Steady-state and EPS

Hydraulic Modelling

Prof. Raziyeh Farmani

r.farmani@exeter.ac.uk

ECMM133
 Outline
• Introduction
• Steady-state hydraulics:
– Hardy-Cross method (incl. example)
– Linear Theory method (incl. example)
– Other methods
– Tree-like networks method
• Extended period simulation
• EPANET2 software demo
 Pressurised Fluid Flow
Energy Components
Small!
Head (m) Symbol EGL
v2/2g

HGL
Elevation head Z

p/rg
Pressure head p/rg
Ground
Velocity head v2/2g E H Small!
Pipe
(Piezometric) H = z + p/rg centreline
Head
Z
Energy E = H+ v2/2g

Datum
 Generalised Pipe Headloss
Equation (SI Units)
n 1
h f  RQ  R Q
n
Q

Formulation R n

Darcy- 0.8106 L 2.00

RDW 
Weisbach gD5

Hazen- 10.648L
10.67 1.852
RHW 
Williams C1.852 D 4.871
 Pipe Network Hydraulics
• Problem: Given network
configuration, L, D and k
(or C or n) for each pipe,
Z and Qd for each node
and at least one Hf for a
fixed head node, find the
unknown Q (or v) in each
pipe and H (or P) at each
node
• Example: 12 unknowns
(7 pipe flow rates Q and 5
nodal heads H)
 Hardy-Cross Method
• Proposed by Hardy and Cross (1936) and
updated by Cornish (1939)
• Mass balance equation for each node i:

Qd,i

Ni i

Q
Q1
m  Qd ,i  0 QNi
m 1 Qm
 Hardy-Cross Method (cont.)
• Energy balance equation for each loop l:

hf,1
Nl

h
j 1
f,j 0
Loop l
hf,Nl

hf,j

Qd
 Hardy-Cross Procedure
1. Identify loops in the network. Set iteration
counter k=1.
2. Assume flows in each pipe (must be
balanced at each node). Assign the flow
signs based on the ‘clockwise convention’.
 Hardy-Cross Procedure (cont.)
3. Calculate the headloss in each pipe.

4. Check the headloss sum for each loop in the

system. If this sum is less than some target
accuracy eH, go to step 7. If not, calculate flow
correction for each loop l (l=1,2,…,Nl):
Ni
 h(f k,i)
Ql( k )  i 1
Ni
h(f k,i)
n
i 1 Qi( k )
 Hardy-Cross Procedure (cont.)
5. Update all pipe flows using the corrections
obtained:

Qi
( k 1)
Q
i
(k )
  Q l
(k )

l i

6. Increase iteration counter k by 1 and go to

step 3.
7. Calculate nodal heads H (and the associated
pressures) from the obtained flows Q.
 Hardy-Cross Example
• Fixed head reservoir
at node 1
• Find flows in all pipes
and heads in all nodes
assuming target
accuracy of eH=0.01 m
• 7 pipes and 6 nodes
• 2 loops
 Hardy-Cross Example: Data

From To Length Diameter k Head/ Demand

Pipe Node Node (m) (mm) (mm) Node Type Elevation (m) (l/s)
1 1 2 530 250 0.1 1 Res 50 -
2 2 3 410 150 0.1 2 Junc 12 10
3 1 4 1200 100 0.1 3 Junc 22 22
4 2 5 630 200 0.1 4 Junc 17 10

5 6 3 1040 100 0.1 5 Junc 25 18

6 4 5 540 100 0.1 6 Junc 20 15

7 5 6 580 150 0.1

Target accuracy: eH=0.01 m
 Hardy-Cross Example:
Assumed Flow Directions

10 l/s 22 l/s

10 l/s
18 l/s 15 l/s
 Hardy-Cross Example:
Assumed Flow Values

50 l/s 22 l/s
10 l/s

25 l/s 20 l/s
20 l/s 2 l/s
10 l/s
18 l/s 15 l/s

15 l/s
17 l/s
 Hardy-Cross Example: Iteration 1

=-(-147.16)/(2*6700.54)*(1000)

