Viscous Drag

• Viscous drag is defined as

the frictional force between an object and a?fluid?which opposes the the motion between the object and the fluid

• Viscous drag is calculated using Stoke’s Law;

F?= 6πηrv

• Where
  • F?= viscous drag (N)
  • η?= coefficient of viscosity of the fluid (N s m?2?or?Pa s)
  • r?= radius of the object (m)
  • v?= velocity of the object (ms?1)

• The viscosity of a fluid can be thought of as its thickness, or how much it resists flowing
  • Fluids with?low viscosity?are?easy?to pour, while those with?high viscosity?are?difficult?to pour

• The?coefficient of viscosity?is a property of the fluid (at a given temperature) that indicates how much it will resist flow
  • The rate of flow?of a fluid is inversely proportional to the coefficient of viscosity

Drag Force at Terminal Velocity

• Terminal velocity is useful when working with Stoke’s Law since at terminal velocity the forces in each direction are balanced

Ws?= Fd?+ U (equation 1)

• Where;
  • Ws?= weight of the sphere
  • Fd?= the drag force (N)
  • U?=?upthrust?(N)

At terminal velocity forces are balanced: W (downwards) = Fd + U (upwards)

• The weight of the sphere is found using volume, density and gravitational force

Ws?=vsρsg

• Where
  • vs?= volume of the sphere (m3)
  • ρs?= density of the sphere (kg m–3)
  • g = gravitational force (N kg?1)

• Recall Stoke’s Law

Fd?= 6πηrvterm?(equation 3)

• Upthrust equals weight of the displaced fluid
  • The?volume of displaced fluid is the same as the volume of the sphere
  • The weight of the fluid is found from volume, density and gravitational force as above

• Substitute equations 2, 3 and 4 into equation 1

• Rearrange to make terminal velocity the subject of the equation

• Finally, cancel out?r?from the top and bottom to find an expression for?terminal velocity?in terms of the?radius of the sphere?and the?coefficient of viscosity

• This final equation shows that terminal velocity is;
  • directly proportional to the square of the radius of the sphere
  • inversely proportional to the viscosity of the fluid

Conditions for Stoke’s Law Equation

• The equation can only be used when certain conditions are met;
  • The flow is laminar
  • The object is small
  • The object is spherical
  • Motion between the sphere and the fluid is at a slow speed

Laminar and Turbulent Flow

• As an object moves through a fluid, or a fluid moves around an object, layers in the fluid are created
• In laminar flow all the layers are moving in the same direction and they do not mix
• This tends to happen for slow moving objects, or slow flowing liquids
• The equation above only applies for laminar flow
• In turbulent flow the layers move in different directions and the layers do mix

Changing Viscosity

• Viscosity is temperature-dependent
  • Liquids?are?less?viscous as temperature increases
  • Gases?get?more?viscous as temperature increases

Step 1: List the known quantities in SI units

• • Radius of the sphere,?rs?= 5.0 mm = 5.0 × 10-3?m
  • Terminal velocity of the sphere,?v?= 0.03 m s-1
  • Viscosity of oil,?η?= 0.3 Pa s
  • Density of oil,?ρf?= 900 kg m?3

Step 2: Sketch a free-body diagram to resolve the forces at constant speed

Ws?= Fd?+ U

Step 3: Calculate the value for viscous drag,?Fd

Fd?= 6πηrv?= 6 × π × 0.3 × 5.0 × 10-3?× 0.03 = 0.008482

Step 4: Write the complete answer to the correct significant figures and include units

• • ?The viscous drag,?Fd??= 8.5 × 10-4?N