/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 In a food-processing plant, hone... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 54

In a food-processing plant, honey is pumped through an annular tube. The tube is \(L=2 \mathrm{m}\) long, with inner and outer radii of \(R_{i}=5 \mathrm{mm}\) and \(R_{o}=25 \mathrm{mm},\) respectively. The applied pressure difference is \(\Delta p=125 \mathrm{kPa}\), and the honey viscosity is \(\mu=5 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2} .\) The theoretical velocity profile for laminar flow through an annulus is: \\[u_{z}(r)=\frac{1}{4 \mu}\left(\frac{\Delta p}{L}\right)\left[R_{i}^{2}-r^{2}-\frac{R_{o}^{2}-R_{i}^{2}}{\ln \left(\frac{R_{q}}{R_{o}}\right)} \cdot \ln \left(\frac{r}{R_{i}}\right)\right]\\] Show that the no-slip condition is satisfied by this expression. Find the location at which the shear stress is zero. Find the viscous forces acting on the inner and outer surfaces, and compare these to the force \(\Delta p \pi\left(R_{o}^{2}-R_{i}^{2}\right) .\) Explain.

Short Answer

Expert verified
By substituting \(r = R_{i}\) and \(r = R_{o}\) into the velocity profile equation and getting zero, it shows that the no-slip condition is satisfied. Solving the derivative of the velocity equation for zero gives the location of zero shear stress. Finally, integrating the shear stress gives the viscous forces acting on the surfaces, and as per the equilibrium condition, these balance out the pressure difference force.

Step by step solution

01

No-Slip Validation

We can validate the no-slip condition, which states that the velocity of the fluid at the wall is zero, by substituting \(r = R_{i}\) and \(r = R_{o}\) into the velocity profile equation: \(u_{z}(r)=\frac{1}{4 \mu}\left(\frac{\Delta p}{L}\right)\left[R_{i}^{2}-r^{2}-\frac{R_{o}^{2}-R_{i}^{2}}{\ln \left(\frac{R_{q}}{R_{o}}\right)} \cdot \ln \left(\frac{r}{R_{i}}\right)\right]\). If these equal zero, the no-slip condition is satisfied.
02

Zero Shear Stress Calculation

To find the location where the shear stress is zero, we take the derivative of the velocity profile equation with respect to r, set it to zero and solve for r. This gives the radius at which the shear stress will be zero in the annulus.
03

Viscous Forces Calculation

Now, calculate the viscous forces acting on the inner and outer surfaces. To find these, you apply the shear stress formula \(\tau = \mu \frac{du}{dr}\), evaluate this at \(r = R_{i}\) and \(r = R_{o}\), and then integrate over the surfaces to obtain the forces. The result has to be the force due to the pressure difference, \(\Delta p \pi(R_{o}^{2}-R_{i}^{2})\), as the net force on a fluid element in steady flow must be zero.

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Most popular questions from this chapter

The cone and plate viscometer shown is an instrument used frequently to characterize non-Newtonian fluids. It consists of a flat plate and a rotating cone with a very obtuse angle (typically \(\theta\) is less than 0.5 degrees). The apex of the cone just touches the plate surface and the liquid to be tested fills the narrow gap formed by the cone and plate. Derive an expression for the shear rate in the liquid that fills the gap in terms of the geometry of the system. Evaluate the torque on the driven cone in terms of the shear stress and geometry of the system.

Crude oil, with specific gravity \(\mathrm{SG}=0.85\) and viscosity \(\mu=2.15 \times 10^{-3}\) lbf \(\cdot \mathrm{s} / \mathrm{ft}^{2},\) flows steadily down a surface inclined \(\theta=45\) degrees below the horizontal in a film of thickness \(h=0.1\) in. The velocity profile is given by \\[u=\frac{\rho g}{\mu}\left(h y-\frac{y^{2}}{2}\right) \sin \theta\\] (Coordinate \(x\) is along the surface and \(y\) is normal to the surface.) Plot the velocity profile. Determine the magnitude and direction of the shear stress that acts on the surface.

A female freestyle ice skater, weighing 100 lbf, glides on one skate at speed \(V=20 \mathrm{ft} / \mathrm{s}\). Her weight is supported by a thin film of liquid water melted from the ice by the pressure of the skate blade. Assume the blade is \(L=11.5\) in. long and \(w=0.125\) in. wide, and that the water film is \(h=0.0000575\) in. thick. Estimate the deceleration of the skater that results from viscous shear in the water film, if end effects are neglected.

For the velocity fields given below, determine: a. whether the flow field is one- , two-, or three-dimensional, and why. b. whether the flow is steady or unsteady, and why. (The quantities \(a\) and \(b\) are constants.) (1) \(\vec{V}=\left[a y^{2} e^{-b t}\right]\) (2) \(\vec{V}=a x^{2} i+b x j+c k\) (3) \(\vec{V}=\)axyi\(-\)byt (4) \(\vec{V}=a x \hat{i}-b y j+c t \hat{k}\) (5) \(\bar{V}=\left[a e^{-\ln x}\right] i+b t^{2}\) (6) \(\vec{V}=a\left(x^{2}+y^{2}\right)^{1 / 2}\left(1 / z^{3}\right) \hat{k}\) (7) \(\vec{V}=(a x+t) i-b y^{2}\) (8) \(\vec{V}=a x^{2} \hat{i}+b x z j+c y k\)

Fluids of viscosities \(\mu_{1}=0.1 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\) and \(\mu_{2}=0.15 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\) are contained between two plates (each plate is \(1 \mathrm{m}^{2}\) in area). The thicknesses are \(h_{1}=0.5 \mathrm{mm}\) and \(h_{2}=0.3 \mathrm{mm},\) respectively. Find the force \(F\) to make the upper plate move at a speed of \(1 \mathrm{m} / \mathrm{s}\). What is the fluid velocity at the interface between the two fluids?
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