Classification of Flow Regimes THE CRITICAL DEPTH CONCEPT Since the flow within the open channels occur due to gravity-slope coupled effect, the velocity and the occurance of the flow type is definitely controlled by the steepness of the channel bottom slope. Hence one should catagorize the slopes based on the discharge ‘Q’ and the cross-section parameters like ‘A’, ‘T’, ‘n’ etc.’ To classify any slope, the required unique parameter is the critical flow depth ‘y cr ’ of that cross-section. This depth is a fictitious water depth within any channel cross-section that gives a useful indication for classifying the flow regimes and hence the slopes. The critical flow depth ‘y cr ’ of any flow for any cross-section can only be obtained using Froude Number approach. Fr 2 = Q 2 T cr gA cr 3 =1
Transcript
Classification of Flow RegimesTHE CRITICAL DEPTH CONCEPT
Since the flow within the open channels occur due to gravity-slopecoupled effect, the velocity and the occurance of the flow type isdefinitely controlled by the steepness of the channel bottomslope. Hence one should catagorize the slopes based on thedischarge ‘Q’ and the cross-section parameters like ‘A’, ‘T’, ‘n’ etc.’
To classify any slope, the required unique parameter is the criticalflow depth ‘ycr’ of that cross-section. This depth is a fictitiouswater depth within any channel cross-section that gives a usefulindication for classifying the flow regimes and hence the slopes.
The critical flow depth ‘ycr’ of any flow for any cross-section can only be obtained using Froude Number approach.
Fr2 =Q2Tcr
gAcr3 = 1
Rearranging the above equation yields,
The Hydraulic depth is defined as D=A/T so the final form of theequation becomes
or
This is basically the Froude Number Fr, which is 1 at critical flow.
For a rectangular channel Dcr=Acr/Tcr=ycr , so the Froude number for critical flow becomes:
HOW TO OBTAIN CRITICAL DEPTH ‘ycr’ FOR ANY CHANNEL CROSS-SECTION
• USE THE DISCHARGE ‘Q’ VALUE THAT IS FLOWING INTHAT CROSS-SECTION (THIS VALUE IS EITHER GIVENOR CAN BE CALCULATED USING MANNING’SEQUATION).
• INSERT THIS DISCHARGE ‘Q’ VALUE INTO FROUDE ‘Fr’EQUATION AND EQUATE IT TO 1.00. WHILEESTABLISHING THIS EQUATION, LET THE CRITICALFLOW DEPTH ‘ycr’ TO BE A VARIABLE. THEN SOLVETHE EQUATION TO GET ‘ycr’. (DEPENDING ON TYPEOF CROSS-SECTION, THE TOP WIDTH ‘T’ MAY NOT BEA CONSTANT VALUE! MAY DEPEND ON ‘ycr’.)
Fr2 =Q2Tcr
gAcr3 = 1
HOW TO OBTAIN CRITICAL SLOPE ‘scr’ FOR ANY SIMPLE – SINGLE CHANNEL CROSS-SECTION
• TO OBTAIN THE CRITICAL SLOPE ‘scr’ FOR ANY REACHFIRST DETERMINE THE CRITICAL DEPTH ‘ycr’.
• THEN INSERT THIS ‘ycr’ IT INTO THE MANNING’SEQUATION BY SUBSTITUTING THE VALUES OF AREA‘Acr’ AND PERIMETER ‘Pcr’ (CALCULATED BASED ONTHE CRITICAL DEPTH VALUE) WITH THE PROPERMANNING’S ROUGHNESS COEFFICIENT ‘n’ OF THATREACH SO THAT THE SAME DISCHARGE ‘Q’ VALUEWILL BE SATISFIED. HENCE, THE SOLUTION GIVES THECRITICAL SLOPE ‘scr’.
Q =Acr
5 3
n.Pcr 2 3 scr
HOW TO OBTAIN CRITICAL SLOPE ‘scr’ FOR ANY SIMPLE – COMPOSITE CHANNEL CROSS-SECTION
• FOR THE COMPOSITE CHANNEL CROSS-SECTION(different ‘n’ values for the perimeter) THEEQUIVALENT ‘nequi’ SHOULD BE DETERMINE BASEDON THE CALCULATED CRITICAL DEPTH ‘ycr’.
