What is (∂u ∕∂x)dx in the expression for mass flow rate?

[Madonmaths]

Hi

I have seen the expression for mass flow rate in one of the problems I am working on. I used to simply apply the expression for calculating the mass flow rate with respect to the position as (ρu + (∂u ∕∂x)dx) dy dz). ρ, u are density and velocity component respectively.

I would like to understand the exact meaning of multiplication of (∂u ∕∂x)dx.
I expect a good explanation as soon as possible.

Thanks in advance.
Madguy

 

[Astronuc]

Hi

I have seen the expression for mass flow rate in one of the problems I am working on. I used to simply apply the expression for calculating the mass flow rate with respect to the position as (ρu + (∂u ∕∂x)dx) dy dz). ρ, u are density and velocity component respectively.

I would like to understand the exact meaning of multiplication of (∂u ∕∂x)dx.
I expect a good explanation as soon as possible.

Thanks in advance.
Madguy

(ρu + (∂u ∕∂x)dx) should be ρ(u + (∂u ∕∂x)dx), if the fluid is incompressible or (ρu + (∂ρu ∕∂x)dx) if ρ = ρ(x), which would imply the fluid is compressible (or changing temperature).

(∂u ∕∂x)dx = du, and ∂u ∕∂x is just the velocity gradient, i.e. u is not constant with x (position or displacement). This could be because the fluid is compressible or the cross-section (normal to flow) is changing.

 

[Madonmaths]

Hi Astronyc

Thanks for your explanations!