Transcript Using MIDUSS 98

Using MIDUSS 98
by
Alan A. Smith
Alan A. Smith Inc.
Dundas, Ontario, Canada
Getting Started
Required first steps:
Confirm that program is authorized
Accept statement of Disclaimer
Select system of units
Define an Output file
Set the time parameters
Selecting Units
Metric/SI or Imperial
Units are all kinematic (no force or mass)
mm/hr, metre/sec, cub.m, hectare-m
or
inch/hr, feet/sec, cub.m, acre-feet
Units cannot be changed after time
parameters have been defined
Output file
Contains a record of all commands, data
and results to allow design session to be
repeated exactly
Can be converted to an Input database
to run in Automatic mode
Should be stored in special folder for
each job
Time Parameters
Time step for
hydrology
Maximum
expected storm
duration
Maximum
expected
hydrograph length
Routing or stability time step t = t/N, N = 1,2,3,...
Other features & options
Show or hide status bar
Include full path in Output
file
Review Output file at any time
Use context sensitive help
Show or hide ‘tool-tips’
Select from ‘Other Options’
Defining a Design Storm
Define the time parameters
Select storm type - 5 options
Enter required parameters
e.g. Depth, duration etc.
Display rainfall as table and graph
Accept the storm
Define a 5-character descriptor
Storm Types Available
Chicago hyetograph
Huff rainfall distribution
Mass rainfall distribution (can be user defined)
Canada AES (Atmospheric Environment Service)
Historic storm (user defined)
Chicago hyetograph
iave
a

c
td  b
Huff Distribution
Huff storm - 1st quadrant
1st quadrant
Huff storm - 2nd quadrant
2nd quadrant
Huff storm - 3rd quadrant
3rd quadrant
Huff storm - 4th quadrant
4th quadrant
Mass Rainfall Distribution
Defined by series of uniformly spaced
vertical coordinates which increase
continuously from 0.0 to 1.0
Various standard distributions for North
America included with MIDUSS 98
User defined distributions are easily
defined using local data
Maximum number of points is unlimited
Mass Rainfall example
Historic Storm
15 values
still zero
Last value
entered by
user
City of Guelph Design
Storms
Return
period
a
(mm/hr)
b
(min)
c
r
2-years
743
6
.7989
0.4
5-years
1593
11
.8789
0.4
10-years
2221
12
.9080
0.4
25-years
3158
15
.9355
0.4
50-years
3886
16
.9495
0.4
100-years
4688
17
.9624
0.4
Estimating Catchment
Runoff
Direct runoff or
Effective rainfall
Storm
Initial abstraction
Losses
Infiltration
Surface depression
storage
Infiltration methods
Soil Conservation Service (SCS)
Curve Number (CN) method
Horton’s equation
‘moving curve’ method
Green & Ampt model
SCS Curve Number method
CN depends on soil type and pre-wetting
1000
S
 10
CN
inches
P(t) = depth of rainfall
25,400
S
 254 mm
CN
( P(t )  I a )
Q(t ) 
( P(t )  S  I a )
2
Q(t) = depth of runoff
Ia
= initial abstraction
S
= potential storage
CN = curve number  100
Horton equation
f capacity  f c   f 0  f c  e
t
K
Green & Ampt model
 MS 
f  K 1 

K 

where
M=moisture deficit
S =suction head
K =hydraulic conductivity
Rainfall-Runoff models (1)
Rainfall
Effective
rainfall
Infiltration Model
Runoff
Losses
Catchment Model
Losses subtracted from rainfall to get effective
rainfall which is then applied to catchment.
Rainfall-Runoff models (2)
Losses and infiltration
calculated along with
runoff as part of Runoff
Model
Rainfall
Runoff
Catchment Model
Losses and
infiltration
Surface
Depression
Storage
Calculating the Runoff (1)
Runoff from pervious and impervious
fractions computed and added together
Flow lengths can
be:(a) equal
(b) proportional
(c) user supplied
Calculating the Runoff (2)
Symmetrical catchment
Area = 2.2 ha
A
22000
L

 47m
2W 263  75  96
Overland flow
length can be
estimated as
area divided by
length of
stream bank
available
for inflow.
96m
One-sided catchment
Area = 2.4 ha
L
A 24000

 125 m
W
192
192m
Calculating the Runoff (3)
Overland flow routing choices:
Combine effective rainfall with:
triangular response function
rectangular response function
single linear reservoir response function
Combine infiltration & other losses with
outflow from idealized inclined plane.
(Similar to SWMM RUNOFF method)
Design of a Pipe
7
101
1
Runoff 102
8
2
3
4
102
9
10
103
11
5
6
In a tree network, each node can have only one outflow
link. Therefore we use the convention that link
numbers are the same as the upstream node number.
Get the Maximum Inflow
If no inflow hydrograph exists the user
can specify a peak flow for the design
Use Hydrograph|Add Runoff to update
Inflow hydrograph
Uniform Flow in Pipes
2
1
M
3
Q
AR S 0 2
n
D2
  sin  
A
8
PD

2
D

y  1  cos 
2
2
Solve for y0 using
3
 2 Q  5
 
f      sin   
Q 
 full 
2
5
0
Critical Depth in Pipes
Solution for Ycr is based on the
minimum energy criterion
A3 Q 2
f y 

0
T
g
T  2 Dy  y 2 

 T 

 2 tan 
 D  2y 
1
Q 2T
1
3
gA
A Trial Pipe Design
Table of feasible
designs for given Q
and ‘n’
Double click on a row
to test trial design
Click [Design] to
get results of partfull flow analysis
Surcharged Pipes
Due to closed top boundary resistance increases
as depth y approaches diameter D.
At y = 0.81963 D
Q = Qfull
1.2
Q/Qfull
1
0.8
When y = 0.93815 D
Q = 1.07571 Qfull.
0.6
0.4
0.2
0
0
0.5
y/D
1
Surcharged Pipes
Energy line
Q > Qfull
Water surface
Q = Qfull
Q < Qfull
MIDUSS 98 assumes uniform flow for
part-full pipes
Exercise 4
Design a pipe to carry 2 c.m/s when running
75% full with a gradient of 0.4% and n =
0.013
Check for surcharged hydraulic grade line if
discharge increases to 3 c.m/s
Channel Design
Channel design based on use of Manning eq. to find
normal depth Yo for a specified discharge.
Q0  f  y0 , Slope, Manning' n' , Geometry
Using Manning eq.
2
1
M
Q0 
A0 R0 3 S 0 2
n
M = 1.49 imperial
1.00 metric
A = flow area
R = hydraulic radius
S = bed slope
Channel Flow Assumptions
Flow is fully developed rough turbulent.
Channel is prismatic, i.e. cross-section is
constant along length.
Flow is uniform, i.e. Sf = S0.
A0, P0, R0 = f(Y0, geometry).
Cross-section is fixed boundary.
Simple Cross-section
T
General Trapezoidal
section can be:
rectangular
trapezoidal
triangular
non-symmetrical
GR
GL
B
y0
Complex Cross-section
Y
Y3
1
9
2
3
7
4
5
10
8
WL
6
X
X3
Datum
Cross-sections can be defined by a set of straight
lines joining up to 50 coordinate pairs. These can
be drawn graphically and edited numerically.
Defining the Discharge
Peak value of current Inflow
hydrograph if one exists.
User specified discharge if no
Inflow hydrograph is defined
Design of a simple channel
Peakand
Display
Plot
flow
table
design
is from
of
details
Depth
current
appear.
- Grade
Inflow
- Velocity
hydrograph
Enter
channel
depth
and
slope,
press
[Design]
Design of a complex channel (1)
Draw section and specify peak flow = 15 c.m/s
Design of a complex channel (2)
Check low flow channel for reduced flow = 1.5 c.m/s
Design of a complex channel (3)
Reduce
Manning
n=0.025
Increase width of low flow channel to 3.5 m
Design of a complex channel (4)
Check modified section for maximum flow of 15 cm/s
Exercise
Design a trapezoidal channel to carry 2 c.m/s with
gradient of 0.3% and n=0.04
Design a channel which includes a low flow channel
to carry maximum flow of 12 c.m/s and low flow of
2 c.m/s. Allow freeboard of 0.3 m. Try for gradient
= 0.3%, n=0.04 for main channel and n=0.02 in
low flow channel
Flood Routing definitions
Q(t)
lag
Peak flow attenuation
Inflow
at x
Recession limb
Rising
limb
Outflow at x+x
tp
time
c t
time t
time t+t
x
Flood Routing methods
Hydraulic
Uses both dynamic and continuity equations
Allows backwater effects to be modelled
Solution advanced by timestep t
Hydrologic
Uses only continuity equation
Cannot model backwater effects
Solution advanced downstream by x
Kinematic Wave Equation
Continuity with no lateral inflow yields:
Q A

