Groundwater Flow Equations
Groundwater Hydraulics
Daene C. McKinney
Summary
• General Groundwater Flow
– Control Volume Analysis
– General Continuity Equation
• Confined Aquifer Flow
– Continuity Equation
– Integrate over vertical dimension
– Transmissivity
– Continuity
– Examples
• Unconfined Aquifer Flow
– Darcy Law
– Continuity Equation
– Examples
Control Volume
• Control volume of dimensions Dx, Dy, Dz
• Completely saturated with a fluid of density r
x
yz
Mass flux in Mass flux out
2
x
x
D

2
x
x
D

2
)( x
x
q
q x
x
D



r
r
2
)( x
x
q
q x
x
D



r
r
xD
x
yD
zD
Control
volume
Mass Flux
• Mass flux = Mass in - Mass out:
mass flux in mass flux out2
x
x
D

2
x
x
D

2
)( x
x
q
q x
x
D



r
r
2
)( x
x
q
q x
x
D



r
r
xD
x
yD
zD
rqx -
¶ rqx( )
¶x
Dx
2
é
ë
ê
ù
û
úDyDz - rqx +
¶ rqx( )
¶x
Dx
2
é
ë
ê
ù
û
úDyDz = -
¶ rqx( )
¶x
DV
Mass fluxMass in Mass out
Mass Flux
• Mass flux =
• Continuity: Mass flux = change of mass
• Fluid mass in the volume:
• Continuity
mass flux in mass flux out2
x
x
D

