Pipe flow

Prandtl's mixing length theory

In turbulent flow, fluid lumps move across the layers and carry momentum with them. Prandtl assumed that a lump keeps its identity over a distance $l$ (the mixing length) and then mixes with the new layer, like the mean free path in kinetic theory.

• Velocity difference between layers a distance $l$ apart: $u' \approx l\,\dfrac{du}{dy}$
• The transverse fluctuation $v'$ is of the same order as $u'$, so $v' \approx l\,\dfrac{du}{dy}$
• Turbulent (Reynolds) shear stress: $\tau_t = -\rho\,\overline{u'v'} = \rho\, l^2 \left(\dfrac{du}{dy}\right)^2$

Near a wall the eddy size grows with distance from the wall, so Prandtl took $l = \kappa y$, where $\kappa \approx 0.4$ is von Karman's constant. Total shear stress is $\tau = \mu\dfrac{du}{dy} + \rho l^2\left(\dfrac{du}{dy}\right)^2$; away from the wall the viscous part is negligible.

Logarithmic velocity distribution in a pipe

Near the wall the shear stress is nearly constant and equal to the wall shear $\tau_0$. Let $u_* = \sqrt{\tau_0/\rho}$ be the shear velocity and $y$ the distance from the wall.

$$\tau_0 = \rho\,\kappa^2 y^2\left(\frac{du}{dy}\right)^2 \;\Rightarrow\; \frac{du}{dy} = \frac{u_*}{\kappa\, y}$$

Integrating:

$$u = \frac{u_*}{\kappa}\ln y + C$$

so velocity varies with the logarithm of the distance from the wall. The constant $C$ is fixed by the wall condition:

• Smooth pipe: $\dfrac{u}{u_*} = 5.75\log_{10}\dfrac{u_* y}{\nu} + 5.5$
• Rough pipe: $\dfrac{u}{u_*} = 5.75\log_{10}\dfrac{y}{k} + 8.5$

Here $\dfrac{1}{\kappa}\ln y = \dfrac{2.303}{0.4}\log_{10} y = 5.75\log_{10} y$. Measured velocities in pipes follow this law over almost the whole cross-section, and for $y = R$ at the axis, $\dfrac{u_{max}-u}{u_*} = 5.75\log_{10}\dfrac{R}{y}$.

Head loss due to sudden expansion

   1          2            Section 1: A1, V1, p1 (small pipe)
  ____       _______       Section 2: A2, V2, p2 (large pipe)
 |    |_____|      
 |  V1   ~~~~ eddies  V2  Control volume between 1 and 2
 |____       _______      (eddy zone pressure = p1 on the annulus)
      |_____|

Take a control volume from section 1 (just before the expansion) to section 2 (where the jet has spread over the full area). Eddies in the corner zone keep the pressure on the annular area $(A_2 - A_1)$ equal to $p_1$. Friction on the short length is neglected.

Momentum equation (flow direction):

$$p_1 A_2 - p_2 A_2 = \rho Q (V_2 - V_1), \qquad Q = A_2 V_2$$ $$\frac{p_1 - p_2}{\gamma} = \frac{V_2(V_2 - V_1)}{g}$$

Energy equation between 1 and 2 (same level), with head loss $h_e$:

$$\frac{p_1}{\gamma} + \frac{V_1^2}{2g} = \frac{p_2}{\gamma} + \frac{V_2^2}{2g} + h_e \;\Rightarrow\; h_e = \frac{p_1-p_2}{\gamma} + \frac{V_1^2 - V_2^2}{2g}$$

Substituting the momentum result:

$$h_e = \frac{V_2^2 - V_1V_2}{g} + \frac{V_1^2 - V_2^2}{2g} = \frac{2V_2^2 - 2V_1V_2 + V_1^2 - V_2^2}{2g}$$ $$\boxed{h_e = \frac{(V_1 - V_2)^2}{2g}}$$

Using continuity $V_2 = V_1A_1/A_2$, this is also $h_e = \left(1 - \dfrac{A_1}{A_2}\right)^2\dfrac{V_1^2}{2g}$.

Definitions

• Hydraulic Gradient Line (HGL) is the line joining the levels to which water would rise in piezometer tubes along the pipe. Its height above datum is the piezometric head $\dfrac{p}{\gamma} + z$.
• Total Energy Line (TEL) (also called EGL) is the line showing the total head $\dfrac{p}{\gamma} + z + \dfrac{V^2}{2g}$. It is above the HGL by the velocity head $\dfrac{V^2}{2g}$.

Features for a real fluid

• TEL always falls in the direction of flow because of head losses. Its slope equals the energy gradient $S_f = h_f/L$.
• In a uniform pipe $V$ is constant, so the HGL is parallel to the TEL, a fixed distance $V^2/2g$ below it.
• Minor losses (entrance, expansion, valve, bend) show as sudden vertical drops of the TEL.
• Where the HGL is below the pipe axis the pressure is below atmospheric (negative); where above, it is positive.
• A pump gives a sudden rise of both lines; a turbine gives a drop.
• At a free outlet to the atmosphere the HGL meets the pipe axis (gauge pressure zero); at a reservoir both lines coincide with the free surface (V = 0).

