Flow fluid about a line source

Problem Statement

Consider the symmetric radial flow of an incompressible, inviscid fluid outward from an infinitely long uniform source, coincident with the z-axis of a cylindrical coordinate system. Fluid is being generated at a volumetric rate Г per unit length of source.

a. Show that the Laplace equation for the velocity potential for this system is:

1r∂∂r (r ∂∅∂ r )=0

b. From the equation find the velocity potential, velocity and pressure as functions of position:

∅=−Γ2π

ln (r ) V r=Γ2 πr

Ρ∞−Ρ=ρΓ 2

8π 2r2

Where Ρ∞ is the value of the modified pressure far away from the source.

c. Discuss the applicability of the results in (b) to the flow about a well drilled into a large body of porous rock.

d. Sketch the flow net of streamlines and equipped liners

Analytical Analysis

Parameters given:

Steady state ∂∂ t

=0

Incompressible fluid i.e. constant density Inviscid fluid (viscosity, μ=0) Volumetric rate per unit length is Г

Flow of fluid is only in the radial direction, thus

vr=vr(r ); vθ=0 ; vz=0

From equation 4.2-1C, Axisymmetric Cylindrical coordinates with no dependence on θ, velocity components for stream function (ѱ) defined as:

vr=+1r∂ψ∂ z

vz=−1r∂∅∂ r

Any vector that has zero curl can be written in gradient of a scalar function such as

v=−∇∅=−i ∂∅∂r

=vr i. Therefore, vr=−∂∅∂r

Stream function ѱ(r) is defined as: vr=+1r∂ψ∂ z

Equating equations (nnnnnnn)

−∂∅∂r

=1r∂ψ∂z

Laplacian function of velocity potential is defined as: ∇2∅=∂2∅∂r 2

Rearranging the above term followed by differentiating with respect to r:

r∂∅∂ r

=−∂ψ∂z

∂∂ r (r ∂∅∂r )= ∂

∂r (−∂ψ∂ z )Recallψ is only a function of r , hence

∂ψ∂ z

=0

Therefore by substituting this relationship into the above equation the Laplace equation is:

∂∂ r (r ∂∅∂r )=0.

Alternatively, we can derive this equation from continuity equation as follows.

The equation for continuity in cylindrical coordinate is given by equation B.4-2:

∂ ρ∂t

+ 1r∂(ρr vr)∂r

+1r∂( ρ vθ)∂θ

+∂(ρ v z)∂ z

=0

vr=vr(r ) and steady state, incompressible fluid.

Recall that vz=0 i.e. no fluid flows along the z-direction since the length of the cylinder is infinite in this

direction. Also vθ=0since there is no circular rotation.

Thus, the continuity equation reduces to 1r∂(ρr vr)∂r

which in vector form is written as ∇ . v=0. Also

v=−∇∅ , therefore, we have;

∇ . v= -(∇ .∇∅ ¿= ∇2∅=0.

For purely radial flow, this laplace equation can be expressed as ∂2∅∂ r2

+ 1r∂∅∂r

=0. This can be written as

total differential since it is a 1D equation. Thus 1rddr (r d∅dr )=0.

Deriving velocity potential, velocity and pressure distribution

1rddr (r d∅d r )=0dd r (r d∅d r )=0Integrating the above term with respect to ‘r’ you will have:

rd∅d r

=C1

Further integration with respect to ‘r’ results into:

ϕ=C1 ln (r )+C2

Recall from velocity potential definition in eqn (YYY) above vr=−d∅d r

Therefore vr=−d∅d r

=−C1r

∴r vr=−C1

Volumetric rate per unit length (QL ) is given by Г

Q=vA=vr∂ A

Q=∬vr r ∂θ∂ z

Therefore QL

=∫0

2π

vr r ∂θ=∫0

2 π

−C1∂θ

QL

=−2π C1

∴C1=−Q2πL

=−Γ2π

Substituting C1 into eqn (PPPP), the velocity equation derived will be:

vr=Γ2πr

Velocity potential ф

vr=−∂∅∂r

=−C1r

Recall from eqn(ppp)

ϕ=C1 ln (r )+C2

Since flow is radially symmetric at r=0, maximum velocity acts at this point, thus

∂vr∂r

=0

∂vr∂r

= −Γ2π r2

=0

Hence

Γ=0 , vr=0

Therefore C1 at r=0 can be computed as:

C1=−Γ2 π

= 02 π

=0

Boundary Conditions:

BC1: r=0 ф=0

Substituting boundary condition to equation(pppp)

0=0 ln (r )+C2

C2=0

ϕ=−Γ2π

ln (r )

Pressure as function of distribution

Equation of motion (Equation B.6-4)

ρ( ∂ vr∂ t +vr∂vr∂ r

+vθr∂vr∂θ

+v z∂ vr∂ z

−vθ2

r )=−∂ p∂r

+μ [ ∂∂r ( 1r ∂∂ r (r vr ))+ 1r2∂2 vr∂θ

2 +∂2 vr∂ z

2 −2

r2

∂2 vθ∂θ ]+ ρ gr

1

Reduced Equation:

ρ vrd vrd r

=−d Pd r

vr=Γ2πr

; ∂vr∂r

= −Γ2π r2

Substituting these terms to the reduced equation above results in:

−d Pdr

=−ρ Γ2

4π 2r3

Integrating above equation

Boundary conditions (BC)

BC1. r=r, P=P

BC2. r=∞, P=P∞

∫P

P∞

∂P=∫r

∞ρΓ 2

4 π2r 3∂r

P∞−P=(− ρΓ 28π 2r2 )r∞

P∞−P=ρΓ 2

8π 2r2