# Gas density in a piston

1. Oct 20, 2008

### stunner5000pt

1. The problem statement, all variables and given/known data
A gas filled pneumatic piston of a strut in a car suspension behaves like a piston apparatus. At one instant the piston is L from the closed end of the cylinder and the gas density is ρ. The piston is moving away from the closed end at v. The gas velocity varies linearly from zero at the closed end to velocity V at the piston. Find the rate of change of gas density at this instant. Also find the average density as a function of time.

2. The attempt at a solution
since there is a linear profile of the velocity, u=kx
u=0,y=0 and
u=V,y=L

we can use continuity equation
$$\frac{d \rho}{dt}=-\rho\frac{du}{dx}$$
$$\frac{d\rho}{dt}=-\rho k$$

For some reason the LaTex is not working...
dp/dt = -p du/dx
dp/dt = -pk

i used p for rho

Does this work till this step solve the first question?

can we just integrate and use the above conditions to get our density as a function of time?

Thank you for your input, it is greatly appreciated

2. Oct 24, 2008

### stunner5000pt

any suggestions??

I just need to know if i'm right or wrong that's all...

3. Oct 25, 2008

### stunner5000pt

$$\frac{d \rho}{dt}=-\rho\frac{du}{dx}$$
$$\frac{d\rho}{dt}=-\rho k$$

when i rearrange this equation i get

$$\int\frac{d\rho}{\rho}=-k\int dt$$

this presents a problem to integrate because I do not have limits the density or the time

all i am given is velocity, density and the distance of the piston to the cylinder head. The resulting expression from the above is

$$\rho\left(t\right)=C\exp\left(-kt\right)$$

Rate of change of hte density is $\frac{d\rho}{dt} =-k\rho$
using u=kx and u=V when x=L we can solve for k.

For the average density as a function of time...
How to solve for C??