2
x
x
D

2
)( x
x
q
q x
x
D



r
r
2
)( x
x
q
q x
x
D



r
r
xD
x
yD
zD
-
¶ rqx( )
¶x
DV
m =frDV
Mass flux change of mass
Aquifer Storage
Water compressibility Aquifer compressibility
Chain rule
Now, put it back into the continuity equation
Continuity Equation
¶
¶x
Kx
¶h
¶x
æ
è
ç
ö
ø
÷+
¶
¶y
Ky
¶h
¶y
æ
è
ç
ö
ø
÷+
¶
¶z
Kz
¶h
¶z
æ
è
ç
ö
ø
÷ = S
¶h
¶t
Confined Aquifer Flow
Horizontal Aquifer Flow
• Most aquifers are thin
compared to horizontal
extent
– Flow is horizontal, qx and qy
– No vertical flow, qz = 0
– Average properties over
aquifer thickness (b)
h(x,y,t)=
1
b
h(x,y,z,t)dz
0
b
ò
Ground surface
Bedrock
Confined aquifer
Qx
K
x
yz
h
Head in confined aquifer
Confining Layer
b
qx(x,y,t)=
1
b
qx(x,y,z,t)dz
0
b
ò Qx = bqx
Aquifer Transmissivity
• Transmissivity (T)
– Discharge through thickness of
aquifer per unit width per unit
head gradient
– Product of conductivity and
thickness
Hydraulic
gradient = 1 m/m
b
1 m
1 m
1 m
Transmissivity, T, volume
of water flowing an area 1
m x b under hydraulic
gradient of 1 m/m
Conductivity, K, volume of water
flowing an area 1 m x 1 m under
hydraulic gradient of 1 m/m
Continuity Equation
• Continuity equation
• Darcy’s Law
• Continuity
-
¶Qx
¶x
= S
¶h
¶t
Qx = -Tx
¶h
¶x
¶
¶x
Tx
¶h
¶x
æ
è
ç
ö
ø
÷ = S
¶h
¶t
Ground surface
Bedrock
Confined aquifer
Qx
K
x
yz
h
Head in confined aquifer
Confining Layer
b
1
r
¶
¶r
r
¶h
¶r
æ
è
ç
ö
ø
÷ =
S
T
¶h
¶t
Radial Coordinates
Example – Horizontal Flow
• Consider steady flow from left to right in a confined aquifer
• Find: Head in the aquifer, h(x)
¶
¶x
T
¶h
¶x
æ
è
ç
ö
ø
÷ = S
¶h
¶t
= 0
T
d2
h
¶x2
= 0
Ground surface
Bedrock
Confined aquifer
Qx
K
x
yz
hB
Confining Layer
b
hA
L
steady flow
h(x)
Example – Horizontal Flow
• L = 1000 m, hA = 100 m, hB = 80 m, K = 20 m/d, f = 0.35
• Find: head, specific discharge, and average velocity
Ground surface
Bedrock
Confined aquifer
Qx
K=2-m/d
x
yz
hB=80m
Confining Layer
b
hA=100m
L=1000m
Unconfined Aquifer Flow
Flow in an Unconfined Aquifer
• Dupuit approximations
– Slope of the water table is small
– Velocities are horizontal
Ground surface
Bedrock
Unconfined aquifer
Water table
Dx
Qx
K
h
x
yz
Qx = qxh = (-K
¶h
¶x
)h
-
¶Qx
¶x
= Sy
¶h
¶t
¶
¶x
Kh
¶h
¶x
æ
è
ç
ö
ø
÷ = Sy
¶h
¶t
Steady Flow in an Unconfined Aquifer
• 1-D flow
• Steady State,
• K = constant
• Find h(x)
¶
¶x
Kh
¶h
¶x
æ
è
ç
ö
ø
÷ = Sy
¶h
¶t
h
FlowhA
hB
Water Table
Ground Surface
Bedrock L
x
Steady Flow in an Unconfined Aquifer
• K = 10-1 cm/sec
• L = 150 m
• hA = 6.5 m
• hB = 4 m
• x = 150 m
• Find h(x), Q
h
FlowhA=6.5m
hB=4m
Water Table
Ground Surface
Bedrock L=150m
x
K=0.1cm/s
Summary
• General Groundwater Flow
– Control Volume Analysis
– General Continuity Equation
• Confined Aquifer Flow
– Continuity Equation
– Integrate over vertical dimension
– Transmissivity
– Continuity
– Examples
• Unconfined Aquifer Flow
– Darcy Law
– Continuity Equation
– Examples
Groundwater Flow Equations
Examples
Example – Horizontal Flow
• Consider steady flow from left to right in a confined aquifer
• Find: Head in the aquifer, h(x)
¶
¶x
T
¶h
¶x
æ
è
ç
ö
ø
÷ = S
¶h
¶t
= 0
T
d2
h
¶x2
= 0
h(x) = hA +
hB - hA
L
x
Ground surface
Bedrock
Confined aquifer
Qx
K
x
yz
hB
Confining Layer
b
hA
L
steady flow
Head in the aquifer
h(x)
Example – Horizontal Flow
• L = 1000 m, hA = 100 m, hB = 80 m, K = 20 m/d, f = 0.35
• Find: head, specific discharge, and average velocity
h(x) = hA +
hB - hA
L
x =100- 0.02x m q = -K
hB - hA
L
= -(20 m/d)
80 -100
1000
= 0.4 m/day
v =
q
f
=1.14 m/day
Ground surface
Bedrock
Confined aquifer
Qx
K=2-m/d
x
yz
hB=80m
Confining Layer
b
hA=100m
L=1000m
Steady Flow in an Unconfined Aquifer
• 1-D flow
• Steady State,
• K = constant
¶
¶x
Kh
¶h
¶x
æ
è
ç
ö
ø
÷ = Sy
¶h
¶t
d
dx
Kh
dh
dx
æ
è
ç
ö
ø
÷ = 0
h2
(x) = hA
2
+(
hB
2
- hA
2
L
)x
h
FlowhA
hB
Water Table
Ground Surface
Bedrock L
x
Q = (-K
dh
dx
)h = -
K
2
dh2
dx
= -
K
2
hB
2
- hA
2
L
æ
è
ç
ö
ø
÷
Steady Flow in an Unconfined Aquifer
• K = 10-1 cm/sec
• L = 150 m
• hA = 6.5 m
• hB = 4 m
• x = 150 m
• Find Q
h
FlowhA=6.5m
hB=4m
Water Table
Ground Surface
Bedrock L=150m
x
Q = -
K
2
hB
2
- hA
2
L
æ
è
ç
ö
ø
÷ = -
86.4 m/d
2
6.52
- 42
150
æ
è
ç
ç
ö
ø
÷
÷
= 7.56 m3
/d /m
K=0.1cm/s

