# Fluid Mechanics/Dimensional Analysis

Dimensional analysis is a mathematical technique used to predict physical parameters that influence the flow in fluid mechanics, heat transfer in thermodynamics, and so forth. The analysis involves the fundamental units of dimensions MLT: mass, length, and time. It is helpful in experimental work because it provides a guide to factors that significantly affect the studied phenomena.
Dimensional analysis is commonly used to determine the relationships between several variables, i.e. to find the force as a function of other variables when an exact functional relationship is unknown. Based on understanding of the problem, we assume a certain functional form.

## Units/Dimensions

The defined units are based on the modern MLT system: mass, length, time. All other quantities can be expressed in terms of these basic units.
For example, Force /area×veilocity gradient MLT^-2/L^2×T^-1 |- |velocity||m/s||= L/T |- |acceleration||m/s²||= L/T² |- |force||kgm/s²||= ML/T² |}
Where L/T, L/T², ML/T², etc. are referred to as the derived units.
Another system for dimensionless analysis is the FLT system, the force, length, time system. In this case, mass ≡ F/a, which makes the units of mass as FT²/L, since acceleration has units of L/T².

## Rayleigh Method

An elementary method for finding a functional relationship with respect to a parameter in interest is the Rayleigh Method, and will be illustrated with an example, using the MLT system.
Say that we are interested in the drag, D, which is a force on a ship. What exactly is the drag a function of? These variables need to be chosen correctly, though selection of such variables depends largely on one's experience in the topic. It is known that drag depends on
| Quantity | Symbol | Dimension |
| Size | l | L |
| Viscosity | μ | M/LT |
| Density | ρ | M/L$^3$ |
| Velocity | V | L/T |
| Gravity | g | L/T$^2$ |
This means that D = f(l,ρ,μV,g) where f is some function.
With the Rayleigh Method, we assume that D=Cl$^a$ρ$^b$μ$^c$V$^d$g$^e$, where C is a dimensionless constant, and a,b,c,d, and e are exponents, whose values are not yet known.
Note that the dimensions of the left side, force, must equal those on the right side. Here, we use only the three independent dimensions for the variables on the right side: M, L, and T.

### Step 1: Setting up the equation

Write the equation in terms of dimensions only, i.e. replace the quantities with their respective units. The equation then becomes
    ${\frac {ML}{T^{2}}}=(L)^{a}\displaystyle \left({\frac {M}{L^{3}}}\right)^{b}\displaystyle \left({\frac {M}{LT}}\right)^{c}\displaystyle \left({\frac {L}{T}}\right)^{d}\displaystyle \left({\frac {L}{T^{2}}}\right)^{e}$
On the left side, we have M¹L¹T$^{-2}$, which is equal to the dimensions on the right side. Therefore, the exponents of the right side must be such that the units are M¹L¹T$^{-2}$

### Step 2: Solving for the exponents

Equate the exponents to each other in terms of their respective fundamental units:
    M: 1 = b + c since M¹ = M$^b$M$^c$
    L: 1 = a - 3b - c + d + e since L¹ = L$^a$L$^{-3b}$L$^{-c}$L$^d$L$^e$
    T: -2 = -c - d - 2e since T$^{-2}$ = T$^{-c}$T$^{-d}$T$^{-2e}$
It is seen that there are three equations, but 5 unknown variables. This means that a complete solution cannot be obtained. Thus, we choose to solve a, b, and d in terms of c and e. These choices are based on experience. Therefore,
| From M: | b = 1 - c | (i) |
| From T: | d = 2 - c - 2e | (ii) |
| From L: | a = 1 + 3b + c - d - e | (iii) |
Solving (i), (ii), and (iii) simultaneously, we obtain
    a = 2 - c + e
Substituting the exponents back into the original equation, we obtain
    D = Cl$^{2+e-c}$ρ$^{1-c}$μ$^c$V$^{2-c-2e}$g$^e$
Collecting like exponents together,
    $D=C\displaystyle \left({\frac {V^{2}}{lg}}\right)^{-e}\displaystyle \left({\frac {Vl\rho }{\mu }}\right)^{-c}{\rho }l^{2}V^{2}$
Which means
    D = Cl$^2$l$^e$l$^{-c}$ρρ$^{-c}$μ$^c$V$^2$V$^{-c}$V$^{-2e}$g$^e$
For the different exponents,
    Terms with exponent of 1: Cρ
    Terms with exponent of 2: l$^2$V$^2$
    Terms with exponent of e: l$^e$V$^{-2e}$g$^e$ = $\displaystyle \left({\frac {lg}{V^{2}}}\right)^{e}=\displaystyle \left({\frac {V^{2}}{lg}}\right)^{-e}$ (iv)
    Terms with exponent of c: l$^{-c}$ρ$^{-c}$μ$^c$V$^{-c}$ = $\displaystyle \left({\frac {l{\rho }V}{\mu }}\right)^{-c}$ (v)
The right sides of (iv) and (v) are known as the dimensionless groups.

