You need to consider all types of fluids. If you consider only incompressible fluids like water, then mass flow rate in - mass flow rate out = 0 (dm/dt=0). If you consider compressible fluid, dm/dt under certain conditions is not 0. In general, it is better to derive the formula based on compressible fluid.
Hey André! Do you remember the definition of derivatives using limits? It's pretty close (compare it and you'll see why using the negative sign). The only thing changing is which variable you are deriving in respect. Therefore, you have a partial derivative. Hope you'll understand!
When there are more than 1 independent variables then the derivative becomes partial derivative... here flow is varying with x,y,z and t(time) so the derivative of u with respect to x will be partial derivative and similarly applied for all cases
IN 2:35 ,I "think not sure" there is a dt that is the amount of time elapsed in : M^. in - M^. out =dM/dt, then will be omitted when dividing by dx dy dz and dt in 4:05 to give the same equation in 4:23 and so on. However, you have shown a clear, intensive, simplified and qualitative derivation.
You considered that velocity is changing in space along the dx, dy and dz but density is not! But finally at the end you put the density inside the divergence , how did you do that ? Density also can change in space.
The cube has some infinitesimally small edges of length dx, dy and dz respectively. So obviously you will have to ADD some the length, width and height of cube to reach the outlet face, right?
Confused. Checking with definitions in a couple popular fluid mechanics textbooks state "incompressible flow is where the density of the fluid remains constant". Could you elaborate on your definition please?
The flow is said to be incompressible when the div(vel)=0 this could happen in 2 cases 1)is when as you said, density=Const w.r.t time and space 2) when the divergence of velocity is zero, which in turn sets the temporal component of density to be zero ...to make it clear take out the continuity equation and then look at the case 1 and 2 that i have given. So in simple words, constant density flows are a special case of incompressible flows.
Or lets look at it this way, constant density flow is of a fluid that can not be compressed or expanded.....incompressible flow encompasses that case and also the case were a compressible/expandable fluid flows in such a way that there is no compression/expansion of the fluid(div(vel)=0)
***** I think there is a slight error in your deduction. First of all the continuity equation is formally given by d(rho)/dt + div(rho.v) = 0. Where rho is the local density (a scalar) and v is the velocity field (a vector function) and the (.) operator is simply multiplication. Now for steady state flows, i.e.: time independent flows, d(rho)/dt = 0, so div(rho.v) = 0. Expanding div(rho.v) one gets div(rho.v) = rho.div(v) + v*grad(rho). Here the (*) operator represents the dot product. Now second of all, getting to my point, one can define incompressible flows mathematically as div(v)=0. However this is only possible if the density of the fluid in the system is constant. This is because for time independent flows div(rho.v) = 0 = rho.div(v) + v*grad(rho) and so if div(v) =0 then rho.div(v) = 0 and consequently v*grad(rho) = 0. If v is not the null vector (which is the case when fluid is flowing) then grad(rho) =0 which implies constant density. However even for time dependent processes one can prove that if div(v) =0 then the fluid density must be constant. Let's prove this. The continuity equation for time dependent flow is given by d(rho)/dt + div(rho.v) = 0 = d(rho)/dt + rho.div(v) + v*grad(rho). If div(v) = 0 then rho.div(v) = 0 and so d(rho)/dt + v*grad(rho) = 0 = D(rho)/Dt, where D(rho)/Dt is the substantial derivative of the density rho. The physical meaning of this derivative is change of the material property (here rho) of a fluid element which follows the flow of the material (or moves with the substance as stated in Bird, Stewart and Lightfoot). If rho in any fluid element does not change as the element flows along with the fluid flow then the density of the fluid must be constant. Hereby again concluding that if div(v) = 0 then it follows that the density is constant.
"In fluid mechanics or more generally continuum mechanics, incompressible flow (isochoric flow) refers to a flow in which the material density is constant within a fluid parcel-an infinitesimal volume that moves with the flow velocity." - en.wikipedia.org/wiki/Incompressible_flow
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if there could be an academy award for the explanation of an equation I'd vote for it till my arms fell off
Incredibly well explained. Thank you for this!
Why does the mass difference is equal to dm /dt? Please elaborate. Thank you?
You need to consider all types of fluids. If you consider only incompressible fluids like water, then mass flow rate in - mass flow rate out = 0 (dm/dt=0). If you consider compressible fluid, dm/dt under certain conditions is not 0. In general, it is better to derive the formula based on compressible fluid.
Is this derivation different from the Reynolds Transport Theorem?
This was excellent!!
excellent explanation!
so much easier to understand than with the vectors stuff my professor uses. thanks!
I didn't understand the 4:42 step. How does it become 3 partial derivates?
Hey André! Do you remember the definition of derivatives using limits? It's pretty close (compare it and you'll see why using the negative sign). The only thing changing is which variable you are deriving in respect. Therefore, you have a partial derivative.
Hope you'll understand!
Perfect
lim dx-> 0 [( f(x) - f(x+dx))/dx] = -df(x)/dx
Thank you clever guy
Brazilian fellow. Just like me!
Isso aí! Feliz em ver gente interessada em estudar nesse nosso país.
