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Chapter 1: Problem 72

For a parallel plate arrangement of the type shown in Fig. 1.5 it is found that when the distance between plates is \(2 \mathrm{mm}\), a shearing stress of 150 Pa develops at the upper plate when it is pulled at a velocity of \(1 \mathrm{m} / \mathrm{s}\). Determine the viscosity of the fluid between the plates. Express your answer in SI units.

Short Answer

Expert verified
The viscosity of the fluid between the plates is 0.3 Pa.s or 0.3 kg/(m.s)

Step by step solution

01

Understand the given values

In this problem, the following values are given: shearing stress (\(τ\)) = 150 Pa, velocity (\(v\)) = 1 m/s, and the distance (\(d\)) between the plates = \(2 \times 10^{-3}\)m (Converted 2 mm to meters)
02

Understand Newton's law of viscosity

According to Newton's law of viscosity, shear stress (\(τ\)) is directly proportional to the velocity gradient in the direction normal to the surface. The formula to calculate shear stress is \(τ =μdv/dx\), where \(μ\) is the dynamic viscosity, \(dv/dx\) is the velocity gradient and \(τ\) is the shear stress.
03

Apply the values and rearrange the formula

The velocity gradient \(dv/dx\) is equal to the change in velocity over the distance between the plates. Since the velocity of the lower plate is zero, \(dv/dx\) equals \(1/(2 \times 10^{-3})\). Substitute the known values into the formula for Newton's law of viscosity and rearrange to solve for \(μ\): \(150 =μ \times 1/(2 \times 10^{-3})\).
04

Calculate the viscosity of the fluid

Solving the equation for \(μ\), we get \(μ = 150 \times 2 \times 10^{-3} = 0.3 \, Pa.s\) or \(0.3 kg/(m.s)\).
05

Final Step: Validate the Units of the Calculated Viscosity

The SI units for dynamic viscosity are Pascals-second (Pa.s), so the final result, 0.3 Pa.s, is consistent with this.

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