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Chapter 9: Problem 54

A viscous liquid is flowing steadily through a pipe of diameter \(D .\) Suppose you replace it by two parallel pipes, each of diameter \(D / 2,\) but the same length as the original pipe. If the pressure difference between the ends of these two pipes is the same as for the original pipe, what is the total rate of flow in the two pipes compared to the original flow rate?

Short Answer

Expert verified
Based on the given information and using the Hagen-Poiseuille equation, we have calculated the flow rates for the original pipe and the two parallel pipes. Through our analysis, we found that the total flow rate in the two parallel pipes is twice the original flow rate. Therefore, the flow rate in the two parallel pipes is 2 times the flow rate in the original pipe.

Step by step solution

01

Calculate the flow rate for the original pipe

First, we need to calculate the flow rate for the original pipe with diameter D and length L. Using the Hagen-Poiseuille equation: $$Q_{original} = \frac{\pi {(D/2)^4} \Delta P}{8 \eta L}$$
02

Calculate the flow rate for each of the new pipes

Next, we need to calculate the flow rate for each of the parallel pipes with diameter D/2 and length L. Using the Hagen-Poiseuille equation for the new pipes: $$Q_{new} = \frac{\pi {((D/2)/2)^4} \Delta P}{8 \eta L}$$ $$Q_{new} = \frac{\pi {(D/4)^4} \Delta P}{8 \eta L}$$
03

Calculate the total flow rate for the two parallel pipes

Now, we have two parallel pipes, so the total flow rate is the sum of the flow rates for each of the new pipes: $$Q_{total} = 2 \times Q_{new} = 2 \times \frac{\pi {(D/4)^4} \Delta P}{8 \eta L}$$
04

Compare the total flow rate with the original flow rate

Finally, we can calculate the ratio of the total flow rate of the parallel pipes to the original pipe flow rate: $$Ratio = \frac{Q_{total}}{Q_{original}}$$ Substitute the values of \(Q_{total}\) and \(Q_{original}\): $$Ratio = \frac{2 \times \frac{\pi {(D/4)^4} \Delta P}{8 \eta L}}{\frac{\pi {(D/2)^4} \Delta P}{8 \eta L}}$$ Simplify the expression: $$Ratio = \frac{2 \times (D / 4)^4}{(D / 2)^4} = \frac{2 \times (D^4/2^8)}{(D^4/2^8)} = 2$$ So, the total flow rate in the two parallel pipes is 2 times the original flow rate.

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