Model is similar with real object required in certain scale ratio. It is
needed to be tested in laboratory with similar condition which is likely to be occurring
in real phenomena. The size of the model might not be smaller than prototype.
PROTOTYPE
Prototype is an actual object that is in its full scale and needed to be
reacting properly in real phenomena for mankind example like: ship, spillway
structure in open channel.
However, scaling effect might affect the characteristic of a real-world
prototype. Factor which lead to scale effect might be air concentration, Earth
characteristics.
Advantages for using similarities.
Performances of
object can be predicted.
Economics and easy
to build, until it’s tested to reach a value.
Non-functional
structure can be measured. Such as dam.
A physical scale model satisfying mechanical similarity is completely
similar to its real-world prototype and involves no scale effects.
Mechanical similarity requires three criteria:
Geometric
similarity
Kinematic
similarity
Dynamic similarity
Geometric similarity
Boundaries between 2 flows are exact dilation (expand) or shrinkages of
each other.
Length, Lp/Lm = Lr
Area, Ap/Am = Ar
Volume, Vp/Vm = Vr
Kinematic similarity
Streamlines between flows are exact dilation or shrinkage of each
other.
Velocity scale ratio, Vr = Vp/Vm
Discharge scale ratio,
Acceleration scale ratio,
Dynamic similarity
Corresponding forces at corresponding locations mutually parallel and
have the same ratio of magnitudes. This ratio is the same for all sets of
corresponding forces.
∑F = Fg + Fp + Fv + Fe + Ft =
Resultant
Therefore: Fg + Fp + Fv + Fe + Ft + Fi= 0
Where Fi=
-resultant force
Where:
Gravitational force
Fg= mg =
ρgL^3
Pressure
Fp= ΔpA =
ΔpL^2
Viscosity
Fv= μ(V/L)A = μ(V/L)L^2= μVL
Elasticity
Fe = EvA = EvL^2
Surface Tension
Ft = σL
Inertia force
Fi= ma = ρL^2v^2 In a general flow field, complete similarity between a model and prototype is achieved only when there is geometric, kinematic and formed dynamic similarity to perform for real life situation.
For further description about dimensionless quantities:
There are a lot of methods can be used to reduce into a smaller number of dimensionless parameters. We will discuss about the two commonly used methods. 6.4.1 Rayleigh’s Method
A basic method to dimensional analysis method and can be simplified to yield dimensionless groups controlling the phenomenon.
The Rayleigh’s method is based on the following steps: 1. First of all, write the functional relationship according to the given data.
2. Now write the equation in terms of aconstant with exponents i.e. powers a, b,c,...
3. With the help of the principle of dimensional homogeneity, find out the values of a, b,
c… by obtaining simultaneous equation and simplify it.
4. Nowsubstitute the values of these exponents in the main equation, and simplify it.
Example 1
Express dimensionless equation for the speed v with a wave pressure travels through a fluid. Consider the physical factors probably influence the speeds are compressibility, E, (or K), density,p and kinematics viscosity, v.
Solution: Write the fundamental dimension (used MLTθ) for all dimensions given
V= [L/T] K= [F/L^2 ]= [M/(LT^2 )] p = [M/L^3 ] v = [L^2/T]
Let write the equation like this: V = CK^a p^b v^d
Inserts fundamental dimension into the equation while C is dimensionless constant.
L/T = (M/(LT^2 ))^a (M/L^3 )^b (L^2/T)^d
To satisfy dimensional homogeneity, net power of each dimension must be identical on both sides of this equation. Thus,
For M: 0 = a + b
For L: 1 = -a + (-3b) + 2d
For T: -1 = -2a – d
a = 1/2 b = - 1/2 d = 0
V = CK^(1/2) p^(-1/2) v^0
Or
V = C √(K/p)
Therefore, wave speed is not affected by the fluid’s kinematics viscosity, v.
6.4.2 The Buckingham pi Theorem
When a large number of variables are involved, Raleigh’s method becomes lengthy. In such circumstances, the Buckingham’s method is useful. This method expressed the variables related to a dimensional homogeneous equation as:
xr: f(xr,xr....x.)
where, the dimension at each section is the same.
Example:
The drag D, which is a force, on a ship. What exactly is the drag a function of?
Say that we have n number of quantities (e.g. 6 quantities, which are D,l,ρ,μ,V, and g) and m number of dimensions (e.g. 3 dimensions, which are M, L, and T). These quantities can be reduced to (n - m) independent dimensionless groups.
Step 1: Setup π groups
For the MLT System, m = 3, so choose A1, A2, and A3 as the repeating variables.
Using the Buckingham π Theorem on the Drag Equation:
f(D, l, ρ, μ, V, g) = 0
Where m = 3, n = 6, so there will be n - m = 3 π groups.
We will select ρ, V, and l as the repeating variables (RV), leaving the remaining quantities as D, μ, and g. Note that if the analysis does not work out, we could always go back and repeat using new RVs. Thus,
Which are all dimensionless quantities, i.e. having units of
Step 2: Determine π groups
For the first π group,
π1 M^0 T^0 L^0=(〖M/L^3
)〗^(x_1
) (L/T)^(y_1 ) (L)^(z_1 ) (ML/T^2 )
Expanding and collecting like units, we can solve for the exponents:
For M: 0 = x1 + 1 ⇒ x1 = -1
For T: 0 = -y1 - 2 ⇒ y1 = -2
For L: 0 = -3x1 + y1 + z1 + 1 ⇒ z1 = 3(-1) - (-2) - 1 = -2
Therefore, we find that the exponents x1, y1, and z1 are -1, -2, and -2 respectively.
This means that the first dimensionless π group, π1, is
= D/(p V^2 l^2)
For the second π group,
π2 M^0 L^0 T^0=(〖M/L^3 )〗^(x_2 ) (〖L/T)〗^(y_2 ) (〖L)〗^(z_2 ) (M/LT)
Solving for the exponents,
For M: x2 + 1 = 0 ⇒ x2 = -1
For T: -y2 - 1 = 0 ⇒ y2 = -1
For L: -3x2 + y2 + z2 - 1 = 0 ⇒ z2 = 1 - (-1) + 3(-1) = -1
Thus,
π_2=ρ^(-1) V^(-1) l^(-1)=μ/( ρVl)=v/Vl
However, we will now invert π2 so that,
π_2 = (Vl/v)^-1 = Reynolds Number
It is permissible to exponentiate any π group, e.g. π-1, π½, π2, etc., to form a new group, as this does not alter the functional form.
For the third π group,
π3 M^0 L^0 T^0=(〖M/L^3 )〗^(x_3 ) (〖L/T)〗^(y_3 ) (〖L)〗^(z_3 ) (L/T^2 )
Solving for the exponents,
For M: x3 = 0 ⇒ x3 = 0
For T: -y3 - 2 = 0 ⇒ y3 = -2
For L: -3x3 + y3 + z3 + 1 = 0 ⇒ z3 = -1 - (-2) = 1
Thus,
π_3=ρ^0 V^(-2) lg=lg/V^2
Raising it to the power of -½,
√(1/π_3 )= V/√lg =Froude Number
Thus, the three π groups can be written together as
f(D/(ρl^2 V^2 ),Re,Fr)=0
