Hydrokinematics

The path of a particle is found by integrating $dx/dt = u$, $dy/dt = v$, $dz/dt = w$ with the initial position at the given time, then eliminating $t$.

Given: $u = 5x$, $v = 15y + 11$, $w = 19t^2$; at $t = 3$ s the particle is at $(4, 6, 2)$.

x-motion

$$\frac{dx}{x} = 5\,dt \;\Rightarrow\; x = 4\,e^{5(t-3)}$$

y-motion

$$\begin{aligned} \frac{dy}{15y + 11} &= dt\\ \ln(15y + 11) &= 15t + C\\ y + \tfrac{11}{15} &= \left(6 + \tfrac{11}{15}\right)e^{15(t-3)} = 6.733\,e^{15(t-3)} \end{aligned}$$

z-motion

$$z = 2 + \frac{19}{3}\left(t^3 - 27\right)$$

Parametric path (parameter $t$):

$$x = 4e^{5(t-3)},\quad y = 6.733e^{15(t-3)} - 0.733,\quad z = 2 + 6.333(t^3 - 27)$$

Eliminating t: from $x$, $e^{5(t-3)} = x/4$ and $t = 3 + \tfrac{1}{5}\ln(x/4)$. Hence

$$y = 6.733\left(\frac{x}{4}\right)^{3} - 0.733 = 0.1052\,x^3 - 0.733$$ $$z = 2 + 6.333\left[\left(3 + \tfrac{1}{5}\ln\tfrac{x}{4}\right)^3 - 27\right]$$

Answer: the path is the curve $y = 0.1052x^3 - 0.733$ in the xy-plane, with $z = 2 + 6.333\,(t^3 - 27)$ and $t = 3 + 0.2\ln(x/4)$. It passes through $(4, 6, 2)$; the particle moves away from the origin as $t$ increases.

Given: $u = Axy$, $A = 2\ \text{m}^{-1}\text{s}^{-1}$, 2-D incompressible, steady, no $w$.

Step 1: find v from continuity

$$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0\;\Rightarrow\;\frac{\partial v}{\partial y} = -Ay\;\Rightarrow\; v = -\frac{Ay^2}{2}$$

(taking $v = 0$ at $y = 0$). At $(2, 1)$:

$$u = 2\times 2\times 1 = 4\ \text{m/s},\qquad v = -\frac{2\times 1}{2} = -1\ \text{m/s}$$

Step 2: acceleration (steady flow, so only the convective part)

$$\begin{aligned} a_x &= u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} = u(Ay) + v(Ax)\\ &= 4(2) + (-1)(4) = 4\ \text{m/s}^2\\ a_y &= u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial y} = 0 + (-1)(-Ay) = (-1)(-2) = 2\ \text{m/s}^2 \end{aligned}$$ $$\vec a = 4\hat i + 2\hat j,\qquad |a| = 4.472\ \text{m/s}^2\ \text{at } 26.6^\circ \text{ above +x}$$

Step 3: radius of curvature. The normal component of acceleration is $a_n = V^2/R$ and $a_n = |\vec V\times\vec a|/|V|$.

$$\begin{aligned} |V| &= \sqrt{4^2 + 1^2} = 4.123\ \text{m/s}\\ |\vec V\times\vec a| &= |u a_y - v a_x| = |4(2) - (-1)(4)| = 12\\ R &= \frac{|V|^3}{|\vec V\times\vec a|} = \frac{4.123^3}{12} = 5.84\ \text{m} \end{aligned}$$

Step 4: streamline.

$$\frac{dy}{dx} = \frac{v}{u} = \frac{-Ay^2/2}{Axy} = -\frac{y}{2x}\;\Rightarrow\; \frac{dy}{y} = -\frac{dx}{2x}\;\Rightarrow\; xy^2 = C$$

Through $(2,1)$: $C = 2$, so $xy^2 = 2\ \text{m}^3$.

 y
 3 |.
 2 |  .
 1.5| .
 1 |    .--->V(4,-1) at (2,1)
   |     ^ .
   |   a(4,2)   .   .
 0.5|            .      .       .
   +----------------------------- x
   0    1    2    3   ...   8

At (2,1) the velocity vector is directed down-right (slope $-0.25$, tangent to the streamline); the acceleration vector $(4, 2)$ points up-right, and its normal component points towards the centre of curvature (the concave side of the streamline).