Loop Pipe Q │v│ Re  hf │hf/Q│ Q

(l/s) (m/s) (-) (-) (m) (s/m2) (l/s)
1 1 50.00 1.02 2.23E+05 0.018 2.04 40.78
4 20.00 0.64 1.12E+05 0.020 1.31 65.51
6 -15.00 1.91 1.68E+05 0.021 -21.52 1434.49
3 -25.00 3.18 2.79E+05 0.021 -128.99 5159.77
Sum: -147.16 6700.54 10.98

2 2 20.00 1.13 1.49E+05 0.020 3.62 180.76

5 -2.00 0.25 2.23E+04 0.028 -0.95 473.18
7 -17.00 0.96 1.27E+05 0.021 -3.75 220.78
4 -20.00 0.64 1.12E+05 0.020 -1.31 65.51
Sum: -2.39 940.23 1.27
 Hardy-Cross Example: Iteration 2

= 20+(10.98-1.27)

= 50+10.98
Loop Pipe Q │v│ Re  hf │hf/Q│ Q
(l/s) (m/s) (-) (-) (m) (s/m2) (l/s)
1 1 60.98 1.24 2.72E+05 0.018 2.98 48.86
4 29.71 0.95 1.66E+05 0.019 2.77 93.14
6 -4.02 0.51 4.49E+04 0.025 -1.77 440.12
3 -14.02 1.78 1.57E+05 0.022 -41.96 2993.23
Sum: -37.98 3575.35 5.31

2 2 21.27 1.20 1.58E+05 0.020 4.07 191.20

5 -0.73 0.09 8.12E+03 0.034 -0.16 214.64
7 -15.73 0.89 1.17E+05 0.021 -3.24 205.89
4 -29.71 0.95 1.66E+05 0.019 -2.77 93.14
Sum: -2.09 704.86 1.48
= -20-(10.98-1.27)
 Hardy-Cross Example: Iteration 3

Loop Pipe Q │v│ Re  hf │hf/Q│ Q

(l/s) (m/s) (-) (-) (m) (s/m2) (l/s)
1 1 66.29 1.35 2.96E+05 0.018 3.50 52.75
4 33.53 1.07 1.87E+05 0.019 3.48 103.91
6 1.29 0.16 1.44E+04 0.030 0.22 173.42
3 -8.71 1.11 9.72E+04 0.022 -16.84 1933.93
Sum: -9.63 2264.01 2.13

2 2 22.76 1.29 1.69E+05 0.020 4.63 203.35

5 0.76 0.10 8.47E+03 0.034 0.17 221.62
7 -14.24 0.81 1.06E+05 0.021 -2.68 188.45
4 -33.53 1.07 1.87E+05 0.019 -3.48 103.91
Sum: -1.37 717.33 0.96
 Hardy-Cross Example: Iteration 6
(final)

Loop Pipe Q │v│ Re  hf │hf/Q│ Q

(l/s) (m/s) (-) (-) (m) (s/m2) (l/s)
1 1 68.73 1.40 3.07E+05 0.018 3.75 54.54
4 34.79 1.11 1.94E+05 0.019 3.74 107.43
6 3.73 0.48 4.17E+04 0.025 1.54 413.23
3 -6.27 0.80 7.00E+04 0.023 -9.03 1440.78
Sum: 0.00 2015.97 0.00

2 2 23.95 1.36 1.78E+05 0.020 5.10 213.05

5 1.95 0.25 2.17E+04 0.028 0.90 462.81
7 -13.05 0.74 9.72E+04 0.021 -2.28 174.44
4 -34.79 1.11 1.94E+05 0.019 -3.74 107.43

