Equation of continuity in euler' s form.

     

 let ρ be the density of the fluid at p(x,y,z)  and let q=(u,v,w)be the velocity. 

Construct a small parallelopiped taking the point P as centre with edges of length h,k and l parallel to  the coordinate axes which is occupied by fluid.

the mass of fluid that flows across the unit area parallel to yz plane through  P in time δt                                                                        =ρuδt

                                                  =f(x,y,z)δt….say

                                                                         

      

we know that ,mass=volume×density                        and here volume=Area×distance                     velocity= →distance=velocity×δt=uδt and area=1 unit               ∴volume=1×uδt=uδt , so mass=uδt×ρ=ρuδt      since ρ and u depend upon the variable of x,y and z so they both can be considered as f(x,y,z)

                      

     let P₁ and P₂  be the projection of P at the faces kl near to and away from the origin respectively where the coordinate of

it' s because P is the centre of the parallelopiped and the length along x axis is h and following figure

             

   due to only projection of P ,P₁&P₂ do not change the coordinate of y and z it only change x coordinate here P₁ lies on faces kl near to origin and P₂  lies on faces kl but away from origin

                      the x coordinate of P₁=ON-MN

                                                        

                 the x coordinate of P₂=ON+NA

                                                

To find the flow across the face kl of the parallelopiped nearest to origin ,take a point          

 on the surface having area dη.dξ at this point 

then the mass of fluid that flows across the area dη.dξ in time δt 

   [∵using Taylor^' s expansion for several variables neglecting higher order]      

    The mass of the fluid that flows across the faces kl near to the origin in x↑direction in time δt, 

The mass of the fluid that flows across the faces kl away from the origin in x↑direction  in time δt, 

 

therefore the increase in mass inside the parallelopiped in time δt due to the flow across these two faces in x↑direction is

similarly the excess of mass of fluid that flow in across the hl faces near to the origin over the mass that flows out across the hl faces away from the origin in y↑direction in time δt 

 and the excess of mass of fluid that flow in across the kh faces near to origin over the mass flows out across the kh faces away from the origin in z↑direction in time δt,

  

hence the total mass of fliud that flows in the parallelopiped over the mass that flows out of it in time δt,

on the other hand ,the initial mass of the fluid inside the parallelopiped=ρhkl 

and the increase in mass of fluid inside the parallelopiped in time δt

 now by principal of continuity,(1)and (2)are equal. i.e

    which is the required equation of continuity in euler' s form.                                                              

Information about fluid

Fluid :-

          Fluid is a substance which is capable to flow.

Classification of fluid :

                                                                  

Isotropic substance

                    Fluid is treated as isotropic substance which implies that physical property(i.e pressure, density etc) is same in all direction.

Anisotropic substance

                    Fluid is treated as anisotropic substance if that  property is not same in all direction.

Newtonian fluid

The fluid which obeys Newtonian relationship between shearing stress and gradient of velocity given by

                        

where  
 is the velocity gradient and τ is shearing stress.

Ideal fluid

Ideal fluid is a Newtonian fluid in which there exist no shearing or tangential stress but exist only normal or direct stress between two contacting layers.

Note that in nature ideal fluid doesn’t occur. They are only as mathematical concept.

Real  fluid

Real fluid is a Newtonian fluid in which there exist both tangential and normal stress between two contacting layers.

A uniform solid cylinder is placed with its axis horizontal on a plane, whose inclination to the horizon is α. Show that the least coefficient of the friction between it and the plane, so that it may roll and not slide, is 1/3 tanα. If the cylinder be hollow and of small thickness, the least value is 1/2 tanα.

Solution: In this figure 1 ,

 a uniform solid cylinder is placed horizontally on the plane BOCD Which makes an angle α with the horizon. Now we make another cross sectional figure from side view only to make our calculation easy.
we get figure 2.

 Let O be the initial point from which the cylinder rolls down. The supposition related to the cylinder are :
                                                       C→centre, M→mass and a→radius
 The plane OD makes an angle α with horizon then we get ∡PCT=α from this following figure and reasons.

 The weight mg act downwardly and it makes an angle α with ck and when we decompose it we get the force along ck is Mgcosα and the force along ct is Mgsinα.
Let CA be the line fixed in the body and the normal line ck in the space and let ∡Ack=θ and F& R be the friction and normal reaction acting on a plane respectively.
 Now, after time t, let the cylinder rolls in x-axis(it is supposed) so that F & Mgsinα act along x axis and R & Mgcosα act along y-axis and covers a distance x from initial point O at k.
 Then arc Ak = Ok = x
                    aθ = x                      [ ∵θ=arc/radius ]
differentiating w.r.to t upto second order
we get (d^2 x)/(dt^2 ) = a(d^2 θ)/(dt^2 )………………..(1)
 We have the equation of motion are M (d^2 x)/(dt^2 ) = Sx …(2)
                                                           M (d^2 y)/(dt^2 ) = Sy …(3)
            and also
 we have
 Mk^2 (d^2 θ)/(dt^2 ) = the forces along x axis in opposite direction × radius = F.a …(4)
From (1) & (2)
Or M a(d^2 θ)/(dt^2 ) = Mgcosα + (-F) [∵ the forces along x axis are mgcosα and (-f) due to acting                                                                    opposite direction]
Multiplying both sides by k^2
                                  Or M a(d^2 θ)/(dt^2 ) k^2 = k^2Mgcosα - k^2 F
                                 Or F(a^2+k^2) = k^2Mgcosα
                                  Or F= (k^2/(k^2+a^2 ))Mgcosα...(5)
From (3)
                         M (d^2 y)/(dt^2 ) = Mgcosα + (-R)    [∵ the forces along x axis are Mgcosα and (-R)                                                                                             due to acting opposite direction]
                    or, 0=Mgcosα -R                      [∵ the body moves only in one direction]
                    or, R=Mgcosα...(6)
Hence from (5) and (6)
        F/R=(k^2/(k^2+a^2 ))tanα
If the cylinder rolls then,
    F/R<m(coefficient of friction)
i.e (k^2/(k^2+a^2 ))tanα<m
here we have the cylinder is solid so k^2=a^2/2  so {(a^2/2)/  (a^2/2+a^2)}tanα<mi.e
                                                        (1/3)tanα<m
and if the cylinder is hollow then k^2=a^2 so, (a^2/a^2+a^2)tanα<m i.e 
                                                        (1/2)tanα<m