Unit 5 | unit 5 laminar flow

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Introduction to fluid mechanics

Unit - 5

Laminar Flow

Q1) Explain laminar flow through circular pipes

A1)

I 2.png

Fig no 1 Circular pipes

Fig: Shows a horizontal circular pipe of radius R, having laminar flow of fluid through it.

Consider a small concentric cylinder of radius r and length dx as a free body.

If is the shear stress, the shear force F is given by

F = x 2 r x dx

Let P be the intensity of pressure at left end and the intensity of pressure at the right end be (

Thus, the force acting on the fluid element are:

The shear force, x 2r x dx on the surface of fluid element.

The pressure force, P x on the left end.

The pressure force, ( on the right end.

For steady flow, the net force on the cylinder must be zero.

[ p x - (] –

- -

Shear stress is zero at the centre of pipe (r = 0) and maximum at the pipe wall given by

I 1.png

Fig no 2 When shear stress is zero

()

From Newton’s Law of viscosity

In this equation, the distance y is measured from the boundary. The radial distance r is related to distance y by the relation.

y = R-r ordinate y = -d r

Comparing two values of

du = ()r.dr

Integrating the above equation w. r. t. ‘r’ we get

u = . r2 + C --------------- (ii)

Where C is the constant of integration and its value is obtained from the boundary conditions.

At r = R, u = 0

0 = . R2 + C or C = . R2

Substituting this value of C in eqn. (ii) we get,

u = . r2 . R2

u = - . (R2 – r2)

Shows that velocity distribution curve is a parabola.

The maximum velocity occurs, at the centre and is given by.

umax = - R2

Q2) Explain annulus and parallel plates?

A2) 1. Flow of viscous fluid between two parallel plates when one plate is moving & other of Rest – Coutts flow

Consider laminar flow between two parallel flat plates located at a distance b apart such that lower plate is at rest & the upper plates moves uniformly with a constant velocity U as shown in fig.

I 4.png

Fig no 3 Parallel plates

A small rectangular element of fluid of length dx, thickness dy& unit width is

Considered as a free body.

The forces acting on the fluid element are

1) The pressure force p dy x 1 on left end

2) The pressure force (p +. d x) d y x1 on the right end.

3) The shear force

4) The shear force [ . dy] dx x 1 on the surface.

For steady & uniform flow the resultant force in the direction of flow is zero.

p. dy – (p + . dx) dy – + (+ . dy) dx = 0

-dx. dy + dy. dx = 0

Dividing by the volume of element dx dy we get

= ------------------- (1)

Above equation shows the interdependence of shear & pressure gradients

According to Newton’s law of viscosity for laminar flow the shear stress

= .

Put in equation (1) we get,

Integrating above equation twine w.r.t. y

u = y2 + C1 y + C2 ------------- (2)

At y = 0, u = 0 & at y = b, u = U

We get C2 = 0 & C1 = - () b

Put values of C1& C2 in eqn. (2)

u = y - (by – y2)

q = = (by – y2)] dy

U. -

The distribution of shear stress across any section may be determined by using

Newton’s Law of viscosity

= u. = [ - () (b – 2y)]

= - (b – 2y)

Both Plates at Rest:

In this case the equations for velocity, discharge q and the shear can be obtained from similar equation for generalizes Coutts flow by putting U = 0

Thus, for flow between two stationary parallel plates

I 5.png

Fig no 4 Both pates are in parallel

u = - . (by – y2)

Discharge per unit width,

q = .

Shear stress . (b –2y)

Q3) Explain stokes law?

A3) The strength of the viscosity in a small area conveying viscous fluid is provided by:

Fd = 6 pi u R v

Where:

Fd is the force of friction - known as Stokes' drag - which operates on a visible link between liquid and particles

μ is strong viscosity (some authors use the symbol η)

R is the radius of the circular object

v is the velocity of the flow relative to an object.

In SI units, F d is given in Newton’s (= kg m s - 2), μ in Pa · s (= kg m-1 s - 1), R in meters, and v in m / s.

Stokes Act makes the following assumptions on the behavior of a particle in a liquid:

Laminar Flow

Round particles

Similar (made in the same way)

Smooth surfaces

The particles do not interfere with each other.

In molecules Stoke's law is used to describe the breadth and depth of Stokes.

The CGS unit of kinematic viscosity was called "stokes" after his work.

Applications

Stokes rule is the base of the fall-sphere viscometer, where the liquid rests on a vertical glass tube. The known size and quantity range is allowed to decrease with this liquid.