Sum: -0.01 957.73 0.01

Hardy-Cross Example: Results

= 50 - 3.75
= 46.25 -12

Pipe Flow hf Node Head Pressure

(l/s) m (m) (m)
1 68.73 3.75 1 50.00 -
2 23.95 5.10 2 46.25 34.25
3 6.27 9.03 3 41.15 19.15
4 34.79 3.74 4 40.97 23.97
5 1.95 0.90 5 42.51 17.51
6 3.73 1.54
6 40.24 20.24
7 13.05 2.28
 Hardy-Cross Example
• Problem with 12
unknowns (7 pipe
flows and 5 nodal
heads) reduced 2
principal unknowns
(loop correction flows)
at each iteration
• Great for doing
manual calculations!
 Another Hardy-Cross Example
• What is the number of loops?
 Another Hardy-Cross Example
(cont.)

Loop 2-4-5: h f ,2  h f ,4  h f ,5  0 (-1)

Pseudo Loop 1-2-3: + h f ,1  h f ,2  h f ,3  hA  hB

Pseudo Loop 1-5-4-3: h f ,1  h f ,5  h f ,4  h f ,3  hA  hB

 Hardy-Cross Method: Summary

• Pros:
– Smaller number of principal unknowns (flow
corrections in loops), hence simpler to use if doing
calculations manually
• Cons:
– Identification of (independent) network loops
(including pseudo loops) may not be so simple in
larger networks
– The above is also not easy to code in a computer
programme
 Linear Theory Method
• Developed by Wood and Charles (1972)
• Equations:
– Mass balance equation for each node I (flow going
out of node is positive):
Ni

Q
j 1
ij  Qd ,i  0

– Energy balance equation for each pipe i-j:

n 1
Hi  H j  Rij Qij Qij  Uij Qij

(where Uij=RijQijn-1)
Linear Theory Method (cont.)
H i( k 1)  H (j k 1)
Hi  H j  Uij Qij  Qij( k 1)  (k )
(1)
U ij

Now the mass balance equation for node i becomes:

Ni
Hi  H j Ni
1 Ni
Hj

j 1 U ij
 Qd ,i  H i 
j 1 U ij

j 1 U ij
 Qd ,i  f ( H i )  0

Newton-Raphson method:

H ( k 1)
H (k )

 
f H i( k )
f H 
i i ' (k )
i
Linear Theory Method (cont.)
H (j k ) Ni

 
Ni Ni

 
1

1
H    ( k )  Qd ,i
(k ) (k ) ' (k )
f H i i (k )
f H i (k )
j 1 U ij j 1 U ij j 1 U ij

Ni
1 Ni
H (j k )
H i
(k )
 (k )
j 1 U ij
  ( k )  Qd ,i
j 1 U ij
H i( k 1)  H i( k )  Ni
1
U
j 1
(k )
ij

Ni
H (j k )
Uj 1
(k )
 Qd ,i
H i( k 1) 
ij
Ni
(2)
1
 (k )
j 1 U ij
 Linear Theory Method Procedure
1. Initialise iteration counter k=0. Initialise nodal heads by
assuming Hi(0) and calculate the corresponding pipe
flows Qij(0). Alternatively, assume Qij(0) (by assuming
velocities vij(0)) and calculate Hi(0). Calculate Uij(0).

2. Update nodal heads by using equation (2).

Ni (k )
H
U
j 1
j
(k )
 Qd ,i
( k 1)

ij
H i Ni
1
 (k )
j 1 U ij
 Linear Theory Method Procedure
3. Update pipe flows by using equation (1).