Answer: $\vec a = (4\hat i + 2\hat j)$ m/s², $|a| = 4.47$ m/s²; $R = 5.84$ m; streamline $xy^2 = 2$.

Velocity field: $u = Bx(1+At) = x(1+0.5t)$, $v = Cy = y$.

Streamlines through (1, 1) at fixed t

$$\frac{dy}{dx} = \frac{y}{x(1+At)}\;\Rightarrow\; \ln y = \frac{1}{1+At}\ln x\;\Rightarrow\; y = x^{1/(1+0.5t)}$$

Points: for $x = 1, 2, 3, 4$: $t=0$: $y = 1, 2, 3, 4$; $t=1$: $y = 1, 1.587, 2.080, 2.520$; $t=2$: $y = 1, 1.414, 1.732, 2.0$.

Streakline

Take the streakline observed at $t = 3$ s, formed by all particles that passed through (1,1) at release times $\tau$ from 0 to 3 s. Integrate the path of a particle released at $\tau$:

$$\frac{dy}{dt} = y \Rightarrow y = e^{(t-\tau)},\qquad \frac{dx}{dt} = x(1+0.5t)\Rightarrow x = \exp\left[(t-\tau) + 0.25(t^2-\tau^2)\right]$$

At $t = 3$:

Eliminating $\tau$ (with $s = \ln y = t-\tau$): $x = y^{1+0.5t}e^{-0.25(\ln y)^2}$ and at $t=3$, $x = y^{2.5}e^{-0.25(\ln y)^2}$.

 y
20 |                                  * (190,20)
   |                       * (108,12)
12 |            
 7 |        * (54.6,7.4)
 4.5| * (24,4.5)
 1 |*--*(1,1)...
   +----------------------------------- x

Comparison: a streamline is the instantaneous curve tangent to velocity, a straight line $y = x$ at $t=0$ that flattens with time ($y = x^{1/(1+0.5t)}$). The streakline at $t = 3$ is much steeper and longer, because particles released earlier have been carried far by the growing $u$ and so do not lie on any one streamline. The flow is unsteady, so streaklines, pathlines and streamlines are different curves; they would coincide only for steady flow.

Answer: streamlines $y = x,\ x^{2/3},\ x^{1/2}$ at $t = 0, 1, 2$ s; streakline from $(1,1)$ to $(190.6, 20.1)$ at $t = 3$ s given by the table above.

Velocity field: $u = x(1+0.5t)$, $v = y$ (A = 0.5 s⁻¹, B = C = 1 s⁻¹).

(i) Pathline of the particle at (1,1) at t = 0

$$\frac{dy}{dt} = y\Rightarrow y = e^{t}$$ $$\frac{dx}{dt} = x(1+0.5t)\Rightarrow \ln x = t + 0.25t^2\Rightarrow x = e^{t+0.25t^2}$$

Eliminating $t = \ln y$:

$$\ln x = \ln y + 0.25(\ln y)^2$$

(ii) Streamlines through (1,1,0)

$$\frac{dy}{dx} = \frac{v}{u} = \frac{y}{x(1+0.5t)}\Rightarrow y = x^{1/(1+0.5t)}$$

 y
 4 |    /  <- t=0 (y = x)
   |   / .-' t=1
 2 |  / .'  t=2
   | /.'
 1 |*-----------------------> x
   1    2    3    4

The pathline starts along the $t=0$ streamline (slope 1 at $t=0$) but bends below it: as time passes the $x$-velocity grows, so the particle crosses successive streamlines that become flatter. The pathline and streamlines differ because the flow is unsteady.