Properly selected, it reaches the velocity terminal, which can be measured in the time it takes to transfer two marks to the tube.

Electronic sensors can be used for opaque liquids. Knowing the final velocity, size and size of the sphere, and the size of this liquid, Stokes law can be used to calculate the viscosity of fluid.

A series of steel ball bearings with different widths are commonly used in old tests to improve mathematical accuracy.

School experiments use glycerin or gold syrup as a liquid, and this process is used industrially to test the viscosity of the liquids used in the processes.

Many school tests often include fluctuations in temperature and / or concentration of materials used to indicate the effects this may have on viscosity. Industrial methods include various oils, and polymer beverages as solutions.

The significance of Stoke's law is reflected in the fact that he played a key role in the research that led to at least three Nobel Prizes.

Stokes Act is important in understanding the swimming of microorganisms and sperm; also, the formation of tiny particles and organisms in water, under gravity.

In the air, the same concept can be used to explain why tiny water droplets (or ice crystals) can remain suspended in the air (like clouds) until they grow to a critical size and begin to fall as rain (or snow and hail).

The same use of mathematics can be made in dissolving fine particles in water or in other liquids.

Terminal velocity of speed falling in a fluid

Fig no 5 Terminal velocity of speed falling in a fluid

At the end (or resolution) of speed, the force exerted by Fg due to the difference between gravity and gravity (both caused by gravity given by:

F g =( ρ p - ρ f ) g 4/ 3 pi R3

With and p and ρf the size of the field and the field of fluid, respectively, and g the speed of gravity. Finding the power balance Fd = Fg and resolving velocity v gives terminal velocity vs. Note that as the maximum force increases as R3 and Stokes' increases as R, the final velocity increases as R2 and therefore differs significantly in particle size as shown below.

If the particle only feels its weight while falling into the visible liquid, then the final pressure is reached when the frictional force and particle strength due to the liquid directly measure the gravitational force. The velocity v (m / s) is given by:

V = 2/ 9 ( ρ p - ρ f ) / μ g R2

Where

g field gravitational force (m / s2)

R is the radius of circular particles (m)

ρ p particle size (kg / m3)

ρf maximum water weight (kg / m3)

μ is the strong viscosity (kg / (m * s)).

Q4) Explain measurements of viscosity

A4) 1. Rotating cylinder method

In this method, Newton’s law of viscosity is used to measure the viscosity of a fluid.

Figure shows a rotating cylinder viscometer.

Fig no 6 Rotating cylinder viscometer

It consists of two concentric cylinders, the annular space between them is filled with the liquid whose viscosity is to be determined.

The outer cylinder is rotated at a constant angular velocity with respect to the inner stationary cylinder.

The torque transmitted by the enclosed liquid to the stationary cylinder is measured by the torsional strain of the restraining spring attached to the top of the inner cylinder.

According to Newton’s law of viscosity,

Since the annular space t = (R2 – R1) is quite small (where R2 and R1 are the radii of the outer and inner cylinders respectively), the velocity gradient

(Where N= the rotational speed of the outer cylinder in rpm)

Shear stress =

Viscous drag = Shear stress × Area

(Where h = height of liquid)

Viscous torque = Viscous drag x radius

= x

Viscous torque must be equal to the torque T exerted by the torquemeter.

Thus, a rotational type viscometer can be calibrated to directly give u for given speed of rotation N.

2. Falling Sphere Method

Falling sphere method of measuring viscosity of a liquid is based on Stokes law.

Figure shows a falling sphere viscometer.

A small spherical ball is released into the liquid to be tested and it accelerate under the gravitational force until it reaches a maximum terminal velocity V when the buoyant and viscous drag forces balance the gravity force.

Fig no 7 Falling sphere viscometer

d = Diameter of the spherical ball.

l = distance travelled by the sphere in viscous liquid.

t = Time taken by the sphere to cover distance l

= Density of sphere

= Density of fluid/liquid

W = weight of sphere

FB = Buoyant force acting on the sphere

FD = Drag force, and

= Dynamic viscosity of the liquid under test.

The force acting on the sphere are-

1) Weight W, acting vertically downward.

2) Buoyant force FB, acting vertically upward

3) Drag force FD, acting vertically upward.

W= volume × Density of sphere × g

FB = weight of liquid displaced

= volume of liquid displaced × density of liquid × g

FD = 3VD

For equilibrium

FD + FB = W

FD = W – FB

3VD = - =

From the above equation the value of dynamic viscosity u can be determined.

3. Redwood viscometer

The redwood viscometer consists of vertical cylindrical oil cup with an orifice in the centre of its base.

