Stream Function Definitions

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U = −T∇h

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Φ=Th

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U = −∇ Th( ) = −∇Φ

DischargePotential

U is aquifer flux. It is a vector. U = qb, where q is flux per unit depth and b is aquifer thickness

For Confined Aquifer

Then

Governing Equations

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divergence U( ) =∇U = 0

Continuity Equation becomes

Zero Divergence implies no sources or sinks

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∇U =∇ −∇Φ( ) =∇ 2Φ = 0

Get LaPlace Equation by substitution into above

Solutions to LaPlace equation are potential functions. When Divergence of a vector field is zero, flux is the gradient of the potential.

Unconfined Aquifer (redefine T)Base of Aquifer is datum, H is saturated thickness. T=KH.

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U = −T∇h = −KH∇H

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U = −∇Φ = −KH∇H

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Φ=KH 2

2

Potentials are Constant HeadsBecause , gradient of potential field gives aquifer flux. Lines along which potential is constant are called Equipotential lines, and are like elevation contours. Darcy velocity may be calculated from the aquifer flux U.

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U = −∇Φ

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q =U

bb is the aquifer thickness for a confined aquifer.

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q =U

H= −

∇Φ

2Φ

K

= − 2K∇Φ1/ 2 = −∇ 2KΦ

Substituting for U and H.

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Φ=KH 2

2 implies H =

2Φ

K. Also U = −∇Φ.

For an Unconfined Aquifer

Parallel VectorsIn both cases (confined, unconfined) the following vector fields are parallel

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v || q || UNote: v is the seepage velocity v = q/n, where n is porosity

Since the divergence is zero, the streamlines cannot cross or join.U is convenient, since its formulation is the same for confined andunconfined aquifers.

Flow along a line lConsider a fluid particle moving along a line l. For each small displacement, dl,

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dl = idx + jdy

Where i and j are unit vectors in the x and y directions, respectively.Since dl is parallel to U, then the cross product must be zero.

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U = iUx + jUyRemembering that:

Then

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U × dl = iUx + jUy( ) × idx + jdy( )

And that:

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i × i = 0 and

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i × j = k

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= Uxdy + Uydx( )k = 0

And

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dx

Ux

=dy

Uy

dx

dydl

Stream FunctionSince must be satisfied along a line l, such a line is

called a streamline or flow line. A mathematical construct called a stream function can describe flow associated with these lines.

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dx

Ux

=dy

Uy

The Stream Function is defined as the function which is constant along a streamline, much as a potential function is constant along an equipotential line.

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Ψ x,y( )

Since is constant along a flow line, then for any dl,

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Ψ x,y( )

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dΨ =∂Ψ

∂xdx +

∂Ψ

∂ydy = 0 Along the streamline

Stream Functions

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dΨ =∂Ψ

∂xdx +

∂Ψ

∂ydy = 0Rearranging: and

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dx

Ux

=dy

Uy

we get

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∂Ψ∂x

dx = −∂Ψ

∂ydy

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Uydx = Uxdyand

from which we can see that

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Ux = −∂Ψ

∂y and Uy =

∂Ψ

∂x

So that if one can find the stream function, one can get the dischargeby differentiation.

Interpreting Stream Functions

Flow Line 1

Flow Line 2

Any Line

t

n

l l is an arbitrary linet is tangent to ln is normal to l

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t =dx

dli +

dy

dlj

n =dy

dli −

dx

dlj

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t • n = 0

What is the flow that crosses l between Flow Lines 1 and 2?For each increment dl:

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dQ = U • ndl = Uxi + Uy j( ) •dy

dli −

dx

dlj

⎛

⎝ ⎜

⎞

⎠ ⎟dl

= Uxdy −Uydx

= −∂Ψ

∂ydy −

∂Ψ

∂xdx

= −dΨ

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Ψ1

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Ψ2

Discharge Between Lines 1, 2If we integrate along line l, between Flow Lines 1 and 2we will get the total flow across the line.

