Log out Go to App
Go to App

Learning Materials

Features

Discover

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

This table gives the terminal speeds of various spheres falling through the same fluid. The spheres all have the same radius. $$\begin{array}{llllllll}\hline m= & 5.0 & 11.3 & 20.0 & 31.3 & 45.0 & 80.0 & (\mathrm{g}) \\\ \hline v_{1}= & 1.0 & 1.5 & 2.0 & 2.5 & 3.0 & 4.0 & (\mathrm{cm} / \mathrm{s}) \\\\\hline\end{array}$$, Is the drag force primarily viscous or turbulent? Explain your reasoning.
A fish uses a swim bladder to change its density so it is equal to that of water, enabling it to remain suspended under water. If a fish has an average density of \(1080 \mathrm{kg} / \mathrm{m}^{3}\) and mass \(10.0 \mathrm{g}\) with the bladder completely deflated, to what volume must the fish inflate the swim bladder in order to remain suspended in seawater of density $1060 \mathrm{kg} / \mathrm{m}^{3} ?$
In an aortic aneurysm, a bulge forms where the walls of the aorta are weakened. If blood flowing through the aorta (radius \(1.0 \mathrm{cm}\) ) enters an aneurysm with a radius of \(3.0 \mathrm{cm},\) how much on average is the blood pressure higher inside the aneurysm than the pressure in the unenlarged part of the aorta? The average flow rate through the aorta is $120 \mathrm{cm}^{3} / \mathrm{s} .$ Assume the blood is non-viscous and the patient is lying down so there is no change in height.
A flat-bottomed barge, loaded with coal, has a mass of $3.0 \times 10^{5} \mathrm{kg} .\( The barge is \)20.0 \mathrm{m}\( long and \)10.0 \mathrm{m}$ wide. It floats in fresh water. What is the depth of the barge below the waterline? (W) tutorial: boat)
Someone steps on your toe, exerting a force of \(500 \mathrm{N}\) on an area of \(1.0 \mathrm{cm}^{2} .\) What is the average pressure on that area in atm?
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Energy Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Magnetism

Read Explanation

Particle Model of Matter

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free