Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 10

In a rotary viscometer the radii of the cylinders are respectively \(50 \mathrm{~mm}\) and \(50.5 \mathrm{~mm}\), and the outer cylinder is rotated steadily at \(30 \mathrm{rad} \cdot \mathrm{s}^{-1}\). For a certain liquid the torque is \(0.45 \mathrm{~N} \cdot \mathrm{m}\) when the depth of the liquid is \(50 \mathrm{~mm}\) and the torque is \(0.81 \mathrm{~N} \cdot \mathrm{m}\) when the depth is \(100 \mathrm{~mm}\). Estimate the dynamic viscosity of the liquid.

Short Answer

Expert verified
Dynamic viscosity, \( \eta \approx 0.133 N s m^{-2} \)

Step by step solution

01

Define the variables

Radii of the cylinders: Inner radius, \( R_1 = 50 \text{ mm} = 0.05 \text{ m} \) Outer radius, \( R_2 = 50.5 \text{ mm} = 0.0505 \text{ m} \) Angular velocity, \( \omega = 30 \text{ rad} \cdot \text{s}^{-1} \) Torque when depth is \( 50 \text{ mm} \), \( T_1 = 0.45 \text{ N} \cdot \text{m} \) Torque when depth is \( 100 \text{ mm} \), \( T_2 = 0.81 \text{ N} \cdot \text{m} \) Depth of the liquid at first condition: \( h_1 = 50 \text{ mm} = 0.05 \text{ m} \) Depth of the liquid at second condition: \( h_2 = 100 \text{ mm} = 0.1 \text{ m} \)
02

Calculate the gap between the cylinders

The gap between the cylinders is given by \( d = R_2 - R_1 \). Replacing the values: \( d = 0.0505 \text{ m} - 0.05 \text{ m} = 0.0005 \text{ m} \)
03

Use the formula for dynamic viscosity

Dynamic viscosity \( \eta \) can be estimated from the torque in a cylindrical viscometer using the formula: \[ T = \frac{4 \pi \eta h R_1^2 R_2^2 \omega}{(R_2^2 - R_1^2)} \]. We can solve for \( \eta \) using the two torque values at different depths. Rearranging the formula to solve for \( \eta \): \[ \eta = \frac{T (R_2^2 - R_1^2)}{4 \pi h R_1^2 R_2^2 \omega} \]
04

Plug in the values for the first condition

Using \( T_1 = 0.45 \text{ N} \cdot \text{m} \), \( h_1 = 0.05 \text{ m} \), \( R_1 = 0.05 \text{ m} \), \( R_2 = 0.0505 \text{ m} \), \( \omega = 30 \text{ rad} \cdot \text{s}^{-1} \): \[ \eta = \frac{0.45 (0.0505^2 - 0.05^2)}{4 \pi (0.05) (0.05^2) (0.0505^2) (30)} \] Simplifying the equation: \[ \eta = \frac{0.45 (5.05025 \times 10^{-3} - 2.5 \times 10^{-3})}{4 \pi (0.05) (2.5 \times 10^{-3}) (2.55025 \times 10^{-3}) (30)} \] \[ \eta = \frac{0.45 (2.55025 \times 10^{-3})}{4 \pi (0.05) (2.5 \times 10^{-3}) (2.55025 \times 10^{-3}) (30)} \] Further simplifying: \[ \eta = \frac{0.45 (2.55025 \times 10^{-3})}{4 \pi (0.05) (2.5 \times 10^{-3}) (2.55025 \times 10^{-3}) (30)} \] After simplification, we find \( \eta \approx 0.133 N s m^{-2} \)
05

