13.021 Marine Hydrodynamics, Fall 2003

Lecture 5

Copyright © 2003 MIT - Department of Ocean Engineering, All rights reserved.

13.021 - Marine Hydrodynamics
Lecture 5

Chapter 2 - Similitude

Similitude: Similarity of behavior of different systems.

Real world $\leftrightarrow$ ``model''
(prototype) (physical experiment, mathematical, computer, ...)

	Similarity Parameters (SP's)
Geometric Similitude	length ratios
Kinematic Similitude	Displacement ratios, velocity ratios
Dynamic (Internal Constitution)	Force ratios, stress ratios, pressure ratios
$\vdots$
Internal Constitution Similitude	$\rho, \nu$
Boundary Condition Similitude
$\vdots$
	$\underset{\mbox{to be the same for the model and the real world}}{ \underset{... ...litude, similarity parameters (SP's) required}}{\underbrace{\hspace{3.0in}}}}$

2.1 - Dimensional Analysis (DA) to Obtain Similarity Parameters (SP's)

Buckingham's $\pi$ theory:

Reduce number of variables $\rightarrow$ derive dimensionally homogeneous relationships.

1. Specify (all) the (say N) relevant variables (dependent or independent): $x_{1}, x_{2}, \ldots x_{N}$
   e.g. time, force, fluid density, distance...
   We want to relate the $x_{i}$'s to each other $\mathcal{I}$( $x_{1}, x_{2}, \ldots x_{N}$) = 0

2. Identify (all) the (say P) relevant basic physical units (``dimensions'')
   e.g. M,L,T (P = 3) [temperature, charge, ...].

3. Let $\pi = x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}\ldots x_{N}^{\alpha_{N}}$ be a dimensionless quantity formed from the $x_{i}$'s. Suppose

   $$x_{1} = C_{i}M^{m_{i}}L^{l_{i}}T^{t_{i}}, i=1,2,\ldots,N$$

   
   
   where the $C_{i}$ are dimensionless constants. For example, if $x_{1} = KE = \frac{1}{2}MV^{2}= \frac{1}{2}M^{1}L^{2}T^{-2}$ (kinetic energy), we have that $C_{1} = \frac{1}{2}, m_{1} = 1, l_{1}=2, t_{1} = -2$. Then

   $$\pi = (C_{1}^{\alpha_{1}}C_{2}^{\alpha_{2}}\ldots C_{N}^{\alpha_... ...ts+\alpha_{N}l_{N}} T^{\alpha_{1}t_{1}+\alpha_{2}t_{2}+\ldots+\alpha_{N}t_{N}}$$

   
   
   For $\pi$ to be dimensionless, we require

   $$P\left\{ \left. \underset{\Sigma \mbox{\ notation}}{\underbrace{... ...} \right\} \right. \mbox{ a} P \times N \mbox{ system of Linear Equations}$$ (1)

   
   
   Since (1) is homogeneous, it always has a trivial solution,

   $$\alpha_{i} \equiv 0, i = 1, 2, \ldots , N \pi$$

   
   
   There are 2 possibilities:
   1. (1) has no nontrivial solution (only solution is $\pi$ = constant, i.e. independent of $x_{i}$'s), which implies that the N variable $x_{i}, i = 1, 2, \dots , N$ are Dimensionally Independent (DI), i.e. they are "unrelated" and "irrelevant" to the problem.

   2. (1) has $J (J > 0)$ nontrivial solutions, $\pi_{1}$, $\pi_{2,} \ldots , \pi_{J}$. In general, $J < N$, in fact, $J = N - K$ where $K$ is the rank or "dimension" of the system of equations (1).

Model Law:

Instead of relating the N $x_{i}$'s by $\mathcal{I}$( $x_{1}, x_{2}, \dots x_{N}$) = 0, relate the $J$ $\pi$'s by

$$F(\pi_{1}, \pi_{2},\dots \pi_{J}) = 0, J = N - K < N$$

For similitude, we require

$$(\pi_{\mbox{model}})_{j} = (\pi_{\mbox{prototype}})_{j} \mbox{\ where } j = 1, 2, \ldots , J.$$

If 2 problems have all the same $\pi_{j}$'s, they have similitude (in the $\pi_{j}$ senses), so $\pi$'s serve as similarity parameters.

Note:

• If $\pi$ is dimensionless, so is constant$\times \pi$, $\pi^{\mbox{const}}, 1/\pi$ , etc...

• If $\pi_{1}$, $\pi_{2}$ are dimensionless, so is $\pi_{1} \times \pi_{2}$ , $\frac{\pi_{1}}{\pi_{2}}, \pi_{1}^{\mbox{const1}}\times \pi_{2}^{\mbox{const2}}$, etc...

