Fluid Mechanics and order of magnitude calculation

  • Thread starter Niles
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Hi

In my lecture notes we making some calculations and all terms [itex]\mathcal O(M^3)[/itex] are to be thrown away. Here M is the Mach number. Now, there is the expression (u denotes the velocity):
[tex]
uu\partial_t \rho \approx \rho_0 uu\nabla u
[/tex]
which in my notes are thrown away because they claim it is [itex]\mathcal O(M^3)[/itex]. But is it really true, I mean the derivative of u will not necessarily be on the same order as Ma, right?
 

olivermsun

Science Advisor
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That's right. The gradient also contains a length scale in each direction. In many cases one simply asserts on physical grounds that du/dx is same order as u (so the flow is sufficiently "smooth") or that there is some characteristic length scale of order one. Does the problem assume M << 1 and also no viscous effects?
 

boneh3ad

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Insights Author
Gold Member
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I'd also postulate that there is some more complicated order-of-magnitude analysis that can be done here a la that done in deriving Prandtl's boundary layer equations, but it would be difficult to carry that out without more information from the OP on what assumptions were made and what the physical situation is.
 

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