Dimensional analysis
- 1 Purpose of dimensional analysis
- 2 Requirements for dimensional analysis
- 3 Let’s try the inviscid flow solution
- 4 Dimensional analysis in open channel flow problems
- 5 Buckingham $\Pi$ theorem
- 5.1 Mathematically speaking
- 5.2 Basic dimensions
- 5.3 Let’s see how it works
- 5.4 Drag problem for a completely immersed cylinder
- 5.5 Drag problem for a completely immersed cylinder (step 1)
- 5.6 Drag problem for a completely immersed cylinder (step 2)
- 5.7 Drag problem for a completely immersed cylinder (step 3)
- 5.8 Drag problem for a completely immersed cylinder (step 4)
- 5.9 Drag problem for a completely immersed cylinder (step 5)
- 5.10 Drag problem for a completely immersed cylinder (step 6)
- 5.11 Drag problem for a completely immersed cylinder (step 7)
- 5.12 Drag problem for a completely immersed cylinder (step 8)
- 6 Homework: Dimensional analysis
1 Purpose of dimensional analysis
To reduce the number of independent variables in open channel flow problems by using a functional relationship between the dependent and independent variables.
We can reduce many variables into a smaller number of dimensionless ratios, such as the Froude number.
2 Requirements for dimensional analysis
Dimensional homogeneity: Every term in an equation should have the same dimensions.
Sometimes, “constants” take some dimensions to make equations dimensionally homogeneous. For example, Manning’s equation:
\begin{equation} Q=\frac{k}{n}AR_h^{2/3}S^{1/2} \end{equation}
3 Let’s try the inviscid flow solution
See Figure 1.3.
\begin{equation} p_\infty+\rho\frac{V_\infty^2}{2}=p+\rho\frac{v^2}{2} \end{equation} where a subscript of $\infty$ indicates the approaching flow.
\begin{equation} p-p_\infty=\rho\frac{V_\infty^2}{2}-\rho\frac{v^2}{2} \end{equation} \begin{equation} \frac{p-p_\infty}{\rho\frac{V_\infty^2}{2}} =1-\frac{\rho\frac{v^2}{2}}{\rho\frac{V_\infty^2}{2}} =1-\left(\frac{v}{V_\infty}\right)^2 =C_p \end{equation} where $C_p$ is a dimensionless pressure coefficient for the inviscid flow solution.
Plugging the solution for the circumferential velocity $v=2V_\infty\sin\theta$ yields \begin{equation} C_p=1-4\sin^2\theta \end{equation}
4 Dimensional analysis in open channel flow problems
Many open channel flow problems need experimental results to evaluate unknown parameters in the theoretical solution or to formulate and solve the governing equations.
For example, calculating the drag force on a circular bridge pier requires the experimental drag coefficient.
5 Buckingham $\Pi$ theorem
A physical process involving $n$ variables in terms of $m$ dimensions can be reduced to a relation between $(n-m)$ dimensionless variables or $\Pi$ terms by choosing $m$ repeating variables. Check the full definition on page 12.
5.1 Mathematically speaking
If a dependent variable $A_1$ can be expressed in terms of $(n-1)$ independent variables as \begin{equation} A_1=f(A_2,A_3,\cdots,A_n), \end{equation}
a functional relation among $(n-m)$ $\Pi$ terms can be formulated as \begin{equation} \phi(\Pi_1,\Pi_2,\cdots,\Pi_{n-m})=0. \end{equation}
5.2 Basic dimensions
- Mass: $M$
- Length: $L$
- Time: $T$
5.3 Let’s see how it works
- $n=5$ with $A_1$, $A_2$, $A_3$, $A_4$, and $A_5$
- $m=3$ with $M$, $L$, and $T$
- How many $\Pi$ terms can be defined by the Buckingham $\Pi$ theorem? $n-m=2$
- How many “repeating” variables do we need? $m=3$
- Let’s repeat $A_2$, $A_3$, and $A_4$
\begin{align} [\Pi_1]&=M^0L^0T^0=[A_2]^{x_1}[A_3]^{y_1}[A_4]^{z_1}[A_1]\\ [\Pi_2]&=M^0L^0T^0=[A_2]^{x_2}[A_3]^{y_2}[A_4]^{z_2}[A_5] \end{align}
Solve this system of equations!
5.4 Drag problem for a completely immersed cylinder
- Dependent variable: Drag force $D$ in $MLT^{-2}$
- Independent variables
- Cylinder diameter $d$ in $L$
- Cylinder length $l_c$ in $L$
- Approach velocity $V_\infty$ in $LT^{-1}$
- Fluid density $\rho$ in $ML^{-3}$
- Fluid viscosity $\mu$ in $ML^{-1}T^{-1}$
\begin{equation} D=f_1(d,l_c,V_\infty,\rho,\mu) \end{equation}
5.5 Drag problem for a completely immersed cylinder (step 1)
$n=6$ and $m=3$: $m=3$ repeating variables and $n-m=3$ $\Pi$ terms
Don’t choose $d$ and $l_c$ as repeating variables because they have the same dimension, forming a $\Pi$ term themselves.
