Exercise 3

Velocity decomposition

We infer on an expression for the flow field given by

\[ v(x) = \begin{pmatrix} 1-y \\ 1 + x \\ 0 \end{pmatrix} \tag{1}\]

1. Decompose the given velocity field by calculating the axial vector of spin tensor \(\mathbf{W}\) and the strain rate tensor \(\mathbf{D}\)

2. Superposition of which two of the following flow fields would lead to the flow field as described in Equation 1.

\(\boxed{\phantom{X}} \text{ A}\)

\(\boxed{\phantom{X}} \text{ B}\)

\(\boxed{\phantom{X}} \text{ C}\)

\(\boxed{\phantom{X}} \text{ D}\)

\[\begin{multline*} \shoveright{} \textbf{Points: 4} \end{multline*}\]

Solution

1. Axial vector of spin tensor \(\mathbf{W}\) \[ \mathbf{w} = \dfrac{1}{2} \nabla \times \mathbf{v} = \dfrac{1}{2} \begin{pmatrix} 0 \\ 0 \\ 1 + 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \]

\[\begin{multline*} \shoveright{} \textbf{Points: 1} \end{multline*}\]

\[ \begin{aligned} \mathbf{D} &= \dfrac{1}{2} (\nabla \mathbf{v} + \nabla \mathbf{v}^T) \\ &= \dfrac{1}{2} \left( \begin{bmatrix} 0 & -1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 1 & 0\\ -1 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} \right) \\ \mathbf{D} &= \mathbf{0} \end{aligned} \]

\[\begin{multline*} \shoveright{} \textbf{Points: 1} \end{multline*}\]

2. Superposition of flow fields A(pure translation) and C(pure rotation)

\[\begin{multline*} \shoveright{} \textbf{Points: 2} \end{multline*}\]

Bernoulli

The figure shows a Venturi tube, which can be used to measure fluid flow. It comprises a cylindrical inlet with cross section \(A_1\) followed by a convergent entrance into a cylindrical throat with cross section \(A_2\) and a divergent outlet. The velocity of the fluid at inlet and the throat is \(\mathbf{v}_1\) and \(\mathbf{v}_2\) respectively. The flowing fluid has density \(\rho_a\) and the liquid in the U-shaped tube has a density \(\rho_f\).

Tasks

Task 1

State the Bernoulli equation and explain to what fluid flow scenarios it can be applied.

Solution

Along a stationary flow streamline 

Task 2

Calculate the velocity of the flowing fluid in terms of the difference in height of the fluid \(\Delta h\) in the U-shaped tube and other given quantities given in the figure.

Solution

To solve the problem, we need to formulate Bernoulli between the different points.

1. Bernoulli between points 1 and 2: These points lie at the same height

\[ p_1 + \frac{1}{2}\,\rho_a\,v_1^2 = p_2 + \frac{1}{2}\,\rho_a\,v_2^2 \]

and thus

\[ p_1 - p_2= \frac{1}{2}\,\rho_a\,\left(\,v_2^2 - v_1^2\right) \]

2. The hydrostatic pressure at the bottom of the tube can be expressed in terms of two heights:

\[ p_{\text{bottom}} = p_1 + \rho_a\,g\,d_1 + \rho_f\, g\,h_1 \quad , \]

and

\[ p_{\text{bottom}} = p_2 + \rho_a\,g\,d_2 + \rho_f\, g\,h_2 \quad . \]

Setting these equations equal yields an expression in terms of \(p_1\), \(p_2\) and the height difference \(\Delta h = h_2 - h_1\)

\[ p_1 - p_2 = \left(\rho_f - \rho_a \right)\, g \,\Delta h \]

Substituting the expression for \(p_1 -p_2\) back into the first expression for \(\Delta p = p_1 - p_2\) gives

\[ \frac{1}{2}\,\rho_a\,\left(\,v_2^2 - v_1^2\right) = \left(\rho_f - \rho_a \right)\, g \,\Delta h \quad . \]

The continuity equation for the constant density fluid yields

\[ v_2\,A_2 = v_1\,A_1 \Rightarrow v_2 = \frac{A_1}{A_2} \,v_1 \quad . \]

Substituting in the above equation yields

\[ \frac{1}{2}\,\rho_a\,\left(\,\left(\frac{A_1}{A_2}\right)^2 - 1\right)\,v_1^2 = \left(\rho_f - \rho_a \right)\,g\,\Delta h \]

which can be rearranged for \(h\) or \(v_1\) as needed.

\[ v_1 = \sqrt{\frac{2\,\left(\rho_f - \rho_a \right)\,g\,\Delta h}{\rho_a\,\left(\,\left(\frac{A_1}{A_2}\right)^2 - 1\right)}} \]