of velocity through the start of the hose and so we can save these of one is equaling the volumetric florid divided by the cross sectional area, which of course would be a pie. Our sub one squared. This is equaling the volumetric flow rate of 40 0.0 leaders per second. This would be divided by pi multiplied by 3.20 times 10 to the negative second meters quantity squared. This is equaling 12.4 meters per second. The velocity through the nozzle would be the volumetric flow rate, which is again constant times pi times the radius of the nozzle squared and so this is equaling again. 40 0.0 leaders per second. This would be divided by pi times 1.50 times 10 to the negative second meters quantity squared. This is equaling 56 0.6 meters per second and we found both of these values Ah, by simply knowing that of course there are 10 to the negative third meters cubed for every one leader. So you're gonna have to use this conversion in order to get the velocity in terms of a court of, of course, meters per second. You can't keep it in leaders. And so you're simply going to multiply the numerator by tend to the negative third meters cubed and again, you can then find the respective velocities. Now we can then find Reynolds number. We can say that. Um four. We're gonna find Reynolds number for the first part of the hose and then through the nozzle so Americans number would be equaling 22 times the density of the fluid times of velocity times the radius of the tube or of the hose at that point divided by the coefficient of viscosity. And so we can say that ends of our one would be equaling two times 1000 kilograms per cubic meter multiplied by the velocity of 12.4 meters per second times the radius of against 3.20 times 10 to the negative second meters. This would be divided by the coefficient of viscosity for water 1.2 times 10 to the negative third Pascal seconds. And so we find that the first Reynolds number is equaling 792,000 15. This is greater than 3000. And so we have. The flow is we have the turbulent flow and so we can say ends up. Our two is equaling two times again. 1000 kilograms per cubic meter multiplied by here, 56.6 meters per second. This would be multiplied by 1.50 times, 10 to the negative second meters and this would be divided. I again 1.2 times 10 to the negative third pass cow seconds. And we have that Reynolds numbers two through the nozzle is equaling. 1,649,000 611. And this is of course, greater than 3000. So here we also have a turbulent flow. That is the end of the solution. Thank you. For what

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