problems with divergence


by Summer2442
Tags: divergence
Summer2442
Summer2442 is offline
#1
Jan16-13, 11:58 AM
P: 8
Hello,

I am new to calculus, and am having problems with divergence, I was reading something to explain the physical interpretation of divergence and i got stuck in the very first part.

it says that if we have a small volume dxdydz at the origin, and that a fluid flowing into this volume from the positive x-direction per unit time, the the rate of flow in is
= ρvx|x=0 = dy dz,
where ρ is the density at (x, y, z), and vx is the velocity of the fluid in the x-direction, what does "|x=0 = dy dz" part mean.

Thanks Alot.
Phys.Org News Partner Science news on Phys.org
Better thermal-imaging lens from waste sulfur
Hackathon team's GoogolPlex gives Siri extra powers
Bright points in Sun's atmosphere mark patterns deep in its interior
Erland
Erland is offline
#2
Jan16-13, 04:59 PM
P: 302
"|x=0" probably means that the function, here ρvx, should be evaluauted at x=0. So they imagine the volume element as a cuboid with one vertex at the origin and the sides as dx, dy, and dz in the posotive directions. One face of the cuboid is then contained in the plane x=0.

The flux into the volume through this face is then ρvxdydz, so something seems wrong in what you wrote anyway.

Btw, you wrote "from the positive x-direction", but I assume that you meant "along the positive x-direction", so that positive vx is directed to the right.
Summer2442
Summer2442 is offline
#3
Jan16-13, 05:57 PM
P: 8
yes what i meant is "along the positive x-direction"

but about the flux into the volume, i agree that it should be 'ρvxdydz' but i am pretty sure this is what the book says 'ρvx|x=0 = dy dz', i think its a mistake.

I have another question please, it then says that the flux out of the opposing face is
ρvx|x=dx dydz
which i understand but then it equates this equation with the following
[ρvx + ∂ (ρvx)/∂x dx] dydz
which i do not understand, how was the partial derivative introduced and why?

thanks,

Erland
Erland is offline
#4
Jan17-13, 12:56 PM
P: 302

problems with divergence


Quote Quote by Summer2442 View Post
yes what i meant is "along the positive x-direction"

but about the flux into the volume, i agree that it should be 'ρvxdydz' but i am pretty sure this is what the book says 'ρvx|x=0 = dy dz', i think its a mistake.
Yes, it is a typo. It should be ρvx|x=0dxdy.

Quote Quote by Summer2442 View Post
I have another question please, it then says that the flux out of the opposing face is
ρvx|x=dx dydz
which i understand but then it equates this equation with the following
[ρvx + ∂ (ρvx)/∂x dx] dydz
which i do not understand, how was the partial derivative introduced and why?

thanks,
The second expression is a linear approximation of the first one.

Recall that (ρvx|x=dx - ρvx|x=0)/dx ≈ ∂(ρvx)/∂x|x=0.
engnr_arsalan
engnr_arsalan is offline
#5
Jan17-13, 01:57 PM
P: 29
i think it is talking about mass flow rate... density X velocity X area=flow rate..or more precisely mass flow rate..it says rho X V X x such that at x=0 area is dy dz..for differential element the area perpendicular to flow in x direction is dydz...
the second thing is taylor's theorem is being applied here..which means when fluid has flowed a length dx its mass flowrate has changed depnding upon dx..that is why patial derivative is introduced here..
Summer2442
Summer2442 is offline
#6
Jan17-13, 05:24 PM
P: 8
Ok I get it now, thanks guys.


Register to reply

Related Discussions
Why the divergence of a diagram when superficial degree of divergence D=0 is Ln(lambd Quantum Physics 2
Divergence Theorem - Confused :s (2 problems) Calculus 4
Can someone double check to see if I did these problems correctly? E-Flux problems Introductory Physics Homework 3