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

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 ($m/s$)

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:

$$D_1^2{V}_1=D_2^2{V}_2$$

Solve for $V_2$:

$$V_2=\approx 7.99\text{m/s}$$

Energy equation

Definitions

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 ($m$)

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:

$$V_2\approx 7.99\text{m/s}$$

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 $\frac{P_2}{\gamma }$:

$$30.58+0.204=\frac{P_2}{\gamma }+3.25\Rightarrow \frac{P_2}{\gamma }=30.784-3.25=27.534$$

Convert back to pressure:

$$P_2=27.534\cdot 9810\approx 270000\text{Pa}=270\text{kPa}$$

Hydraulic grade line and Energy line

Hydraulic gradient (grade line)

Definitions

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

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 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)

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$L$, we can use following tricks

$$R=\frac{A}{P}=\frac{\pi D^2/4}{\pi D}=\frac{D}{4}$$

$$S=\frac{h_f}{L}$$

and

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

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:

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

Solve for $h_f$:

$$h_f=0.552\text{m}$$

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 $ft/s$)
• $C$ = 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 $ft$)
• $K_1$= 0.849 for SI units, 1.318 for USCS units

Similarly, to compute head loss (h_f) over a pipe length$L$, we can use following tricks

$$R=\frac{A}{P}=\frac{\pi D^2/4}{\pi D}=\frac{D}{4}$$

$$S=\frac{h_f}{L}$$

and

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

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:

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

Solve for $h_f$:

$$h_f\approx 0.0189\text{m}$$

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 $0.1$ respectively.

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