Answer: pathline $\ln x = \ln y + 0.25(\ln y)^2$ (parametric $x = e^{t+0.25t^2}$, $y = e^t$); streamlines $y = x$, $y = x^{2/3}$, $y = x^{1/2}$ at $t = 0, 1, 2$ s.

Given: $\psi = -6Ax - 8Ay$ with $A = 1$ m/s, so $\psi = -6x - 8y$. Using $u = \partial\psi/\partial y$ and $v = -\partial\psi/\partial x$.

(i) Streamlines

• $\psi = 0$: $-6x - 8y = 0\Rightarrow y = -0.75x$
• $\psi = 8$: $-6x - 8y = 8\Rightarrow y = -0.75x - 1$

Both are parallel straight lines of slope $-0.75$ ($-36.9^\circ$ to the x-axis). The second line cuts the y-axis at $y = -1$ m and the x-axis at $x = -1.333$ m.

 y
   |   \  
   |    \ psi=0  (y = -0.75x)
   |     \
---+------\------------ x
   |\      \
   | \ psi=8  (y = -0.75x - 1)

(ii) Velocity at (0, 0)

$$u = \frac{\partial\psi}{\partial y} = -8A = -8\ \text{m/s},\qquad v = -\frac{\partial\psi}{\partial x} = 6A = 6\ \text{m/s}$$ $$|V| = \sqrt{8^2 + 6^2} = 10\ \text{m/s},\qquad \theta = \tan^{-1}\frac{6}{-8} = 143.1^\circ\ \text{from +x}$$

The velocity is uniform everywhere (the same at every point) and parallel to the streamlines.

(iii) Flow rate between (1, 1) and (4, 1)

$$\psi_{(1,1)} = -6 - 8 = -14\ \text{m}^2/\text{s},\qquad \psi_{(4,1)} = -24 - 8 = -32\ \text{m}^2/\text{s}$$ $$q = |\psi_2 - \psi_1| = |-32 - (-14)| = 18\ \text{m}^2/\text{s per metre depth}$$

Answer: (i) $y = -0.75x$ and $y = -0.75x - 1$; (ii) $\vec V = (-8\hat i + 6\hat j)$ m/s, $10$ m/s at $143.1^\circ$ to +x; (iii) $q = 18$ m²/s per unit width.

Given: $u = A/x$, $v = Ay/x^2$, $A = 2$ m²/s.

(i) Streamline through (1, 3)

$$\frac{dy}{dx} = \frac{v}{u} = \frac{Ay/x^2}{A/x} = \frac{y}{x}\;\Rightarrow\;\ln y = \ln x + \ln C\;\Rightarrow\; y = Cx$$

At $(1, 3)$: $C = 3$.

Streamline: $y = 3x$ (a straight line through the origin).

(ii) Time from x = 1 m to x = 3 m

For a particle, $u = dx/dt$:

$$\frac{dx}{dt} = \frac{A}{x}\;\Rightarrow\; x\,dx = A\,dt$$ $$t = \frac{1}{A}\int_1^3 x\,dx = \frac{x_2^2 - x_1^2}{2A} = \frac{9 - 1}{2\times 2} = 2\ \text{s}$$

Answer: (i) $y = 3x$; (ii) $t = 2.0$ s.

Given: $\vec V = \hat i\,A/x + \hat j\,Ay/x^2$ with $A = 2$ m²/s.

Streamline through (1, 3)

The streamline satisfies $dy/dx = v/u$:

$$\frac{dy}{dx} = \frac{Ay/x^2}{A/x} = \frac{y}{x}\;\Rightarrow\;\int\frac{dy}{y} = \int\frac{dx}{x}\;\Rightarrow\; y = Cx$$

Using $(x, y) = (1, 3)$: $C = 3$, so the streamline is $y = 3x$.