When there are more than 1 independent variables then the derivative becomes partial derivative... here flow is varying with x,y,z and t(time) so the derivative of u with respect to x will be partial derivative and similarly applied for all cases
Excellent video. Clear and simple.
IN 2:35 ,I "think not sure" there is a dt that is the amount of time elapsed in : M^. in - M^. out =dM/dt, then will be omitted when dividing by dx dy dz and dt in 4:05 to give the same equation in 4:23 and so on.
However, you have shown a clear, intensive, simplified and qualitative derivation.
how did d(rho)/dt become partial derivative(rho)/partial derivative t
Well, it was never supposed to be a total derivative. Might've put the total derivative by mistake
that is why we come here youtube.. thank you very much it is very well explanation
Omg, what a perfect expanation,Thank you!
you use partial or complete derivative in continuity equation??
Thank you sir..That was soo smooth..👍
You considered that velocity is changing in space along the dx, dy and dz but density is not! But finally at the end you put the density inside the divergence , how did you do that ? Density also can change in space.
He considered density as a constant only in incompressible flows, yes density also can change in space but only in compressible flows
The best explanation !!
Awesome explanation
2019!!!! excellent explanation!
Can someone give me a proof in the lagrangian frame of reference?
crystal clear explanation. thank you
I didn't understand how in de mass outlet eqn in the x direction the velocity is at x +dx
The cube has some infinitesimally small edges of length dx, dy and dz respectively. So obviously you will have to ADD some the length, width and height of cube to reach the outlet face, right?
very good explanation!!
i cannot understand how mass equals to mass/time. 1:40
Great point. The dot over the m is there to indicate that it is a mass flow per unit time.
@@LearnChemE thanks!
Wow, incredible video. Thank you.
Great Video! Thank you!
thank u sir for excellent presentation
2:56 "u" shouldnt be volumetric flow it should be velocity!
u dy dz=dV/dt, i think that's the reason he said "density times volumetric flow is equal to mass flow".
Perfect video !!
Great explanation 😆 Thanks a lot
Amazing video thank you
Volume
i want Derivation of the Continuity Equation in cylindrical coordinate
just replace cubic dimensions by cylindrical dimensions and u will get your result..
Energy equation derivation plz
Thank you!
Excellent
Well done!!!!!
Please start making transport videos :-)
enlightening
Thanks
nice
amazing
Incompressible flow does not mean the density is constant.Please rectify that......
Confused. Checking with definitions in a couple popular fluid mechanics textbooks state "incompressible flow is where the density of the fluid remains constant". Could you elaborate on your definition please?
The flow is said to be incompressible when the div(vel)=0 this could happen in 2 cases 1)is when as you said, density=Const w.r.t time and space 2) when the divergence of velocity is zero, which in turn sets the temporal component of density to be zero ...to make it clear take out the continuity equation and then look at the case 1 and 2 that i have given. So in simple words, constant density flows are a special case of incompressible flows.
Or lets look at it this way, constant density flow is of a fluid that can not be compressed or expanded.....incompressible flow encompasses that case and also the case were a compressible/expandable fluid flows in such a way that there is no compression/expansion of the fluid(div(vel)=0)
*****
I think there is a slight error in your deduction. First of all the continuity equation is formally given by d(rho)/dt + div(rho.v) = 0. Where rho is the local density (a scalar) and v is the velocity field (a vector function) and the (.) operator is simply multiplication. Now for steady state flows, i.e.: time independent flows, d(rho)/dt = 0, so div(rho.v) = 0. Expanding div(rho.v) one gets div(rho.v) = rho.div(v) + v*grad(rho). Here the (*) operator represents the dot product. Now second of all, getting to my point, one can define incompressible flows mathematically as div(v)=0. However this is only possible if the density of the fluid in the system is constant. This is because for time independent flows div(rho.v) = 0 = rho.div(v) + v*grad(rho) and so if div(v) =0 then rho.div(v) = 0 and consequently v*grad(rho) = 0. If v is not the null vector (which is the case when fluid is flowing) then grad(rho) =0 which implies constant density.
However even for time dependent processes one can prove that if div(v) =0 then the fluid density must be constant. Let's prove this. The continuity equation for time dependent flow is given by d(rho)/dt + div(rho.v) = 0 = d(rho)/dt + rho.div(v) + v*grad(rho). If div(v) = 0 then rho.div(v) = 0 and so d(rho)/dt + v*grad(rho) = 0 = D(rho)/Dt, where D(rho)/Dt is the substantial derivative of the density rho. The physical meaning of this derivative is change of the material property (here rho) of a fluid element which follows the flow of the material (or moves with the substance as stated in Bird, Stewart and Lightfoot). If rho in any fluid element does not change as the element flows along with the fluid flow then the density of the fluid must be constant.
Hereby again concluding that if div(v) = 0 then it follows that the density is constant.
"In fluid mechanics or more generally continuum mechanics, incompressible flow (isochoric flow) refers to a flow in which the material density is constant within a fluid parcel-an infinitesimal volume that moves with the flow velocity."
- en.wikipedia.org/wiki/Incompressible_flow
thanks a alot!
dp/dt+ del(pv)=0
Worst style of writing 😒
You haven't seen handwriting from Americans it seems.
enlightening