05 groundwater flow equations

  • 1.
    Groundwater Flow Equations GroundwaterHydraulics Daene C. McKinney
  • 2.
    Summary • General GroundwaterFlow – Control Volume Analysis – General Continuity Equation • Confined Aquifer Flow – Continuity Equation – Integrate over vertical dimension – Transmissivity – Continuity – Examples • Unconfined Aquifer Flow – Darcy Law – Continuity Equation – Examples
  • 3.
    Control Volume • Controlvolume of dimensions Dx, Dy, Dz • Completely saturated with a fluid of density r x yz Mass flux in Mass flux out 2 x x D  2 x x D  2 )( x x q q x x D    r r 2 )( x x q q x x D    r r xD x yD zD Control volume
  • 4.
    Mass Flux • Massflux = Mass in - Mass out: mass flux in mass flux out2 x x D  2 x x D  2 )( x x q q x x D    r r 2 )( x x q q x x D    r r xD x yD zD rqx - ¶ rqx( ) ¶x Dx 2 é ë ê ù û úDyDz - rqx + ¶ rqx( ) ¶x Dx 2 é ë ê ù û úDyDz = - ¶ rqx( ) ¶x DV Mass fluxMass in Mass out
  • 5.
    Mass Flux • Massflux = • Continuity: Mass flux = change of mass • Fluid mass in the volume: • Continuity mass flux in mass flux out2 x x D  2 x x D  2 )( x x q q x x D    r r 2 )( x x q q x x D    r r xD x yD zD - ¶ rqx( ) ¶x DV m =frDV Mass flux change of mass
  • 6.
    Aquifer Storage Water compressibilityAquifer compressibility Chain rule Now, put it back into the continuity equation
  • 7.
  • 8.
  • 9.
    Horizontal Aquifer Flow •Most aquifers are thin compared to horizontal extent – Flow is horizontal, qx and qy – No vertical flow, qz = 0 – Average properties over aquifer thickness (b) h(x,y,t)= 1 b h(x,y,z,t)dz 0 b ò Ground surface Bedrock Confined aquifer Qx K x yz h Head in confined aquifer Confining Layer b qx(x,y,t)= 1 b qx(x,y,z,t)dz 0 b ò Qx = bqx
  • 10.
    Aquifer Transmissivity • Transmissivity(T) – Discharge through thickness of aquifer per unit width per unit head gradient – Product of conductivity and thickness Hydraulic gradient = 1 m/m b 1 m 1 m 1 m Transmissivity, T, volume of water flowing an area 1 m x b under hydraulic gradient of 1 m/m Conductivity, K, volume of water flowing an area 1 m x 1 m under hydraulic gradient of 1 m/m
  • 11.
    Continuity Equation • Continuityequation • Darcy’s Law • Continuity - ¶Qx ¶x = S ¶h ¶t Qx = -Tx ¶h ¶x ¶ ¶x Tx ¶h ¶x æ è ç ö ø ÷ = S ¶h ¶t Ground surface Bedrock Confined aquifer Qx K x yz h Head in confined aquifer Confining Layer b 1 r ¶ ¶r r ¶h ¶r æ è ç ö ø ÷ = S T ¶h ¶t Radial Coordinates
  • 12.
    Example – HorizontalFlow • Consider steady flow from left to right in a confined aquifer • Find: Head in the aquifer, h(x) ¶ ¶x T ¶h ¶x æ è ç ö ø ÷ = S ¶h ¶t = 0 T d2 h ¶x2 = 0 Ground surface Bedrock Confined aquifer Qx K x yz hB Confining Layer b hA L steady flow h(x)
