MM303 FLUID MECHANICS I PROBLEM SET 1 (CHAPTER 2)

FALL 2018

1) For the velocity fields given below, determine:

i) Whether the flow field is one-, two-, or three-dimensional, and why.

ii) Whether the flow is steady or unsteady, and why

a) 2 btV ax e i

b) 2bxV ae i bx j

c) 2V ax t i by j

d) V axy i byztj

NOTE: The quantities a and b are constants.

2) The velocity field V=axyi+by2j, where a=2 1/m s, b=-6 1/m s, and the coordinates are measured in

meters. Calculate the velocity at the point (2, ½). Find an expressionfor for the streamline passing

through this point. Plot several streamlines in the first quadrant, including the one passes through the

point (x,y)=(2,1/2)

Solution:

At point (2, ½), the velocity components are u=axy=2×2×1/2=2 m/s

v=by2=-6×(1/2)2= -3/2 m/s

Steamline is tangent to the velocity vector. Hence, for streamline, 2dy v by by

dx u axy ax

Separating the variables, dy bdx

y ax

Integrating, we get, / 3ln( ) ln( ) b ab

y x c y Cx y Cxa

The streamline passing through point (2,1/2) is obtained as 3 3

3

1 1 42 2 4

2 2C C C y

x

3) Velocity field of a flow is given as V=Axi+2Ayj, where A=2 s-1. Verify that the parametric equations for particle motion are xp=c1eAt and yp=c2e2At. Obtain the equation for the pathline of the particle located at the point (x,y)=(2,2) at time t=0. Compare the pathline with the streamline through the same point.

Solution:

For pathlines 1 1

1 1ln( ) At c c At At

p

dx dx dxu Ax Adt x At c x e x e e C e

dt dt x

2 22 2 2

2 22 2 ln( ) 2 At c c At At

p

dy dy dyv Ay Adt y At c y e y e e C e

dt dt y

Hence the parametric equations of the motion are 1

Atx C e and 2

2

Aty C e

Equation of the pathlines can be obtained by combining the parametric equations as follows:

1

1

At At xx C e e

C subtituting this into the following equation, we get,

2

2 2

2 2

1

At xy C e y C y Cx

C

At point (2,2), 2 2

2 1

2 2

yC

x .

The path line of the particle at point (2,2) at time t=0 is 21

2y x

For streamline, 2 2dy v Ay y

dx u Ax x

Separating the variable and integrating, we get,

2

2

2int ln( ) 2ln( ) ln

dy dx yegrating y x c c y Cx

y x x

At point (2,2), 2 2

2 1

2 2

yC

x .

The steamline passing through point(2,2) is 21

2y x

The streamline passing through point (2,2) and the pathine that started at point (2,2) coincide because the flow

is steady.

4) Velocity field of a flow is given by V=ai+bxj, where a= 3 m/s and b= 2 s-1. Coordinates are measured in

meters.

a) Obtain the equation for the streamline passing through point (2, 4).

b) At t=5 s, what are the coordinates of the particle that passed through point (0, 4) at time t=0 s?

c) What conclusion can you draw about the pathline, streamline and streakline for this flow?

5) Crude oil, with specific gravity SG=0.85 and viscosity =0.1 Ns/m2, flows steadily down a surface inclined

=450 in a film of thickness h=2.5 mm. The velocity profile is given by the expression below. (Coordinates

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.

𝑢 =𝜌𝑔

𝜇(ℎ𝑦 −

𝑦2

2) 𝑠𝑖𝑛𝜃

6) A block weighing 45 N and having dimensions 250 mm on each edge is pulled up on an inclined surface

on which there is a film of SAE 10W oil at 370C. If the speed of the block is 0.6 m/s and the oil film is 0.025

mm thick, find the force required to pull the block. Assume that velocity profile in the oil film is linear.

The surface is inclined at an angle of 250 from the horizontal.

7) A concentric cylinder viscometer is driven by a falling

mass M=0.20 kg connected by a cord and pulley to

the inner cylinder, as shown. The liquid to be tested

fills the annular gap of width a=0.4mm and height

H=160 mm. After a brief starting transient, the mass

falls at constant speed Vm=60 mm/s. Develop an

algebraic expression for the viscosity of the liquid in

the device in terms of M, g, Vm, r, R, a and H. Evaluate

the viscosity of the liquid. Note: r=50 mm, R=100

mm.

8. At a depth of 7.5 km in the ocean, the pressure is 75 Mpa. Assume a specific weight of 10 KN/m3 at the

surface and an average bulk modulus of elasticity of 2.5 GPa for that pressure range. Find

(a) the change in specific volume between the surface and 7.5 km depth,

(b) the specific volume at 7.5 km

(c) the specific weight at 7.5 km.