Uniform pipe (constant diameter)

 reservoir          entrance loss
 TEL=HGL ---.
 (surface)   \_
              `-._  TEL (slope = hf/L)
        0.5V²/2g  `-._
               ___  `-._
          HGL ..  `-.__ `-._
        V²/2g gap const `-._`-._
 pipe axis ======================== outlet

TEL and HGL are two parallel straight sloping lines; the gap is constant at $V^2/2g$.

Non-uniform pipe (diameter changes from $D_1$ to $D_2 > D_1$)

 TEL ---.\_
         \ `-._ steeper (small pipe, high V)
          \     `-._
           |        `-.|  <- drop (V1-V2)²/2g
           |  HGL        |`-._ flatter slope
           | rises?      |     `-._ TEL
   ==========[small D1]==[large D2]=======

• In the smaller pipe the velocity head is large, so the HGL is far below the TEL and the slope $h_f/L$ is steeper.
• At the sudden expansion the TEL drops by $(V_1 - V_2)^2/2g$; as $V$ falls, the velocity head $V^2/2g$ falls and the HGL can rise (pressure recovery) even though energy is lost.
• In the larger pipe the TEL is flatter and the gap between the lines is smaller.
• In a gradual contraction/expansion the lines change slope smoothly instead of jumping.

Given. Reservoir level 8 m above the pipe axis (datum = pipe axis). Pipe 1: $D_1 = 0.15$ m, $L_1 = 25$ m. Pipe 2: $D_2 = 0.3$ m, $L_2 = 15$ m, $f = 0.01$. The pipe discharges freely, so the outlet velocity head stays as kinetic energy (no exit loss term, the HGL ends at the axis).

Velocities

Continuity: $V_2 = V_1\left(\dfrac{D_1}{D_2}\right)^2 = V_1/4$.

Energy equation from the reservoir surface to the outlet (all losses):

$$H = 0.5\frac{V_1^2}{2g} + f\frac{L_1}{D_1}\frac{V_1^2}{2g} + \frac{(V_1-V_2)^2}{2g} + f\frac{L_2}{D_2}\frac{V_2^2}{2g} + \frac{V_2^2}{2g}$$ $$8 = \frac{V_1^2}{2g}\left[0.5 + 0.01\left(\frac{25}{0.15}\right) + (1-0.25)^2 + 0.01\left(\frac{15}{0.3}\right)(0.25)^2 + (0.25)^2\right] = 2.8229\,\frac{V_1^2}{2g}$$ $$V_1 = \sqrt{\frac{2(9.81)(8)}{2.8229}} = 7.457\ \text{m/s},\qquad V_2 = 1.864\ \text{m/s}$$

Velocity heads: $V_1^2/2g = 2.834$ m, $V_2^2/2g = 0.177$ m. Discharge $= 131.8$ l/s.

Head losses

The total equals the 8 m available head.

Levels of TEL and HGL (m above the pipe axis)

 8.0 ___                        TEL
     |  |\ 0.5V²/2g
     |  | \__ TEL falls steeply in pipe 1 (small D)
     |  |    `-.__
     |  |  HGL     `-.  | drop (V1-V2)²/2g
     |  |  (well below) |`.__ TEL falls gently in pipe 2
     |  |               |    `-.__ HGL
 0 --|__|===[D1=0.15]===[D2=0.3]===> free outlet

Draw to scale: TEL: 8.0, 6.58 (entrance drop), 1.86 (end of pipe 1), 0.27 (after expansion drop), 0.18 (outlet). HGL is lower by 2.83 m in pipe 1 and by 0.18 m in pipe 2, and meets the pipe axis at the outlet. The HGL rises slightly at the expansion (from -0.97 to 0.09 m) because the velocity head falls.

Answer: $V_1 = 7.46$ m/s, $V_2 = 1.86$ m/s, TEL ends 0.177 m above the axis at the outlet and the HGL ends at the axis.

Reservoir A (higher) feeds reservoir B through three pipes in series, with $d_1 < d_3 < d_2$. Both reservoir surfaces are at rest, so HGL and EGL start at the water surface of A and end at the water surface of B.

Rules used

• EGL falls in the direction of flow: slope $= f V^2/(2gd)$, so it is steepest in the smallest pipe 1 and flattest in pipe 2.
• HGL is below the EGL by $V^2/2g$; the gap is largest in pipe 1 (highest velocity) and smallest in pipe 2.
• Sudden drops of the EGL: entrance loss $0.5V_1^2/2g$; expansion from $d_1$ to $d_2$: $(V_1 - V_2)^2/2g$; contraction from $d_2$ to $d_3$: $0.5\left(1-\dfrac{A_3}{A_2}\right)V_3^2/2g$ (approximately); exit loss into B: $V_3^2/2g$.
• At the exit, the whole velocity head is lost, so the EGL falls to the surface of B and both lines end at the water level of B.