Q =Acr
5 3
nequi.Pcr 2 3 scr
CRITICAL SLOPE ‘scr’ FOR COMPOUND – COMPOSITE CHANNELS
• ONCE THE CRITICAL DEPTH ‘ycr’ IS OBTAINED, THE CRITICALSLOPE ‘scr’ FOR COMPOUND CHANNELS IS DETERMINED BYSUPER IMPOSING TECHNIQUE.
• THERE ARE SEVERAL SOLUTION APPROCHES (METHODS)SUGGESTED BY DIFFERENT RESERCHERS. AMONG THEM 3SOLUTIONS APPROACHES WILL BE DETAILED..
CRITICAL SLOPE ‘scr’ FOR COMPOUND – COMPOSITE CHANNELS
SOLUTION APPROACH # 1
INSERTING THE ‘ycr’ INTO THE MANNING’S EQUATION BYSUBSTITUTING THE VALUES OF AREA ‘Acr’ AND PERIMETER ‘Pcr’FOR EACH PART OF THE COMPOUND CHANNEL APPROPRIATELYWITH THEIR PROPER EQUIVALENT MANNING’S ROUGHNESSCOEFFICIENTS ‘nequi-cr’ (calculated based on the critical depth‘ycr’ value) AND EQUATING IT TO THE TOTAL DISCHARGE ‘QTotal’VALUE WILL GIVE THE CRITICAL SLOPE ‘scr’.
QTotal =Acr
5 3
nequi−cr. Pcr 2 3
Part 1
+Acr
5 3
nequi−cr. Pcr 2 3
Part 2
+⋯ scr
CRITICAL SLOPE ‘scr’ FOR COMPOUND CHANNELS
SOLUTION APPROACH # 2
SIMPLY EQUATING THE PARTS SEPERATELY BASED ON THEIRRELATIVE DISCHARGES, CRITICAL SLOPE FOR EACH PART ISDETERMINED. THE EQUIVALENT CRITICAL SLOPE FOR THEWHOLE CHANNEL CROSS-SECTON IS OBTAINED(APPROXIMATELY) BY APPLYING THE GEOMETRIC AVERAGE (for2 parts use square-root, 3 parts use cubic root, 4 parts 4th
root...) METHOD.
QPart 1 =Acr
5 3
nequi. Pcr 2 3
Part 1
scr−part1
QPart 2 =Acr
5 3
nequi. Pcr 2 3
Part 2
scr−part2
scr−equivalent =2 scr−part 1. scr−part 2
CRITICAL SLOPE ‘scr’ FOR COMPOUND CHANNELS
SOLUTION APPROACH # 3
CONSIDERING THE GIVEN COMPOUND CHANNEL CROSS-SECTION AS IF IT IS A SINGLE CROSS-SECTION ANDUSING THE RELEVANT PARAMETERS ACCORDINGLY BYINSERTING THE CRITICAL DEPTH VALUE (even for the‘nequi-cr total’ based on ‘ycr’), A REPRESENTATIVE SINGLECRITICAL SLOPE VALUE CAN BE CALCULATED.
QTotal =Acr −total
5 3
nequi−cr total. Pcr−total 2 3
scr
Question 8:1 For the given symmetrical trapezoidal channel cross-section of side slopes 1:2 (V:H) having a uniform flow depth yn=2.00 m. Its Mannings roughness coefficient is n=0.016 and the longitudinal slope s0= 0.001. Find:a) the discharge passing from this cross-section ‘Q’?b) the critical depth ycr?c) the flow regime?d) the critical slope?
A concrete lined trapezoidal channel with steady and uniform flow has a normal depth
of 2.0 m. The base width is 5.0 m and both of the side slopes are equal 1 V: 2 H. Manning’s n
can be taken as 0.015 and the bed slope sb=0.001. Calculate the discharge and the mean
(average) velocity vav of this open channel.
