Viscous Effects

Reynolds' experiment

Osborne Reynolds (1883) studied the nature of flow in a glass tube by injecting a thin filament of dye into flowing water.

  water tank
 +-----------+      dye
 |           |       |
 |           |   ____v___
 |           |==|  tube  |=====> to valve
 +-----------+  |_________|

Observations

1. At low velocity, the dye moves as a straight, thin line parallel to the tube axis: laminar flow (layers slide over each other).
2. When the velocity is increased, the dye line begins to wave: transition.
3. At high velocity, the dye mixes completely with the water, showing eddies and random motion: turbulent flow.

Reynolds found that the change depends on the dimensionless group:

$$Re = \frac{\rho V D}{\mu} = \frac{V D}{\nu} = \frac{\text{inertia force}}{\text{viscous force}}$$

Critical values (pipe flow)

• $Re < 2000$: laminar
• $2000 < Re < 4000$: transition (unstable)
• $Re > 4000$: turbulent

The exact values depend on the disturbances present; the lower critical value of about 2000 is the commonly used limit.

Laminar vs turbulent flow

  Laminar          Turbulent
  |   .--          |  .----
  |  /             | |
  | |              | |
  |  \             | |
  |   '--          |  '----

Assumptions

Steady, laminar, fully developed flow of a Newtonian incompressible fluid between two fixed wide plates, gap $h$. Flow is in the $x$-direction; $y$ is measured from the lower plate. There is no body force effect other than hydrostatic, and $dp/dx$ is constant.

   y ^ ---------------------------  upper plate (u = 0)
     |        ______
     |  h   /        \   u(y)
     |      \________/
     +---------------------------> x   lower plate (u = 0)

Force balance on an element

Consider a fluid element of length $dx$, thickness $dy$ and unit width. Net force in $x$ is zero:

$$p\,dy - \left(p + \frac{dp}{dx}dx\right)dy - \tau\,dx + \left(\tau + \frac{d\tau}{dy}dy\right)dx = 0$$ $$\frac{d\tau}{dy} = \frac{dp}{dx}$$

With $\tau = \mu\,du/dy$:

$$\mu\frac{d^2u}{dy^2} = \frac{dp}{dx}$$

Velocity distribution

Integrating twice, $u = \dfrac{1}{2\mu}\dfrac{dp}{dx}y^2 + C_1y + C_2$. Boundary conditions $u = 0$ at $y = 0$ and at $y = h$ give $C_2 = 0$ and $C_1 = -\dfrac{h}{2\mu}\dfrac{dp}{dx}$.

$$u = -\frac{1}{2\mu}\frac{dp}{dx}\left(hy - y^2\right)$$

The profile is parabolic. The pressure falls in the flow direction, so $dp/dx < 0$ and $u > 0$.

Shear stress distribution

$$\tau = \mu\frac{du}{dy} = -\frac{dp}{dx}\left(\frac{h}{2} - y\right)$$

It is zero at the centre line and maximum at the walls: $\tau_w = \dfrac{h}{2}\left|\dfrac{dp}{dx}\right|$ (linear variation).

Maximum velocity

At $y = h/2$:

$$u_{max} = -\frac{h^2}{8\mu}\frac{dp}{dx}$$

Discharge and average velocity

$$q = \int_0^h u\,dy = -\frac{1}{2\mu}\frac{dp}{dx}\left[\frac{hy^2}{2} - \frac{y^3}{3}\right]_0^h = -\frac{h^3}{12\mu}\frac{dp}{dx}$$ $$\bar u = \frac{q}{h} = -\frac{h^2}{12\mu}\frac{dp}{dx} = \frac23 u_{max}$$

Pressure drop

For a length $L$, $dp/dx = -\Delta p/L$:

$$\Delta p = \frac{12\mu \bar u L}{h^2} = \frac{12\mu q L}{h^3}$$

The head loss is $h_f = \Delta p/\rho g = \dfrac{12\mu\bar u L}{\rho g h^2}$.

Given data

$\mu = 0.12\ \text{Pa s}$, $\rho = 900\ \text{kg/m}^3$, $h = 0.01\ \text{m}$, $L = 5\ \text{m}$, $\Delta p = 20\,000\ \text{Pa}$, $b = 1\ \text{m}$.

Pressure gradient magnitude: $\left|\dfrac{dp}{dx}\right| = \dfrac{20\,000}{5} = 4000\ \text{Pa/m}$

(a) Maximum velocity

$$u_{max} = \frac{h^2}{8\mu}\left|\frac{dp}{dx}\right| = \frac{(0.01)^2 \times 4000}{8 \times 0.12} = 0.4167\ \text{m/s}$$

(b) Discharge

Per unit width:

$$q = \frac{h^3}{12\mu}\left|\frac{dp}{dx}\right| = \frac{(0.01)^3 \times 4000}{12 \times 0.12} = 2.778\times10^{-3}\ \text{m}^2/\text{s}$$

For $b = 1\ \text{m}$: $Q = 2.778\times10^{-3}\ \text{m}^3/\text{s} = 2.78\ \text{L/s}$.