  • 13.
    Example – HorizontalFlow • L = 1000 m, hA = 100 m, hB = 80 m, K = 20 m/d, f = 0.35 • Find: head, specific discharge, and average velocity Ground surface Bedrock Confined aquifer Qx K=2-m/d x yz hB=80m Confining Layer b hA=100m L=1000m
  • 14.
  • 15.
    Flow in anUnconfined Aquifer • Dupuit approximations – Slope of the water table is small – Velocities are horizontal Ground surface Bedrock Unconfined aquifer Water table Dx Qx K h x yz Qx = qxh = (-K ¶h ¶x )h - ¶Qx ¶x = Sy ¶h ¶t ¶ ¶x Kh ¶h ¶x æ è ç ö ø ÷ = Sy ¶h ¶t
  • 16.
    Steady Flow inan Unconfined Aquifer • 1-D flow • Steady State, • K = constant • Find h(x) ¶ ¶x Kh ¶h ¶x æ è ç ö ø ÷ = Sy ¶h ¶t h FlowhA hB Water Table Ground Surface Bedrock L x
  • 17.
    Steady Flow inan Unconfined Aquifer • K = 10-1 cm/sec • L = 150 m • hA = 6.5 m • hB = 4 m • x = 150 m • Find h(x), Q h FlowhA=6.5m hB=4m Water Table Ground Surface Bedrock L=150m x K=0.1cm/s
  • 18.
    Summary • General GroundwaterFlow – Control Volume Analysis – General Continuity Equation • Confined Aquifer Flow – Continuity Equation – Integrate over vertical dimension – Transmissivity – Continuity – Examples • Unconfined Aquifer Flow – Darcy Law – Continuity Equation – Examples
  • 19.
  • 20.
    Example – HorizontalFlow • Consider steady flow from left to right in a confined aquifer • Find: Head in the aquifer, h(x) ¶ ¶x T ¶h ¶x æ è ç ö ø ÷ = S ¶h ¶t = 0 T d2 h ¶x2 = 0 h(x) = hA + hB - hA L x Ground surface Bedrock Confined aquifer Qx K x yz hB Confining Layer b hA L steady flow Head in the aquifer h(x)
  • 21.
    Example – HorizontalFlow • L = 1000 m, hA = 100 m, hB = 80 m, K = 20 m/d, f = 0.35 • Find: head, specific discharge, and average velocity h(x) = hA + hB - hA L x =100- 0.02x m q = -K hB - hA L = -(20 m/d) 80 -100 1000 = 0.4 m/day v = q f =1.14 m/day Ground surface Bedrock Confined aquifer Qx K=2-m/d x yz hB=80m Confining Layer b hA=100m L=1000m
  • 22.
    Steady Flow inan Unconfined Aquifer • 1-D flow • Steady State, • K = constant ¶ ¶x Kh ¶h ¶x æ è ç ö ø ÷ = Sy ¶h ¶t d dx Kh dh dx æ è ç ö ø ÷ = 0 h2 (x) = hA 2 +( hB 2 - hA 2 L )x h FlowhA hB Water Table Ground Surface Bedrock L x Q = (-K dh dx )h = - K 2 dh2 dx = - K 2 hB 2 - hA 2 L æ è ç ö ø ÷
  • 23.
    Steady Flow inan Unconfined Aquifer • K = 10-1 cm/sec • L = 150 m • hA = 6.5 m • hB = 4 m • x = 150 m • Find Q h FlowhA=6.5m hB=4m Water Table Ground Surface Bedrock L=150m x Q = - K 2 hB 2 - hA 2 L æ è ç ö ø ÷ = - 86.4 m/d 2 6.52 - 42 150 æ è ç ç ö ø ÷ ÷ = 7.56 m3 /d /m K=0.1cm/s