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Q1−2 = dQ =Ψ1

Ψ2

∫ −dΨ = −ΨΨ1

Ψ2

∫Ψ1

Ψ2

= − Ψ2 − Ψ1( ) = Ψ1 − Ψ2

This is true even if K or T is heterogeneous

Conjugate FunctionsWe have already shown that:

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∇2Φ = 0 and since

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Ux = −∂Φ

∂x; Uy = −

∂Φ

∂y

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U = −∇Φ

The rotation of the discharge potential field may be calculated viathe Curl as follows:

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∇×U =∂Uy

∂x−

∂Ux

∂y

⎛

⎝ ⎜

⎞

⎠ ⎟k =

∂

∂x−

∂Φ

∂y

⎛

⎝ ⎜

⎞

⎠ ⎟−

∂

∂y−

∂Φ

∂x

⎛

⎝ ⎜

⎞

⎠ ⎟

⎛

⎝ ⎜

⎞

⎠ ⎟k = 0

Since the Curl of U is zero, the flow field is irrotational.

Conjugate Functions

Since the Curl of U is zero, the flow field is irrotational. Doing the same calculation, but using the relationship between the Stream Function and U, we get:

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∇×U =∂Uy

∂x−

∂Ux

∂y

⎛

⎝ ⎜

⎞

⎠ ⎟k =

∂

∂x

∂Ψ

∂x

⎛

⎝ ⎜

⎞

⎠ ⎟−

∂

∂y−

∂Ψ

∂y

⎛

⎝ ⎜

⎞

⎠ ⎟

⎛

⎝ ⎜

⎞

⎠ ⎟k = ∇ 2Ψ( )k = 0

From this, we see that , because we know thatU is irrotational.

Conjugate Functions

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∇2Ψ = 0

Thus, the Potential Function and Stream Function both satisfyLaPlace’s Equation.

By definition, flow lines are parallel to streamlines (lines of constant stream function value), and perpendicular to lines ofconstant potential.

Thus, the streamlines and potential lines are also perpendicular.

If and Then

SuperpositionOne special property of solutions to the LaPlace Equation is that it is linear. Thus, solutions to the equation may be added together, and the sum of solutions will also be a solution.

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∇2Ψ1 = 0

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∇2Ψ 2 = 0

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∇2 Ψ1 + Ψ 2( ) = 0

This implies that if one can find a solution for uniform flow and fora point source or sink, then they can be added together to get a solution for (uniform flow) + (source) + (sink)

Main Equations - Aquifer Flux

In Cartesian Coordinates:

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Ux = −∂Φ

∂x= −

∂Ψ

∂y

Uy = −∂Φ

∂y=

∂Ψ

∂x

In Polar Coordinates:

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Ur = −∂Φ

∂r= −

1

r

∂Ψ

∂θ

Uθ = −1

r

∂Φ

∂θ=

∂Ψ

∂rThese are the Cauchy-Riemann Equations.

Uniform FlowFor a uniform flow rate U (L2/T) at an angle with respect to the x-axis:

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Ux = −∂Φ

∂x= U cosα

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Uy = −∂Φ

∂y= U sinα

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Φ=Φ0 −U x cosα + y sinα( )

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dΦ =∂Φ

∂xdx +

∂Φ

∂ydy

= −Uxdx −Uydy

= − U cosα( )dx − U sinα( )dy

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Ψ=Ψ0 −U y cosα − x sinα( )

Radial Flow SourceFor an injection well at the origin, flow Q across any circle withradius r is equal to due to continuity.

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Ur =Q

2πr= −

∂Φ

∂r= −

1

r

∂Ψ

∂θr

Circumference=2r

U = Flow per unit length

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Uθ = −1

r

∂Φ

∂θ=

∂Ψ

∂r= 0

There is no rotational flow:

Q

Develop Φ and Ψ Lines

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∂Φ∂r

= −Q

2πr

Φ = Φ0 −Q

2πln r( )

−1

r

∂Ψ

∂θ=

Q

2πr∂Ψ

∂θ= −

Q

2π

Ψ = Ψ0 −Q

2πθ

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Ur =Q

2πr= −

∂Φ

∂r= −

1

r

∂Ψ

∂θ

Separate

IntegratingEquipotential

Streamline

Φand Ψ Lines for Capture Zone Theory

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Combine a uniform field and a well

Φ = Φ0 −Ux +Q

2πln r( )

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Ψ=Q

2πθ −Uy

or =Q

2πtan−1 y

x−Uy