 A __                                           
    |\ entrance                 
    | \___ EGL                              
    |     `-.__ (steep, d1)                  
    |  HGL     `-.   | expansion drop        
    |  (gap     `-.__|`-.__ (flat, d2)     
    |   V1²/2g)         |  `-.__|contraction
    |                    HGL     |`-.__ (d3)
    |==[d1]=======[d2 big]=======[d3]=====> 
                                       exit drop
                                          \__ B

• The HGL rises after the expansion (velocity falls) and falls again after the contraction (velocity rises).
• Total of all drops (friction + minor losses) equals the difference of the two reservoir levels, $H = h_{f1} + h_{f2} + h_{f3} + h_{entry} + h_{exp} + h_{contr} + h_{exit}$.

Criterion. A pipe is hydrodynamically smooth when the roughness is buried in the laminar sub-layer, i.e. when the roughness Reynolds number (Nikuradse) satisfies

$$\frac{u_* k}{\nu} \le 4$$

with $u_* = V\sqrt{f/8}$ and $k = 0.15$ mm.

Step 1: friction factor of a smooth pipe at $Re = 2\times10^5$. Prandtl's smooth-pipe law $\dfrac{1}{\sqrt f} = 2\log_{10}(Re\sqrt f) - 0.8$ gives (by iteration)

$$f = 0.01564, \qquad \sqrt{f/8} = 0.04421$$

Step 2: express the criterion in terms of $D$. Since $Re = VD/\nu$,

$$\frac{u_* k}{\nu} = \frac{V\sqrt{f/8}\,k}{\nu} = Re\,\sqrt{\frac{f}{8}}\,\frac{k}{D} \le 4$$ $$D \ge \frac{Re\,k\sqrt{f/8}}{4} = \frac{(2\times10^5)(0.00015)(0.04421)}{4} = 0.332\ \text{m}$$

Answer: Minimum diameter $\approx 0.332$ m (about 330 mm). Any smaller pipe at this Reynolds number is not smooth. (Using the limit $u_*k/\nu = 5$ instead of 4 gives 0.265 m.)

Given. $\rho = 900$ kg/m³, $\nu = 0.0002$ m²/s so $\mu = \rho\nu = 0.18$ Pa s; $d = 0.06$ m, $L = 10$ m, inclination 40°. At section 1: $p_1 = 350\,000$ Pa, $z_1 = 0$. At section 2: $p_2 = 250\,000$ Pa, $z_2 = L\sin 40^\circ = 6.428$ m.

Direction of flow

Flow goes from high to low piezometric pressure $p^* = p + \rho g z$, not from high to low $p$.

$$p_1^* = 350\,000 + 0 = 350\,000\ \text{Pa}, \qquad p_2^* = 250\,000 + 900(9.81)(6.428) = 306752\ \text{Pa}$$

Since $p_1^* > p_2^*$, the flow is from 1 to 2, i.e. upward, as required. The driving pressure drop is

$$\Delta p^* = p_1^* - p_2^* = 43248\ \text{Pa}$$

Flow rate (Hagen-Poiseuille, laminar)

$$Q = \frac{\pi d^4\,\Delta p^*}{128\,\mu L} = \frac{\pi (0.06)^4 (43248)}{128(0.18)(10)} = 0.007643\ \text{m}^3/\text{s} = 7.64\ \text{l/s}$$

Check that the flow is laminar

$$V = \frac{Q}{\pi d^2/4} = 2.703\ \text{m/s},\qquad Re = \frac{Vd}{\nu} = \frac{2.703(0.06)}{0.0002} = 811 < 2000$$

So the laminar assumption is valid.

Answer: The flow is upward; $Q = 0.007643$ m³/s ($7.64$ l/s), $Re \approx 811$.

Loss due to sudden enlargement

  1                2
 _______          _________    Pipe area A1 -> A2 (A2 > A1)
        |  ~ ~ ~ |             Eddies form in the corner;
  V1,p1 |  eddy  |  V2,p2      the pressure on the annular face
 _______|  zone  |_________    equals p1.

Consider section 1 (just before the enlargement) and section 2 (downstream, where the flow has spread over the full area $A_2$). Assume the pipe is horizontal and neglect wall friction between 1 and 2. The pressure on the annular shoulder is taken as $p_1$ (eddy zone).

Momentum equation:

$$p_1A_1 + p_1(A_2 - A_1) - p_2A_2 = \rho Q (V_2 - V_1), \qquad Q = A_2V_2$$ $$\frac{p_1 - p_2}{\gamma} = \frac{V_2(V_2 - V_1)}{g} \quad \cdots (1)$$

Energy equation (head loss $h_e$):

$$\frac{p_1}{\gamma} + \frac{V_1^2}{2g} = \frac{p_2}{\gamma} + \frac{V_2^2}{2g} + h_e \quad\Rightarrow\quad h_e = \frac{p_1-p_2}{\gamma} + \frac{V_1^2 - V_2^2}{2g}$$