0
x t
Q
t+ t
t
A
Q+Q
x
For quasi-uniform flow:
Q dQ
Q  f (WL) so that

A dA
Substitute and separate variables to get wave eq.
Q
dA  Q 
1 Q



x
dQ  t 
c t
where c = dQ/dA is wave celerity
or
Q 1 Q

0
x c t
Space-Time Coordinates
Time t
a x
3
8
5
Flow Q4
unknown
4
6
Nucleus
t
b t
1
x
7
2
Distance x
Continuity Around the Nucleus
Q Q6  Q5

x
x
Q Q8  Q7

t
t
Q5  b Q3  1  b Q1
Q6  b Q4  1  b Q2
Q7  a Q1  1  a Q2
3
8
5
1
4
6
7
2
adx
Q8  a Q3  1  a Q4
Q 1 Q

0
x c t
 c t 

Q6  Q5   Q8  Q7  0
 x 
bdt
Generalized Muskingum equation
c t 
Let   
and get Q4=f(Q1 , Q2 , Q3)


x


 b Q4   1  b  Q2   b Q3   1  b  Q1 
a Q3  1  a  Q4  a Q1  1  a  Q2  0
Collecting terms,
C4Q4  C1Q1  C2Q2  C3Q3
where C1  a  1  b 
C2  1  a   1  b 
C3  a  b 
C4  1  a   b 
Setting b = 0.5 yields

X  t 2 K
 1  X  t 2 K
  X  t 2 K
 1  X  t 2 K
Deriving the Diffusion equation
Non-centered finite difference
scheme creates a numerical error
or
Q 1 Q

  0
x c t
Q 1 Q x
 2Q
1  2a   2b  1  2  0


x c t
2
x
Convert the Wave equation to a
Diffusion equation
Diffusion coefficient is
related to channel conveyance
Q 1 Q
 2Q

D 2
x c t
x
  2Q
Q 1 Q 
K3


 2
2

dK
x c t

x
2Q
dh 

Determine weighting coefficients
Compare the two
equations for the
diffusion coeff. D
Q
D
2s f dQ
1  2a   2b  1
f(a,b,D)=0 leads to multiple
sets of (a,b) coordinates
for any value of D.

dh
x
1  2a   2 b  1 

2
Q
h f dQ
where h f  s f x
dh
Q
a  1
2h f dQ
; b  0.5
dh
Numerical Stability Criteria
Condition for numerical stability is
Unstable
a
b
1a
 c t 