Time for a particle to move from x = 1 m to x = 2 m

Along its path $dx/dt = u = A/x$, so $x\,dx = A\,dt$:

$$t = \int_1^2\frac{x}{A}dx = \frac{x_2^2 - x_1^2}{2A} = \frac{4 - 1}{2\times 2} = 0.75\ \text{s}$$

(The flow is steady, so the particle path and the streamline coincide; the particle moves along $y = 3x$, from $(1, 3)$ to $(2, 6)$.)

Answer: streamline $y = 3x$; time $= 0.75$ s.

A pathline is the track of one particle; a streakline is the line joining all particles that passed through the injection point earlier. Here the flow is unsteady (velocity changes at $t = 8$ s). The injection point is the origin, and the streak is observed at $t = 16$ s.

Velocity: $0\le t<8$: $(u, v) = (0, 2)$ m/s; $8\le t\le 16$: $(2, -2)$ m/s.

Pathline (particle at origin at t = 0)

• 0 to 8 s: moves $\Delta y = 2\times 8 = 16$ m, from $(0, 0)$ to $(0, 16)$.
• 8 to 16 s: moves $\Delta x = 2\times 8 = 16$ m, $\Delta y = -16$ m, from $(0, 16)$ to $(16, 0)$.

Streakline at t = 16 s

For a particle released at $\tau$:

• Released in $0\le\tau\le 8$: moves up for $(8-\tau)$ s, then diagonally for 8 s:

$$x = 16,\qquad y = 2(8-\tau) - 16 = -2\tau$$

So these particles lie on the vertical line $x = 16$ from $y = 0$ ($\tau=0$) to $y = -16$ ($\tau=8$).

• Released in $8\le\tau\le 16$: move diagonally for $(16-\tau)$ s:

$$x = 2(16-\tau),\qquad y = -2(16-\tau)$$

So they lie on the line $y = -x$ from $(16, -16)$ ($\tau=8$) to the origin ($\tau=16$).

 y
16 |*(0,16)
   | \         pathline: (0,0)->(0,16)->(16,0)
   |  \
   |   \
 0 *----\------*(16,0) ----> x
   0\    \     |
     \         |   streakline: (0,0) -> (16,-16)
      \        |   then vertical up to (16,0)
       \       |
-16     ------*(16,-16)

Answer: pathline is (0,0) to (0,16) to (16,0); streakline at $t = 16$ s is (0,0) to (16,-16) (straight line) and then vertically up the line $x = 16$ to (16,0). The two curves are different because the flow is unsteady. Scale: 1 division = 4 m on both axes.

Given: $\psi = x^2 + y^2$, with $u = \partial\psi/\partial y$, $v = -\partial\psi/\partial x$.

Streamlines

A streamline is $\psi =$ constant, so

$$x^2 + y^2 = C = r^2$$

These are concentric circles about the origin with radius $r = \sqrt{\psi}$. For $\psi = 1, 4, 9, 16, 25$ the radii are $1, 2, 3, 4, 5$ m; the streamlines get closer as $r$ increases, so velocity increases outward.

        y
        |   . - ' ' - .
      .' .-'''''''-. '.
     / .'   .---.   '. \
 ---|-|----|--O--|----|-|--- x
     \ '.   '---'   .' /
      '. '-.......-' .'
        |  ' - . - '

Velocity at (3, 4)

$$u = \frac{\partial\psi}{\partial y} = 2y = 8\ \text{m/s},\qquad v = -\frac{\partial\psi}{\partial x} = -2x = -6\ \text{m/s}$$ $$|V| = \sqrt{8^2 + 6^2} = 10\ \text{m/s},\qquad \theta = \tan^{-1}\frac{-6}{8} = -36.87^\circ$$

The velocity is $10$ m/s at $36.9^\circ$ below the +x axis. It is tangent to the circle $r = 5$ m through $(3, 4)$ (because $\psi = 25$ there) and the flow is clockwise. In general $|V| = 2r$ (a rotational flow like a rigid-body rotation).

Answer: concentric circles $x^2 + y^2 = \psi$; at (3, 4) the velocity is $(8\hat i - 6\hat j)$ m/s, i.e. 10 m/s at $-36.9^\circ$ to the x-axis.

