Fluid Mechanics/Dimensional Analysis< Fluid Mechanics
Dimensionless 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.
Dimensionless 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.
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.
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.
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
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 equationEdit
Write the equation in terms of dimensions only, i.e. replace the quantities with their respective units. The equation then becomes
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 exponentsEdit
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 = 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 = (iv)
- Terms with exponent of c: l-cρ-cμcV-c = (v)
The right sides of (iv) and (v) are known as the dimensionless groups.
Step 3: Determining the dimensionless groupsEdit
Note that e and c are unknown. Consider the following cases:
- If e = 1 then (iv) becomes
- If e = -1 then (iv) becomes
- If c = 1 then (v) becomes
- If c = -1 then (v) becomes
- Where ν is the kinematic viscosity of the fluid.
And so on for different exponents. It turns out that:
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 -½ 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
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.
The Rayleigh Method has limitations because of the premise that an exponential relationship exists between the variables.
The Buckingham π Theorem/MethodEdit
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.
- A1 = f(A2, A3, A4, ... , An)
where Ax are quantities such as drag, length, and so forth, as mentioned under the n number of quantities, and f implies the functional relationship between A1 and the other quantities.
Then re-arranging, we obtain
- 0 = f(A2, A3, A4, ... , An) - A1 = f(A1, A2, A3, A4, ... , An)
Which can be further reduced, using the Buckingham π Theorem, to obtain
- 0 = f(π1, π2, ... , πn-m)
Forming π GroupsEdit
For each π group, take m of the quantities, Ax, 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
- π1 = A1x1A2y1A3z1A4
- π2 = A1x2A2y2A3z2A5
- πn-m = A1xn-mA2yn-mA3zn-mAn
which are all dimensionless quantities.
Step 1: Setup π groupsEdit
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,
- π1 = ρx1Vy1lz1D
- π2 = ρx2Vy2lz2μ
- π3 = ρx3Vy3lz3g
Which are all dimensionless quantities, i.e. having units of M0L0T0
Step 2: Determine π groupsEdit
For the first π group,
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
- π1 = ρ-1V-2l-2D =
For the second π group,
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
However, we will now invert π2 so that
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,
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
Raising it to the power of -½,
Thus, the three π groups can be written together as
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 GroupsEdit
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:
|V||velocity of flow|
|ρ||density of fluid|
|K or Ev||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: