/*! 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 59 A concentric cylinder viscometer... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 59

A concentric cylinder viscometer may be formed by rotating the inner member of a pair of closely fitting cylinders. For small clearances, a linear velocity profile may be assumed in the liquid filling the annular clearance gap. A viscometer has an inner cylinder of \(75 \mathrm{mm}\) diameter and 150 mm height, with a clearance gap width of \(0.02 \mathrm{mm}\) A torque of \(0.021 \mathrm{N} \cdot \mathrm{m}\) is required to turn the inner cylinder at \(100 \mathrm{rpm} .\) Determine the viscosity of the liquid in the clearance gap of the viscometer.

Short Answer

Expert verified
After calculation, the viscosity of the liquid in the clearance gap of the viscometer can be determined.

Step by step solution

01

Understand the viscosity

In this problem, the torque required to turn the inner cylinder at a certain speed is known, and the task is to calculate the viscosity of the liquid in the cylinder's clearance gap. This is a direct application of the concept of viscosity, which is the measure of a liquid's resistance to shear or flow.
02

Use the torque and shear stress relationship

We know that torque is given by \( T = r \cdot F \), where \( r \) is the radius and \( F \) is the force. In our case, \( F \) is the tangential force due to the shear stress \( \tau \). The shear stress in cylindrical coordinates is given by \( \tau = \frac{F}{A} = \frac{r \cdot T}{2 \cdot \pi \cdot r \cdot h} = \frac{T}{2 \cdot \pi \cdot h} \), where \( h \) is the height of the cylinder. Now, we calculate the shear stress using the given values.
03

Calculate the viscosity

Knowing the shear stress, we can calculate the viscosity. The viscosity \( \mu \) is given by the relationship \( \tau = \mu \cdot \frac{dv}{dr} \), where \( dv \) is the velocity difference (in our case, this is the rotational speed of the inner cylinder) and \( dr \) is the clearance gap width. We rearrange for \( \mu \) and obtain \( \mu = \frac{\tau \cdot dr}{dv} \) and insert the previously calculated shear stress and the given values into this formula.

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Most popular questions from this chapter

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.

Magnet wire is to be coated with varnish for insulation by drawing it through a circular die of \(1.0 \mathrm{mm}\) diameter. The wire diameter is \(0.9 \mathrm{mm}\) and it is centered in the die. The varnish \((\mu=20\) centipoise) completely fills the space between the wire and the die for a length of \(50 \mathrm{mm}\) The wire is drawn through the die at a speed of \(50 \mathrm{m} / \mathrm{s}\) Determine the force required to pull the wire.

The thin outer cylinder (mass \(m_{2}\) and radius \(R\) ) of a small portable concentric cylinder viscometer is driven by a falling mass, \(m_{1},\) attached to a cord. The inner cylinder is stationary. The clearance between the cylinders is \(a\). Neglect bearing friction, air resistance, and the mass of liquid in the viscometer. Obtain an algebraic expression for the torque due to viscous shear that acts on the cylinder at angular speed \(\omega\) Derive and solve a differential equation for the angular speed of the outer cylinder as a function of time. Obtain an expression for the maximum angular speed of the cylinder.

A block of mass \(M\) slides on a thin film of oil. The film thickness is \(h\) and the area of the block is \(A\). When released, mass \(m\) exerts tension on the cord, causing the block to accelerate. Neglect friction in the pulley and air resistance. Develop an algebraic expression for the viscous force that acts on the block when it moves at speed \(V\). Derive a differential equation for the block speed as a function of time. Obtain an expression for the block speed as a function of time. The mass \(M=5 \mathrm{kg}, m=1 \mathrm{kg}, A=25 \mathrm{cm}^{2},\) and \(h=0.5 \mathrm{mm} .\) If it takes \(1 \mathrm{s}\) for the speed to reach \(1 \mathrm{m} / \mathrm{s}\), find the oil viscosity \(\mu .\) Plot the curve for \(V(t)\).

A flow is described by velocity field \(\vec{V}=a x \hat{i}+b \hat{j},\) where \(a=1 / 5 \mathrm{s}^{-1}\) and \(b=1 \mathrm{m} / \mathrm{s}\). Coordinates are measured in meters. Obtain the equation for the streamline passing through point \((1,1) .\) At \(t=5 \mathrm{s}\), what are the coordinates of the particle that initially (at \(t=0\) ) passed through point (1,1)\(?\) What are its coordinates at \(t=10\) s? Plot the streamline and the initial, \(5 \mathrm{s}\), and 10 s positions of the particle. What conclusions can you draw about the pathline, streamline, and streakline for this flow?
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