What is the difference between Lagrangian and Eulerian Methods?
Solution:
|
Lagrangian Method |
Eulerian Method |
|
In the Lagrangian method, we analyse a fluid flow by
assuming that the fluid is composed of a very large number of particles whose
motion must be described. |
In the Eulerian method, the properties of a flow
field are described as functions of space coordinates and time. |
|
In this method, we Identify (or label) the material
of the fluid and track (or follow) it as it moves, and monitor change in its
properties (which may be velocity, temperature, density, mass, or
concentration, etc.) in the flow field. |
In this method, we identify (or label) a certain fixed location in the flow field and follow a change in its property, as different materials pass through that location. In such a case, the following the property, say the temperature is recorded by the sensor. |
 |
| Water Droplet |
In a 450 bend, a rectangular air duct of 1 m2 cross-sectional area is gradually reduced to 0.5 m2 area. Find the magnitude and direction of the force required to hold the duct in position if the velocity of flow at the 1 m2 section is 10 m/s and pressure are 2.943 N/cm2. Take the density of air as 1.16 kg/m3.
Solution:
Area at section (1), A1 = 1 m2
Area at section (2), A2 = 0.5 m2
Velocity at section (1), V1 = 10 m/s
Pressure at section (1),p1 = 2.943 N/cm2 = 2.943 x ((10)
^4) N/m2 = 29430 N/m2
Density of air, ρ = 1.16 kg/m3
Applying the continuity equation at sections (1) and (2)
A1V1 = A2V2
V2 = A1V1/A2 = (1/0.5) * 10 = 20 m/s
Discharge
Q = A1V1 = 1 x 10 = 10 m3/s
Applying Bernoulli's equation at sections (1) and (2)
[p1/ρg
+ ((V1) ^2) / 2g] = [p2/ρg +
((V2) ^2) / 2g]
[(2.943 * ((10) ^4)) / (1.16 * 9.81)] + [((10) ^2)/ (2 *
9.81)] = [p2 / ρg +
((20) ^2) / (2*9.81)]
p2/ρg =
[(2.943 * ((10) ^4)) / (1.16 * 9.81)] + [((10) ^2) / (2 * 9.81)] – [((20) ^2) /
(2 * 9.81)]
p2/ρg=
2586.2 + 5.0968 - 20.387 = 2570.90 m
p2 = 2570.90 * 1.16 * 9.81 = 29255.8 N
Force along x-axis, Fx = ρQ [V1x- V2x] + (P1A1)x + (P2A2)x
where A1x = 10 m/s, V2x = V2 cos45° = 20 * 0.7071
(p1A1)x = p1A1 = 29430 * 1 = 29430 N
and(p2A2)x = - p2A2 cos45° = - 29255.8 * 0.5 *0.7071
Fx = 1.16 x 10[10 - 20 * 0.7071] + 29430 * 1 - 29255.8 *
0.5 * 0.7071
= - 48.04 + 29430 - 10343.37 = 0 - 19038.59 N
Similarly force along y-axis, Fy = ρQ [V1y - V2y] + (p1A1)y +
(p2A2)y
where V1y = 0, V2y = V2 sin45° = 20 *0.7071 = 14.142
(p1A1)y = 0 and (p2A2)y = - p2A2 sin45° = - 29255.8 *0.5
*0.7071 = - 10343.37
Fy = 1.16 x 10 [0 - 14.142] + 0-10343.37
= - 164.05 - 10343.37 = - 10507.42 N
Resultant force, FR= √ [((Fx) ^2) + ((Fy) ^2)] = √ [((19038.6) ^2) + ((10507.42) ^2) = 21746.6
N
The direction of FR with the x-axis is-
tanθ=
Fy/Fx =10507.42/19038.6 = 0.5519
θ=
28° 53'
How will you obtain Bernoulli’s equation from Euler’s equation of motion along a streamline? Write assumptions of Bernoulli’s equation.
Solution:
Using Euler’s Equation:
dp/P +gdz + vdv = 0
On Integrating this we get,
p/P +gz + v^2/2 = constant
On dividing by g we get,
p/Pg + z + v^2/2g = constant
And the above equation is called Bernoulli’s Equation.
Assumptions of Bernoulli’s Equations are as follows
1. The flow must be steady, i.e. the flow
parameters at any point shall not change with time,
2. The flow must be incompressible – even though pressure varies, the density must
remain constant along a streamline;
3.
The friction by viscous forces must be negligible.