### Step 3: Determining the dimensionless groups

Note that e and c are unknown. Consider the following cases:
    If e = 1 then (iv) becomes $\displaystyle \left({\frac {lg}{V^{2}}}\right)$
    If e = -1 then (iv) becomes $\displaystyle \left({\frac {V^{2}}{lg}}\right)$
    If c = 1 then (v) becomes $\displaystyle \left({\frac {\mu }{l{\rho }V}}\right)$
    If c = -1 then (v) becomes $\displaystyle \left({\frac {l{\rho }V}{\mu }}\right)=\displaystyle \left({\frac {lV}{\nu }}\right)$
       Where ν is the kinematic viscosity of the fluid.
And so on for different exponents. It turns out that:
    ${Reynolds\ Number}\equiv {\frac {Vl}{\nu }}=N_{R}=Re$
    ${Froude\ Number}\equiv \displaystyle \left({\frac {V^{2}}{lg}}\right)^{\frac {1}{2}}={\frac {V}{\sqrt {lg}}}=N_{F}=Fr$
Where N$_R$ or Re and N$_F$ or Fr are the usual notations for the Reynolds and Froude Numbers respectively. Such dimensionless groups keep reoccurring throughout Fluid Mechanics and other fields.
Choosing exponents of -1 for c and -½ for e, which result in the Reynolds and Froude Numbers respectively, we obtain
    D = g(Fr, Re)ρl$^2$V$^2$
    Where g(Fr, Re) is a dimensionless function
This can also be written as
    ${\frac {D}{{\rho }l^{2}V^{2}}}=g(Fr,Re)$
Which is a dimensionless quantity, and a function of only 2 variables instead of 5. This dimensionless quantity turns out to be the drag coefficient, C$_D$.
    $C_{D}\equiv {\frac {D}{{\rho }l^{2}V^{2}}}$

### Notes

The Rayleigh Method has limitations because of the premise that an exponential relationship exists between the variables.

## The Buckingham π Theorem/Method

This method will be illustrated by the same example as that for Rayleigh Method, the drag on a ship.
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, such as Re and Fr.
Say that
    A$_1$ = f(A$_2$, A$_3$, A$_4$, ... , A$_n$)
where A$_x$ are quantities such as drag, length, and so forth, as mentioned under the n number of quantities, and f implies the functional relationship between A$_1$ and the other quantities.
Then re-arranging, we obtain
    0 = f(A$_2$, A$_3$, A$_4$, ... , A$_n$) - A$_1$ = f(A$_1$, A$_2$, A$_3$, A$_4$, ... , A$_n$)
Which can be further reduced, using the Buckingham π Theorem, to obtain
    0 = f(π$_1$, π$_2$, ... , π$_{n-m}$)

### Forming π Groups

For each π group, take m of the quantities, A$_x$, known as m repeating variables, and one of the other remaining variables. Note that experience dictates which quantities make the best repeating variables.
The π groups, in general form, would then be
    π$_1$ = A$_1$$^{x1}$A$_2$$^{y1}$A$_3$$^{z1}$A$_4$
    π$_2$ = A$_1$$^{x2}$A$_2$$^{y2}$A$_3$$^{z2}$A$_5$
    π$_{n-m}$ = A$_1$$^{xn-m}$A$_2$$^{yn-m}$A$_3$$^{zn-m}$A$_n$
which are all dimensionless quantities.

### Step 1: Setup π groups

For the MLT System, m = 3, so choose A$_1$, A$_2$, and A$_3$ 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,
    π$_1$ = ρ$^{x1}$V$^{y1}$l$^{z1}$D
    π$_2$ = ρ$^{x2}$V$^{y2}$l$^{z2}$μ
    π$_3$ = ρ$^{x3}$V$^{y3}$l$^{z3}$g
Which are all dimensionless quantities, i.e. having units of M$^0$L$^0$T$^0$

