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Concrete is pumped from a cement mixer to the place it is being laid, instead of being carried in wheelbarrows. The flow rate is 200.0 L/min through a 50.0-m-long, 8.00-cm-diameter hose, and the pressure at the pump is $8.00 \times 10^{6} \mathrm{N} / \mathrm{m}^{2} .$ (a) Calculate the resistance of the hose. (b) What is the viscosity of the concrete, assuming the flow is laminar? (c) How much power is being supplied, assuming the point of use is at the same level as the pump? You may neglect the power supplied to increase the concrete's velocity.

(a) $R=2.4 \times 10^{9} \mathrm{N} \mathrm{sm}^{-5}$(b) $\eta=48.3 \mathrm{N} \mathrm{sm}^{-2}$(c) power $=26.6 \mathrm{kW}$

Physics 101 Mechanics

Chapter 12

Fluid Dynamics and Its Biological and Medical Applications

Fluid Mechanics

University of Washington

University of Winnipeg

McMaster University

Lectures

04:16

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So for card A to find the resistance in the pipe. We can say that the volumetric Flory is gonna be equaling the pressure. Sub two minus pressure. Someone to a difference in pressure divided by the resistance. And so the resistance would be equaling. Piece of two minus piece of one, divided by the volumetric floor eight Q and so this would be equaling. We know that piece of one is gonna be zero Nunes Pierre. Ah, square meter. And so we can say that this is gonna be 8.0 times 10 to the sixth Newtons per square meter, divided by the boy metric florid of 200 leaders for a minute and we're going to then then divide this by now The multiply this by tend to the negative third cubic meters for every 60.0 seconds. And we're dividing this entire term, of course, for everyone leader per minute. And so we can then find the resistance are this is equaling 2.40 times 10 to 1/9. This would be Newton seconds per meter, D'oh per meter to the fifth power. This would be our answer for the resistance for part a and then for part B ah, we can use process hes equation to find the viscosity And so we have that are no the resistance is equaling eight times new the Scott the coefficient of viscosity times that of the length divided by pi are to the fourth power. And so the viscosity would be equaling. Hi are to the fourth power times the resistance are divided by eight times the length. And so we can, uh, we know that then we can solve for the resistance rather than the coefficient of viscosity by plugging in the resistance. And so the coefficient of viscosity would be the resistance 2.40 times 10 to the ninth Newton seconds per meter to the fifth power times pi times the radius of point 04 00 meters. So we're simply dividing the diameter by two raised to the fourth power. This would be divided by eight times the length of 50.0 meters and so we find that the coefficient of viscosity would be equaling 48.3 Newton seconds per square meter. This would be our final answer for the coefficient of viscosity for part B and then four part See, we can say that he supplied power. P would be equaling the pressure at 0.2 multiplied by the volumetric flow rate and so this would be equaling 8.0 times tend to be sixth Newtons per square meter. This would be multiplied by weekend again, 200 litres per minute and then again convert. So this would be 10 to the negative third cubic meters per every 60 seconds per every one leader for a minute. And so we can then say that the power the supplied power would be 2.67 times 10 to the fourth units. Of course, wants this would be our final answer for the supply power for part C. That is the end of the solution. Thank you for one.

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