Let’s choose $d$, $V_\infty$, and $\rho$ as repeating variables.
\begin{align} [\Pi_1]&=M^0L^0T^0=[d]^{x_1}[V_\infty]^{y_1}[\rho]^{z_1}[D]\\ [\Pi_2]&=M^0L^0T^0=[d]^{x_2}[V_\infty]^{y_2}[\rho]^{z_2}[l_c]\\ [\Pi_3]&=M^0L^0T^0=[d]^{x_3}[V_\infty]^{y_3}[\rho]^{z_3}[\mu] \end{align}
5.6 Drag problem for a completely immersed cylinder (step 2)
Plug the fundamental dimensions.
\begin{align} [\Pi_1]&=M^0L^0T^0=L^{x_1}(LT^{-1})^{y_1}(ML^{-3})^{z_1}(MLT^{-2})\\ [\Pi_2]&=M^0L^0T^0=L^{x_2}(LT^{-1})^{y_2}(ML^{-3})^{z_2}L\\ [\Pi_3]&=M^0L^0T^0=L^{x_3}(LT^{-1})^{y_3}(ML^{-3})^{z_3}(ML^{-1}T^{-1}) \end{align}
5.7 Drag problem for a completely immersed cylinder (step 3)
A little algebra...
\begin{align} [\Pi_1]&=M^0L^0T^0=M^{z_1+1}L^{x_1+y_1-3z_1+1}T^{-y_1-2}\\ [\Pi_2]&=M^0L^0T^0=M^{z_2}L^{x_2+y_2-3z_2+1}T^{-y_2}\\ [\Pi_3]&=M^0L^0T^0=M^{z_3+1}L^{x_3+y_3-3z_3-1}T^{-y_3-1} \end{align}
5.8 Drag problem for a completely immersed cylinder (step 4)
Solve these systems for $x_i$, $y_i$, and $z_i$ for all $i\in[1,3]$.
\begin{equation} \left\{\begin{aligned} z_1+1&=0\\ x_1+y_1-3z_1+1&=0\\ -y_1-2&=0\\ \end{aligned}\right. \end{equation}
\begin{equation} \left\{\begin{aligned} z_2&=0\\ x_2+y_2-3z_2+1&=0\\ -y_2&=0\\ \end{aligned}\right. \end{equation}
\begin{equation} \left\{\begin{aligned} z_3+1&=0\\ x_3+y_3-3z_3-1&=0\\ -y_3-1&=0 \end{aligned}\right. \end{equation}
Each system of equations has three unknown variables and three equations (solvable).
5.9 Drag problem for a completely immersed cylinder (step 5)
Check your answers!
\begin{equation} \left\{\begin{aligned} x_1&=-2\\ y_1&=-2\\ z_1&=-1 \end{aligned}\right. \end{equation}
\begin{equation} \left\{\begin{aligned} x_2&=-1\\ y_2&=0\\ z_2&=0 \end{aligned}\right. \end{equation}
\begin{equation} \left\{\begin{aligned} x_3&=-1\\ y_3&=-1\\ z_3&=-1 \end{aligned}\right. \end{equation}
5.10 Drag problem for a completely immersed cylinder (step 6)
Plug the found exponents into the three $\Pi$ terms in step 1.
\begin{align} [\Pi_1]&=M^0L^0T^0=[d]^{-2}[V_\infty]^{-2}[\rho]^{-1}[D]\\ [\Pi_2]&=M^0L^0T^0=[d]^{-1}[V_\infty]^{0}[\rho]^{0}[l_c]\\ [\Pi_3]&=M^0L^0T^0=[d]^{-1}[V_\infty]^{-1}[\rho]^{-1}[\mu] \end{align}
5.11 Drag problem for a completely immersed cylinder (step 7)
Strip out the brackets.
\begin{align} d^{-2}V_\infty^{-2}\rho^{-1}D&=\frac{D}{\rho d^2V_\infty^2}&=C_1\\ d^{-1}V_\infty^{0}\rho^{0}l_c&=\frac{l_c}{d}&=C_2\\ d^{-1}V_\infty^{-1}\rho^{-1}\mu&=\frac{\mu}{\rho V_\infty d}&=C_3 \end{align} where $C_i$ is a dimensionless constant for all $i\in[1,3]$.
We found three dimensionless ratios.
5.12 Drag problem for a completely immersed cylinder (step 8)
The Reynolds number $\mathbf{Re}=\frac{1}{C_3}=\frac{\rho V_\infty d}{\mu}$ is still dimensionless.
The result is given by \begin{equation} \frac{D}{\rho d^2V_\infty^2}=f_2\left(\frac{l_c}{d}, \mathbf{Re}\right) \end{equation} where $\frac{D}{\rho d^2V_\infty^2}$ is the dimensionless drag ratio in terms of the Reynolds number and the ratio of cylinder length to diameter.
6 Homework: Dimensional analysis
- Exercise 1.13
- Exercise 1.14
Show your full work.