### Step 2: Determine π groups

For the first π group,
    π$_1$ $M^{0}L^{0}T^{0}=\displaystyle \left({\frac {M}{L^{3}}}\right)^{x_{1}}\displaystyle \left({\frac {L}{T}}\right)^{y_{1}}(L)^{z_{1}}\displaystyle \left({\frac {ML}{T^{2}}}\right)$
Expanding and collecting like units, we can solve for the exponents:
    For M: 0 = x$_1$ + 1 ⇒ x$_1$ = -1
    For T: 0 = -y$_1$ - 2 ⇒ y$_1$ = -2
    For L: 0 = -3x$_1$ + y$_1$ + z$_1$ + 1 ⇒ z$_1$ = 3(-1) - (-2) - 1 = -2
Therefore, we find that the exponents x$_1$, y$_1$, and z$_1$ are -1, -2, and -2 respectively. This means that the first dimensionless π group, π$_1$, is
    π$_1$ = ρ$^{-1}$V$^{-2}$l$^{-2}$D = ${\frac {D}{{\rho }V^{2}l^{2}}}$
For the second π group,
    π$_2$ $M^{0}L^{0}T^{0}=\displaystyle \left({\frac {M}{L^{3}}}\right)^{x_{2}}\displaystyle \left({\frac {L}{T}}\right)^{y_{2}}(L)^{z_{2}}\displaystyle \left({\frac {M}{LT}}\right)$
Solving for the exponents,
    For M: x$_2$ + 1 = 0 ⇒ x$_2$ = -1
    For T: -y$_2$ - 1 = 0 ⇒ y$_2$ = -1
    For L: -3x$_2$ + y$_2$ + z$_2$ - 1 = 0 ⇒ z$_2$ = 1 - (-1) + 3(-1) = -1
Thus,
    $\pi _{2}=\rho ^{-1}V^{-1}l^{-1}\mu ={\frac {\mu }{{\rho }Vl}}={\frac {\nu }{Vl}}$
However, we will now invert π$_2$ so that
    $\pi _{2}={\frac {Vl}{\nu }}=Reynolds\ Number$
It is permissible to exponentiate any π group, e.g. π$^{−1}$, π$^½$, π$^2$, etc., to form a new group, as this does not alter the functional form.
For the third π group,
    π$_3$ $M^{0}L^{0}T^{0}=\displaystyle \left({\frac {M}{L^{3}}}\right)^{x_{3}}\displaystyle \left({\frac {L}{T}}\right)^{y_{3}}(L)^{z_{3}}\displaystyle \left({\frac {L}{T^{2}}}\right)$
Solving for the exponents,
    For M: x$_3$ = 0 ⇒ x$_3$ = 0
    For T: -y$_3$ - 2 = 0 ⇒ y$_3$ = -2
    For L: -3x$_3$ + y$_3$ + z$_3$ + 1 = 0 ⇒ z$_3$ = -1 - (-2) = 1
Thus,
    $\pi _{3}=\rho ^{0}V^{-2}lg={\frac {lg}{V^{2}}}$
Raising it to the power of -½,
    ${\sqrt {\frac {1}{\pi _{3}}}}={\frac {V}{\sqrt {lg}}}=Froude\ Number$
Thus, the three π groups can be written together as
    $f\displaystyle \left({\frac {D}{{\rho }l^{2}V^{2}}},Re,Fr\right)=0$
Finally,
    ${\frac {D}{{\rho }l^{2}V^{2}}}=f(Re,Fr)$
Note that this is the same result as obtained with the Rayleigh Method, but with the Buckingham π Method, we did not have to assume a functional dependence.

## Common π Groups

Using the Buckingham π Theorem, we will now examine the π groups which appear most frequently in fluid dynamics. Most fluid flow situations depend on the following quantities:
| l | length |
| D | diameter |
| ε | surface roughness |
| V | velocity of flow |
| ρ | density of fluid |
| Δp | pressure drop |
| g | gravity |
| μ | absolute/dynamic viscosity |
| σ | surface tension |
| K or E$_v$ | Compressibility/Bulk Modulus |
There are 10 quantities, n = 10, and 3 dimensions, m = 3, so this gives n - m = 7 π groups. Choosing V, ρ, and l as the repeating variables, performing the Buckingham π analysis, and using different exponents for some π groups, we obtain the following π groups, which are common in the study of Fluid Mechanics:
$\pi _{1}={\frac {{\Delta }p}{{\rho }V^{2}}}={\text{Pressure Coefficient}}=C_{P}$ $\pi _{2}={\frac {V}{\sqrt {gl}}}={\text{Froude Number}}=Fr$ $\pi _{3}={\frac {Vl\rho }{\mu }}={\text{Reynolds Number}}=Re$ $\pi _{4}={\frac {V^{2}l\rho }{\sigma }}={\text{Weber Number}}=We$ $\pi _{5}={\frac {V}{\sqrt {\frac {K}{\rho }}}}={\text{Mach Number}}=Ma$ $\pi _{6}={\frac {l}{D}}={\text{Length:Diameter Ratio}}$ $\pi _{7}={\frac {\epsilon }{D}}={\text{Relative Roughness}}$