anonymous
  • anonymous
ques
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1440220440808:dw| This is a volume element of in an imcompressible fluid flowing at a distance x,y,z from the respective axes In the book it is given coordinates of P as x,y,z but shouldn't they be (x+dx),(y+dy),(z+dz) or (dx),(dy),(dz) if we consider the element at origin Let \[\vec V=V_{x}\hat i+V_{y}\hat j+V_{z}\hat k\] be the velocity of fluid at a point P(x,y,z) Mass of fluid flowing through face ABCD in unit time is \[=V_{x}(dy.dz)\] Now the problem is right in starting, this equation is dimensionally incorrect, the left side is having a unit of mass but right side is jut made up of velocity and length elements
anonymous
  • anonymous
Maybe what they mean by P(x,y,z) is \[\vec V \equiv \vec V(x,y,z)\] A vector point function \[\vec V=V_{x}(x,y,z)\hat i+V_{y}(x,y,z)\hat j+V_{z}(x,y,z)\hat k\] But using a notation like this makes u think the coordinates of P are x,y,z but that's impossible if the length is dx,dy,dz how can it have coordinates x,y,z It must have either (dx,dy,0) or (x+dx,y+dy,0) Depending on where we taking our volume element (point P lies in the x-y plane, so z=0)
anonymous
  • anonymous
@ganeshie8 @IrishBoy123

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ganeshie8
  • ganeshie8
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anonymous
  • anonymous
,
IrishBoy123
  • IrishBoy123
In the Gospel of Mary, V is defined as \(\vec V = \rho \vec v\) with units \(\frac{kg}{m^2 \ s}\) and so flow rate through "unit area" of surface with normal \(\hat n\) is \(\vec V \bullet \hat n\) which has units \(\frac{kg}{m^2 \ s}.\frac{m^2}{1}=\frac{kg}{s}\) http://www.amazon.co.uk/Mathematical-Methods-Physical-Sciences-Mary/dp/0471365807 it's only 3-4 pages so i'll copy them if you are using a different text. [and as it will come with a recommendation to buy, i reckon the copyright lawyers will forgive me.!]
anonymous
  • anonymous
and how does \[V_{x}((x+dx),y,z)=V_{x}+\frac{\partial V_{x}}{\partial x}.dx\]
IrishBoy123
  • IrishBoy123
does this help?
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anonymous
  • anonymous
thanks, that was a lot helpful

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