- #1
benny_91
- 47
- 5
Hello!
While studying from the book 'Fluid Mechanics' by Cengel I came across the section which explains the behavior of fluids acting like a rigid body when the cylindrical container which contains the liquid rotates with a uniform angular velocity. Without much explanation the author states the governing equations using cylindrical co-ordinate system which are as follows:
$$\frac{\partial P}{\partial r}=\rho r \omega^{2} , \frac{\partial P}{\partial \theta}=0 ,
\frac{\partial P}{\partial z}=-\rho g$$
But when I tried to derive the first equation myself I ended up with something else. Here is my attempt to derive it. I am not including the diagram of the container so I request you to imagine the shape of the fluid element while I explain it. Let us assume that the density of the liquid is constant throughout.
As stated above, the variation in pressure with respect to angle in cylindrical co-ordinate system is zero. Hence let us imagine a cylindrical fluid element of radius 'r', thickness 'dr' and height 'dz'. Now we know that the pressure is the same at regions having same elevation and radius. Let the pressure be denoted by 'P'. Now we can do the force balance considering the pressure forces and the centrifugal force.
$$(P+\frac{\partial P}{\partial r}\frac{dr}{2})2\pi(r+\frac{dr}{2})dz - (P-\frac{\partial P}{\partial r}\frac{dr}{2})2\pi(r-\frac{dr}{2})dz = 2\pi r dr dz \rho r \omega^2$$
On simplifying this we get,
$$P dr+r\frac{\partial P}{\partial r} dr = \rho r^2 \omega^2 dr$$
Further simplification yields,
$$P+r\frac{\partial P}{\partial r} = \rho r^2 \omega^2$$
Finally this simplifies to
$$\frac{\partial}{\partial r}(Pr) = \rho r^2 \omega^2$$
This final result that we get has considerable difference from the one that is given in the book.
Please help me to find out where I have gone wrong in the derivation.
Thank you!
While studying from the book 'Fluid Mechanics' by Cengel I came across the section which explains the behavior of fluids acting like a rigid body when the cylindrical container which contains the liquid rotates with a uniform angular velocity. Without much explanation the author states the governing equations using cylindrical co-ordinate system which are as follows:
$$\frac{\partial P}{\partial r}=\rho r \omega^{2} , \frac{\partial P}{\partial \theta}=0 ,
\frac{\partial P}{\partial z}=-\rho g$$
But when I tried to derive the first equation myself I ended up with something else. Here is my attempt to derive it. I am not including the diagram of the container so I request you to imagine the shape of the fluid element while I explain it. Let us assume that the density of the liquid is constant throughout.
As stated above, the variation in pressure with respect to angle in cylindrical co-ordinate system is zero. Hence let us imagine a cylindrical fluid element of radius 'r', thickness 'dr' and height 'dz'. Now we know that the pressure is the same at regions having same elevation and radius. Let the pressure be denoted by 'P'. Now we can do the force balance considering the pressure forces and the centrifugal force.
$$(P+\frac{\partial P}{\partial r}\frac{dr}{2})2\pi(r+\frac{dr}{2})dz - (P-\frac{\partial P}{\partial r}\frac{dr}{2})2\pi(r-\frac{dr}{2})dz = 2\pi r dr dz \rho r \omega^2$$
On simplifying this we get,
$$P dr+r\frac{\partial P}{\partial r} dr = \rho r^2 \omega^2 dr$$
Further simplification yields,
$$P+r\frac{\partial P}{\partial r} = \rho r^2 \omega^2$$
Finally this simplifies to
$$\frac{\partial}{\partial r}(Pr) = \rho r^2 \omega^2$$
This final result that we get has considerable difference from the one that is given in the book.
Please help me to find out where I have gone wrong in the derivation.
Thank you!
Last edited:
You need to be careful: what is the shape of this element? Is the only radial pressure coming from those two sides?