Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 94

Fluid with kinematic viscosity \(\nu\) flows down an inclined circular pipe of length \(\ell\) and diameter \(D\) with flow rate \(Q\). Find the vertical drop per unit length of the pipe so that the pressure drop \(\left(p_{1}-p_{2}\right)\) is zero for laminar flow.

Short Answer

Expert verified
The vertical drop per unit length of the pipe is 0.

Step by step solution

01

Apply Hagen-Poiseuille law

The Hagen-Poiseuille law relates the pressure drop across a fluid flowing through a tube to the flow rate, the tube length and diameter, and the fluid viscosity. It is given by \(p_{1}-p_{2}=\frac{8 \nu l Q}{\pi D^{4}}\). Given \(p_{1}-p_{2} = 0\), the equation reduces to \(0=\frac{8 \nu l Q}{\pi D^{4}}\)
02

Decipher Hagen-Poiseuille law result

For Step 1's result to be true in a mathematical context, the flow rate \(Q\) must be zero (indicating no flow) or it implies that other factors influencing the flow are extreme enough to counteract the effect of the flow rate on pressure drop.
03

Bernoulli's Equation and Continuity Equation

The Bernoulli's equation is expressed as \(p_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = p_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2\). Since pressure drop is zero, \(p_1 = p_2\). The height \(h_1\) is \(l\) and \(h_2 = 0\). Given the flow is steady, the continuity equation \(A_1v_1 = A_2v_2 = Q\), holds, where \(A\) is the cross-sectional area of the pipe, and \(v\) is the speed of the fluid. Simplifying these gives: \(v = \frac{Q}{\pi(D/2)^2}\)
04

Simplify Given and Derived Values

Substitute the derived \(v\) value and given \(p_1 = p_2\) into the Bernoulli's equation. This provides \(\frac{1}{2}\rho (\frac{Q}{\pi(D/2)^2})^2 + \rho gl = \frac{1}{2}\rho (\frac{Q}{\pi(D/2)^2})^2\). Simplifying gives \(g = 0\) or \(l = 0\), indicating the pipe is horizontal, or no vertical drop occurs.
05

Calculate the Vertical Drop

The vertical drop per unit length is given by the height difference \(l\) by length of the pipe. Hence the vertical drop is \(l = 0\).

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

Two fixed, horizontal, parallel plates are spaced 0.4 in. apart. A viscous liquid $\left(\mu=8 \times 10^{-3} 16 \cdot \mathrm{s} / \mathrm{ft}^{2}, 3 G=0.9\right)$ flows between the plates with a mean velocity of 0.5 fls. The flow is laminar. Determine the pressure drop per unit length in the direction of flow. What is the maximum velocity in the channel?

Two immiscible, incompressible, viscous fluids having the same densities but different viscosities are contained between two infinite, horizontal, parallel plates (Fig. \(P 6.80\) ). The bottom plate is fixed, and the upper plate moves with a constant velocity \(U\) Determine the velocity at the interface. Express your answer in terms of \(U, \mu_{1},\) and \(\mu_{2}\). The mo:ion of the fluid is caused entirely by the movement of the upper plate; that is, there is no pressure gradient in the \(x\) direction. The fluid velocity and shearing stress are continuous across the interface between the two fluids. Assume laminar flow:

Blood flows at volume rate \(Q\) in a circular tube of radius \(R\). The blood cells concentrate and flow near the center of the tube. while the cell-free fluid (plasma) flows in the outer region. The center core of radius \(R_{c}\) has a viscosity \(\mu_{c}\) and the plasma has a viscosity \(\mu_{p}\). Assume laminar, fully developed flow for both the core and plasma flows and show that an "apparent" viscosity is defined by $$\mu_{\mathrm{app}}=\frac{\pi R^{4} \Delta p}{8 L Q}$$ is given by $$\mu_{\mathrm{app}}=\frac{\mu_{p}}{1-\left(R_{c} / R\right)^{4}\left(1-\mu_{p} / \mu_{c}\right)}$$

Verify that the momentum correction factor \(\beta\) for fully developed, laminar flow in a circular tube is \(4 / 3\).

The three components of velocity in a flow field are given by $$\begin{aligned}u &=x^{2}+y^{2}+z^{2} \\\v &=x y+y z+z^{2} \\\w &=-3 x z-z^{2} / 2+4\end{aligned}$$ (a) Determine the volumetric dilatation rate and interpret the results. (b) Determine an expression for the rotation vector. Is this an irrotational flow field?
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Thermodynamics

Read Explanation

Electromagnetism

Read Explanation

Solid State Physics

Read Explanation

Translational Dynamics

Read Explanation

Physics of Motion

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