# Fluid Mechanics/Dimensional Analysis

(Redirected from Fluid Mechanics/Ch4)

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/L3 Velocity V L/T Gravity g L/T2

This means that D = f(l,ρ,μV,g) where f is some function.

With the Rayleigh Method, we assume that D=ClaρbμcVdge, 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 = MbMc
L: 1 = a - 3b - c + d + e since L¹ = LaL-3bL-cLdLe
T: -2 = -c - d - 2e since T-2 = T-cT-dT-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 = Cl2+e-cρ1-cμcV2-c-2ege

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 = Cl2lel-cρρ-cμcV2V-cV-2ege

For the different exponents,

Terms with exponent of 1: Cρ
Terms with exponent of 2: l2V2
Terms with exponent of e: leV-2ege = $\displaystyle \left({\frac {lg}{V^{2}}}\right)^{e}=\displaystyle \left({\frac {V^{2}}{lg}}\right)^{-e}$ (iv)
Terms with exponent of c: l-cρ-cμcV-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 NR or Re and NF 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 for e, which result in the Reynolds and Froude Numbers respectively, we obtain

D = g(Fr, Re)ρl2V2
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, CD.

$C_{D}\equiv {\frac {D}{{\rho }l^{2}V^{2}}}$ 