Mass and energy conservation

This chapter covers the following:

• Continuity equation
• Energy equation
• Hydraulic grade line and energy line
• Head loss due to flow
• Minor losses in pipe fittings, contractions, and expansions

Continuity equation

Definitions

Continuity equation

The continuity equation is based on the conservation of mass law. It states that for an incompressible fluid flowing in a closed conduit (pipe), the mass flow rate must remain constant from one cross-section to another.

For incompressible flow, a constant mass flow rate means a constant volumetric flow rate. That gives the continuity relationship:

$$Q=A_1{V}_1=A_2{V}_2\Rightarrow D_1^2{V}_1=D_2^2{V}_2$$

Where:

• $Q$ = volumetric flow rate ($m^3/s$)
• $A_1,A_2$ = cross-sectional areas at sections 1 and 2 ($m^2$)
• $V_1,V_2$ = fluid velocities at sections 1 and 2 ($m/s$)
• $D_1,D_2$ = diameter of circular pipe at sections 1 and 2 ($m$)

Example

Water flows through a pipe that narrows from a diameter of 0.30 m to 0.15 m. If the velocity in the larger section is 2 m/s, what is the velocity in the narrower section?

Given:

• $D_1=0.30\text{m},V_1=2\text{m/s}$
• $D_2=0.15\text{m}$

Solution:

Apply the continuity equation and solve for $V_2$:

$$V_2={\left(\frac{D_1}{D_2}\right)}^2{V}_1={\left(\frac{0.30}{0.15}\right)}^2\cdot 2=4\cdot 2=8\text{m/s}$$

Energy equation

Definitions

Energy equation

The energy equation is derived from the conservation of energy law and applies to steady, incompressible flow along a streamline.

The energy equation (often introduced as Bernoulli’s equation with pumps, turbines, and losses) balances energy per unit weight between two sections:

$$\frac{P_1}{\gamma }+\frac{V_1^2}{2g}+z_1+h_p-h_t=\frac{P_2}{\gamma }+\frac{V_2^2}{2g}+z_2+h_L$$

Where:

• $P$ = pressure ($Pa$)
• $\gamma$ = specific weight of fluid ($N/m^3$)
• $V$ = velocity ($m/s$)
• $z$ = elevation head ($m$)
• $h_L$ = head loss ($m$)
• $h_p$ = head added by the pump into the system ($m$)
• $h_t$ = head taken by the turbine from the system ($m$)

Exam tip: Keep pressures on a consistent basis - if $P_1$ is given as gauge pressure, $P_2$ must come out gauge as well; don’t mix gauge and absolute values in the same equation. It also helps to carry a few extra decimal places through the intermediate steps, since rounding too early can shift the final answer.

Example

Water flows through a horizontal pipe that narrows from 0.30 m to 0.15 m in diameter. If the pressure in the wider section is 300 kPa and velocity is 2 m/s, find the pressure in the narrower section.

Given:

• $D_1=0.30\text{m},V_1=2\text{m/s},P_1=300\text{kPa}$
• $D_2=0.15\text{m},z_1=z_2$ (horizontal pipe)

From the previous example, we know:

$$V_2=8\text{m/s}$$

Apply Bernoulli (ignoring elevation):

$$\frac{P_1}{\gamma }+\frac{V_1^2}{2g}=\frac{P_2}{\gamma }+\frac{V_2^2}{2g}$$

Substitute the given values and solve for $\frac{P_2}{\gamma }$:

$$30.58+0.204=\frac{P_2}{\gamma }+3.262\Rightarrow \frac{P_2}{\gamma }=30.784-3.262=27.522$$

Convert back to pressure:

$$P_2=27.522\cdot 9810\approx 270000\text{Pa}=270\text{kPa}$$

Hydraulic grade line and energy line

Hydraulic gradient (grade line)

Definitions

Hydraulic grade line

The hydraulic grade line represents the sum of pressure head and elevation head at any point in a pipe.

The hydraulic grade line (HGL) tracks how the pressure head plus elevation head changes along a pipe. A practical way to picture it is with piezometers: if you installed piezometer tubes along the pipe, the HGL would connect the water levels in those tubes.

$$HGL=z+\frac{p}{\gamma }$$

Energy gradient (grade line)

Definitions

Energy grade line

The energy grade line represents the total head, including velocity head.