H i( k 1)  H (j k 1)
Qij( k 1)  (k )
U ij

4. Increase k by 1 and check for convergence, e.g.:

max H i( k 1)  H i( k )  e H
i
If converged, stop. If not, go back to step 2.
 Linear Theory Example: Data

From To Length Diameter Head/ Demand

Pipe Node Node (m) (mm) k (mm) Node Type Elevation (m) (l/s)
1 1 2 530 250 0.1 1 Res 50 -
2 2 3 410 150 0.1 2 Junc 12 10
3 1 4 1200 100 0.1 3 Junc 22 22
4 2 5 630 200 0.1 4 Junc 17 10

5 6 3 1040 100 0.1 5 Junc 25 18

6 4 5 540 100 0.1 6 Junc 20 15

7 5 6 580 150 0.1

Target accuracy: eH=0.01 m
 LT Method Example:
Assumed Flow Directions

22 l/s
10 l/s

10 l/s
18 l/s 15 l/s
 LT Method Example: Iteration 0
Pipe v Q Re  R U hf
(-) (m/s) (l/s) (-) (-) (s2/m5) (s/m2) (m)
1 1.00 49.09 2.19E+05 0.018 8.17E+02 40.11 1.97
2 1.00 17.67 1.32E+05 0.020 9.15E+03 161.62 2.86
3 1.00 7.85 8.77E+04 0.023 2.24E+05 1762.34 13.84
4 1.00 31.42 1.75E+05 0.019 3.12E+03 97.95 3.08
5 1.00 7.85 8.77E+04 0.023 1.94E+05 1527.36 12.00
6 1.00 7.85 8.77E+04 0.023 1.01E+05 793.05 6.23
7 1.00 17.67 1.32E+05 0.020 1.29E+04 228.63 4.04

Node Head
(-) (m)
1 50.00
2 48.03
3 45.18
4 36.16
=50.00-1.97
5 44.95
6 40.91
 LT Method Example: Iteration 1
Ni
H (j k )
U
j 1
(k )
 Qd ,i
H i( k 1) 
ij
Ni
1 Node Head Error =48.03-47.79
 (k )
j 1 U ij (-) (m) (m)
Calculated 1 50.00 0.00
using 2 47.79 0.24
equation 3 44.13 1.04
(2), i.e. by 4 41.05 4.89
using H(0) 5 43.99 0.97
for nodes 1,
6 42.00 1.09
3 and 5, U(0)
for pipes 1, Max: 4.89
2 and 4 and
Pipe Q v Re  R U
demand at
node 2 (-) (l/s) (m/s) (-) (-) (s /m5)
2
(s/m2)
1 55.12 1.12 2.46E+05 0.018 8.08E+02 44.56
2 22.61 1.28 1.68E+05 0.020 8.94E+03 202.16
Calculated 3 5.08 0.65 5.67E+04 0.024 2.36E+05 1197.61
using
4 38.83 1.24 2.17E+05 0.019 3.06E+03 118.76
equation (1),
5 -1.40 0.18 1.56E+04 0.030 2.54E+05 355.12
i.e. by using
H(1) for 6 -3.70 0.47 4.13E+04 0.025 1.11E+05 410.15
nodes 1 and 7 8.68 0.49 6.47E+04 0.022 1.41E+04 122.25
2 and U(0) for
pipe 1 ( k 1)
H i( k 1)  H (j k 1)
Q ij 
U ij( k )
LT Method Example: Iteration 2
Node Head Error
(-) (m) (m)
1 50.00 0.00
2 47.50 0.29
3 42.85 1.28
4 42.46 1.41
5 43.49 0.49
6 42.66 0.66
Max: 1.41

Pipe Q v Re  R U
(-) (l/s) (m/s) (-) (-) (s /m5)
2
(s/m2)
1 56.19 1.14 2.51E+05 0.018 8.07E+02 45.35
2 22.96 1.30 1.71E+05 0.020 8.93E+03 205.00
3 6.29 0.80 7.03E+04 0.023 2.30E+05 1446.00
4 33.71 1.07 1.88E+05 0.019 3.10E+03 104.39
5 -0.55 0.07 6.15E+03 0.037 3.17E+05 174.78
6 -2.51 0.32 2.80E+04 0.026 1.18E+05 295.86
7 6.82 0.39 5.08E+04 0.023 1.46E+04 99.50
LT Method Example: Iteration 32
Node Head Error
(-) (m) (m)
1 50.00 0.00
2 46.28 0.00
3 41.20 0.01
4 41.03 0.00
5 42.58 0.01
6 40.31 0.00
Max: 0.01