Substituting (1):

$$h_e = \frac{V_2^2 - V_1V_2}{g} + \frac{V_1^2 - V_2^2}{2g} = \frac{V_1^2 - 2V_1V_2 + V_2^2}{2g}$$ $$\boxed{h_e = \frac{(V_1 - V_2)^2}{2g} = \left(1 - \frac{A_1}{A_2}\right)^2\frac{V_1^2}{2g}}$$

Loss due to sudden contraction (deduced)

  1      c         2       Jet contracts to the vena contracta c
 ________                  (area Cc A2), then expands to fill
          \ ~~~ _          the smaller pipe A2.
  V1       \___|___ V2     Loss is only in the expansion c -> 2.
 ________/  Vc

In a sudden contraction from $A_1$ to $A_2$ the stream contracts to a vena contracta of area $A_c = C_c A_2$, then expands suddenly to fill the pipe $A_2$. The only significant loss is this expansion from $c$ to 2. Applying the enlargement result between $c$ and 2:

$$h_c = \frac{(V_c - V_2)^2}{2g}, \qquad V_c = \frac{A_2V_2}{A_c} = \frac{V_2}{C_c}$$ $$\boxed{h_c = \left(\frac{1}{C_c} - 1\right)^2\frac{V_2^2}{2g} = k_c\frac{V_2^2}{2g}}$$

With $C_c \approx 0.62$, $k_c = (1/0.62 - 1)^2 \approx 0.375$; a common value used is $k_c \approx 0.5$ for a large area ratio.

Consider steady laminar flow in a horizontal circular pipe of radius $R$. Take a coaxial cylinder of fluid of radius $r$ and length $L$.

        <-- tau (on surface)
   p1 ->  ____________  <- p2
         |   cylinder |  radius r, length L
   ------|------------|-------- axis
         |____________|

Force balance (no acceleration): pressure force on the ends = shear force on the curved surface.

$$(p_1 - p_2)\pi r^2 = \tau\,(2\pi r L) \quad\Rightarrow\quad \tau = \frac{(p_1 - p_2)}{L}\frac{r}{2} = -\frac{dp}{dx}\frac{r}{2}$$

So the shear stress varies linearly from zero at the axis to $\tau_0$ at the wall.

Newton's law of viscosity. With $y = R - r$, $\tau = \mu\dfrac{du}{dy} = -\mu\dfrac{du}{dr}$:

$$-\mu\frac{du}{dr} = -\frac{dp}{dx}\frac{r}{2} \quad\Rightarrow\quad \frac{du}{dr} = \frac{1}{2\mu}\frac{dp}{dx}\,r$$

Integrate:

$$u = \frac{1}{4\mu}\frac{dp}{dx}\,r^2 + C$$

Boundary condition (no slip): $u = 0$ at $r = R$, so $C = -\dfrac{1}{4\mu}\dfrac{dp}{dx}R^2$.

$$\boxed{u = -\frac{1}{4\mu}\frac{dp}{dx}\left(R^2 - r^2\right)}$$

This is the equation of a parabola in $r$ ($u$ falls as $r^2$), with its maximum at the axis:

$$u_{max} = -\frac{1}{4\mu}\frac{dp}{dx}R^2, \qquad \frac{u}{u_{max}} = 1 - \frac{r^2}{R^2}$$

The mean velocity is $V = u_{max}/2$, so the velocity distribution is parabolic. (For an inclined pipe, replace $p$ by the piezometric pressure $p + \gamma z$.)

Criterion. Flow is fully turbulent (hydrodynamically rough) when the roughness Reynolds number (Nikuradse) is

$$\frac{u_* k}{\nu} \ge 100$$

Take $\nu = 1.0\times10^{-6}$ m²/s (water at about 20 °C), $k = 1$ mm $= 0.001$ m, $D = 1.2$ m.

Step 1: rough-pipe friction factor (von Karman-Prandtl, $D/k = 1200$):

$$\frac{1}{\sqrt f} = 2\log_{10}\frac{D}{k} + 1.14 \;\Rightarrow\; f = 0.01877, \qquad \sqrt{f/8} = 0.04844$$

Step 2: shear velocity at the limit

$$u_* = \frac{100\,\nu}{k} = \frac{100(1.0\times10^{-6})}{0.001} = 0.100\ \text{m/s}$$

Step 3: mean velocity

$$V = \frac{u_*}{\sqrt{f/8}} = \frac{0.100}{0.04844} = 2.06\ \text{m/s}$$

Answer: The flow becomes fully turbulent (rough) for $V \ge 2.06$ m/s, about 2 m/s. (If the limit $u_*k/\nu = 70$ is used, the velocity is 1.45 m/s.)