(c) Mean velocity

$$\bar u = \frac{q}{h} = \frac{2.778\times10^{-3}}{0.01} = 0.2778\ \text{m/s} \quad (= \tfrac23 u_{max})$$

(d) Wall shear stress

$$\tau_w = \frac{h}{2}\left|\frac{dp}{dx}\right| = 0.005 \times 4000 = 20\ \text{Pa}$$

(e) Velocity at $y = 2.5\ \text{mm}$

$$u = \frac{1}{2\mu}\left|\frac{dp}{dx}\right|(hy - y^2) = \frac{4000}{0.24}\left(0.01 \times 0.0025 - 0.0025^2\right) = 0.3125\ \text{m/s}$$

(f) Head loss

$$h_f = \frac{\Delta p}{\rho g} = \frac{20\,000}{900 \times 9.81} = 2.27\ \text{m of oil}$$

Check for laminar flow

Taking the plate gap as the characteristic length:

$$Re = \frac{\rho \bar u h}{\mu} = \frac{900 \times 0.2778 \times 0.01}{0.12} = 20.8 \ll 2000$$

So the flow is laminar.

Answer: $u_{max} = 0.417\ \text{m/s}$, $Q = 2.78\ \text{L/s}$, $\bar u = 0.278\ \text{m/s}$, $\tau_w = 20\ \text{Pa}$, $u(2.5\ \text{mm}) = 0.3125\ \text{m/s}$, $h_f = 2.27\ \text{m}$

Assumptions

Steady, fully developed, laminar, incompressible flow of a Newtonian fluid in a horizontal circular pipe of radius $R$ and length $L$. Velocity is axial only and depends on $r$ only.

        r
        ^   _____________________________
   R    |  /      ->  ->   ->
        | /   ->  ->  ->   ->   -> u(r)
   0 ---+--------------------------------->  x
        | \   ->  ->  ->   ->
        |  \_______________________________
           p1                       p2

Force balance on a cylinder of radius r

Pressure force $= (p_1 - p_2)\pi r^2 = \Delta p\,\pi r^2$. Shear force on the cylindrical surface $= \tau\,2\pi r L$. For equilibrium:

$$\tau = \frac{\Delta p\, r}{2L}$$

This is linear in $r$: zero at the axis, maximum at the wall.

Velocity distribution

With $\tau = -\mu\,du/dr$ (velocity decreases as $r$ increases):

$$\frac{du}{dr} = -\frac{\Delta p}{2\mu L}r$$

Integrating, with $u = 0$ at $r = R$:

$$u = \frac{\Delta p}{4\mu L}\left(R^2 - r^2\right)$$

The profile is a paraboloid.

Maximum velocity (at $r = 0$)

$$u_{max} = \frac{\Delta p R^2}{4\mu L}$$

Discharge

$$Q = \int_0^R u\,2\pi r\,dr = \frac{\pi \Delta p}{2\mu L}\left[\frac{R^4}{2} - \frac{R^4}{4}\right] = \frac{\pi \Delta p R^4}{8\mu L} = \frac{\pi \Delta p D^4}{128\mu L}$$

This is the Hagen-Poiseuille equation.

Average velocity

$$\bar V = \frac{Q}{\pi R^2} = \frac{\Delta p R^2}{8\mu L} = \frac{u_{max}}{2}$$

Wall shear stress

$$\tau_w = \frac{\Delta p R}{2L} = \frac{\Delta p D}{4L} = \frac{8\mu\bar V}{D}$$

Head loss

$$h_f = \frac{\Delta p}{\rho g} = \frac{32\mu \bar V L}{\rho g D^2}$$

Friction factor

Equating with the Darcy-Weisbach equation $h_f = \dfrac{f L \bar V^2}{2 g D}$:

$$f = \frac{64\mu}{\rho \bar V D} = \frac{64}{Re}$$

The result is valid only for laminar flow, $Re < 2000$. The head loss is directly proportional to the mean velocity and independent of the pipe roughness.

Given data

$\rho = 850\ \text{kg/m}^3$, $\mu = 0.08\ \text{Pa s}$, $D = 0.05\ \text{m}$, $R = 0.025\ \text{m}$, $L = 100\ \text{m}$, $Q = 1.5\times10^{-3}\ \text{m}^3/\text{s}$.

Mean velocity and Reynolds number

$$\begin{aligned} A &= \frac{\pi}{4}(0.05)^2 = 1.9635\times10^{-3}\ \text{m}^2 \\ \bar V &= \frac{Q}{A} = \frac{1.5\times10^{-3}}{1.9635\times10^{-3}} = 0.764\ \text{m/s} \\ Re &= \frac{\rho \bar V D}{\mu} = \frac{850 \times 0.764 \times 0.05}{0.08} = 406 \end{aligned}$$

$Re = 406 < 2000$, so the flow is laminar.