The energy grade line (EGL) includes everything in the HGL plus the velocity head. In other words, it represents the total head above a horizontal datum.

The difference between the EGL and the HGL is the velocity head term $\frac{v^2}{2g}$.

$$EGL=z+\frac{p}{\gamma }+\frac{v^2}{2g}$$

Head loss due to flow

Definitions

Head loss

Head loss is the reduction in total mechanical energy due to friction and turbulence as fluid flows through a pipe.

In fluid mechanics, head loss is the drop in the fluid’s total mechanical energy (head) as it moves through a pipe or conduit. The main causes are friction along the pipe wall and turbulence in the flow.

Types of head loss

There are two primary types of head loss:

1. Major head loss - due to friction in straight pipe sections.
2. Minor head loss - due to fittings, bends, valves, entrances, exits, etc.

Major head loss (Darcy-Weisbach equation)

A common way to compute friction loss in straight pipe flow is the Darcy-Weisbach equation:

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

Where:

• $h_f$ = head loss ($m$)
• $f$ = Darcy friction factor
• $L$ = length of the pipe ($m$)
• $D$ = diameter of the pipe ($m$)
• $v$ = average velocity of fluid ($m/s$)

Example

Problem: Water flows through a 50 m long, 0.1 m diameter pipe with a velocity of 2 m/s. The Darcy friction factor is 0.02. Calculate the major head loss.

Given:

• $L=50\text{m}$
• $D=0.1\text{m}$
• $v=2\text{m/s}$
• $f=0.02$

Solution:

Using the Darcy-Weisbach equation:

$$h_f=0.02\cdot \frac{50}{0.1}\cdot \frac{2^2}{2\cdot 9.81}$$

$$h_f=2.04\text{m}$$

Answer: The major head loss is 2.04 m.

Head loss due to flow (Manning’s and Hazen-Williams equations)

Two other widely used empirical equations to estimate head loss are:

• Manning’s equation (typically used for open channel flow and full-flowing circular pipes)
• Hazen-Williams equation (commonly used for pressurized pipe flow with water)

Watch the units: Manning’s $K$ (1.486 for USCS units, 1.0 for SI units) and Hazen-Williams $K_1$ (1.318 for USCS units, 0.849 for SI units) must match the unit system of the given data - using the wrong $K$ silently produces a plausible-looking wrong answer. All examples on this page use SI units.

Manning’s equation

Manning’s equation estimates the velocity of flow in an open channel or full-flowing pipe:

$$Q=\frac{K}{n}A{R}^{2/3}{S}^{1/2}$$

$$V=\frac{K}{n}{R}^{2/3}{S}^{1/2}$$

Where:

• $V$ = velocity of flow ($m/s$ or $ft/s$)
• $n$ = Manning’s roughness coefficient
• $R$ = hydraulic radius ($m$ or $ft$), $R=\frac{A}{P}$
• $S$ = slope of the energy grade line
• $A$ = cross-sectional area of flow ($m^2$ or $f{t}^2$)
• $P$ = wetted perimeter ($m$ or $ft$)
• $K$ = 1.486 for USCS units, 1.0 for SI units

To compute head loss ($h_f$) over a pipe length $L$ for a full-flowing circular pipe, it helps to rewrite $R$, $S$, and $A$ in terms of the pipe diameter and length. These relations are reused in the Hazen-Williams example below as well:

$$R=\frac{A}{P}=\frac{\pi D^2/4}{\pi D}=\frac{D}{4},S=\frac{h_f}{L},A=\frac{\pi }{4}{D}^2$$

Example

A 300 mm diameter concrete pipe ($n=0.013$) is flowing full. Flow velocity is 1.5 $m/s$, and the pipe length is 200 m.

$$V=\frac{1}{n}{\left(\frac{D}{4}\right)}^{2/3}{\left(\frac{h_f}{L}\right)}^{1/2}$$

Substitute the given values:

$$1.5=\frac{1}{0.013}{\left(\frac{0.3}{4}\right)}^{2/3}{\left(\frac{h_f}{200}\right)}^{1/2}$$