Pipe Q
(-) (l/s)
1 68.38
2 23.90
3 6.27
4 34.64
5 -1.92
6 -3.70
7 12.99
 LT Example: Results

Pipe Flow Node Head Pressure

(l/s) (m) (m)
1 68.38 1 50.00 -
2 23.90 2 46.28 34.28
3 6.27 3 41.20 19.20
4 34.64 4 41.03 24.03
5 1.92 5 42.58 17.58
6 3.70 6 40.31 20.31
7 12.99
 LT Method: Summary

• Pros:
– Relatively easy to code in a computer
programme (no need to identify loops, etc.)
• Cons:
– Larger number of unknowns than in the
Hardy-Cross method
– Requires larger number of iterations to
converge
 Todini & Pilati (1987) Method
• Global Gradient Algorithm (GGA)
• Based on the Newton-Raphson technique
• Equations:
– Mass balance for nodes and
– Energy balance pipes (i.e. links)
• Used in Epanet2 software (Rossman
2000)
 GGA: System Equations
 A11  A12  Q   A10H 0 
         
    
 A 21  0   H    q 
Node inflow
where: assumed positive
QT  [Q1 , Q2 , , QNp ] unknown pipe flows

HT  [ H1 , H 2 , , H Nj ] unknown nodal heads

HT0  [ H N j 1 , H N j 2 , , H Nn ] known nodal heads

qT  [q1 , q2 , , qNn ] given nodal demands

Np = number of network pipes (i.e. links)
Nj = number of network junctions (nodes with unknown head)
Nn = number of network nodes
Nn-Nj= number of nodes with known heads
GGA: System Equations (cont.)

 A11  A12  Q   A10H 0 

          (1)
    
 A 21  0   H    q 

n 1
A11  i, i   R Qij (diagonal matrix)

Topological (incidence) matrix:

1, if flow of pipe ij leaves node i

A12  i, j    0,if pipe ij is not connected to node i
1, if pipe ij is not connected to node i

A12   A12 A10  A12  A T21
Topological Matrix: Example

Node 2 3 4 5 6 1 Pipe
22 l/s 1 0 0 0 0 1 1
10 l/s
 1 1 0 0 0 
0 2

0 0 1 0 0 1 3
 
10 l/s
A12   1
18 l/s 15 l/s
0 0 1 0 0 4
0 1 0 0 1 0  5
Assumed flow directions!  
0 0 1 1 0 0  6
0 0 1 1 0  7
 0
 System of Equations (1): Example
10 l/s
22 l/s

10 l/s
18 l/s 15 l/s

 R1 Q1
n 1
0 0 0 0 0 0 1 0 0 0
0
 n 1
  Q1   50 
 0 R2 Q2 0 0 0 0 0 1 1 0 0 0  Q   0 
 n 1  2   
 0 0 R3 Q3 0 0 0 0 0 0 1 0 0   Q3   50 
 n 1    
 0 0 0 R4 Q4 0 0 0 1 0 0 1 0   Q4   0 
 n 1    
 0 0 0 0 R5 Q5 0 0 0 1 0 0 1  Q5   0 
 1 1 0   6   
n 1 Q 0 
0 0 0 0 0 R6 Q6 0 0 0
  Q   0 
 0 1 1     
n 1 7
0 0 0 0 0 0 R7 Q7 0 0
   H 2   0.010 
 1 1 0 0 0   
0.022 
1 0 0 0 0 0 0
H3
 0 1 0 0 1 0 0 0 0 0 0 0    
   H 2   0.010 
 0 0 1 0 0 1 0 0 0 0 0 0    
   H 5 0.018 
 0 0 0 1 0 1 1 0 0 0 0 0   
 H 6   0.015
 0 0 0 0 1 ECMM123
0 1 0 0 0 0 0 
 GGA Method
Differentiate system of equations (1) with respect to unknowns Q and H:

