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Chapter 2: Problem 66

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.

Short Answer

Expert verified
The expression for the torque needed to turn the disk is \(T = r^2 * \pi * h * \mu * du/dy\). The device could not be used to measure the viscosity of a non-Newtonian fluid because their viscosity changes with the shear rate, making the relationship between shear stress and shear rate non-linear.

Step by step solution

01

Obtain the Shear Stress Equation

We start by using the basic equation for shear stress. Shear stress (\(\tau\)) is given by the formula \(\tau = \mu * du/dy\) where \(\mu\) is the dynamic viscosity, \(du\) is the change in tangential velocity between the two disks, and \(dy\) is the change in height.
02

Relate Shear Stress to Torque

The relationship between the torque (\(T\)) applied and shear stress can be given by the formula \(T = r^2 * \pi * h * \tau\). Where \(\tau\) is the shear stress, \(r\) is the radius of the disk, \(\pi\) is the constant \(Pi\), and \(h\) is the height or distance between the two disks.
03

Substitute Shear Stress into Torque Equation

Substitute the equation for shear stress from Step 1 into the torque equation from Step 2: \(T = r^2 * \pi * h * \mu * du/dy\)
04

Consider Non-Newtonian Fluids

Regarding the possibility of using this device to measure the viscosity of a non-Newtonian fluid, the answer would be no. This is because non-Newtonian fluids do not have a constant viscosity. The relationship between shear stress and shear rate is not linear for these fluids, thus the derivation above for torque would not apply.

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Most popular questions from this chapter

A block weighing 10 lbf and having dimensions 10 in. on cach edge is pulled up an inclined surface on which there is a film of SAE \(10 \mathrm{W}\) oil at \(100^{\circ} \mathrm{F}\). If the speed of the block is \(2 \mathrm{ft} / \mathrm{s}\) and the oil film is 0.001 in. thick, find the force required to pull the block. Assume the velocity distribution in the oil film is linear. The surface is inclined at an angle of \(25^{\circ}\) from the horizontal.

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