Pipe hydraulics

This chapter covers the following:

• Reynolds number and moody diagram,
• Velocity and shear stress distribution in a pipe
• Pump characteristics
• Power and efficiency equations
• Scaling laws for pumps
• Pipes in series and parallel
• Pipe network problems

Reynolds number and Moody diagram

Reynolds number

Definitions

The Reynolds number is a dimensionless parameter used to characterize the flow regime in a pipe.

The Reynolds number is a dimensionless quantity that helps you identify the flow regime in a pipe:

$$\text{Re}=\frac{\rho VD}{\mu }=\frac{VD}{\nu }$$

Where:

• $\rho$: fluid density ($kg/m^3$)
• $V$: average flow velocity ($m/s$)
• $D$: pipe diameter ($m$)
• $\mu$: dynamic viscosity ($Pa\cdot s$)
• $\nu$: kinematic viscosity ($m^2/s$)

Flow regimes for pipe flow:

• Laminar: $\text{Re}<2100$
• Transitional: $2100<\text{Re}<10000$
• Turbulent: $\text{Re}>10000$

Moody diagram

The Moody diagram summarizes how the Darcy friction factor depends on:

• Reynolds number ($Re$)
• Friction factor ($f$)
• Relative roughness $\frac{ϵ}{D}$

For laminar flow, the friction factor is:

$$f=\frac{64}{\text{Re}}$$

Velocity and shear stress distribution in a pipe

Velocity distribution (laminar flow)

For fully developed laminar flow in a circular pipe, the velocity profile is parabolic:

$$v(r)=V_{\text{max}}\left(1-\frac{r^2}{R^2}\right)$$

Where:

• $r$: radial distance from the center
• $R$: pipe radius
• $V_{max}$: maximum velocity at the centerline
• $V_{avg}=\frac{1}{2}{V}_{max}$

Shear stress distribution

In fully developed laminar pipe flow, shear stress varies linearly with radius:

$$\tau (r)={\tau }_w\frac{r}{R}$$

Where ${\tau }_w$ is the wall shear stress.

Pump characteristics

Pumps add energy to fluids. Their performance is commonly described using these curves:

• Head vs Flow rate curve
• Power vs Flow rate
• Efficiency vs Flow rate

Total dynamic head

Total dynamic head combines pressure head, velocity head, and elevation head changes across the pump:

$$H=\frac{p_{\text{out}}-p_{\text{in}}}{\rho g}+\frac{V_{\text{out}}^2-V_{\text{in}}^2}{2g}+z_{\text{out}}-z_{\text{in}}$$

Net positive suction head available (${\text{NPSH}}_A$)

Net positive suction head available is written as:

$${\text{NPSH}}_A=H_{pa}+H_s-\sum h_L-H_{vp}=\frac{P_{\text{inlet}}}{\rho g}+\frac{v_{\text{inlet}}^2}{2g}-\frac{P_{\text{vapor}}}{\rho g}$$

where:

• $H_{pa}$ = atmospheric pressure head on the surface of the liquid in the sump ($ft$ or $m$)
• $H_s$ = static suction head of liquid (height of the surface of the liquid above the centerline of the pump impeller) ($ft$ or $m$)
• $\sum h_L$ = total friction losses in the suction line ($ft$ or $m$)
• $H_{vp}$ = vapor pressure head of the liquid at the operating temperature ($ft$ or $m$)
• $v$ = fluid velocity at pump inlet
• $P_{vapor}$ = fluid vapor pressure at pump inlet
• $\rho$ = fluid density

Power and efficiency equations

Hydraulic power (water horsepower):

Hydraulic power is the rate at which the pump adds energy to the fluid:

$$P_{hyd}=\gamma QH$$

Where:

• $Q$: flow rate ($m^3/s$)
• $H$: total head ($m$)

Pump efficiency:

Pump efficiency compares hydraulic power delivered to the fluid with the input (brake) power:

$$\eta =\frac{P_{\text{hyd}}}{P_{\text{input}}}$$

Where:

• $\eta$: efficiency
• $P_{input}$: input (brake) power to the pump

Example:

Given:

• Flow rate $Q=0.05{\text{m}}^3/\text{s}$
• Head $H=20\text{m}$
• Fluid: water $\rho =1000{\text{kg/m}}^3$
• Input power $P_{\text{input}}=15\text{kW}$

Hydraulic power

$$P_{hyd}=\gamma QH=1000\times 9.81\times 0.05\times 20=9810\text{W}=9.81\text{kW}$$

Efficiency

$$\eta =\frac{P_{hyd}}{P_{input}}=\frac{9.81}{15}=0.654=65.4\%$$

Scaling laws for pumps (affinity laws)