1 b
 x 
Limits for x and t
2D
For b = 0.5 2a  1 
x
and
c t
2a 
 21  a 
x
2 D c t
1

or x  2 D  c t
x
x
For very long channels, route hydrograph over multiple
sub-reaches of length x=Length/N, N = 2,3,4...
From parts 1 & 2
Limits for x and t
2D
For b = 0.5 2a  1 
x
and
c t
2a 
 21  a 
x
2 D c t
1

or x  2 D  c t
x
x
For very long channels, route hydrograph over multiple
sub-reaches of length x=Length/N, N=2,3,4...
From parts 1 & 2
From parts 2 & 3
c t
2D
 2 1 
x
x
or
t 
x  2 D
c
For very short channels, use routing time-step equal to
sub-multiple of hydrology time step, t=t/N, N=2,3,4...
MIDUSS 98 Route Command
MIDUSS 98 Route Command
Details of last
conduit design
are displayed
Changes to x
or t reported
for information
User can change
computed X or K
values
Estimated
values of
weighting
coefficients
Results of Route command
Calculating celerity
Q
2
Design of a Detention Pond
Volume
WL
Discharge
Q(t)
Inflow
Peak outflow is
on recession
limb of inflow.
Outflow
Time
Types of Detention Pond
‘In-line’ storage reservoir with outflow control
device to reduce peak flow
‘Off-line storage reservoir with connection above
normal hydraulic grade line
On-site storage on parking lots or below ground in
oversized storm sewers or trench
On rooftops of proposed new commercial buildings
Theory of Reservoir Routing
QI2
QO2 = ?
Law of
Continuity
Inflow
QI1
Outflow
t
QO1
Inflow = Outflow + Rate of change of storage
Assume:(1) Storage depends only on outflow
(2) Reservoir surface is horizontal
(3) Water surface elev. is function of outflow
Theory of Reservoir Routing (2)
Inflow = Outflow +
Rate of change of storage
QI1  QI 2 QO1  QO2 S 2  S1


2
2
t
 2S 2
  2S1

QI1  QI 2  
 QO2   
 QO1  2QO1
 t
  t

f(QO)
f (QO2)  f (QO1)  QI1  QI 2  2QO1
QI1 + QI2 - 2QO1
Outflow QO
Outflow Orifice Controls
Submerged orifice
Q  Cc


d 2 2g H  2 d
3
4
Ccd

for
H d
H
d
Non-submerged orifice
5
H
Q  f  Cc g d 2
d 
for H  d
1.57
where
H
H 
f    0.496 
d 
d 
H 
 0.04 
d 
0.5
H
d
Outflow Weir Controls
Rectangular weir
Qcr  Cd B g ycr
3
2
H
Ycr
2
ycr  H
3
Triangular weir
Qcr  Cd
4
ycr  H
5
g
m 52
y
2 2 cr
Ycr
H
Storage Models
MIDUSS 98 provides 4 tools to assist in
defining the depth-storage relation.
“Rectangular” reservoir or pond
Oversized storm sewers
Wedge shaped storage (parking lots)
Rooftop storage
Rectangular Pond storage
Aj+1 = Lj+1 x Bj+1
Lj+1
Am
m
Lj
Aj = Lj x Bj
H
V  A j  4 Am  A j 1 
6
Aj  L j  B j
Am  L j  m H B j  m H 
A j 1  L j  2m H B j  2m H 
H
Aspect ratio
R = L/B
For irregularly shaped
ponds the aspect ratio R
is defined by:
2
Perim
eter
f ( R)  4 R  4  8 
R
Area
Oversized Storm Sewers
Weir & orifice
outflow control
D
S0
WL
IL
Datum
Wedge shaped Storage
Parking lot storage
created by restricting
capacity of catch basins
V

