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

Exam tip: The equations in this chapter - Reynolds number, the Moody diagram friction factor, Darcy-Weisbach, the pump affinity laws, and the Hardy Cross correction - all appear in the FE Reference Handbook’s fluid mechanics section. The tested skill is locating and correctly applying the right equation under time pressure, not memorizing it.

Reynolds number and Moody diagram

Reynolds number

Definitions

Reynolds number

A dimensionless parameter used to characterize the flow regime (laminar, transitional, or turbulent) in a pipe.

You calculate the Reynolds number as:

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

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: $\text{Re}>10000$

Example: Reynolds number calculation

Water ($\nu =1.0\times 10^{-6}{\text{m}}^2/\text{s}$) flows at $V=2\text{m/s}$ through a pipe with $D=0.05\text{m}$.

$$\text{Re}=\frac{VD}{\nu }=\frac{2\times 0.05}{1.0\times 10^{-6}}=100,000$$

Answer: $\text{Re}=100,000$, so the flow is turbulent.

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:

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

For turbulent flow, the exam typically hands you the Moody diagram itself: you locate the Reynolds number on the horizontal axis, trace up to the curve matching the pipe’s relative roughness, and read the friction factor off the vertical axis.

Example: reading a friction factor off the Moody diagram

Water flows through a pipe with $\text{Re}=5\times 10^5$ and relative roughness $ϵ/D=0.001$.

Locate $\text{Re}=5\times 10^5$ on the horizontal axis, trace up to the curve labeled $ϵ/D=0.001$, and read across to the friction factor axis.

Answer: $f\approx 0.02$

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:

$${\text{NPSH}}_A=H_{pa}+H_s-\sum h_L-H_{vp}$$

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 $m$)

When $P_{\text{inlet}}$ is the actual absolute pressure measured at the pump suction - which already reflects the atmospheric pressure, static suction head, and friction losses upstream - ${\text{NPSH}}_A$ can equivalently be expressed directly in terms of the pressure and velocity at the pump inlet:

$${\text{NPSH}}_A=\frac{P_{\text{inlet}}}{\rho g}+\frac{v_{\text{inlet}}^2}{2g}-\frac{P_{\text{vapor}}}{\rho g}$$

Watch out: $P_{\text{inlet}}$ and $P_{\text{vapor}}$ must both be absolute pressures - mixing a gauge value for one with an absolute value for the other gives a wrong (and often negative) ${\text{NPSH}}_A$. This same discipline applies anywhere you combine pressure and head terms: keep every term in one unit system (SI or USCS) throughout a calculation, converting up front if the problem gives you a mix.

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: pump hydraulic power and efficiency

Given (SI units):

• 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 $P_{\text{input}}=15\text{kW}$

Specific weight: $\gamma =\rho g=1000\times 9.81=9810{\text{N/m}}^3$

Hydraulic power:

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

Efficiency:

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

Answer: $\eta =65.4\%$

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.

The Handbook prints the head group as $\frac{H}{N^2{D}^2}$; written as a true dimensionless group it is $\frac{gH}{N^2{D}^2}$. Because $g$ is the same at both conditions, it cancels, so either form gives $H_2=H_1{\left(\frac{N_2}{N_1}\right)}^2{\left(\frac{D_2}{D_1}\right)}^2$.

Example: pump affinity law scaling

A pump operating at $N_1=1750\text{rpm}$ delivers $Q_1=500\text{gpm}$ at $H_1=100\text{ft}$. If the pump speed increases to $N_2=2000\text{rpm}$ with the same impeller diameter, find the new flow rate and head.

Since $D$ is unchanged, ${\left(\frac{Q}{N{D}^3}\right)}_2={\left(\frac{Q}{N{D}^3}\right)}_1$ reduces to $Q_2=Q_1\cdot \frac{N_2}{N_1}$:

$$Q_2=500\times \frac{2000}{1750}=571\text{gpm}$$

Similarly, ${\left(\frac{H}{N^2{D}^2}\right)}_2={\left(\frac{H}{N^2{D}^2}\right)}_1$ reduces to $H_2=H_1\cdot {\left(\frac{N_2}{N_1}\right)}^2$:

$$H_2=100\times {\left(\frac{2000}{1750}\right)}^2=131\text{ft}$$

Answer: $Q_2=571\text{gpm}$, $H_2=131\text{ft}$

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: head loss for pipes in series

Three pipes are connected in series with the following properties (SI units):

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 the total head loss.

Velocity (the same in every pipe, since $D$ doesn’t change):

$$V=\frac{Q}{A}=\frac{0.2}{\frac{\pi (0.3{)}^2}{4}}=\frac{0.2}{0.0707}=2.83\text{m/s}$$

Head loss in each pipe, from $h_f=f\cdot \frac{L}{D}\cdot \frac{V^2}{2g}$:

$$h_{f1}=0.02\times \frac{100}{0.3}\times \frac{(2.83{)}^2}{2(9.81)}=2.72\text{m}$$

$$h_{f2}=0.02\times \frac{200}{0.3}\times \frac{(2.83{)}^2}{2(9.81)}=5.44\text{m}$$

$$h_{f3}=0.02\times \frac{150}{0.3}\times \frac{(2.83{)}^2}{2(9.81)}=4.08\text{m}$$

Answer: $H_f=2.72+5.44+4.08=12.24\text{m}$

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: discharge through pipes in parallel

Two pipes run in parallel between two tanks with a head difference of $10\text{m}$ (SI units):

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

Calculate the discharge through each pipe.

Velocity in each pipe, solving $h_f=f\cdot \frac{L}{D}\cdot \frac{V^2}{2g}=10\text{m}$ for $V$:

$$V_1=\sqrt{\frac{10\times 2(9.81)}{0.02\times (100/0.2)}}=\sqrt{\frac{196.2}{10}}=4.43\text{m/s}$$

$$V_2=\sqrt{\frac{10\times 2(9.81)}{0.02\times (100/0.3)}}=\sqrt{\frac{196.2}{6.67}}=5.42\text{m/s}$$

Discharge in each pipe:

$$Q_1=A_1{V}_1=\frac{\pi (0.2{)}^2}{4}\times 4.43=0.139{\text{m}}^3/\text{s}$$

$$Q_2=A_2{V}_2=\frac{\pi (0.3{)}^2}{4}\times 5.42=0.383{\text{m}}^3/\text{s}$$

Answer: $Q=Q_1+Q_2=0.139+0.383=0.522{\text{m}}^3/\text{s}$

Pipe network problems

Flow rules in pipe networks

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

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

2. 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.

The head loss in each pipe follows a head-loss law of the form $h_f=r{Q}^n$, where $r\propto \frac{fL}{D^5}$ is a resistance term and $n$ is the exponent in that law. For the Darcy-Weisbach equation used elsewhere in this chapter, $h_f=f\frac{L}{D}\frac{V^2}{2g}$ is proportional to $Q^2$, so $n=2$.

Correction formula:

$$\Delta Q=-\frac{\sum (r{Q}^n)}{n\cdot \sum (r\cdot |Q{|}^{n-1})}$$

Where:

• $\Delta Q$: correction to assumed flow
• $r$: resistance term, $r\propto \frac{fL}{D^5}$
• $Q$: flow in the pipe (signed, based on assumed direction)
• $n$: exponent in the head-loss law $h_f=r{Q}^n$; $n=2$ for Darcy-Weisbach

Steps:

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