The orifice can be closed by a ball.

Fig no 8 Redwood viscometer

A hook pointing upward serves as a guide mark for filling the oil.

The cylindrical cup is surrounded by the water bath.

The water bath maintains the temperature of the oil to be tested at constant temperature.

The oil is heated by heating the water bath by means of an immersed electric heater in the water bath.

The provision is made for stirring the water, to maintain the uniform temperature in the water bath and to place the thermometer to record the temperature of oil and water bath.

This viscometer is used to determine the kinematic viscosity of the oil. From the kinematic viscosity the dynamic viscosity is determined

Q5) Show that for a steady laminar flow through a circular pipe mean velocity of flow occurs at a radial distance of 0.707R from the centre of pipe where R is radius of pipe.

A5) Velocity distribution in laminar flow

Average velocity ü

Radius at u= u. put in Equation (1)

Q6) An oil of viscosity 15 poise flows through a pipe of diameter 150 mm at a velocity of 2.5 m/s. If the specific gravity of oil is 0.9, find:

(1) Pressure gradient in direction of flow

(2) Share stress at the wall of pipe

(3) Reynolds number of flow of oil.

(4) Velocity at a distance of 50 mm from wall.

A6)

Given:

Viscosity = 15 poise = 1.5 N s/m²,

Diameter D = 150 mm= 0.15 m

Velocity 13 m/s,

Specific gravity, 0.9

To find

2) 3) Re 4) Vat r = 0.05 m

Radius of pipe R

Pressure gradient in the direction of flow

Using relation

(2) Shear stress at the wall of pipe,

)

3] Reynolds’s number

Laminar flow

(4) Velocity at a distance of 50 mm from wall

Q7) A laminar flow is taking place in a pipe of diameter 0.2m. The maximum velocity is 1.5m/s. Find the mean velocity and the radius at which this occurs also calculate the velocity at 0.04m from the wall of the pipe.

A7)

Given:

D= 0.2 m

Umax= 1.5 m/s

Y= 0.04 m

Radius of pipe R=-0.1 m

Mean velocity u= Umax/2 = 0.75 m/s

Velocity at 0.04 m from pipe wall

Put the value

The Velocity at y = 0.04 m is 0.96 m/s

Q8) Heated compound in the plastic state is forced through the die is an injection molding process under a pressure of 1 MPa. The diameter and length of the die being 8 mm and 3 m long respectively. If the mean velocity of flow is 0.48 m/s, calculate the viscosity of the plastic compound the heated state. Assume laminar flow

A8)

Given:

P1-P2 = 1 Mpa= 1x10^6 N/m^2

D= 8 mm= 0.008 m.

L= 3 m

u = 0.48 m/sec

To find:

The viscosity of plastic compound is 1.388 Nam.

Q9) An oil of density 917 kg/m' is being pumped in a 15 cm diameter pipe. The discharge is measured as 850 L/min. The drop in pressure in a stretch of 800 m of pipeline, both ends of which are at the same elevation, is measured as 95 kPa. Estimate the absolute viscosity of the oil.

A9)

Given:

P1-p2 =

D= 15 cm = 0.15 m

L= 800 m

Q= 850 lit/min = 0.0142 m2/sec

Find: Absolute viscosity

Average velocity

Using relation of pressure difference

The absolute Viscosity is 0.1039 N.s/m^2

Q10) An oil having a viscosity of 10 poise flows through a pipe of 500 mm diameter at a rate of 500 lit per second. Find the type of flow occurring in the pipe, if the specific gravity of all is 0.9. May 2014 3 Marks

A10)

Given

D= 500 mm = 0.5 m

Q= 500 Liter per second-0.5 m/sec,

S=0.9

To find: Type of fluid

Reynolds number

The flow is laminar.

Q11) What power will be required per km length of pipe line to overcome viscous resistance to flow of crude oil of viscous 1.9 poise through a horizontal pipe of 10 cm diameter at the rate of 650 liters per hour? Take specific gravity of oil = 0.85.

A11)

Given:

L.= 1000 m.

= 1.9 poise = 0.19 N.s/m²

D=10 cm=0.1 m

Q=650 liter/hour = 1.806x10^-4 m^3/sec

S = 0.85

Average velocity

To find:

P Pressure head,

Power

Power required to overcome viscous resistance is 2.526

Q12) Crude oil R.D. 0.9 is pumped through a smooth horizontal oil is 2.5 stokes. Differential pressure head between two ends of the pipe is 16.31 m of oil. Assuming the flow of oil is to be laminar, Find:

(1) Rate of flow of oil through pipe.