Set-up. Pipe of radius $R$ inclined at angle $\alpha$ below the horizontal; the fluid flows down the pipe under gravity alone, so there is no externally imposed pressure gradient: $\dfrac{\partial p}{\partial x} = 0$ (the pressure along the pipe is atmospheric/constant). Flow is steady, laminar, fully developed, so $u = u(r)$ only, and $v_r = v_\theta = 0$.

        x (down the pipe)
     ___ \
    /     \  alpha
   / pipe  \
  /_________\ ____ horizontal
  gravity component along x: g sin(alpha)

Momentum equation along $x$ (Navier-Stokes in cylindrical coordinates, reduced for this case):

$$0 = \rho g\sin\alpha + \mu\,\frac{1}{r}\frac{d}{dr}\left(r\frac{du}{dr}\right)$$

(Equivalent force balance on a cylinder of radius $r$: weight component $\rho g \sin\alpha\,\pi r^2 L$ is balanced by the shear force $-\mu\dfrac{du}{dr}\,2\pi r L$.)

Integrate.

$$\frac{d}{dr}\left(r\frac{du}{dr}\right) = -\frac{\rho g\sin\alpha}{\mu}\,r$$ $$r\frac{du}{dr} = -\frac{\rho g\sin\alpha}{2\mu}\,r^2 + C_1$$

$C_1 = 0$, since $du/dr$ must be finite at $r = 0$. Then

$$\frac{du}{dr} = -\frac{\rho g\sin\alpha}{2\mu}\,r \quad\Rightarrow\quad u = -\frac{\rho g\sin\alpha}{4\mu}\,r^2 + C_2$$

Boundary condition. No slip at the wall: $u = 0$ at $r = R$, so $C_2 = \dfrac{\rho g\sin\alpha}{4\mu}R^2$.

$$\boxed{u(r) = \frac{\rho g\sin\alpha}{4\mu}\left(R^2 - r^2\right)}$$

The profile is parabolic with $u_{max} = \dfrac{\rho g\sin\alpha}{4\mu}R^2$ at the axis. With $R = D/2$, $u = \dfrac{\rho g\sin\alpha}{16\mu}\left(D^2 - 4r^2\right)$. The mean velocity is $V = u_{max}/2 = \dfrac{\rho g D^2\sin\alpha}{32\mu}$.

The type of boundary is decided by the roughness Reynolds number $Re_* = \dfrac{u_* k}{\nu}$ (Nikuradse): smooth $Re_* < 4$, transition $4 \le Re_* \le 100$, rough $Re_* > 100$.

Data. $D = 0.2$ m, $V = 5$ m/s, $k = 0.5$ mm, $f = 0.02$, $\rho = 800$ kg/m³. The viscosity is not given, so it is assumed that for kerosene at about 20 °C, $\mu \approx 1.9\times10^{-3}$ Pa s, giving $\nu = \mu/\rho = 2.375\times10^{-6}$ m²/s.

Shear velocity

$$u_* = V\sqrt{\frac{f}{8}} = 5\sqrt{\frac{0.02}{8}} = 0.250\ \text{m/s}$$

Roughness Reynolds number

$$Re_* = \frac{u_*k}{\nu} = \frac{(0.250)(0.0005)}{2.375\times10^{-6}} = 52.6$$

Check: laminar sub-layer thickness $\delta' = \dfrac{11.6\nu}{u_*} = 0.110$ mm, which is smaller than $k = 0.5$ mm, but $k/\delta' \approx 4.5$ lies in the transition range (0.25 to 6).

Since $4 < Re_* < 100$ (about 53), the pipe is neither smooth nor fully rough.

Answer: The pipe behaves in the transition regime ($Re = 421053$, $k/D = 0.0025$, $Re_* \approx 53$). The conclusion is the same for any realistic kerosene viscosity ($\nu$ between 1.9 and 2.8 $\times10^{-6}$ m²/s).

Data. $\rho = 850$ kg/m³ (s.g. 0.85), $\mu = 0.01$ Pa s, $D = 0.12$ m, $L = 50$ m, slope 20°, outlet pressure 1 bar $= 10^5$ Pa, $Q = 0.25$ m³/min $= 0.004167$ m³/s. The fluid is viscous, so Hagen-Poiseuille (laminar) flow is assumed.

(i) Pressure needed if there were no slope

Mean velocity and Reynolds number:

$$V = \frac{Q}{\pi D^2/4} = 0.3684\ \text{m/s}, \qquad Re = \frac{\rho V D}{\mu} = 3758$$

(This is above 2000, but in the critical zone; the laminar formula is used because the fluid is described as viscous and the flow is steady and smooth. A turbulent estimate would give a friction drop of the order of 1 kPa, which is still negligible against the 142 kPa static term, so the overall pressure is almost unchanged.)

Friction pressure drop (Hagen-Poiseuille):

$$\Delta p_f = \frac{128\mu L Q}{\pi D^4} = \frac{128(0.01)(50)(0.004167)}{\pi(0.12)^4} = 409.3\ \text{Pa}$$

Pressure at the inlet without slope: $P = 10^5 + 409.3 = 100409.3$ Pa $\approx 100.41$ kPa.