D  A12  dQ  dE 

           (2)
    
 A 21  0  dH   dq 
where:
n 1
D  k , k   nR Qij
dQ  Q ( k )  Q( k 1)
dH  H ( k )  H ( k 1)
dE  A11Q( k )  A12 H ( k )  A10 H 0
dq  A 21Q ( k )  q
 GGA Method (cont.)
D  A12  dQ  dE 
          
    
 A 21  0  dH   dq 

Solution to the above system of equations:

1
dQ   D A12  dE
    
     
dH   A 21 0   dq 
 GGA Method (cont.)
The inversion of the system matrix can be performed
by taking into account its partitioned structure:
1
 D A12   B11 B12 
   
   
 A 21 0  B 21 B 22 
This gives:

B 11  D 1  D 1 A 12 A 21D 1 A 12 
1
A 21D 1
1

B 12  D A 12 A 21D A 12 1
1


B 21  A 21D 1 A 12 
1
A 21D 1

B 22   A 21D 1 A 12 1
 GGA Method (cont.)
The solution of (2) now becomes:

dQ  B11dE  B12dq
dH  B 21dE  B 22dq

Therefore:
  A 21D A12  A 21D1  A11Q ( k )  A12 H ( k )  A10 H 0 
( k 1) 1 1
dH  H (k )
H

  A 21D A12  A  q
1 1 (k )
21Q
dQ  Q( k )  Q( k 1)  D1  A11Q ( k )  A12 H ( k )  A10 H 0 

D1 A12  A 21D1 A12  A 21D1  A11Q ( k )  A12 H ( k )  A10 H 0 

1

 D A12  A 21D A12  A Q(k )  q 

1 1 1
21
 GGA Method (cont.)
This gives:

  A 21D A12  A  q    A 21D A12  A 21D1A12 H ( k )

( k 1) 1 1 1 1
H H (k )
21Q
(k )

  A 21D A12  A 21D A11Q   A 21D A12  A 21D1 A10 H 0

1 1 1 (k ) 1 1

A   
1
 D1
A  A I  D 1
A Q (k )
 q  A D 1
A10 H 0 
21 12  21 11 21

Q ( k 1)  Q ( k )  D1 A11Q ( k )  D1 A10 H 0  D1 A12 H ( k 1)

  I  D1 A11  Q( k )  D1  A12 H ( k 1)  A10 H 0 
 GGA Equations (cont.)
Let us denote:

A   A 21D A12  1

1 1
F  A 21Q (k )
 q  A 21D A11Q (k )
 A 21D A10 H 0

Finally:

H( k 1)  A1F (3)

Q ( k 1)
Q(k )
D1
A
11Q
(k )
 A12 H ( k 1)
 A10 H0  (4)
 GGA Procedure
1. Initialise iteration counter k=0. Initialise Q(0) (by
e.g. assuming velocities v(0)) and calculate
H(0).
2. Update nodal heads by using equation (3).
3. Update pipe flows by using equation (4).
4. Increase k by 1 and check for convergence,
e.g.:
max Q i
( k 1)
Q
i
(k )
 eQ
i

If converged, stop. If not, go back to step 2.

 GGA Method: Summary
• Pros:
– Computationally efficient
– Numerically stable
– Can handle the change in status of system
components without changing the structure of
the linearised system matrix
• Cons:
– Bit more difficult to code than the LT method
 Pressure-driven Hydraulic Model
Qd
• Demand assumed to Wagner et al. (1988)
be a function of Qd,req
pressure
• Different formulations
of Qd(P) exist
• More realistic but also KIWA (1993)
more complicated to
model Pmin
P
Pmax
• Slows down the
convergence of the
steady-state solver
 Device Modelling: Pumps
• Impart energy to a fluid H
thereby raising its
hydraulic head
• Can be modelled as links
Q
• Principal data:
– pump curve(s)
– start and end nodes (if i j
modelled as a link)
• Fixed vs variable speed
H j  Hi  H Q 
 Device Modelling: Valves
• Different control valves
exist:
– PRV, PSV, FCV, TCV
– Other i j
• Principal data:
– Valve headloss
curve/equation Hi  H j  H  Q, 
– Valve specific data (e.g.
various settings, etc.)
– Start and end nodes (if
modelled as a link)
 Tree-like Pipe Systems
A QB
QD