Solution:

Pressure at 7.5km (P2) = 75 Mpa = 75x106 N/m2

Specific weight at the surface (𝛾)= 10 KN/m3 = 10x1000 = 10000 N/m3

Bulk modulus of elasticity at the surface (Ev) = 2.5 GPa = 2.5x109 N/m2

a) Density at the surface (𝜌1 ) = 10000/9.81 = 1019.4 kg/m3

Specific volume at the surface (vs1)= 1/𝜌1 =1/1019.4=0.000981 m3/kg

Bulk modulus in terms of specific volume is

Ev = −∆𝑃

∆𝑉𝑠𝑉𝑠1

2.5 × 109 = −7.5 × 106 − 0

∆𝑉𝑠0.000981

∆𝑉𝑠 = −0.0000294 𝑚3/𝑘𝑔

b) vs2 = vs1 +∆𝑉𝑠 = 0.000981-0.0000294 = 0.000951 m3 /kg

c) Density at 7.5 km (𝜌2 ) = 1/𝑣𝑠2 = 1051.5 kg/m3

𝛾2 = 𝜌2𝑔 = 1051.5 × 9.81 = 10315 𝑁/𝑚3

9. In a fluid the velocity measured at a distance of 75 mm from the boundary is 1.125 m/s. The fluid has a

dynamic viscosity of 0.048 Ns/m2 and a specific gravity (SG) of 0.9. What is the velocity gradient and the shear

stress at the boundary assuming a linear velocity distribution. Also calculate the kinematic viscosity.

Solution:

Assumptions:

The fluid is Newtonian

The velocity profile is linear

Change in velocity (u) =1.125-0 = 1.125 m/s

Change in distance (y) = 75 – 0 = 75 mm = 75/1000 m = 0.075 m

Dynamic viscosity (µ) = 0.048 Ns/m2

specific gravity (SG)= 0.9

velocity gradient (du/dy) = ?

Shear stress (𝝉 ) = ?

Kinematic viscosity (𝒗) = ?

Velocity gradient on the boundary can be calculated as

du/dy= u/y=1.125/0.075=15 1/s

𝜏 = 𝜇𝑑𝑢

𝑑𝑦= 0.048 × 15 = 0.72 𝑁/𝑚2

Densitiy of fluid= SGx𝜌𝑤𝑎𝑡𝑒𝑟=0.9x1000=900 kg/m3

𝑣 =𝜇

𝜌=

0.048

900= 5.3𝑥10−5 𝑚2/𝑠

10) A piston of weight 90 N slides in a lubricated pipe. The clearance between piston and pipe is 0.025 mm. If

the piston decelerates at a rate of 0.6 m/s2 when its speed is 0.5 m/s, what is the viscosity of the oil?

Solution:

Assumptions:

- The oil is Newtonian

- Velocity of the oil in the gap is linear.

Weight of piston (W) = 90 N

Clearance (y)= 0.025 mm = 0.000025 m

Deceleration (a) = 0.6 m/s2

Change in velocity (u) = 0.5 m/s

Diameter of piston (D) = 125 mm = 0.125m

Length of piston (L) = 130mm = 0.13 m

Viscosity of oil (µ) = ?

Frictional force acting on piston (F) = Shear stress(τ) at the piston surface x surface area of piston(A)

𝐹 = 𝜇𝑑𝑢

𝑑𝑦(𝜋𝐷𝐿) = 𝜇 ×

0.5

0.000025(𝜋𝑥0.125𝑥0𝑥13) = 1021𝑥02𝜇

Note: Since velocity profile in the gap is linear, on the piston surface du/dy=u/y

Applying Newtons secon la to the piston, we can write

𝑊 − 𝐹 =𝑊

𝑔𝑎

90 − 10210.02𝜇 =90

9.81(−0.6)

𝜇 = 0.094 𝑁𝑠/𝑚2

11) A flat plate 0.3 m2 in area moves horizantally through oil between two large fixed parallel walls. The

distance between the walls is 10 cm. If the velocity of the moving plate is 0.6 m/s and the oil has a kinematic

viscosity of 𝟎. 𝟒𝟓 × 𝟏𝟎−𝟒 𝒎𝟐/𝒔 and specific gravity 0.8, calculate the drag force when

(a) the plate is 2.5 cm from one of the walls and (y1=2.5).

(b) the plate is equidistant from both the walls(y1=y2).

Solution

Assumptions

- The oil is Newtonian.

- The velocity profile between the moving plate and the stationary walls is linear.

a) y1 = 2.5 cm = 0.025 m

y2 = 10-2.5 = 7.5 cm = 0.075 m

Total force (F) = Force on side1 (F1) + Force on side2 (F2) = τ1 A + τ2 A

b) If the plate is equidistant, y1 = y2 = y = 10/2 = 5 cm = 0.05 m

Total force (F) = Force on side1 (F1) + Force on side2 (F2) = τ1 A + τ2 A = (τ1 + τ2) A

12) The tip of glass tube with an internal diameter of 2 mm is immersed to a depth of 1.5 cm into a liquid of

specific gravity 0.85. Air is forced into the tube to form a spherical bubble just at the lower end of the tube.

Estimate surface tension of liquid if the pressure in the bubble is 200 Pa.

Solution:

Radius of bubble (r) = 1 mm = 0.001m

Pressure inside bubble (Pi) = 200 Pa

Depth of liquid (h) = 1.5 cm = 0.015 m

Specific weight of liquid= SGx𝛾𝑤𝑎𝑡𝑒𝑟 = 0.85x9810 = 8338.5 N

Surface Tension (𝜎)=?

Pressure outside the bubble (Po) = x h= 8338.4x0.015 = 125.07 Pa

The difference between the pressure in the bubbleoutside of the bubble (P) = 200-125.07 = 74.93 Pa

The force actin on the surface of the bubble due this pressure difference is balanced by the surface tension force.

Hence we can write

𝑑𝑃 =2𝜎

𝑟

𝜎 =𝑑𝑃𝑟

2= 0.0375 𝑁/𝑚