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+---
+author: Akbar Rahman
+date: \today
+title: MMME2047 // Dimensional Analysis
+tags: [ mmme2047, uni, fluid_dynamics, dimensional_analysis ]
+uuid: 
+---
+
+In lab tests it is not always possible to use the actual scale of the prototype,
+actual flow speed, or actual fluid.
+In these cases a model is used.
+
+Dimensional analysis allows:
+
+- experiments to be performed on a scaled models or using different fluids
+- reduced number of parameters (and therefore tests), and also
+- tests to have greater generality
+
+
+# Introduction Physical Similarity
+
+How to make sure a prototype and a scale model are physically similar
+
+- Geometrical similarity --- all dimensions in all three coordinates have the same
+  linear scale ratioi.
+
+  This includes surface roughness (e.g. a 10x smaller model have 10x smaller roughness)
+- Kinematic similarity ---velocities at corresponding points in the two flows are in the same direction and related by a constant scale factor in magnitude
+- Dynamic similarity --- requires that the magnitude ratio of any two forces in one system must be the same as the magnitude ratio of the corresponding forces in the other system
+
+Kinematic and Dynamic similarity are ensued by equality of the governing
+nondimensional parameters.
+
+# Dimensions and Units
+
+There are four basic dimensions for fluid dynamics:
+
+- Mass (sometimes replaced by a force)
+- Length
+- Time
+- Temperature
+
+A nondimensional (dimensionless) group does not have any dimension or units, such
+as the Reynolds number:
+
+$$\text{Re} = \frac{\rho U x}{\mu)$$
+
+# Example of Dimensional Analysis
+
+![](./images/dim_anal_example.png)
+
+In order to maintain a constant $\omega$ a certain torque $T$ is required to overcome
+the shear stress exerted by the fluid on the surface of the rotating cone.
+
+The experiment wants to answer: How are $T$ and $\omega$ related?
+
+Doing this by testing the parameters individually would take too long as there are
+several: $\omega$, $D$, $\rho$, $\mu$.
+Doing 10 experiments per parameter would take $10^4$ experiments to get a full idea.
+This would take ages.
+
+This is where dimensional analysis helps as it means we only need 2 nondimensional groups
+to relate the 4 parameters: momentum coefficient, $C_m$, and Reynolds number, $Re$.
+
+$$C_m = \frac{T}{\frac12 \rho \omega^2D^5} \rightarow C_m = g(\text{Re})$$
+
+This suggests that the parameters can be reduced to a single parameters - the rotating
+Reynolds number.
+This drastically reduces the number of experiments required.
+
+When you plot the data in a dimensional format, you would see multiple curves,
+but when you plot them in a nondimensional format, you will observe that they fall
+onto one curve:
+
+![](./images/nondim-format.png)
+
+This verifies the relationship $C_m = g(\text{Re})$ and our dimensional analysis.
+
+You can use this nondimensional graph to find the behaviour for any value of $C_m$ and
+$\text{Re}$ within the range of your experiment.
+As always, you should be careful when extrapolating.
+
+If nondimensional points do not fall on the same line, there may be a parameters that
+are left out:
+
+- Thickness of the gap between stator and rotor
+- Cone angle
+- Surface roughness
+- Fluid heating due to viscous dissipation
+
+# Buckingham Pi Theorem
+
+If a physical process is fully described by $n$ variables and $k$ dimensions, then
+$m = n - k$ dimensionless groups are sufficient to describe the process.
+
+For example, say $n = 5$ and $k = 3$ such that $v_1 = f(v_2, v_3, v_4, v_5)$.
+The theorem says that $m = 5 -2$ --- two nondimensional groups should be sufficient to
+describe the process with a functional relationship: $\Pi_1 = g(\Pi_2)$.
+
+Then pick the two variables that are of the most interest, e.g. $v_1$ and $v_5$, and
+use the other three to form the dimensionless groups.
+
+It is important that these three do not themselves form a dimensionless group.
+
+$\Pi_1$, $\Pi_2$ are then formed by the variable we chose and power products of the
+three others.:
+
+$$\Pi_1 = v_1 v_2^av_3^bv_4^c$$
+
+$$\Pi_2 = v_5 v_2^dv_3^ev_4^f$$
+
+The exponents are obtained by knowing the groups are dimensionless:
+
+$$[\Pi_1] = [v_1] [v_2]^a[v_3]^b[v_4]^c = [M]^0[L]^0[T]^0$$
+
+$$[\Pi_2] = [v_5] [v_2]^d[v_3]^e[v_4]^f = [M]^0[L]^0[T]^0$$
+
+where $M$, $L$, and $T$ are units of mass, length, and time respectively.
+
+The two equations results in some simple simultaneous equations to solve to find the
+coefficients $a$, $b$, $c$, $d$, $e$, $f$.
+
+