(2) Power required maintaining the flow.

Also check whether the flow is actually laminar or not pipe 400 m long. 100 mm diameter, kinematics viscosity of

A12)

Given:

RD= 0.9

L= 400 m

D= 100 mm= 0.1 m

V= 2.5 stokes

Density of oil, p=S x P=0.9x1000= 900 kg/m

Dynamic viscosity, u=p v=900x25x10 =0 225 N-x/m

Average velocity by Hagen Poiseuille's formula,

Discharge, Q=ū× A = 0.5 × x 0.1² = 3.927 x 10³ m³/sec

Power, P = pg HQ

= 900×9.81×16.31 x 3.927x10-³

= 565.49 watt

Power required maintaining the flow is 565.49 watt. or How, 2000

Reynolds’s number for given flow is,

The actual flow is laminar

Q13) Two horizontal plates kept at 100 mm apart have laminar flow of oil of viscosity 1.5 N.s/m between them. The maximum velocity of flow is 2 m/s. Find:

(1) Discharge per meter width

(2) Shear stress at the plate

(3) Velocity gradient at the plate

(4) Pressure difference between two points 15 m apart.

A13)

Given: t = 100 mm = 0.1 m, U = 2 m/s, μ = 5 N-s/m², L=15m, b=1m

(i) Relation between ū and Umax,

(ii) Discharge per meter width,

Pressure gradient use Equation

Shear stress at the plate,

(iii) Pressure difference between two points

(iv) Velocity gradient of plates,

Q14) Two parallel plates kept 20 mm apart have laminar flow of oil between them maximum velocity of 2 m/s. Calculate:

(1) Discharge per meter width

(i) Shear stress at the plate

(iii) The difference in pressure between two points 20 m apart

(iv) Velocity gradient of plates

(v) Velocity at 0.02 mm from the plate. Assume viscosity of oil = 2.453 N-s/m Dec. 2004, 10 Marks

A14)

Given:

t = 20 mm = 0.2 m. Um=2 m/sc, = 2.453 N-s/m²,

L = 20 m, b= 1 m

(iii) Relation between ū and Umax,

(iv) Discharge per meter width,

Pressure gradient use Equation

Shear stress at the plate,

(iii) Pressure difference between two points

(iv) Velocity gradient of plates,

(v) Velocity at any point,

y-20 mm = 0.02

Velocity at 20 mm from plate 0.7198 m/sec

Q15) Oil of relative density 0.92 and dynamic viscosity 1.05 poise flowing between two parallel fixed plates, 12 mm apart, has mean velocity of 1.4 m/s. Calculate:

1) The maximum velocity

2) The boundary shear stress

3) Velocity and shear stress at a distance of 2 mm from one of the plate

4) Loss of head over a distance 25 m. Dec. 2012, 10 Marks

A15)

Given: R.D=0.92. = 1.05 poise = 0.105 Pa-s

t = 12mm=0.012=14m

1) Maximum velocity.

2) Boundary shear stress

Now for pressure gradient, use Equation

3) a) Velocity at a distance 2 mm from plate, use Equation

b) Shear stress at y = 2 mm

4) Loss of head over a distance 25 m

Q16) Glycerin of specific gravity of 1.28 and viscosity 8.07 poise flows between two large parallel plates kept 1.5 cm apart. The rate of flow is 4.4 m per hour meter width of plates. Determine the maximum velocity, maximum shear strops, pressure gradient and Reynolds number.

A16)

Given:

S= 1.28

8.07 poise = 0.807 N-s/m',

t= 1.5 cm-0.015 m

Q=- 4.4 m per hour per meter width of plate

Discharge per meter width= mean velocity. Area

Pressure gradient,

Maximum shear stress

Reynolds Number

Q17) Two parallel plates are kept 3 mm apart. For a steady laminar flow of oil between these plates, there was a pressure drop of 10 kN/m² per meter length of plates. If the viscosity of the oil is 5 x 10 N-s/m, determine the discharge per meter width of plates, maximum shear and maximum velocity of flow.

A17)

Given

t = 3 mm = 3x10 m.

p1, -p2 = 10 kN/m²

L = 1m,

5x10¹ N./m²

Drop of pressure in a length of 1 m, use equation

Discharge per meter width,

Maximum shear stress,

Maximum velocity,

Q18) A masonry wall of a water tank is 0.9 m thick. At the bottom a crack of thickness 0.3 mm and 600 mm wide has developed and the crack extends to the entire thickness of the wall. If the tank contains 4 m of water above the crack and the other crack is at atmospheric pressure, estimate the leakage volume per day from the crack. Kinematic viscosity of water = 0.01 stokes.