g
18

1
2

 g1 g 2  g 2 H 3
2
R1
Ponding depth H
g1
3 ft/ 0.92 m
Typical depth of exit pipe
below rim elevation
Roof top Storage
L/2
H
L/2
Roof slope S0
H
Linear Discharge weir
Q = K.H
e.g. Q = 24 litres/min/25mm head
Vol = f(H, L S0)
On-Site Storage Control
Commercial developments may have a percentage of
impervious areas of 85% or more. On-site storage is
often preferred to centralized storage for cost sharing,
quality control and spill control. Methods include:-
Rooftop storage
Parking lot storage
Underground storage
Schematic of Commercial Site
Parking and Roads
65%
Pervious
5%
Parking
95% impervious
1
2
Roads
90% impervious
Total Width W
3
Roads
33% B
Roof storage
75%total roof area
Building
footprint
30%
Schematic of Commercial Site
Parking and Roads
65%
Pervious
5%
Parking
95% impervious
1
2
Roads
90% impervious
Total Width W
3
Roads
33% B
Roof storage
75%total roof area
Building
footprint
30%
Example of On Site control
1
Total
area
10 ha
2
Roof 3.00
Parking
4.33
Parking &
roads 6.50
Imperv. 95% 4.11
Pervious 5% 0.22
Imperv. 90% 1.95
Roads
2.17
Pervious 10% 0.22
Grass 0.50
0.50
3
Part 3 pervious area
Part 3 impervious area
Part 3 total area
0.72
1.95
73%
2.67 imperv
Example of On Site control
Model Rooftop storage
Roof area
hectares
Store area
hectares
Area/drain
sq.metre
Drain flow
L/min/25mm
Roof slope
gH:1V
3.000
2.250
450.0
24.000
200.00
Outflow from Rooftop
Parking Lot storage
100.5
Inlet
Control
Device
100.0
99.0
Define Wedge storage
Wedge
invert
Grade 1
g1H:1V
Grade2
g2H:1V
Angle
subtended
Number
of wedges
100.00
60.00
120.00
90.00
67.00
Outflow from Parking storage
Parking lot storage (2)
Volume
Rim elevation
Discharge
Catch basin
Invert level
Rim capacity
Compare
outflow
with and
without
on-site
storage
0.889
0.492
Working with Files
Topics discussed
Types of files
Commands that use files
Storage arrays that interact with files
Naming a file
File formats
Types of Files
Type
Typical Name
Purpose
Output
Default.out
Records commands, data, results
Input data
Miduss.Mdb
Database for Automatic use
Hyetograph
5year.stm
Rainfall intensities
Hydrograph
Pond flow.hyd
Flowrates
Rainfall dist.
Huff1.mrd
Fractions of total rain depth
Diversion
Div12345.hyd
Flow removed from inflow
Junction
Hyd12345.JNC
Flow accumulated at junction
Help
Miduss98.hlp
Help system
Log
Miduss.log
Error reporting
Commands that use Files
File Type
Output
Related Command
Input
Most commands except Windows, Help, Show
(Table, QGraph),
Automatic (Create, Edit, Run)
Hyetograph
Hydrograph (FileI_O)
Hydrograph
Hydrograph (FileI_O)
Mass Rain Dist.
Storm (Huff, Mass distribution)
Diversion
Design (Diversion)
Junction
Hydrograph (Combine, Confluence)
Rules for File Names
Long names allowed. More than 11
characters used by DOS “nnnnnnnn.eee”
Names can include spaces, periods, e.g.
“Pond Inflow.Pre.005hyd”
Only 11 illegal characters, e.g.
“
\
/
:
*
?
<
>
|
Storage arrays that use Files
Array Name
Size
Contents
RainTemp()
NRain
Temporary storm until accepted by user
Rain()
NRain
Defined design storm
RainEffI()
NRain
Effective rainfall on Impervious areas
RainEffP()
NRain
Effective rainfall on Pervious areas
OvHyd()
MaxHyd
Total runoff hydrograph
OvHydI()
MaxHyd
Runoff from Impervious areas
OvHydP()
MaxHyd
Runoff from Pervious areas
Inflow()
MaxHyd
Inflow to pipe, channel, pond etc.
Outflow()
MaxHyd
Outflow from pipe, channel, pond etc.
TempHyd()
MaxHyd
Temporary junction or diverted flow
BkupHyd()
MaxHyd
Backup storage to allow ‘Undo’ command
Hydrograph File Formats
Record
Content
Example
1
Description
“Hydrograph #1…”
2
File type
4
3
Peak value
“0.432”
4
Gravity
9.81
5
Time step
5
6
No. of values
36
7
Value 1
0.000
8
Value 2
3.456E-03
:
:
:
N+6
Value N
4.567E-04
The FileI_O Command
The FileI_O Command
Select Read
or Write
Choose Rainfall
hyetograph or
flow hydrograph
Select type of
hyetograph or
hydrograph
If file is to be
created, enter
the name here
If file exists
(‘Read’) pick file
from this list
Pick drive from
drop down list
Define type of
file or “All files”
Navigate to folder
where file is found or
is to be created
Exfiltration Trench
Perforated distribution pipe
Outflow Control Device
Outflow Q
Exfiltration X
Water table
Idealized diagram of exfiltration trench
Purpose of Exfiltration
Encourage return of storm runoff to the
groundwater
Reduce the hydraulic load on the minor
(e.g. piped) system
Improve quality of runoff by removal of some
particulate matter
Reduce thermal impact on the runoff
Split inflow hydrograph into two components
of Outflow and Groundwater recharge
Basic Theory
Inflow = Outflow + Exfiltration + Change of storage
I1  I 2 Q1  Q2 X 1  X 2 V2  V1



