20-09-2016, 09:36 AM
1455219717-Hhmga1.docx (Size: 11.76 KB / Downloads: 6)
Q1. WHAT ARE THE TYPES OF SIMILARITIES? EXPLAIN IN BRIEF.
1) Ans:
The types of similarities are as follows:
a. Dynamical similarity: In fluid mechanics, dynamic similarity refers to the phenomenon that when there are two geometrically similar vessels (same shape, different sizes) with the same boundary conditions (ex. No-slip, center-line velocity) and the same Reynolds and Womersley numbers, then the fluid flows will be identical. Ratios of all forces acting on corresponding fluid particles and boundary surfaces in the two systems are constant.
b. Kinematic similarity: Fluid flow of both the model and real application must undergo similar time rates of change motions. (Fluid streamlines are similar).In this similarity, the ratio of acceleration and velocity of model and prototype is constant.
c. Geometric similarity: If two objects have the same shape, they are said to be geometrically similar. By definition, the ratio of any two linear dimensions of one object will be same for any geometrically similar object.
Lm/Lp=Bm/Bp=Hm/Hp.
If this condition is verified, then the bodies are said to be geometric similar. Lm/Lp is called the scale ratio.
Where, L= length, B=breadth and H=height, m=model and p=prototype
Q2. DERIVE THE REYNOLD’S NO.
2) Ans:
Reynolds Number is the ratio of inertial force by viscous force. Here, Inertial force=m*v/t2
= (ρ*l3*v)/t2
= ρ*l2*v2
where, d =density of the fluid
l= length
v= velocity of fluid flowing
Again, Viscous Force= μ *dv/dl*a
= μ *v/l*l2
=μ vl
where, μ = dynamic viscosity
dv/dl= velocity gradient
Hence, reynolds Number is the ratio of inertial force to the viscous force, then
=(ρ*l2*v2)/ (μ *v*l)
= ρlv/μ which is Reynolds Number
Q3. EXPLAIN AND DERIVE:-
1)Euler’s no.
2)Weber’s no.
3)Mach’s no.
4)Froude’s no.
3) Ans:
a. Euler's number: It is the ratio of pressure force to the inertial force.
Pressure force=Pl2
Inertial force= ρ*l2*v2
Hence, Euler's number= pressure force/inertial force
=P/ ρ*v2
Uses:
• In turbo machines
• In lift and drag
b. Froude's Number: It is the ratio of square root of inertial force to the gravitational force.
Here, inertial force= ρ*l2*v2
Gravitational force= m*g
= (ρ*l3)*g
Hence, = (inertial force/gravitational force)1/2
= {(ρ*l2*v2)/(ρ*l3*g)}1/2
=v/(lg)1/2
Uses:
• In floating submarines
• In open channel
• In floating bodies
c. Weber number (Wr): It is the ratio of inertial force to the surface tension force.
Here, inertial force= ρ*l2*v2
Surface tension force= T*l
Hence, Wr= inertial force/ surface tension
= (ρ*l2*v2)/ ( T*l)
Uses:
• Capillary effect
• Boiling and conduction
• liquid jets
• where surface tension is predominant
Q4. WHAT IS DIMENSIONAL ANALYSIS. EXPLAIN.
4) Ans:
Dimensional analysis is the analysis of the relationships between different physical quantities by identifying their fundamental analysis (such as length, mass and time, and electric charge and units of measure (such as miles vs. kilometers, or pounds vs. kilograms vs. grams) and tracking these dimensions as calculations or comparisons are performed. Physical quantities that are commensurable have the same dimension; if they have different dimensions, they are incommensurable. For example, it is meaningless to ask whether a kilogram is less, the same, or more than an hour.
Any physically meaningful equation will have the same dimensions on the left and right sides, a property known as "dimensional homogeneity". Checking this is a common application of dimensional analysis. Dimensional analysis is also routinely used as a check on the plausibility of derived equations and computations. It is generally used to categorize types of physical quantities and units based on their relationship to or dependence on other units.
Uses:
• To deduce physical relations
• To convert one unit of one system to other.
• To check whether the formula is dimensionally correct or not.
Q5. GIVE ONE EXAMPLE OF RAYLEIGH’S METHOD.
5) Ans:
Show that the resistance (F) to the motion of a sphere of diameter (D) is moving with the uniform velocity (v) to a level fluid of density (d) and viscosity (μ) is given by
Solution:
F= f(v, D, d, μ)
F= K. da. vb. Dc. μd
[MLT-2]= K[ML-3] a[LT-1]b[L]c[ML-1T-1]d
=k Ma+d L-3a+b+c-d T-b-d
On equating the above right and left hand sides, we get,
a+d=1
b+d=2
-3a+b+c-d=1
On solving these equations in terms of d, we get
F=k. d 1-d. v 2-d. D 2-d. μ dHence, F= d v2 D2 f( μ/ dvD)