Mass flow rate

Mass flow rate

The mass flow rate is the amount of mass that flows per second along a given streamline. The mass flow rate can be used if the velocity of a flowing fluid needs to be calculated at different points along a streamline with a variable diameter, this is often the case when the velocity in 1 point is known and the pressure is unknown. The mass flow rate follows equation (1) as shown below:

Where $\dot{m}$ $[\frac{kg}{m}]$ is equal to the mass flow rate, $\rho$ $[\frac{kg}{m^3}]$ to the density, $v$ $[\frac{m}{s}]$ to the velocity and $A$ $[m^2]$ to the cross sectional area of the pipe.

Assumptions for the equation

Equation (1) uses the same assumptions as the equation in Bernoulli's principle.

• The stream is steady, the velocity and density at any point can't change over time.
• The flow is incompressible, the density must have the same value at any point of the stream.
• Friction is negligible

Circular cross section

Since most pipes have a circular cross section, the cross sectional area $A$ is equal to $\frac{1}{4}\pi*D^2$. If this expression is substituted for the cross sectional area into equation (1), which stays constant along a streamline, it means that the velocity will increase with a decrease of the diameter, as long as the density remains the same. Because the diameter $D$ has a power of $2$, when the diameter becomes $3$ times smaller, the velocity will become $3^{2}$ larger.

Example problem

In the figure below a pipe is seen with a variable diameter. Let's say the velocity of the stream in point B needs to be calculated, the equation for mass flow rate can be used to do that.

The following values are known:

• $v_A = 10[\frac{m}{s}]$
• $D_A=2*D_B$

Figure 1. Visualization of a streamline along a pipe with a variable diameter

Equation (1) is used for 2 points along the streamline, where 1 point is point B and the other point is a point of which all values in equation (1) are known:

$\rho v_AA_A = \rho v_BA_B$ $A$ can be rewritten to $\frac{1}{4}\pi D^2$ so the new equation becomes:

$\rho v_A\frac{1}{4}\pi D^2_A = \rho v_B\frac{1}{4}\pi D^2_B$

Since $\rho, \frac{1}{4} \pi$ are on both sides of the equation they cancel out:

$v_AD^2_A = v_BD^2_B$

$v_A$ and $(\frac{D_B}{D_A})^2$ are already known thus the equation can be solved for $v_B$:

$v_B = v_A*(\frac{D_B}{2D_B})^2$

$v_B = 10*\frac{1}{4}=2.5\frac{m}{s}$

Now is shown how mass flow rate can be used to perform calculations on streamlines to calculate the velocity of a flow. The next reading give more information about streamlines.