Derive the Continuity Equation in Spherical Coordinates

Consider a two-dimensional incompressible flow field. The two dimensional polar coordinates are 'r' and θ angle subtended by elements is 'dθ' . Consider a fluid elements ABCD between radius 'r' and (r + dr) dθ .

The components of velocity in radial direction Vr ad tangential direction Vθ.

Assume thickness of element perpendicular to the plane of paper is unity.


Consider flow in radial direction:

Mass if fluid entering the face AB during the time dt,

Fluid influx = Density x Velocity in r-direction x Area x Time

ρ . Vr (rdθ x 1) dt             Thickness = 1

Mass fo fluid leaving the face CD during the same time dt is,

Fluid efflux  = ρ ( Vr + ∂ vr / ∂r dr ) (r + dr) dθ . dt

Mass accumulated in element = Fluid influs-Fluid efflux

ρVr . rdθ . dt - [ ∂vr rdθdt + ρ ∂vr / ∂r . r. drdθ . dt + ρVr . dr . d . dt + ρ dvr / dr .  d2 rdθdt ]

Neglect d2r is too small,

=  - [ ρVr. dr. dθ . dt + ρ dV / dr . r. dr . dθ . dt

=  - ρ [ Vr + dVr / dr ] dr . dθ .dt

=  - ρ [ Vr / r + dV / dr ] rdr . dθ .dt

This is written in this form because is the volume f elemtn.
Consider flow in tangential direction,

= [Mass of fluid entering BC- Mass of fluid leaving CD] dt

=  [ ρ Vθ . dr x - ρ (Vθ + ∂Vθ / ∂θ . dθ) dr ] dt

=  [ ρ Vθ drdt - ρ Vθ drdt - ∂Vθ / ∂θ . dθ. drdt ]

=  -  ∂Vθ / ∂θ ρdθ . drdt

=  - ∂ / ∂θ  Vθ rdθ . drdy / r    ( Multiplying and divide by r)

=  - 1/r Vθ / ∂θ (rdr/ dθ / dt)

According law of conservation of mass, "The total gain mass be equal to the rate of change of luid mass in element ABCD".

Total gain of mass = Mass accumulate due to radial + Mass accumulate due to tangential

=  - [ ρVr + ρ dV / dr ] drdθ . dt + [ ρ ∂Vθ  / ∂θ . dθ. dr .dt ]

But, mass of fluid element =  ρ x volume of fluid element.

= ρ [ rdθ . drdt ]

Rate of increases of fluid mass in element with time,

=  ∂ / ∂t [ ρ rdθ + (r+dr) d / 2  dr  dt

=   ∂ / ∂t ρ ( rdθ / 2 + rdθ / 2 + drdθ / 2) drdt

dr x dθ  = 0

By law of conservation,

∂ / ∂t   ( ρ. r. dθ. dr) dt  = - [ ρVr + ρr dVr / dr. r + ρ ∂Vθ / ∂θ ] dr. dθ .dt

Consider dt = 1,

∂ / ∂t   ( ρ. r. dθ. dr) + [ ρVr  + r dVr / dr. r + ρ ∂Vθ / ∂θ ] dr. dθ .dt = 0

For steady flows         ∂ / ∂t  = 0

[ ρVr + ρ ∂ / ∂r (Vr) r + ∂ / ∂θ ] dr. dθ = 0

For incompressible flow, ρ  = constant.

Vr + ∂ / ∂r (Vr) + ∂ / ∂θ . (Vθ)  = 0

Or

∂ / ∂r (r. Vr) + ∂  / ∂ . Vθ = 0

Which is the equation of continuity in polar co-ordinates for two dimensional, steady incompressible flow.

For more help in Continuity Equation in a Polar Form click the button below to submit your homework assignment
θ ρ   ∂

gregorpaped1989.blogspot.com

Source: https://www.tutorhelpdesk.com/homeworkhelp/Fluid-Mechanics-/Continuity-Equation-in-a-Polar-Form-Assignment-Help.html

0 Response to "Derive the Continuity Equation in Spherical Coordinates"

Post a Comment

Iklan Atas Artikel

Iklan Tengah Artikel 1

Iklan Tengah Artikel 2

Iklan Bawah Artikel