A18)

Given:

L= 0.9 m, t=0.3 mm = 3 x 10 m.

Width, b= 600 mm = 0.6 m h, H = 4 m of water above crack.

v=0.01 stokes = 1 x 10 m²/sec

Dynamic viscosity,

μ = p. v = 1000x1x10* = 0.001 N-s/m²

Using drop of pressure head formula,

Rate of flow

Q = Au= (width x thickness) u

= (0.6×3×10) x 0.327 = 5.886 x 10³ m³/sec

Leakage volume per day,

Q 3600C24 = 5.886 x 10³ x 3600x24

= 5.085 m²/day

Leakage from crack is 5.085 m'/day.

Q19) A small air bubble of diameter 1 mm rises with a steady state velocity of 1.5 cm/sec through an oil. Estimate the viscosity of the oil if its specific gravity is 0.9.

A19)

Given:

d=1 mm = 1 x 10 m, U = 1.5 cm/sec = 0.015 m/sec.

S = 0.9

p = sp=0.9 x 1000 = 900 kg/m³

If weight of bubble is neglected, the drag force and the buoyancy force must keep the bubble in equilibrium.

Drag force = Buoyant force

Check Reynolds number

The expression for drag force is valid only for Reynolds’s number less than unity.

Q20) A viscometer consisting of a long vertical tank containing a fluid at the desired ambient conditions and a small sphere is employed for determining the velocity of an oil. If a steel sphere (S=0.8) of 1 cm diameter falls with a constant velocity of 5 cm/s through a lubricating oil (S = 0.9). calculate the velocity of the oil.

A20)

Given: 5, -0.8, d=1 cm = 1x 10¹ m. U-5 cm/s 0.05 m/s, S₁-0.9

Density of sphere

Density of fluid (oil)

= 0.9 x 1000=900 kg/m

Viscosity for falling sphere is given by

Check Reynolds number

0.058 <1 .. OK.

Q21) In a torsional viscometer, the outer cylinder of 150.5 mm diameter is rotated by turning shaft at a constant speed of 100 rpm. Owing to viscosity, the liquid under test transmits a torque of 540 Nm to the inner cylinder of 150 mm diameter, which is suspended by torsion wire fixed at its upper end. If the liquid is 130 deep, find the viscosity

A21)

Given: Inner diameter, D, 150 mm, Outer diameter,

D₁ = 150.5 mm

Number of rotation, N-100 rpm,

Torque, T-540 Nm

h+t= 130

R₁-150/2-75 mm - 0.075 m

R2 = 150./ 2= 75.25 mm=0.07525 m

h = 130-t=130-0.5= 129.5 mm

For torsional viscometer, viscosity is

The viscosity of fluid is 37.36 N.s/m².

Q22) In a rotating cylinder viscometer, the radii of the cylinders are 32 mm and 30 mm and the outer cylinder is rotated steadily at 200 rpm. For certain liquid filled in the annular space to a depth 80 mm, the torque produced on the inner cylinder is 0.9 x 10 Nm. Calculate viscosity.

A22)

Given:

Radius of inner cylinder, R, = 32 mm = 0.032 m.

Radius of outer cylinder, R₂ = 30 mm = 0.03 m,

Height of liquid, h=== 80 mm = 0.08 m,

Speed of cylinder, N = 200 rpm,

Torque T = 0.9 x 10 Nm

For rotating cylinder, viscosity is

The viscosity of fluid is 5.94 x10* Ns/m².

Q23) The viscosity of an oil specific gravity 0.8 is measured by a capillary tube of diameter 40 mm. The difference of pressure head between two point 1.2 m apart is 0.3 m of water. The weight of oil collected in a measuring tank is 400 N in 100 seconds. Find the viscosity of oil.

A23)

Given: Sp. gr. of oil, S = 0.8,

Wt. Density, p = Sxp = 0.8 x 1000 = 800 kg/m³

Diameter of tube, D = 40 mm = 0.04 m,

Length of tube, L = 1.2 m

Pressure head difference, h = 0.3 m of water,

Time, t= 100 sec., Weight of oil, W = 400 N

Discharge Q = Wt. of oil collected/sec /Weight density

=400/100 /800

= 0.509 x 10³ m³/sec

Viscosity by capillary tube viscometer is given by

:. The viscosity of oil is 0.242 N-s/m².

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