Continuity Equation for Fluids in Different Coordinate Systems

Last edited: 2024-11-12 17:40:30

The continuity equation for fluids can be applied to problems like calculating the outlet velocity of a pipe with varying diameters. It can also be expressed in different coordinate systems such as the cylindrical and spherical coordinate systems. Let us explore the properties of the different formulas for the continuity equation for fluids.

Continuity Equation For Steady Flow

Let us assume a fluid moves in a steady flow through a pipe as in the figure above, meaning the mass flow ($\dot{m}$) into $A_1$ is the same as the mass flow out of $A_2$. This leads to the following formula:

$$\dot{m} = \rho_1 v_1 A_1 = \rho_2 v_2 A_2,$$

where $v_1$ and $v_2$ are the inlet and outlet velocities respectively. $A_1$ is the area of the inlet and $A_2$ is the area of the outlet. $\rho_1$ is the density of the fluid at the inlet and $\rho_2$ is the density of fluid at the outlet. To understand the formula you can think of the velocity $v$ as a height per second and $A$ as a base area so when the two multiply you get a volume per second. $\rho$ is the density so that gives us mass per second, aka the mass flow.

Continuity Equation Compressible Flow

If the fluid is compressible meaning that the density changes when the pressure changes then the equation above should be used since when the area, as an example, decreases the density and/or velocity have to increase to maintain the equality in the formula.

Continuity Equation Incompressible Flow

If the fluid is incompressible, which water essentially is, then density is constant ($\rho_1 = \rho_2$) so we can cancel it out on both sides and get:

$$v_1 A_1 = v_2 A_2.$$

So if the area decreases the velocity has to increase. You most likely have experienced this effect when putting your thumb over the end of a water hose. When you do that the area decreases and you get a high velocity flow exiting the hose.

Continuity Equation in Differential Form

The continuity equation in differential form for fluids looks like this:

$$\frac{\partial \rho}{\partial t} = - \nabla \cdot (\rho \vec{u}),$$

where $\rho$ is the density, $t$ is time, and $\vec{u}$ the flow velocity vector field. If you would like a derivation of this formula then check out our post on the continuity equation in electromagnetism and just replace the charge density $\rho$ with the fluid density and the current density $\vec{j}$ with $\rho \vec{u}$.

Continuity Equation in Cartesian Coordinates

The continuity equation in cartesian coordinates $(x,y,z)$ is

$$\frac{\partial \rho}{\partial t} + \frac{\partial (\rho u)}{\partial x} + \frac{\partial (\rho v)}{\partial y} + \frac{\partial (\rho w)}{\partial z} = 0,$$

where $u$, $v$, and $w$ are the flow velocity in the $x$-, $y$-, and $z$-direction respectively.

Continuity Equation in Cartesian Coordinates With Incompressible Flow

If the fluid is incompressible, $\rho$ is constant then the formula becomes:

$$\rho \left(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} \right) = 0,$$

since $\frac{\partial \rho}{\partial t}=0$

Continuity Equation in Cylindrical Coordinates

The continuity equation for cylindrical coordinates $(r,\theta,z)$ is

$$\frac{\partial \rho}{\partial t} + \frac{1}{r} \frac{\partial(\rho r u_r)}{\partial r} + \frac{1}{r} \frac{\partial(\rho u_\theta)}{\partial \theta} + \frac{\partial(\rho u_z)}{\partial z} = 0,$$

where $u_r$, $u_\theta$, and $u_z$ are the flow velocity in the $r$-, $\theta$-, and $z$-direction respectively.

Continuity Equation in Cylindrical Coordinates With Incompressible Flow

If the fluid is incompressible, $\rho$ is constant then the formula becomes:

$$\rho \left( \frac{1}{r} \frac{\partial(r u_r)}{\partial r} + \frac{1}{r} \frac{\partial(u_\theta)}{\partial \theta} + \frac{\partial u_z}{\partial z} \right) = 0.$$

Continuity Equation in Spherical Coordinates

The continuity equation for spherical coordinates $(r,\theta,\varphi)$ is

$$\frac{\partial \rho}{\partial t} + \frac{1}{r^2} \frac{\partial(\rho r^2 u_r)}{\partial r} + \frac{1}{r \sin{\theta}} \frac{\partial(\rho u_\theta \sin{\theta})}{\partial \theta} + \frac{1}{r \sin{\theta}} \frac{\partial(\rho u_\varphi)}{\partial \varphi} = 0,$$

where $u_r$, $u_\theta$, and $u_\varphi$ are the flow velocity in the $r$-, $\theta$-, and $\varphi$-direction respectively.

Continuity Equation in Spherical Coordinates With Incompressible Flow

If the fluid is incompressible, $\rho$ is constant then the formula becomes:

$$\rho \left( \frac{1}{r^2} \frac{\partial (r^2 u_r)}{\partial r} + \frac{1}{r \sin{\theta}} \frac{\partial(u_\theta \sin{\theta})}{\partial \theta} + \frac{1}{r \sin{\theta}} \frac{\partial(u_\varphi)}{\partial \varphi} \right) = 0.$$