Fluid Mechanics (01 1X07)
Assignment 1 (Introduction: Fluid Properties)
1. ………………….Theoretical…………………….
2. ………………….Theoretical…………………….
3. If 5 m3 of certain oil weighs 45 kN calculate the specific weight, specific gravity and
mass density of the oil.
Solution :
Given data: Volume = 5 m3
Weight = 45 kN
Answer:
; 0.917;
4. A plate (2m x 2m), 0.25 mm distant apart from a fixed plate, moves at 40 cm/s and
requires a force of 1 N. Determine the dynamic viscosity of the fluid in between the
plates.
Solution :
Given data: Change of velocity,
Distance between the plates,
Contact area A = 2x2 = 4 m2
Force required, F = 1 N
Now,
Shear stress,
= F/A = 0.25N/m2
And,
Answer:
5. A tape of 0.015 cm thick and 1.00 cm wide is to be drawn through a gap with a clearance
of 0.01cm on each side. A lubricant of dynamic viscosity 0.021 Ns/m2 completely fills
the gap for a length of 80 cm along the tape. If the tape can withstand a maximum tensile
force of 7.5 N calculate the maximum speed with which it can be drawn through the gap.
Fig. Tape sliding with in lubricant
Solution :
Given data: Dynamic viscosity 0.021 Ns/m2
Clearance dy = 0.01 cm
Contact area
maximum tension the tape can with stand =7.5 N
Shear stress, t = F/A
Maximum shear stress the tape can withstand, = 0.467185 kN/m2
Also shear stress,
Substituting the value of
Answer: 2.23 m/s.
in the above equation, we get
6. Find the pressure inside a water droplet having diameter of 0.5 mm at 200 C if the
outside pressure is 1.03N/cm 2and the surface tension of water at that temperature is
0.0736 N/m.
Solution :
Given Data: -
Pressure inside the droplet,
Answer:7. Compare the capillary rise of water and mercury in a glass tube of 2 mm diameter at
200 C. Given that the surface tension of water and mercury at 200 C are 0.0736 N/m and
0.051N/m respectively. Contact angles of water and mercury are 00 and
1300 respectively.
Solution :
Given data: Surface tension of water, sw = 0.0736 N/m
And surface tension mercury, sm =0.051N/m
Capillary rise in a tube
For mercury
and
Note that the negative sign indicates capillary depression.
For water specific weight
The
Answer: -
and
15mm rise and 6.68mm depression.
Fluid Mechanics (01 1X07)
Assignment 2 (Hydrostatics)
1. ………………….Theoretical…………………….
2. ………………….Theoretical…………………….
3. Two pipes on the same elevation convey water and oil of specific gravity 0.88
respectively. They are connected by a U-tube manometer with the manometric liquid
having a specific gravity of 1.25. If the manometric liquid in the limb connecting the
water pipe is 2 m higher than the other find the pressure difference in two pipes.
Solution :
Given data:
Height difference = 2 m
Height difference = 2 m
Specific gravity of oil s = 0.88
Specific gravity of manometric liquid s = 1.25
Equating pressure head at section (A-A)
Substituing h = 5 m and density of water 998.2 kg/m3 we have P A -P B = 10791
Answer:10791 Pa
4. A large iceberg floating in sea water is of cubical shape and its specific gravity is 0.9 If
20 cm proportion of the iceberg is above the sea surface, determine the volume of the
iceberg if specific gravity of sea water is 1.025
Solution :
Let the side of the cubical iceberg be h.
Total volume of the iceberg = h 3
volume of the submerged portion is = ( h -20) x h 2
Now,
For flotation, weight of the iceberg = weight of the displaced water
The side of the iceberg is 164 cm.
Thus the volume of the iceberg is 4.41m3
Answer: 4.41m 3
5. A wooden cylinder of length L and diameter L/2 is floating on water with its axis
vertical. Find the metacentric height if the specific gravity of wood is 0.6
Solution :
Given data: Length of the cylinder = L
Diameter of the cylinder = L/2
Specific gravity of wood = 0.6
For flotation weight of the cylinder should be equal to weight of the water
displaced.
Now if the depth of immersion is h,
Weight of water displaced = Weight of the cylinder
h = 0.6
Metacentric height
Answer: - 0.1739 L m. (The negative sign indicates the body is in unstable
equilibrium.)
6. A hollow cylinder closed in both end, of outside diameter 1.5 m and length of 3.8 m and
specific weight 75 kN per cubic meter floats just in stable equilibrium condition. Find
the thickness of the cylinder if the sea water has a specific weight of 10 kN per cubic
meter.
Solution :
Given data : Outside diameter 1.5 m
Length L = 3.8 m
Specific weight 75 kN/m 3
Let the thickness t and immersion depth h .
For flotation
Weight of water displaced = weight of the cylinder
h = 91 t
For the cylinder to be in equilibrium
Solving for t we have t = 0.0409 or 0.000829m
Answer:- t = 0.83 mm
Fluid Mechanics (01 1X07)
Assignment 3 (Kinematics of Fluid Flow)
1. ………………….Theoretical…………………….
2. A velocity field is defined by u = 2y2, v = 3x, w = 0. At point (1, 2, 0), compute:
a) velocity,
b) local acceleration and
c) convective acceleration.
Solution :
Given velocity field, u = 2y 2 ; v = 3x; w = 0 so,
a) Thus,
b) Now from the above equation we can observe that
which implies the local acceleration is zero.
c) Also from the above equation we have the acceleration component as follow
3. Calculate the velocity of water flowing through a pipe system in which the flow first
enters a pipe of 0.2 m diameter and releases to another pipe of 0.1 m diameter. The water
flows at a rate of 2200 litre per hour.
Solution :
Given data: Flow rate Q = 2200 lph
Velocity of water at the first pipe
If A 2 and V 2 are the cross sections and velocity in the second pipe
Using continuity equation,
Substituting the values
Answer:
4. The x- component of the velocity field for a two dimensional fluid flow over a plate is
given by u = 3y – y 2 .Where y is measured from the surface in a perpendicular direction.
Verify if the flow is rotational. If so find the rotational velocity at (3, 2).
5. The two scalar components of velocity field are given for two flow system. Find the
third component of the velocity. For case:
a) u=x 3 +2y 2 +z 3 and v = -x 2 y – yz - xy
b) u= log (y 2 + z 2) and w= log (x 2 +y 2).
Solution :
Case I: Velocity field should satisfy the continuity equation. For a three dimensional
flow
the
equation
is
The partial derivatives of u and v are
Substituting
Integrating
Answer: -
these
into
the
continuity
we
equation
have,
Case
II:
As
similar
to
the
case
I
we
can
derive
Integration
of
the
above
equation
gives v =
f(x,z)
Since u and w are symmetric function of x, y and z , v should also be a symmetric
function.
So,
6. a) Check whether ϕ = (x 2 - y 2 + y) represents the velocity potential function for 2dimensional irrotational flow.
b) If, it does then determine the stream function.