What is the boundary layer? Explain with a sketch development of boundary layer over a smooth plate.
Solution:
Boundary-Layer:-
A very thin layer of fluid, called the boundary layer, in the immediate neighbourhood of the solid boundary where the variation of velocity from zero at the solid boundary to free-stream velocity in the direction normal to the boundary takes place.
The boundary layer is the locus of the point at which the velocity of fluid particles become 0.99 of U∞ (i.e., 99% of the free stream
velocity).
In the fluid flow, the very first layer adheres to the
surface. This is called No-Slip Condition (where u=0). As moving away from the
surface, the velocity of fluid increases. Hence, there is a relative velocity in
the fluid. The value of shear stress in the boundary layer is
Τ=
µ*(du/dy)
But beyond the boundary layer du/dy becomes zero. Hence, no
shear stress is present.
Discuss the effect of pressure gradient on boundary layer separation.
Solution:
The effect of the pressure gradient (dp/dx) on the boundary layer separation can be explained by considering the flow over the curved surface ABCSD as shown in the figure below.
In the region ABC of the curved surface, the velocity
increases because the area of flow decreases. It means that in this region, the
flow gets accelerated.
The pressure gradient (dp/dx) is negative in this region
because due to the increase of the velocity, the pressure decreases in the direction of the flow.
The entire boundary layer moves forward as long as (dp/dx)
< zero, as shown in the figure.
In the region CSD of the curved surface, the pressure is
minimum at point C.
Along the region CSD of the curved surface, the velocity
of flow along the direction of fluid decreases because the area of a flow
increases.
Due to the decrease of velocity, the pressure increases
and hence dp/dx is positive i.e. dp/dx>0.
The velocity of the layer adjacent to the solid surface
along the length of the solid surface goes on decreasing as the kinetic energy
of the layer is used to overcome the frictional resistance of the surface.
Hence, the combined effect of positive pressure gradient
and surface resistance reduces the momentum of the fluid.
A stage comes, when the momentum of the fluid is unable
to overcome the surface resistance and the boundary layer starts to separate
from the surface at point S. Downstream the point S, the flow is taking
place in the reverse direction and the velocity gradient becomes negative.
A wooden block of rectangular section 1.25 m wide, 2m deep and 4 m long floats horizontally in seawater. If the specific weight of wood is 0.64 and water weight 1.025 kgf/m3. Find the volume of the water displaced and the position of the centre of buoyancy.
Solution:
Rectangular section
b= 1.25 m
h= 2m
l= 4m
Sb= 0.64
ρ=
1000 kg/m3 (Density of water)
Water = 1.025 kgf/m3
=1.025*9.81 = 10.1 N/m3 = ρ*g
Weight of water displaced = Weight of body
Weight of wooden block = ρ*g*Vfd
ρB =
0.64 * 1000
= 640 kg/m3
Weight of wooden block = ρB * V
= 640*2*1.25*4
= 6400*9.81 N
= 62784 N
Vfd= Volume of fluid displaced
Ρ*g*Vfd
= 62784
Vfd = 62784 / (1000*9.81)
Vfd = 6.4 m3
Volume of fluid displaced = 6.4 m3
So, height of wooden block submerged = 6.4 / (1.25*4) =
1.28 m
y= 1.28 m
Centre of buoyancy from bottom = 1.28/2 = 0.64 m
A venturimeter having a diameter of 75mm at the throat and 150mm diameter at the enlarged end is installed in a horizontal pipeline 150mm in diameter carrying oil of specific gravity 0.9. The difference of pressure between the enlarged end and the throat recorded by a U-tube differential manometer is 175mm of Mercury. Determine the discharge through the pipe. Assume the coefficient of discharge of the water (Cd) as 0.97.
Solution:
The discharge through the venturimeter is given by
Q = Cd [a1a2 /√ (((a1) ^2) - ((a2) ^2))]* √ (2gh)
Cd =0.97
d1 =150mm= 0.15m
a1 =(π/4)*((0.15)) ^2) = 0.0177m2
d2 =75mm= 0.075m
a2 = (π/4)*((0.075)) ^2) = 0.0044m2
x= 175mm = 0.175m
h =x[(sh/sp) -1] = 0.175[(13.6/0.9)-1] = 2.469m
By substitution, we get
Q=0.97[0.0177*0.0044 /√ (((0.0177) ^2) - ((0.0044)
^2)]*√(2*9.81*2.469)
Q= 0.03067 m3 /sec
= 30.67 lit/sec