Is there an equation depicting a compressible fluid flowing into a cone, increasing in pressure and velocity as the cone narrows? I'm trying to write an equation which shows a compressible fluid entering a cone with radius R and length L, where the pressure (density) and velocity of the compressible fluid increases as the exit radius R' narrows, a huge bonus would be given the mass of the fluid to also be able to calculate the overall mass of the fluid based on the volume of the cone.
Thank you so much in advance. Also, this is not for homework. It's for a pet project I'm working on.
 A: You don't need to create an equation out of nothing. What you're asking can be combined by a couple of equations. Anyway,  there is a  concept mistake in your question, and it is pressure can't increase when velocity increases for an incompressible fluid in laminar flow. The equation which relates the fluid's velocity to the section area of the tube it is flowing in can be derived from the continuity equation for fluids. A fluid moving in a tube with transversal section $A_1$ with velocity $v_1$ and then the tube narrows to $A_2$, the fluid will have another velocity $v_2$. This four quantities are then related by
$$A_1v_1=A_2v_2$$
For a cone, the area of the transversal section will be $A=\pi r^2$, where $r$ is the radius at a given instant. Pressure comes into play in the Bernouilli equation for fluids
$$\frac{1}{2}\rho v^2+\rho g z+P=\mbox{constant}$$
where $\rho$ is density,  $P$ is the pressure and $z$ is the height. This can be formulated with respect to two points; if they are at the same height then the $\rho g z$ term cancels out:
$$\frac{1}{2}\rho v_1^2+P_1=\frac{1}{2}\rho v_2^2+P_2$$
You can combine this last equation with the first one to relate $v_1$ and $v_2$ and obtain the principle I was talking to you in the beginning. When the area decreases, so does the pressure, but velocity increases. I hope this answers your question.
A: 
Assume a constant volumetric throughput $Q$, of an incompressible fluid, then the continuity equation holds:
$$\frac{v_1}{A_2}=\frac{v_2}{A_1}$$
where the $A$ are the cross sections and $v$ the flow-speeds.
Now we can apply Bernoulli's principle:
$$p_1+\frac{\rho v_1^2}{2}+\rho gz_1=p_2+\frac{\rho v_2^2}{2}+\rho gz_2$$
with $p$ the pressure and $z$ the height.
Assuming $\rho g(z_1-z_2) \approx 0$, then:
$$p_2-p_1=\frac{\rho(v_1^2-v_2^2)}{2}$$
A: I hope this answer does not come too late.
There are indeed a few relationships, assuming subsonic, frictionless, isentropic, uniform, steady and ideal gas flow. Also, if you are getting your air from the atmosphere, the stagnation properties $p_0$, $\rho_0$ and $T_0$ will be those of the atmospheric conditions. Assuming constant specific heat ratio, as well ($k = 1.4$).
The first relevant equation, relates the cross-section area with velocity:
$\frac{dv}{v} = -\frac{dA}{A} \frac{1}{1 - \mathrm{Ma}^2}$
($v$ is velocity, and $\mathrm{Ma}$ is the Mach number, this is, the ratio between velocity and the speed of sound)
Temperature, pressure and density can be obtained by:
$\frac{T}{T_0} = \frac{1}{1 + ((k - 1) / 2) \mathrm{Ma}^2}$
$\frac{p}{p_0} = (\frac{1}{1 + ((k - 1) / 2) \mathrm{Ma}^2})^{k/(k-1)}$
$\frac{\rho}{\rho_0} = (\frac{1}{1 + ((k - 1) / 2) \mathrm{Ma}^2})^{1/(k-1)}$
Please notice how these equations have, implicit, the ideal gas relationship:
$\rho = \frac{p M}{R T}$
(It is possible you use a definition of $R$ equivalent to my $R/M$, which would render $\rho = \frac{p}{R T}$.)
Notice also, the relationship:
$p / \rho^k =$ constant
Now, it all depends on what measurements and/or estimations you can have for your system. Don't forget that for steady flow, the mass flow rate, $\dot{m} = \rho A v$, will have the same value on any cross-section. (Note that, for incompressible flow, this equation is easily converted to the $A_1 v_1 = A_2 v_2$, that the other answers mention.)
If any of the assumptions of this answer do not apply to you, I am afraid you will need help from someone else - it will be beyond my knowledge (I am a low Mach person). If you can access a Fluid Mechanics text book (undergrad is probably enough), you will find further info on this.
Two remarks:

*

*You mention increasing pressure, however, from the equations above, when area decreases, velocity increases and, with increasing velocity, pressure decreases.


*To compute the total mass in the cone you can obtain, from the presented equations, a relationship between density and cross-section area. With that, you can integrate and obtain the mass in the system.
(I've got the $\mathrm{Ma}$ equations from Munson, Fundamentals of Fluid Mechanics.)
