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I'm trying to figure out how the mathematics of conservation of mass in compressible flow works for a simple setup. I posted this problem in hw physics section, and the conversation turned to the physics model and appears to have gone kaput. This was supposed to be a mathematics question (but its difficult to untangle completely from the physics), so it went to the wrong forum.
In the top position the system is static. The mass of air in the pipe at ##t=0## is ##M_i##, then the piston starts to advance pushing air out of the outlet ##o##. The density ##\rho## of the flow across the remainder of the tube will vary in the coordinate ##u##, and time ##t##. the origin of ##u## is to ride along with the piston and will always range from ##0 \leq u \leq l-x(t)##.
The properties ( density and velocity) are distributed uniformly across any particular section ( I'm ignoring realities of what the flow properties look like in transverse directions)
With the subscript ##o## referring to the properties at the fixed location of the outlet. At any fixed time ##t## to get the mass inside the control volume I get that:
$$ M_{cv} = \int_0^{u(t)} A~ \rho (u,t)~du $$
Also, at any time ##t## the rate of change of mass in the control volume is:
$$ \frac{d}{dt}M_{cv} = - A \rho_o(t) v_o(t) $$
So it seems (to me) like I can combine those two equations:
$$ \frac{d}{dt} \left( \int_0^{u(t)} A~ \rho (u,t) ~du \right) = - A \rho_o(t) v_o(t) $$
Now, ##u(t)## ( the upper limit of integration on the LHS) at any fixed time ##t## can be found from the position of the piston w.r.t. to origin:
$$ u(t) = l- x(t) = l - \int_0^t \dot x ~dt $$
My question is:
If I haven't already committed mathematical heresy, is there any way to get ## \rho (u,t)## as a function of all the other variables? It seems like the variable of ##u## is trapped in the integral in some kind of partial integration ( I'm out of my depth on the proper nomenclature), because the variable of the integration is not itself a function of time, just the upper limit. I'm all twisted up on this...
Thanks for any help untangling this mess I'm making.
In the top position the system is static. The mass of air in the pipe at ##t=0## is ##M_i##, then the piston starts to advance pushing air out of the outlet ##o##. The density ##\rho## of the flow across the remainder of the tube will vary in the coordinate ##u##, and time ##t##. the origin of ##u## is to ride along with the piston and will always range from ##0 \leq u \leq l-x(t)##.
With the subscript ##o## referring to the properties at the fixed location of the outlet. At any fixed time ##t## to get the mass inside the control volume I get that:
$$ M_{cv} = \int_0^{u(t)} A~ \rho (u,t)~du $$
Also, at any time ##t## the rate of change of mass in the control volume is:
$$ \frac{d}{dt}M_{cv} = - A \rho_o(t) v_o(t) $$
So it seems (to me) like I can combine those two equations:
$$ \frac{d}{dt} \left( \int_0^{u(t)} A~ \rho (u,t) ~du \right) = - A \rho_o(t) v_o(t) $$
Now, ##u(t)## ( the upper limit of integration on the LHS) at any fixed time ##t## can be found from the position of the piston w.r.t. to origin:
$$ u(t) = l- x(t) = l - \int_0^t \dot x ~dt $$
My question is:
If I haven't already committed mathematical heresy, is there any way to get ## \rho (u,t)## as a function of all the other variables? It seems like the variable of ##u## is trapped in the integral in some kind of partial integration ( I'm out of my depth on the proper nomenclature), because the variable of the integration is not itself a function of time, just the upper limit. I'm all twisted up on this...
Thanks for any help untangling this mess I'm making.