Momentum Flux: Quick Question (Product Rule)

andrew.c
Messages
46
Reaction score
0

Homework Statement


Heya,
I can expand the term on the LHs and get the term on the RHS no problem :)
But,
I don't understand how the bottom line is arrived at; I think its basically just a backward engineering of the product rule, but I can't get it!

Any help would be useful
 

Attachments

  • momentum flux.JPG
    momentum flux.JPG
    15.1 KB · Views: 450
Physics news on Phys.org
[tex]\frac{\partial(\rho u u)}{\partial x} = \rho u \frac{\partial u}{\partial x} + u \frac{\partial (\rho u)}{\partial x}.[/tex]
 
Thanks
That was easier than I thought!
 

Similar threads

Replies
23
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K