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Chapter 2: Problem 40

The velocity distribution for laminar flow between parallel plates is given by \\[\frac{u}{u_{\max }}=1-\left(\frac{2 y}{h}\right)^{2}\\] where \(h\) is the distance separating the plates and the origin is placed midway between the plates. Consider a flow of water at \(15^{\circ} \mathrm{C},\) with \(u_{\max }=0.10 \mathrm{m} / \mathrm{s}\) and \(h=0.1 \mathrm{mm} .\) Calculate the shear stress on the upper plate and give its direction. Sketch the variation of shear stress across the channel.

Short Answer

Expert verified
The shear stress on the upper plate is calculated to be \(\tau= 22.76 N/m^2\). The direction is in the flow direction (along the plate). The shear stress distribution is linear with the maximum at the centerline and zero at plates.

Step by step solution

01

Understand the formula and hypothese

The velocity distribution for laminar flow between parallel plates is given by \[\frac{u}{u_{max}} = 1 - \left(\frac{2y}{h}\right)^2\] where \(u\) is the velocity at point \(y\), \(h\) is the distance between the plates and \(y=0\) is set to be the channel mid-point. This parabolic velocity profile is characteristic for fully developed laminar flow between parallel plates. A maximum velocity \(u_{max}\) is achieved in the middle of the channel.
02

Understand the concept of shear stress

Shear stress \(\tau\) in a fluid flowing between parallel plates can be calculated by \[\frac{du}{dy} = \frac{d}{dy}\left(u_{max}\left(1 - \left(\frac{2y}{h}\right)^2\right)\right)\] taking into account the no-slip condition, meaning that the fluid in immediate contact with the plate is at rest.
03

Plug in values and calculate shear stress

Substitute the given values into the equation:\n\(u_{max}=0.10ms^{-1}\), \(h=0.1mm=0.1x10^{-3}m\). The fluid is water, and the dynamic viscosity \(\mu\) of water at 15掳C is approximately \(1.138x10^{-3} Pas\). Considering the upper plate where \(y=h/2\), the shear stress is hence \(\tau=\mu \times \frac{du}{dy}(\frac{h}{2})\). Solving the equation will give the shear stress.
04

Determine the Direction of the Shear Stress

The direction of the shear stress on upper plate is in the direction of the fluid flow, which is along the plate (parallel to the flow direction) since the fluid adjacent to the upper plate is moving backwards due to the no-slip condition.
05

Sketch the Shear Stress Distribution

The shear stress distribution is linear in a fully developed laminar flow. At the centreline of the channel (y = 0) the shear stress is at maximum. Moving towards the plate surfaces, the shear stress gradually reduces to zero at the wall (y = 卤 h/2). It forms a straight line from the center to the plate.

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Most popular questions from this chapter

A proposal has been made to use a pair of parallel disks to measure the viscosity of a liquid sample. The upper disk rotates at height \(h\) above the lower disk. The viscosity of the liquid in the gap is to be calculated from measurements of the torque needed to turn the upper disk steadily. Obtain an algebraic expression for the torque needed to turn the disk. Could we use this device to measure the viscosity of a nonNewtonian fluid? Explain.

A flow is described by the velocity field \(\vec{V}=(A x+B)\) is \((-A y) j,\) where \(A=10\) ft/s/ft and \(B=20\) ft/s. Plot a few streamlines in the \(x y\) plane, including the one that passes through the point \((x, y)=(1,2)\).

A velocity field is given by \(\vec{V}=a y t \hat{i}-b x \hat{j},\) where \(a=1 \mathrm{s}^{-2}\) and \(b=4 s^{-1}\). Find the equation of the streamlines at any time \(t\) Plot several streamlines at \(t=0\) s, \(t=1\) s, and \(t=20\) s.

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) \(\vec{V}=\left[(a x+t) e^{\log y}\right]\) (2) \(\vec{V}=(a x-b y)\) (3) \(\vec{V}=a x \vec{i}+\left[e^{\ln x}\right]\) (4) \(\vec{V}=a x \hat{i}+b x^{2} j+a x k\) (5) \(\vec{V}=a x \hat{i}+\left[e^{b r}\right] \dot{j}\) (6) \(\vec{V}=a x \hat{i}+b x^{2} \hat{j}+a y k\) (7) \(\vec{V}=a x \hat{i}+\left[e^{\ln }\right] \hat{j}+a y \hat{k}\) (8) \(\vec{V}=a x i+\left[e^{b y}\right] j+a z \hat{k}\)

A viscous liquid is sheared between two parallel disks; the upper disk rotates and the lower one is fixed. The velocity field between the disks is given by \(\vec{V}=\hat{e}_{0} r \omega z / h\). (The origin of coordinates is located at the center of the lower disk; the upper disk is located at \(z=h .\) ) What are the dimensions of this velocity field? Does this velocity field satisfy appropriate physical boundary conditions? What are they?
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