(a) Maximum velocity

$$u_{max} = 2\bar V = 2 \times 0.764 = 1.528\ \text{m/s}$$

(b) Pressure drop and head loss

From Hagen-Poiseuille, $\Delta p = \dfrac{32\mu \bar V L}{D^2}$:

$$\Delta p = \frac{32 \times 0.08 \times 0.764 \times 100}{(0.05)^2} = 78\,228\ \text{Pa} \approx 78.2\ \text{kPa}$$ $$h_f = \frac{\Delta p}{\rho g} = \frac{78\,228}{850 \times 9.81} = 9.38\ \text{m of oil}$$

(c) Wall shear stress

$$\tau_w = \frac{\Delta p D}{4L} = \frac{78\,228 \times 0.05}{4 \times 100} = 9.78\ \text{Pa}$$

(d) Friction factor

$$f = \frac{64}{Re} = \frac{64}{406} = 0.158$$

Check: $h_f = f\dfrac{L}{D}\dfrac{\bar V^2}{2g} = 0.158 \times 2000 \times 0.02975 = 9.38\ \text{m}$. This agrees.

(e) Radius where $u = \bar V$

Since $u = u_{max}(1 - r^2/R^2) = \bar V = u_{max}/2$:

$$r = \frac{R}{\sqrt2} = \frac{25}{1.414} = 17.7\ \text{mm}$$

(f) Velocity at $r = 15\ \text{mm}$

$$u = 1.528\left[1 - \left(\frac{15}{25}\right)^2\right] = 1.528 \times 0.64 = 0.978\ \text{m/s}$$

(g) Power to overcome friction

$$P = Q\,\Delta p = 1.5\times10^{-3} \times 78\,228 = 117.3\ \text{W}$$

Answer: $Re = 406$ (laminar); $u_{max} = 1.53\ \text{m/s}$; $\Delta p = 78.2\ \text{kPa}$ ($h_f = 9.38\ \text{m}$); $\tau_w = 9.78\ \text{Pa}$; $f = 0.158$; $r = 17.7\ \text{mm}$; $u = 0.978\ \text{m/s}$; $P = 117\ \text{W}$

(a) Velocity distribution

For steady, fully developed laminar flow, the x-momentum (Navier-Stokes) equation reduces to

$$\mu\frac{d^2u}{dy^2} = \frac{dp}{dx}$$

Integrating twice: $u = \dfrac{1}{2\mu}\dfrac{dp}{dx}y^2 + C_1y + C_2$.

Boundary conditions: $u = 0$ at $y = 0$, and $u = U$ at $y = h$. Then $C_2 = 0$ and $C_1 = \dfrac{U}{h} - \dfrac{h}{2\mu}\dfrac{dp}{dx}$.

$$u = \frac{U y}{h} - \frac{1}{2\mu}\frac{dp}{dx}\left(hy - y^2\right)$$

The first term is the Couette (linear) part due to the moving plate, and the second is the Poiseuille (parabolic) part due to the pressure gradient.

Discharge per unit width:

$$q = \int_0^h u\,dy = \frac{Uh}{2} - \frac{h^3}{12\mu}\frac{dp}{dx}$$

(b) Effect of pressure gradient

Let $P = -\dfrac{h^2}{2\mu U}\dfrac{dp}{dx}$.

  dp/dx = 0     dp/dx < 0     dp/dx > 0 (large)
      /|           _.|             |
     / |         .'  |            _|
    /  |        /    |          <-'/
   /___|       /_____|         /___|

Shear stress: $\tau = \mu\dfrac{du}{dy} = \mu\dfrac{U}{h} - \left(\dfrac{h}{2} - y\right)\dfrac{dp}{dx}$

At the moving plate ($y = h$): $\tau_h = \dfrac{\mu U}{h} + \dfrac{h}{2}\dfrac{dp}{dx}$

Numerical

$\mu U/h = 0.1 \times 3/0.004 = 75\ \text{Pa}$.

• For $dp/dx = 0$: $\tau_h = 75\ \text{Pa}$.
• For $dp/dx = -10\,000\ \text{Pa/m}$:

$$\tau_h = 75 + \frac{0.004}{2}(-10\,000) = 75 - 20 = 55\ \text{Pa}$$

(The shear stress on the fixed plate would be $75 + 20 = 95\ \text{Pa}$.) The discharge per unit width rises from $Uh/2 = 0.006\ \text{m}^2/\text{s}$ to $0.006 + \dfrac{(0.004)^3 \times 10\,000}{12 \times 0.1} = 0.00653\ \text{m}^2/\text{s}$.

Answer: $\tau_h = 75\ \text{Pa}$ for $dp/dx = 0$, and $55\ \text{Pa}$ for $dp/dx = -10\ \text{kPa/m}$