The affinity laws relate the performance of similar pumps (or the same pump under different operating conditions):

$${\left(\frac{Q}{N{D}^3}\right)}_2={\left(\frac{Q}{N{D}^3}\right)}_1$$

$${\left(\frac{\dot{m}}{\rho N{D}^3}\right)}_2={\left(\frac{\dot{m}}{\rho N{D}^3}\right)}_1$$

$${\left(\frac{H}{N^2{D}^2}\right)}_2={\left(\frac{H}{N^2{D}^2}\right)}_1$$

$${\left(\frac{P}{\rho N^2{D}^2}\right)}_2={\left(\frac{P}{\rho N^2{D}^2}\right)}_1$$

$${\left(\frac{\dot{W}}{\rho N^3{D}^5}\right)}_2={\left(\frac{\dot{W}}{\rho N^3{D}^5}\right)}_1$$

where:

• $Q$ = volumetric flow rate
• $\dot{m}$ = mass flow rate
• $H$ = head
• $P$ = pressure rise
• $\dot{W}$ = power
• $\rho$ = fluid density
• $N$ = rotational speed
• $D$ = impeller diameter

Subscripts 1 and 2 refer to different but similar machines or to different operating conditions of the same machine.

Pipes in series and parallel

Pipes in series

Pipes are said to be in series if they are connected end-to-end, so the flow rate is the same through every pipe.

Main characteristics

• Same discharge (Q) through each pipe.
• Head loss adds up over the entire length.

Total Head Loss

If three pipes with head losses $h_{f1},h_{f2},h_{f3}$ are in series:

$$H_f=h_{f1}+h_{f2}+h_{f3}$$

Using the Darcy-Weisbach equation:

$$h_f=f\cdot \frac{L}{D}\cdot \frac{V^2}{2g}$$

Where:

• $f$: Darcy friction factor
• $L$: Length of pipe
• $D$: Diameter of pipe
• $V$: Velocity of flow

Example:

Three pipes are connected in series with the following properties:

Pipe	Length (m)	Diameter (m)	Friction factor
1	100	0.3	0.02
2	200	0.3	0.02
3	150	0.3	0.02

If $Q=0.2m^3/s$, calculate total head loss.

Solution:

1. Calculate velocity:

   $$V=\frac{Q}{A}=\frac{Q}{\frac{\pi D^2}{4}}$$

2. Apply Darcy-Weisbach for each pipe.

3. Sum all head losses.

Pipes in parallel

Pipes are said to be in parallel if they connect the same two points, so the total flow splits among the branches.

Main characteristics

• Same head loss in each branch.
• Total discharge is the sum of individual discharges.

If:

$$Q=Q_1+Q_2+Q_3$$

and:

$$h_{f1}=h_{f2}=h_{f3}$$

Then each pipe must satisfy:

$$h_f=f_i\cdot \frac{L_i}{D_i}\cdot \frac{V_i^2}{2g}$$

and:

$$Q_i=A_i\cdot V_i$$

Example:

Two pipes are in parallel between two tanks with a head difference of 10 m.

Pipe	Length (m)	Diameter (m)	Friction factor
1	100	0.2	0.02
2	100	0.3	0.02

Calculate discharge through each pipe.

Solution:

1. Assume head loss $h_f=10\text{m}$ (same for both).

2. Use Darcy-Weisbach to find $V_1,V_2$.

3. Compute:

   $$Q_1=A_1{V}_1,Q_2=A_2{V}_2,Q=Q_1+Q_2$$

Pipe network problems

Flow rules in pipe networks

1. At any junction, the total inflow must be equal to the total outflow.

$$\sum Q_{\text{in}}=\sum Q_{\text{out}}$$

1. The loss of head due to flow in a clockwise direction around a loop must be equal to the loss of head due to flow in a counterclockwise direction.

$$\sum h_{\text{clockwise}}=\sum h_{\text{counterclockwise}}$$

Hardy cross method (loop method)

Pipe networks involve a combination of series and parallel pipes. The Hardy Cross method is often used to solve these networks by iteratively correcting assumed flow rates.

Assumptions

• Initial guess for flow in each loop.
• Continuity and energy conservation laws are used.

Correction Formula:

$$\Delta Q=-\frac{\sum (h_f\cdot r)}{n\cdot \sum (r\cdot |Q|)}$$

Where:

• $\Delta Q$: correction to assumed flow
• $h_f$: head loss in each pipe
• $r$: resistance term $=\frac{fL}{D^5}$
• $Q$: flow in the pipe

Steps:

1. Assume flow in each loop.
2. Compute head losses.
3. Apply correction $\Delta Q$.
4. Repeat until convergence.