Given: $u = 2x^3$, $v = -6x^2y$.

Check of continuity

$$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 6x^2 - 6x^2 = 0$$

so the flow is incompressible and a stream function exists.

Equation of streamline

$$\frac{dy}{dx} = \frac{v}{u} = \frac{-6x^2y}{2x^3} = -\frac{3y}{x}$$ $$\frac{dy}{y} = -3\frac{dx}{x}\;\Rightarrow\; \ln y = -3\ln x + \ln C\;\Rightarrow\; x^3y = C$$

Stream function ψ

$u = \partial\psi/\partial y = 2x^3\Rightarrow\psi = 2x^3y + f(x)$.

$v = -\partial\psi/\partial x = -6x^2y - f'(x)$. Comparing with $v = -6x^2y$ gives $f'(x) = 0$.

$$\psi = 2x^3y\ (+\text{constant})$$

This agrees with the streamline $x^3y = C$.

Velocity potential φ

A velocity potential exists only for irrotational flow, i.e. when $\partial v/\partial x = \partial u/\partial y$.

$$\frac{\partial v}{\partial x} = -12xy,\qquad \frac{\partial u}{\partial y} = 0$$

The vorticity $\omega_z = \tfrac12(\partial v/\partial x - \partial u/\partial y) = -6xy \ne 0$, so the flow is rotational.

Answer: streamline $x^3y = C$; $\psi = 2x^3y$; $\phi$ does not exist (flow is rotational, $\omega_z = -6xy$).

A flow is irrotational if the vorticity (twice the rotation) is zero everywhere: $\omega_z = \dfrac{1}{2}\left(\dfrac{\partial v}{\partial x} - \dfrac{\partial u}{\partial y}\right) = 0$.

Given: $u = \dfrac{1}{2\mu}\dfrac{dP}{dx}(y^2 - hy)$, $v = 0$.

$$\frac{\partial v}{\partial x} = 0,\qquad \frac{\partial u}{\partial y} = \frac{1}{2\mu}\frac{dP}{dx}(2y - h)$$ $$\omega_z = \frac{1}{2}\left(0 - \frac{1}{2\mu}\frac{dP}{dx}(2y - h)\right) = -\frac{1}{4\mu}\frac{dP}{dx}(2y - h)$$

The vorticity is $\zeta = 2\omega_z = -\dfrac{1}{2\mu}\dfrac{dP}{dx}(2y - h)$.

Since $dP/dx$ is a non-zero constant, $\omega_z$ is zero only at the mid-plane $y = h/2$, and non-zero elsewhere (positive near the bottom plate, negative near the top plate, because $dP/dx<0$).

Answer: the flow is rotational, because $\omega_z \neq 0$ everywhere except along the centreline $y = h/2$. The rotation arises from viscous shear (velocity gradient $\partial u/\partial y$), which is largest at the walls.

A stagnation point is a point where the velocity is zero, i.e. both $u = 0$ and $v = 0$ at the same time.

Given: $u = 1 + 2.5x + y$, $v = -0.5 - 3x - 2.5y$.

Set both components equal to zero:

$$\begin{aligned} 1 + 2.5x + y &= 0 \quad (1)\\ -0.5 - 3x - 2.5y &= 0 \quad (2) \end{aligned}$$

From (1): $y = -1 - 2.5x$. Substituting in (2):

$$\begin{aligned} -0.5 - 3x - 2.5(-1 - 2.5x) &= 0\\ -0.5 - 3x + 2.5 + 6.25x &= 0\\ 2 + 3.25x &= 0\\ x &= -0.6154\ \text{m} \end{aligned}$$ $$y = -1 - 2.5(-0.6154) = 0.5385\ \text{m}$$

Check: $u = 1 - 1.5385 + 0.5385 = 0$ and $v = -0.5 + 1.8462 - 1.3462 = 0$.

Answer: yes, there is one stagnation point, at $(x, y) = (-0.615\ \text{m},\ 0.538\ \text{m})$.