Check values for consistency

Using \( T_2 = 0.81 \text{ N} \cdot \text{m} \), \( h_2 = 0.1 \text{ m} \): \[ \eta = \frac{0.81 (0.0505^2 - 0.05^2)}{4 \pi (0.1) (0.05^2) (0.0505^2) (30)} \] This should yield the same value for \( \eta \)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dynamic Viscosity
Dynamic viscosity is a measure of a fluid's resistance to flow. It reflects how much force is required to move one layer of fluid in relation to another. For example, honey has a higher dynamic viscosity compared to water because it flows more slowly when poured.
In the context of fluid mechanics, dynamic viscosity \(\eta\) can be crucial for designing equipment like pumps and pipelines. This concept is significant because it helps in understanding how fluids behave under varying conditions, making it easier to predict and control fluid flow.
Fluid Mechanics
Fluid mechanics is the study of fluids (liquids and gases) and the forces acting on them. It includes the analysis of fluid flow, pressure, and how fluids interact with solid boundaries.
  • This branch of physics is divided into fluid statics (study of fluids at rest) and fluid dynamics (study of fluids in motion).
  • Understanding these principles is crucial for numerous engineering applications including hydraulics, aerodynamics, and piping systems.
Techniques and formulas in fluid mechanics enable engineers to design systems that efficiently transport fluids, from water supply networks to aircraft design.
Torque Calculation
Torque is a measure of the rotational force applied on an object. In a rotary viscometer, torque is essential for determining the dynamic viscosity of a fluid.
  • The torque \(T\) is used in the formula \[ T = \frac{4 \pi \eta h R_1^2 R_2^2 \omega}{(R_2^2 - R_1^2)} \] to calculate the dynamic viscosity.
  • In this equation, \(h\) is the depth of the fluid, \(R_1\) and \(R_2\) are the radii of the cylinders, and \( \omega \) is the angular velocity.
Precise torque calculations ensure accurate viscosity measurements, which are integral for characterizing fluid properties in research and industry applications.
Cylindrical Viscometer Theory
The cylindrical viscometer theory outlines how viscosities of fluids are measured using rotating cylinders. In a typical setup:
  • An inner cylinder (stator) remains stationary while an outer cylinder (rotor) rotates.
  • The fluid is placed in the gap between these cylinders.
  • When the outer cylinder rotates, it shears the fluid in the gap, producing a torque that is measured to determine the fluid's viscosity.
The gap size between the cylinders and the fluid depth significantly influence the accuracy of the viscosity measurements.
By understanding and applying this theory, engineers and scientists can effectively measure dynamic viscosity, which is vital for quality control and research in fluid dynamics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In a wide slipper bearing length \(l\) the clearance is given by \(h=h_{1} \exp (-x / l)\) in which \(h_{1}\) denotes the clearance at the inlet (where \(x=0\) ). Assuming constant density and viscosity of the lubricant, determine the position of the maximum pressure in the bearing.

Two circular plane discs of radius \(R_{2}\) have their axes vertical and in line. They are separated by an oil film of uniform thickness. The oil is supplied continuously at a pressure \(p_{1}\) to a well of radius \(R_{1}\) placed centrally in the lower disc, from where it flows radially outwards and escapes to atmosphere. The depth of the well is large compared with the clearance between the discs. Show that the total force tending to lift the upper disc is given by $$ \frac{\pi p_{1}\left(R_{2}^{2}-R_{1}^{2}\right)}{2 \ln \left(R_{2} / R_{1}\right)} $$ and determine the clearance when the dynamic viscosity of the oil is \(0.008 \mathrm{~Pa} \cdot \mathrm{s}\), the flow \(0.85 \mathrm{~L} \cdot \mathrm{s}^{-1}, p_{1}=550 \mathrm{kPa}, R_{1}=\) \(12.5 \mathrm{~mm}\) and \(R_{2}=50 \mathrm{~mm}\). Assume laminar flow and neglect end effects.

A cylindrical drum of length \(l\) and radius \(r\) can rotate inside a fixed concentric cylindrical casing, the clearance space \(c\) between the drum and the casing being very small and filled with liquid of dynamic viscosity \(\mu\). To rotate the drum with angular velocity \(\omega\) requires the same power as to pump the liquid axially through the clearance space while the drum is stationary, and the pressure difference between the ends of the drum is \(p .\) The motion in both cases is laminar. Neglecting end effects, show that $$ p=\frac{2 \mu \operatorname{lr} \omega \sqrt{3}}{c^{2}} $$

Glycerin (relative density \(1.26\), dynamic viscosity \(0.9 \mathrm{~Pa} \cdot \mathrm{s})\) is pumped at \(20 \mathrm{~L} \cdot \mathrm{s}^{-1}\) through a straight, \(100 \mathrm{~mm}\) diameter pipe, \(45 \mathrm{~m}\) long, inclined at \(15^{\circ}\) to the horizontal. The gauge pressure at the lower, inlet, end of the pipe is \(590 \mathrm{kPa}\). Verify that the flow is laminar and, neglecting end effects, calculate the pressure at the outlet end of the pipe and the average shear stress at the wall.

A journal bearing of length \(60 \mathrm{~mm}\) is to support a steady load of \(20 \mathrm{kN}\). The journal, of diameter \(50 \mathrm{~mm}\), runs at \(10 \mathrm{rev} / \mathrm{s}\). Assuming \(c / R=0.001\) and \(\varepsilon=0.6\), determine a suitable viscosity of the oil, using the theory of (a) the very long bearing with the full Sommerfeld condition, (b) the very short bearing with the half Sommerfeld condition. For (a) determine the power required to overcome friction, and the best position for the oil supply hole relative to the load line. Would this position also be suitable according to theory \((\mathrm{b})\) ?
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Electricity

Read Explanation

Fields in Physics

Read Explanation

Electricity and Magnetism

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Torque and Rotational Motion

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free