Pressure needed on the actual slope

Extra pressure to lift the oil through the height $L\sin 20^\circ = 17.10$ m:

$$\Delta p_z = \rho g L \sin 20^\circ = 850(9.81)(50)(0.3420) = 142597\ \text{Pa}$$ $$P = 10^5 + 409.3 + 142597 = 243006\ \text{Pa} \approx 243.0\ \text{kPa}$$

(ii) Number of 8 cm pipes in parallel

At the same driving pressure the friction drop per pipe is the same, and in laminar flow $Q \propto D^4$ (the slope term is the same for any pipe size). Flow per pipe ratio:

$$\frac{Q_8}{Q_{12}} = \left(\frac{8}{12}\right)^4 = 0.1975$$ $$n = \frac{Q_{12}}{Q_8} = \left(\frac{12}{8}\right)^4 = 5.0625 \;\Rightarrow\; n = 6 \text{ pipes (rounded up)}$$

Check with 6 pipes: each carries $Q/6 = 0.000694$ m³/s, $V = 0.138$ m/s, $Re = 939$ (laminar), and the friction drop $345.4$ Pa is less than 409.3 Pa, so the driving pressure is not exceeded.

Answer: (i) $P \approx 100.41$ kPa without slope ($\Delta p_f = 409$ Pa); on the 20° slope $P \approx 243.0$ kPa; (ii) 6 pipes of 8 cm diameter.

Description

Moody's chart is a graph of the Darcy-Weisbach friction factor $f$ against the Reynolds number $Re$ for flow in circular pipes, with a family of curves for different relative roughness $k/D$ (equivalent sand roughness / diameter). Both axes are logarithmic: $Re$ (from about $10^3$ to $10^8$) on the horizontal axis, $f$ (about 0.008 to 0.1) on the vertical axis, and the relative roughness is marked on the right-hand scale.

 f
 0.1 |\ laminar     critical     k/D curves
     | \  f=64/Re   zone    _________ 0.05
 0.05|  \          |      __/   ______ 0.01
     |   \_        |  ___/ ___/ ______ 0.001
 0.02|     \______ |_/ ___/   _______  0.0001
     |  smooth pipe curve ___/  (rough: flat lines)
 0.01|                   transition|  fully rough
     +--------------------------------- Re (log)
     10^3   10^4    10^5   10^6    10^7

Regions of the chart:

Basis of the chart

1. Laminar flow: Hagen-Poiseuille law gives $f = 64/Re$.
2. Smooth pipes: Blasius ($f = 0.316/Re^{0.25}$, $Re < 10^5$) and Prandtl's universal law $\dfrac{1}{\sqrt f} = 2\log_{10}(Re\sqrt f) - 0.8$.
3. Rough pipes: von Karman-Prandtl law from Nikuradse's sand-roughened pipe experiments, $\dfrac{1}{\sqrt f} = 2\log_{10}\dfrac{D}{k} + 1.14$.
4. Transition region: the Colebrook-White equation, which joins the smooth and rough laws:

$$\frac{1}{\sqrt f} = -2\log_{10}\left(\frac{k/D}{3.7} + \frac{2.51}{Re\sqrt f}\right)$$

Moody plotted this equation using the equivalent roughness of commercial pipes (Nikuradse's artificial roughness corrected to commercial pipe behaviour).

Uses

• To find $f$ for a given $Re$ and $k/D$, then the head loss $h_f = \dfrac{fLV^2}{2gD}$.
• To identify the flow regime (laminar, smooth, transition, rough) of a pipe.
• To solve the three types of pipe problems (find $h_f$, find $Q$, find $D$), by iteration.
• To get the equivalent roughness $k$ of an old or commercial pipe from test data.
• To compare materials and judge how much capacity is lost as a pipe ages and becomes rough.

Data. $D = 0.1$ m, $L = 60$ m, inclination 10°, $\rho = 900$ kg/m³ (R.D. 0.9), $\nu = 120$ mm²/s $= 1.2\times10^{-4}$ m²/s, outlet gauge pressure $p_2 = 120$ kPa, maximum mean wall shear $\tau_0 = 200$ Pa.

Which flow regime?

Mean wall shear in terms of $f$: $\tau_0 = \dfrac{f\rho V^2}{8}$.

• If laminar: $\tau_0 = \dfrac{8\mu V}{D}$, i.e. $V = \dfrac{\tau_0 D}{8\mu} = \dfrac{200(0.1)}{8(900)(1.2\times10^{-4})} = 23.15$ m/s, giving $Re = 19290$, which is not laminar. So the flow is turbulent.
• Turbulent: $f = 0.32\,Re^{-1/4}$ (valid since $Re < 10^5$).