Simplify the constant term and isolate the square-root term:

$$1.5=76.92\cdot 0.1778\cdot {\left(\frac{h_f}{200}\right)}^{1/2}=13.68\cdot {\left(\frac{h_f}{200}\right)}^{1/2}$$

$${\left(\frac{h_f}{200}\right)}^{1/2}=\frac{1.5}{13.68}\approx 0.1097$$

Square both sides and solve for $h_f$:

$$h_f\approx (0.1097{)}^2\cdot 200\approx 2.41\text{m}$$

Hazen-Williams equation

Hazen-Williams equation is used for water flow in pressurized pipes:

$$V=K_1\cdot C\cdot R^{0.63}\cdot S^{0.54}$$

$$Q=K_1\cdot C\cdot A\cdot R^{0.63}\cdot S^{0.54}$$

Where:

• $V$ = velocity of flow ($m/s$ or $ft/s$)
• $C$ = Hazen-Williams roughness coefficient (C-factor)
• $R$ = hydraulic radius ($m$ or $ft$), $R=\frac{A}{P}$
• $S$ = slope of the energy grade line
• $A$ = cross-sectional area of flow ($m^2$ or $f{t}^2$)
• $P$ = wetted perimeter ($m$ or $ft$)
• $K_1$ = 0.849 for SI units, 1.318 for USCS units

The same relations for $R$, $S$, and $A$ from the Manning’s equation section above apply here as well.

Example

Water flows through a 150 mm diameter PVC pipe ($C=140$) at a rate of 0.03 $m^3/s$ over 100 m.

$$Q=K_1\cdot C\cdot \left(\frac{\pi }{4}{D}^2\right)\cdot {\left(\frac{D}{4}\right)}^{0.63}\cdot {\left(\frac{h_f}{100}\right)}^{0.54}$$

Substitute the given values:

$$0.03=0.849\cdot 140\cdot \left(\frac{\pi }{4}0.15^2\right)\cdot {\left(\frac{0.15}{4}\right)}^{0.63}\cdot {\left(\frac{h_f}{100}\right)}^{0.54}$$

Simplify the constant term and isolate ${\left(\frac{h_f}{100}\right)}^{0.54}$:

$$0.03=0.2655\cdot {\left(\frac{h_f}{100}\right)}^{0.54}\Rightarrow {\left(\frac{h_f}{100}\right)}^{0.54}=\frac{0.03}{0.2655}\approx 0.1130$$

Since the exponent on the right is $0.54$, raise both sides to the power $\frac{1}{0.54}$ to solve for $h_f$:

$$\frac{h_f}{100}=(0.1130{)}^{1/0.54}\approx 0.0176\Rightarrow h_f\approx 1.76\text{m}$$

This is consistent with the standard SI Hazen-Williams head-loss form, $h_f=\frac{10.67\cdot L\cdot Q^{1.852}}{C^{1.852}\cdot D^{4.87}}$, which gives the same result.

Minor losses in pipe fittings, contractions, and expansions

Minor losses are calculated using the formula:

$$h_m=K_m\cdot \frac{v^2}{2g}$$

Where:

• $h_m$ = minor head loss (m)
• $K_m$ = loss coefficient (depends on fitting type)
• $v$ = velocity of fluid (m/s)

For sharp exit, protruding pipe entrance, sharp entrance, and round entrance the value of $K_m$ will be $1.0$, $0.8$, $0.5$, and $0.04$ respectively.

Total head loss $h_L$ will be major loss $h_f$ plus $h_m$, therefore, $h_L=h_f+h_m$.

Example

Water flows at 3 m/s through a 100 m long, 0.2 m diameter pipe with a Darcy friction factor of 0.02. The pipe has a sharp entrance ($K_m=0.5$). Find the total head loss.

Given:

• $L=100\text{m},D=0.2\text{m},v=3\text{m/s},f=0.02,K_m=0.5$

Solution:

First, find the major head loss:

$$h_f=f\cdot \frac{L}{D}\cdot \frac{v^2}{2g}=0.02\cdot \frac{100}{0.2}\cdot \frac{3^2}{2\cdot 9.81}=10\cdot 0.459\approx 4.59\text{m}$$

Then find the minor head loss from the entrance, using the same $\frac{v^2}{2g}$ term:

$$h_m=K_m\cdot \frac{v^2}{2g}=0.5\cdot 0.459\approx 0.23\text{m}$$

Add the two together for the total head loss:

$$h_L=h_f+h_m=4.59+0.23\approx 4.82\text{m}$$

Answer: The total head loss is 4.82 m.