D1,L1,k1 Q1 D Q2 B
hA Q3 D2,L2,k2
QC
D3,L3,k3
C

• Problem: Given reservoir A water level,

consumptions at nodes B, C & D and all
pipe characteristics (diameter, length and
roughness), find flows in pipes 1-3 and
heads at nodes B, C & D.
 Tree-like Pipe Systems (cont.)
EGL≈HGL
A QB
QD

D1,L1,k1 Q1 D Q2 B
hA Q3 D2,L2,k2
QC
D3,L3,k3
C

• Solution:
1. Phase I (upstream direction):
a) Calculate flows in all pipes: Q3=QC, Q2=QB and Q1=QD+QC+QB
b) Calculate headloss in each pipe hfi (i=1,2,3) using e.g. D-W
equation
2. Phase II (downstream direction):
a) Calculate nodal heads: hD=hA-hf1, hB=hD-hf2 and hC=hD-hf3
 Extended Period Simulation (EPS)
Modelling
• Adds time dimension to the analysis
• Modelled as a sequence of steady-state
conditions in the networks
• Initial conditions:
– Water level in all tanks at t=0
– Status of all automatically regulated devices at t=0
• Boundary conditions:
– Water level with time at all reservoirs (i.e. sources)
– Status of time controlled devices
 General EPS
Computational Algorithm
1. Initialise simulation time t=0. Find heads and
flows in the network given the initial conditions.
2. Increase t by t and update reservoir water
levels and statuses of time controlled devices by
using the boundary conditions. Update tank
water levels by using the flow rates from the
previous time step.
3. Solve network hydraulics for unknown junction
heads, pipe/link flows and statuses of
automatically regulated devices.
4. If t>tmax stop. Otherwise, return to step 2.
 EPS: Tank Level Updating
Differential equation:
dV (t )
 Qin (t )  Qout (t )
z dt
Zmax
Initial condition: V (t0 )  V0
zt+t Boundary condition: Vmin  V (t )  Vmax
zt
V(z) Solution using the Euler’s (first-order)
Zmin
Qin Qout explicit approximation:
Vt t  Vt   Qin,t  Qout ,t  t t=0,1,2,…

Water level zt+t determined from Vt+t

by using the tank volume curve V(z)
 EPS: Simple Operational Controls
• Changes the status or setting of a link
based on:
– Water level in a tank
– Pressure at a junction
– Simulation time
– Example: LINK 12 CLOSED IF NODE 23
ABOVE 20 (shuts down pump 12 when level
in tank 23 exceeds 20 m)
EPS: Rule-based Operational Controls
• Allow link status and settings to be based on a
combination of conditions that might exist in the network
after an initial hydraulic state of the system is computed.
• Example (shuts down a pump and opens a by-pass pipe
when the level in a tank exceeds a certain value and
does the opposite when the level is below another
value):
RULE 1:
IF TANK 1 LEVEL ABOVE 19.1 THEN PUMP 335 STATUS IS
CLOSED AND PIPE 330 STATUS IS OPEN
RULE 2:
IF TANK 1 LEVEL BELOW 17.1 THEN PUMP 335 STATUS IS OPEN
AND PIPE 330 STATUS IS CLOSED
WDS Hydraulic Modelling Software
• Commercial:
– WaterCAD
– H20Net/InfoWater
– MIKE Net
– SynerGEE
– Other
• Epanet2 (US EPA) Epanet2
 Summary
• Introduction
• Hardy-Cross method
– Example
• Linear Theory method
– Example
• Other methods
• Other hydraulic models