(b) Flow rate

$$\tau_0 = \frac{\rho V^2}{8}(0.32)\left(\frac{\nu}{VD}\right)^{1/4} = 0.04\,\rho\left(\frac{\nu}{D}\right)^{1/4} V^{7/4}$$ $$200 = 0.04(900)\left(\frac{1.2\times10^{-4}}{0.1}\right)^{1/4} V^{1.75} = 6.700\,V^{1.75}$$ $$V = \left(\frac{200}{6.700}\right)^{1/1.75} = 6.963\ \text{m/s}$$

Check: $Re = \dfrac{6.963(0.1)}{1.2\times10^{-4}} = 5803 < 10^5$, so the formula applies; $f = 0.32Re^{-1/4} = 0.03666$ and $\dfrac{f\rho V^2}{8} = 200.0$ Pa.

$$Q = \frac{\pi}{4}D^2V = \frac{\pi}{4}(0.1)^2(6.963) = 0.05469\ \text{m}^3/\text{s} = 54.69\ \text{l/s}$$

(a) Maximum inlet pressure

Force balance on the liquid in the pipe along the incline (steady, uniform flow):

$$(p_1 - p_2)\frac{\pi D^2}{4} = \tau_0\,\pi D L + \rho g\,\frac{\pi D^2}{4}L\sin\theta$$ $$p_1 - p_2 = \frac{4\tau_0L}{D} + \rho g L\sin\theta = \frac{4(200)(60)}{0.1} + 900(9.81)(60)\sin10^\circ$$ $$p_1 - p_2 = 480000 + 91988 = 571988\ \text{Pa}$$ $$p_1 = 120\,000 + 480000 + 91988 = 691988\ \text{Pa} \approx 692\ \text{kPa (gauge)}$$

Answer: (a) $p_{1,max} \approx 692$ kPa gauge; (b) $Q = 54.7$ l/s ($V = 6.96$ m/s).

Data. $D = 0.5$ m, $L = 100$ m, water with $\nu = 1.0\times10^{-6}$ m²/s. Area $A = 0.19635$ m².

Friction factor from the test

$f = \dfrac{h_f\,2gD}{LV^2}$ with $V = Q/A$:

At high flows $f$ levels off at about 0.0193, which means the pipe is fully rough there.

(i) Equivalent sand-grain roughness

Use the highest flow ($Q$ = 800 l/s, $Re = 2037183$, $f = 0.01932$) in the Colebrook-White equation:

$$\frac{1}{\sqrt f} = -2\log_{10}\left(\frac{k/D}{3.7} + \frac{2.51}{Re\sqrt f}\right)$$

Solving, $\dfrac{k}{D} = 0.000903$, so

$$k = 0.000903(500) = 0.45\ \text{mm}$$

Check: Colebrook with this $k$ gives $f$ = 0.0218, 0.0198, 0.0195 and 0.0193 for the four flows, which agree with the test values.

(ii) Maximum discharge for smooth behaviour

The pipe is smooth while $\dfrac{u_*k}{\nu} \le 4$:

$$u_* = \frac{4\nu}{k} = \frac{4(10^{-6})}{0.45\times10^{-3}} = 0.0089\ \text{m/s}$$

With smooth-pipe $f$ (Prandtl law) at the limiting $Re$ (found by iteration, $Re \approx 92619$, $f = 0.0183$):

$$V = \frac{u_*}{\sqrt{f/8}} = 0.185\ \text{m/s},\qquad Q = AV = 36.4\ \text{l/s}$$

(iii) Discharge for fully rough behaviour

The pipe is fully rough when $\dfrac{u_*k}{\nu} \ge 100$, i.e. $u_* = \dfrac{100\nu}{k} = 0.2214$ m/s. For the rough pipe, $\dfrac{1}{\sqrt f} = 2\log_{10}\dfrac{D}{k} + 1.14$ gives $f = 0.01914$:

$$V = \frac{u_*}{\sqrt{f/8}} = 4.526\ \text{m/s}, \qquad Q = 889\ \text{l/s}$$

This is the limit of the rough regime: for $Q$ above about 889 l/s (it is a minimum, not a maximum, flow for rough behaviour) the pipe is fully rough and $f$ no longer depends on $Re$.

Answer: (i) $k \approx 0.45$ mm; (ii) smooth for $Q \lesssim 36$ l/s; (iii) fully rough for $Q \gtrsim 889$ l/s (transition in between).

Control volume. Take the fluid in the pipe between the inlet section 1 (uniform velocity $V$, pressure $p_1$) and the section 2 (velocity $u = 2V(1 - r^2/R^2)$, pressure $p_2$). The pipe is horizontal. Let $F$ be the friction force exerted by the wall on the fluid (equal and opposite to the friction force of the fluid on the wall).

Continuity check. Mean velocity at section 2:

$$\bar u = \frac{1}{\pi R^2}\int_0^R 2V\left(1 - \frac{r^2}{R^2}\right)2\pi r\,dr = \frac{4V}{R^2}\left[\frac{R^2}{2} - \frac{R^2}{4}\right] = V$$

So the discharge is the same at both sections.

Momentum flux at section 2. The momentum flux is $\rho\int u^2\,dA$, since $u$ is not uniform:

$$\int_0^R u^2\,2\pi r\,dr = 4V^2\cdot 2\pi\int_0^R\left(1 - \frac{r^2}{R^2}\right)^2 r\,dr = 8\pi V^2\left[\frac{R^2}{2} - \frac{R^2}{2} + \frac{R^2}{6}\right] = \frac{4}{3}\pi R^2V^2$$

(using $\int_0^R (r - 2r^3/R^2 + r^5/R^4)\,dr = R^2/2 - R^2/2 + R^2/6 = R^2/6$).

