• Newton’s law of viscosity,• Pressure and temperature dependence of

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• Pressure and temperature dependence of viscosity

viscosity,

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Viscosity• Viscosity is a property that represents the

internal resistance of a fluid to motion.

• Ethanol

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• Water• Honey• Sludge• Slurry• Pastes

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Newton’s Law of Viscosity

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Velocity Gradient= ∆U/∆Y

Velocity gradient

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Newton’s Law of Viscosity�Shear stress acts tangentially to the

surface (F=tangential force).

F∆X

ForceArea

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F A

area tangential

force shear=τ

ForceArea

Newton’s Law of Viscosity

• Newton’s law of viscosity states that“Shear stress is directly proportional to velocity gradient”

duτ α duτ µ=

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µ = Viscosity of the fluidUnit of µKg/m.sPoisePa.s

1 Poise = 1g/cm. s

du

dyτ α du

dyτ µ=

Role of Viscosity

• Statics– Fluids at rest have no relative motion between

layers of fluid and thus du/dy = 0– Therefore the shear stress is _____ and is zerozero

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– Therefore the shear stress is _____ and is independent of the fluid viscosity

• Flows– Fluid viscosity is very important when the fluid

is moving

zerozero

Kinematic viscosity ν

• The ratio µ / ρ appears in many equations. • Kinematic viscosity ν (pronounced: new)

ρµν =

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ρν =

• Gases: Viscosity increases with increasing temperature, why?

• Liquids: Viscosity decreases with increasing temperature, why?

Kinematic viscosity

• Units

• m2/s• Stokes =cm2/s

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• Stokes =cm /s

Temperature and pressure dependence of viscosity

Viscosity of Newtonian fluids depends only on

temperature and pressure

)PP(exp TTE

exp)P,T( oo

0 −β

−∆µ=µ

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)PP(exp T TR

exp)P,T( oo

o0 −β

µ=µ

Whereµo:viscosity at To and Po (reference temperature and pressure)

∆E: activation energy for flowR: gas constantβ:material property [m2/N]

Temperature and pressure dependency of viscosity

• Viscosity will also change with pressure -but under normal conditions this change is negligible in gasses

• High pressure can also change the

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• High pressure can also change the viscosity of a liquid. As pressure increases the relative movement of molecules requires more energy hence viscosity increases

Kinematic Viscosity of Water vs. Temp

Temp (oC) Viscosity (m2/s)

0 1.79 x 10-3

10 1.31 x 10-3

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10 1.31 x 10

20 1.00 x 10-3

30 7.97x 10--4

40 6.5 x 10-4

50 5.55 x 10-4

ElementAtoms of the same type

MPD/FFO/Lect_3 GasLiquid

Liquid Atoms can move around but are attracted together

MPD/FFO/Lect_3Cold liquid Hot liquid

Gases Large spaces between atoms

MPD/FFO/Lect_3Cold gas Hot gas

Gases fill the whole space

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Liquids fill containers from the bottom

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Next Lecture

• Numerical based on viscosity of fluids

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• Come with calculator

Examples

[1] A plate 0.025 mm apart from a fixed plate moves at 60 cm/s and requires force of 2 N/m2 to maintain the speed. Determine the fluid viscosity between the plates.fluid viscosity between the plates.

Answer: 8.33 x10 -5 Pa.s

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[2] A flat plate of area 1.5 x 106 mm2 is pulled with the speed of 0.4 m/s relative to another plate located at a distance 0.15 mm apart from it. Find the force and power required to maintain the speed, if the fluid separating them having viscosity as 1 Pa.s. (Hint: Power = F.u)as 1 Pa.s. (Hint: Power = F.u)

Answer: F = 4000 N and Power =1600 Watts

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[3] The space between two flat parallel plates is filled with an oil. Each side of the plate is 60 cm. The thickness of the oil plate is 12.5 mm. The upper plate which moves at 2.55 m/s requires a force of 98.1 N to maintain the speed. Determine:

(a) Dynamic viscosity of oil in cP(a) Dynamic viscosity of oil in cP(b) Kinematic viscosity of oil in Stokes if the

specific gravity of the oil is 0.95Answer: (a) 1363.5 cP (b) 14.35 Stokes

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[4] The dynamic viscosity of an oil used for lubrication between a shaft and a sleeve is 6 Poise. The shaft is of diameter 0.4 m and rotates at 190 rpm. Calculate the power lost in the bearing for the sleeve length of 90 mm. The thickness of the film length of 90 mm. The thickness of the film is 1.5 mm. (Hint: u =ЛDN/60, T = FD/2, Power = 2ЛNT/60)

Answer: 716.48 watts

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[5] The velocity distribution for a flow over a flat plate is given by::

u= (¾)y-y2

Where,u = velocity (m/s)y = distance above the plate (m)

Determine shear stress at y = 0.15 m. Take dynamic viscosity of the fluid as 8.6 Poise.

Answer: 0.3875 N/m2

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[6] If the velocity profile over a flat plate is parabolic with a vortex 20 cm from the plate, where the velocity is 120 cm/s. Calculate the velocity gradient and shear Calculate the velocity gradient and shear stress at a distance of 0, 10 and 20 cm from the plate. Take viscosity of oil as 8.5 Poise. (Hint: parabolic velocity profile u = ay2 + by +c)

Answer: 10.2, 5.1 and 0 Pa.s

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Atoms

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Molecules - Atoms bonded together

Mixture Different atoms

Element One type of atom

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not bonded together

Compound Different atoms bonded together (molecules)

CompoundDifferent atoms in fixed proportions

MPD/FFO/Lect_3 1Y 1R 1B 2R 1G1Y 1G

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