Fluid flow and the continuity equation
Flow rates
There are three closely related ways to express the amount of fluid flowing in a system per unit time. The first one is the volume flow rate ($Q$) which expresses the volume of fluid flowing through a system in given amount of time. For example, a household faucet has a volume flow rate of about 2 gallons per minute. If you close the drain in your sink and leave your faucet open for one minute your sink will contain about 2 gallons of water.
The volume flow rate through a section of pipe can be related to the pipe’s cross-sectional area ($A$) and the average flow velocity ($v$) by the relation
\[Q=A\times v\]Two quantities closely related to the volume flow rate are the weight flow rate ($W$) and the mass flow rate ($M$) which represent the weight or mass of the fluid flowing in the system per unit time.
| Symbol | Name | Definition | SI | USCS |
|---|---|---|---|---|
| $Q$ | Volume flow rate | $Q=Av$ | $\m^3/\s$ | $\ft^3/\s$ |
| $W$ | Weight flow rate | $W=\gamma Q$ $W=\gamma A Q$ | $\N/\s$ | $\lb/\s$ |
| $M$ | Mass flow rate | $M=\rho Q$ $M=\rho A V$ | $\kg/\s$ | $\slug/\s$ |
The standard units for volume flow rates are $\m^3/\s$ and $\ft^3/\s$ but are quite large. We will therefore more frequently use liters per minute (L/min) and gallons per minute (gpm). Some useful conversion factors that are available on our table of conversion factors include the following:
\[\frac{\text{449 gal/min}}{\ft^3/\s}\qquad \frac{\text{60,000 L/min}}{\m^3/\text{s}}\qquad \frac{\text{3.785 L/min}}{\text{gal/min}}\qquad\]A 2½-inch discharge outlet on a fire hydrant has a flow rate of 500 gpm. Convert this flow rate to L/min and to $\m^3/\s$. Then estimate the average flow velocity in units of $\ft/\s$.
First, convert the 500 gpm to L/min using the volume flow rate conversion factors listed above,
\[500~\cancel{\text{gal/min}}\times \left(\frac{\text{3.785 L/min}}{\cancel{\text{gal/min}}}\right)=1,892.5~\text{L/min}\]Next, convert this result into $\m^3/\s$,
\[1,892.5~\cancel{\text{L/min}}\times\left(\frac{\text{m}^3/\s}{60,000~\cancel{\text{L/min}}}\right)=0.0315~\m^3/\s\]In order to find the average flow velocity we make use of the definintion of the volume flow rate, $Q=Av$, and solve this for the flow velocity
\[v=\frac{Q}{A}\]Since we want the flow velocity in units of $\ft/\s$ first convert the volume flow rate to $\ft^3/\s$:
\[500~\cancel{\text{gal/min}}\times \frac{\ft^3/\s}{449~\cancel{\text{gal/min}}}=1.114~\ft^3/\s\]The cross sectional area of the 2½-inch outlet is
\[A=\frac{\pi}{4}\left(0.208~\ft\right)^2 =0.0341~\ft^2\]The average flow velocity is therefore
\[v=\frac{Q}{A}=\frac{1.114~\ft^3/s}{0.0341~\ft^2}=33~\ft/\s\]Continuity equation
The continuity equation is a consequence of the conservation of mass. Consider the pipe system shown below. If the flow is steady (velocity, pressure and density do not change with time) the mass of fluid that enters the pipe from the left must equal the mass of fluid that exits from the right.
We therefore must have that
\[M_1=M_2\]using the definition that $M=\rho A v$ we have that
\[\rho_1 A_1 v_1=\rho_2 A_2 v_2\]If the fluid is incompressible then $\rho_1 = \rho_2$ and the density cancels from each side of the above expression. We are therefore left with the
Continuity equation for steady incompressible flow
\[A_1 v_1=A_2 v_2\]In the system shown below the inside diameter of the pipe at section 1 is 50 mm and 100 mm at section 2. Water at room temperature is flowing with an average velocity of 8 m/s at section 1. What is the velocity at section 2?
To solve this problem we first need to recognize that the volume flow rate is constant throughout the pipe and can therefore use the continuity equation for steady incompressible flow
\[A_1 v_1=A_2 v_2\]Dividing both sides for $v_2$ we find
\[v_2=\frac{A_1}{A_2} v_1\]The ratio of areas in the above expression is
\[\frac{A_1}{A_2}=\frac{\cancel{\frac{\pi}{4}}(50~\mm)^2}{\cancel{\frac{\pi}{4}}(100~\mm)^2}=\frac{1}{4}=0.25\]Notice that the factor of $\pi/4$ cancels out in the numerator and denominator. We therefore have
\[v_2=0.25 v_1=0.25\times(8~\m/\s)=2~\m/\s\]Commercially available piping
First, the difference between pipe and tubes. Tubing is generally used for structural purposes and therefore tends to be more expensive. The sizing is typically specified by the tube’s outer diameter and wall thickness. Piping is used to transports fluids (gases or liquids). Sizing for pipes is based on a nominal diameter and schedule number. In many instances the terms piping and tubing are used interchangeably. This class will be interested in piping.
ASTM A53 standard
The ASTM A53 standard is the most common standard for carbon steel pipes. It will also be the standard that we use most frequently, if not exclusively, in this class. Here are two steel pipe data tables that we will use very often and therefore worth bookmarking:
Schedule 40 Steel Pipe Data Table
Schedule 80 Steel Pipe Data Table
Here are some additional references in case you come across other types of commercially available pipes:
PVC and CPVC Pipes Schedule 40 & 80
Copper, Wrought iron, Cast iron
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