Momentum flux at section 1 (uniform): $\rho V^2\,\pi R^2$.

Momentum equation (flow direction). Net force = rate of change of momentum flux:

$$p_1\pi R^2 - p_2\pi R^2 - F = \rho\left(\frac{4}{3}\pi R^2V^2 - \pi R^2V^2\right) = \frac{\rho \pi R^2V^2}{3}$$ $$\boxed{F = \pi R^2\left(p_1 - p_2 - \frac{\rho V^2}{3}\right)}$$

This is the friction force of the wall on the fluid, and by Newton's third law it is also the friction drag on the pipe walls from the inlet to the section. The term $\rho V^2/3$ is the pressure drop used to build up the parabolic profile (increase in momentum flux), not lost by friction.

Idea. A pipe has the minimum energy loss when it is hydrodynamically smooth, since then $f$ depends only on $Re$ and the roughness adds nothing. The pipe stays smooth as long as the roughness is buried in the laminar sub-layer:

$$\frac{u_*k}{\nu} \le 4 \quad\Rightarrow\quad k \le \frac{4\nu}{u_*}$$

Data. $D = 1$ m, $Q = 1.0$ m³/s, $\nu = 1.0\times10^{-6}$ m²/s (water at about 20 °C).

Step 1: velocity and Reynolds number

$$V = \frac{Q}{\pi D^2/4} = 1.2732\ \text{m/s},\qquad Re = \frac{VD}{\nu} = 1273240$$

Step 2: smooth-pipe friction factor (Prandtl: $\dfrac{1}{\sqrt f} = 2\log_{10}(Re\sqrt f) - 0.8$)

$$f = 0.01118$$

Step 3: shear velocity

$$u_* = V\sqrt{\frac{f}{8}} = 1.2732\sqrt{\frac{0.01118}{8}} = 0.04759\ \text{m/s}$$

Step 4: permissible roughness

$$k \le \frac{4\nu}{u_*} = \frac{4(1.0\times10^{-6})}{0.04759} = 0.0840\ \text{mm}$$

Answer: The surface roughness height must not exceed about $0.084$ mm (0.08 mm) for minimum loss. With the looser limit $u_*k/\nu \le 5$, the value is 0.105 mm.

Data. $\nu = 0.6\times10^{-6}$ m²/s, $Q = 0.8$ m³/s, $D = 0.5$ m, same absolute roughness $k$ for the new pipe.

$$V = \frac{4Q}{\pi D^2} = 4.074\ \text{m/s},\qquad Re = \frac{VD}{\nu} = 3395305 > 10^6$$

So $f \propto (k/D)^{1/3}$ may be used.

Part 1: Diameter for half the pumping cost

Pumping cost is proportional to the power $P = \rho g Q h_f$. With $Q$ fixed, cost $\propto h_f$ (neglecting all losses except friction).

$$h_f = \frac{fLV^2}{2gD}, \qquad V = \frac{4Q}{\pi D^2} \;\Rightarrow\; h_f \propto \frac{f}{D^5}$$

For constant $k$: $f \propto \left(\dfrac{k}{D}\right)^{1/3} \propto D^{-1/3}$, so

$$h_f \propto D^{-1/3}\,D^{-5} = D^{-16/3}$$

Cost halved: $\left(\dfrac{D'}{D}\right)^{-16/3} = \dfrac{1}{2}$ gives

$$\frac{D'}{D} = 2^{3/16} = 1.1388 \;\Rightarrow\; D' = 0.5(1.1388) = 0.569\ \text{m} \approx 569\ \text{mm}$$

Check: new velocity $V' = 3.142$ m/s and $Re' = 2981506 > 10^6$, so the assumption holds.

Part 2: n equal pipes in parallel at the same Reynolds number

The same $Re = VD/\nu$ for the same fluid means $V'D' = VD$. Each of the $n$ pipes carries $Q/n$:

$$\frac{Q}{n} = \frac{\pi}{4}D'^2V' \;\Rightarrow\; \frac{\pi}{4}D^2V = n\frac{\pi}{4}D'^2V' = n\frac{\pi}{4}D'\,(VD)\;\Rightarrow\; D' = \frac{D}{n},\quad V' = nV$$

Relative roughness rises to $k/D' = n(k/D)$, so $f' = n^{1/3}f$. Head loss of each pipe (all pipes have the same $h_f$ in parallel):

$$h_f' = \frac{f'LV'^2}{2gD'} = \frac{n^{1/3}f\,L\,n^2V^2}{2g(D/n)} = n^{3+1/3}h_f = n^{10/3}h_f$$

The discharge is the same, so the power and the running cost increase by the factor

$$\boxed{\frac{\text{cost}'}{\text{cost}} = n^{10/3}}$$

For example, two pipes ($n = 2$) cost $2^{10/3} = 10.1$ times as much. Using one large pipe is much cheaper than several small ones.

Answer: $D' \approx 569$ mm (about 570 mm); running cost rises by $n^{10/3}$ for $n$ parallel pipes.