2
2
2
t
I
V
X
 2V2
  2V1

I1  I 2  
 Q2  X 2   
 Q1  X 1   2Q1  2 X 1
 t
  t

I1  I 2  f V2 , Q2 , X 2   f V1 , Q1 , X1   2Q1  2 X1
f (Q2 )  f Q1   2Q1  2 X1  I1  I 2
Q
Trench Cross-section
Topwidth T
For laminar flow:
Filter
Q
 q  KS f
A
Clear stone
Height H
P + y/2
yeff
Invert
elevation IL
B
P=IL-G
Water table elevation G
where
Sf 
IL  G  y
2
IL  G
y
 1
2IL  G 
K = hydraulic
conductivity
Define Trench parameters
Data window is opened by using Geometry/Trench
menu command from Trench Design form.
Define Trench parameters
Define outflow control
[Compute]
Plot V,Q=f(H)
Defining the Trench Pipes
Main storm sewer is
solid 450 mm pipe
with invert =100.85
Two 200 mm diam.
perforated pipes are
plugged at down stream
end to distribute inflow
along trench
Press [Compute]
to update volume
table
Pipes positioned
graphically with
clearance shown.
Position refined
by editing table.
Results of Routing
The Etobicoke Trench
What is Automatic mode?
In Manual mode all commands and required
data are entered by the user. These
commands, data and main results are
copied to the Output file. This
information allows the design session to be
repeated.
In Automatic mode, commands and data
are read from an input file with no entry
required from user.
Reasons for Automatic mode
User can complete a design in two or
more sessions
Repeat a design with a different storm
Revise and compare the design of one
or more components
Add or insert commands in Manual mode
to change hydrology simulation or
design
Files used in Automatic mode
‘Output’ is the
file created in a
previous Manual
session.
Automatic
Create Miduss.Mdb
Edit Miduss.Mdb
Run Miduss.Mdb
‘New Output’ is modified
output file created during
Automatic design session.
‘Miduss.Mdb’ is a
database file created by
the Create Miduss.Mdb
command.
Structure of the Database
Index
Command
Parameter value
Description, data or
results
Advantages of a Database
Direct access to records speeds processing
Database file can be ‘bound’ to a grid
File can be viewed, edited and used as input
source simultaneously
Setting a command as a negative number
causes a continuous Automatic run to stop and
revert to step-by-step EDIT mode or switch to
Manual mode
Automatic processing can re-start where it was
halted
Steps to Run Automatic mode
Create the input database Miduss.Mdb
using the Creat Miduss.Mdb command
Review and/or Edit the database - e.g.
change Command numbers to negative
value to halt processing.
Use the Run Miduss.Mdb command to
process the file in any of three modes
EDIT, STEP or RUN
Using the Control Panel
The Control Panel is displayed when
the Run Miduss.Mdb command is used
RUN starts
continuous
processing of
the data
STEP executes
commands one
by one without
any chance to
modify data
EDIT executes next command and lets
you alter data and [Accept] the result