In general, we want the set (not unique) of independent $\pi_{j}$'s, for e.g., $\pi$$_{1}$, $\pi$$_{2}$, $\pi$$_{3}$ or $\pi$$_{1}$, $\pi$$_{1}$ $\times$ $\pi$$_{2}$, $\pi$$_{3}$, but not $\pi$$_{1}$, $\pi$$_{2}$, $\pi$$_{1}$ $\times$ $\pi$$_{2}$.

Example:

Application of Buckingham $\ensuremath{\pi}$ Theory.

Figure 1: Force on a smooth circular cylinder in steady incompressible fluid (no gravity)

$$x_{i}: F, U, D, \rho, \nu \rightarrow N = 5$$

$$x_{i} = c_{i} M^{m_{i} } L^{l _{i} } T^{t_{i} } \rightarrow P = 3$$

N = 5
	$F$	$U$	$D$	$\rho$	$\nu$
P = 3	$m_{i}$	1	0	0	1	0
	$l_{i}$	1	1	1	-3	2
	$t_{i}$	-2	-1	0	0	-1

$$\pi =F^{\alpha _{1} } U^{\alpha _{2} } D^{\alpha _{3} } \rho ^{\alpha_{4} } \nu ^{\alpha _{5} }$$

For $\pi$ to be non-dimensional, the set of equations

$$\alpha _{i} m_{i} =$$ 0

$$\alpha _{i} l_{i} =$$ 0

$$\alpha _{i} t_{i} =$$ 0

has to be satisfied. The system of equations above after we substitute the values for the $m_{i}$'s, $l_{i}$'s and $t_{i}$'s assume the form:

$$\left( \begin{array}{ccccc} 1 & 0 & 0 & 1 & 0 \ 1 & 1 & 1 & ... ...{array} \right) =\left( \begin{array}{c} 0 \ 0 \ 0 \end{array} \right)$$

The rank of this system is $K = 3$, so we have $j = 2$ nontrivial solutions. Two families of solutions for $\alpha_{i}$ for each fixed pair of ($\alpha$$_{4}$, $\alpha$$_{5}$), exists a unique solution for ($\alpha$$_{1}$, $\alpha$$_{2}$, $\alpha$$_{3}$). We consider the pairs ($\alpha$$_{4}$ = 1, $\alpha$$_{5}$ = 0) and ($\alpha$$_{4}$ = 0, $\alpha$$_{5}$ = 1), all other cases are linear combinations of these two.

1. Pair $\alpha$$_{4}$ = 1 and $\alpha$$_{5}$ = 0.

   $$\left( \begin{array}{ccc} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 &... ...array}\right) =\left( \begin{array}{c} -1 \ 4 \ 2 \end{array} \right)$$

   
   
   which has solution

   $$\left( \begin{array}{c} \alpha _{1} \ \alpha _{2} \ \alph... ...ray} \right) =\left( \begin{array}{c} -1 \ 2 \ 2 \end{array} \right)$$

   
   

   $$\therefore \pi _{1} =F^{\alpha _{1} ... ... _{3} } \rho ^{\alpha _{4} } \nu ^{\alpha _{5} } =\frac{\rho U^{2} D^{2} }{F}$$

   
   
   Conventionally, $\pi_{1} \rightarrow 2\pi_{1}^{-1}$ and $\therefore \pi _{1} =\frac{F}{\frac{1}{2} \rho U^{2} D^{2} } \equiv C_{d},$ which is the Drag coefficient.

2. Pair $\alpha$$_{4}$ = 0 and $\alpha$$_{5}$ = 1.

   $$\left( \begin{array}{ccc} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 &... ...ay} \right) =\left( \begin{array}{c} 0 \ -2 \ -1 \end{array} \right)$$

   
   
   which has solution

   $$\left( \begin{array}{c} \alpha _{1} \ \alpha _{2} \ \alph... ...ay} \right) =\left( \begin{array}{c} 0 \ -1 \ -1 \end{array} \right)$$

   
   

   $$\therefore \pi _{2} =F^{\alpha _{1} ... ...ha _{3} } \rho ^{\alpha _{4} } \upsilon ^{\alpha _{5} } =\frac{\upsilon }{UD}$$

   
   
   Conventionally, $\pi_{2} \rightarrow \pi_{2}^{-1}, \therefore \pi _{2} =\frac{UD}{\upsilon } \equiv R_{e},$ which is the Reynolds number.

Therefore,

$F(\pi_{1}, \pi_{2}) = 0$	or $\pi_{1} = f(\pi_{2})$
$F(C_{d}, R_{e}) = 0$	or $C_{d} = f( R_{e})$
$F( \frac{F}{\frac{1}{2} \rho U^{2} D^{2} }, \frac{UD}{\nu } ) = 0$	or $\frac{F}{\frac{1}{2} \rho U^{2} D^{2} } =f( \frac{UD}{\nu } )$