5.00 m
2
1 2.00 m
a) Q = 5∗2 + 2∗ 4 2 ∗2 5 3
0.016∗ 5+ 22+42+ 22+42 2 3 0.001 = 42.176 m3/s
b) Fr2 =Q2T
gA3= 1.0
42.1762∗(5+2ycr+2ycr)
9.81∗ 5ycr+2ycr
2
2+
2ycr2
2
3 = 1.0
ycr = 1.561 m
Note that as flow depth changes T also changes.
c) Fr2 =Q2T
gA3
Fr2 =42.1762∗[5+ 2∗2 + 2∗2 ]
9.81∗ (5∗2)+2∗22
2+
2∗22
2
3 = 0.404 < 1.0
SUB-CRITICAL REGIME
d) To obtain a critical slope insert the critical depth based values (P, A & Q) into Mannings eqn.
scr= 0.00263 given slope 0.001 (smaller than critical) Mild slope
Previous Exam Question:
For the below given single-composite channel symmetric cross-sections, write the
correct equation of:
a) the discharge ‘Q’
b) the critical Froude number ‘Fr’
Longitudunal slope is also called REACH
Beginning of the reach is called upstream and
End of the reach is called downstream.
How to Express the Longitudinal Slope
1- As decimal notation: S = 0.0015
2- As percentage: S = % 0.15
3- As degree angle: θ = 0.086° S = tan(0.086)
4- As topographic elevation: S = ΔzA-B/L
S = (139.72-137.39)/1553.333=0.0015
8.1 Classification of SlopesSince the flow within the open channels occur due to gravity-slope effect, the velocity and the occurance of the flow type is definitely controlled by the steepness of the channel bottom slope.
Procedure for finding the flow regime of any reach:
1- if the given data is enough, directly determine the Froude Number value where:
Fr < 1 flow regime is sub-criticalFr > 1 flow regime is super critical
2- if the given data is not enough, then, obtain the critical flow depth ycr
by using Fr2 =Q2T
gA3= 1
Then;i- either compare the critical flow depth ycr with uniform flow depth yn:
ycr < yn the flow regime is sub-criticalycr > yn the flow regime is super critical
ii- or subsititute the critical flow depth ycr to Mannings equation and obtain the critical slope s0cr then compare it with given slope s0:
s0 < s0cr slope is MILD expected uniform flow regime is sub-criticals0 > s0cr slope is STEEP expected uniform flow regime is super critical
Example 8.2:A rectangular channel of bottom width 4.0 m carries an
average discharge Q = 3 m3/sec. If the Manning’s roughnesscoefficient is n = 0.019 and the average bottom slope of thechannel is 0.004 (=0.4% = 0.2292°),a) determine yn & ycr ,b) classify the slope,c) find the regime of this flow (Fr),d) draw the reach by showing these depths.
Example 8.3:A rectangular channel of bottom width 4.0 m carries an
average discharge Q = 3 m3/sec. If the Manning’s roughnesscoefficient is n = 0.019 and the average bottom slope of the channelis 0.02, determinea) determine yn & ycr ,b) classify the slopec) find the regime of this flow (Fr)d) draw the reach by showing the depths.
In nature, several successive longitudunal channels ofdifferent slopes and varying cross-sections are forming theopen channel systems. Hence, the uniform flow depthdefinitely varies along the flow direction and non-uniformflow zone(s) form(s) along some places.For any channel: bed slope and/or cross-section and/or any obstacle within the channelcauses the flow to be non-uniform where the flow may be gradually varied or rapidly varied type.
Hence for any channel cross-section once the slopes and/or flow regimes occurs, different water surface profiles observed.
Non-uniform Flow Regions
Gradually Varied Flow (GVF)no rapid fluctuations of water and the slope is small, no discontinuites or zigzags along the water surface.Occurs due slope and/or cross-sectional variations.In this case the regime of the flow does not change...
Rapidly Varied Flow (RVF)the water depth along the flow direction varies incomperatively short distance; discontinuities and zigzags occurs along the water surface.Hydraulic jump is a good example...
In this case the regime of the flow changes...
For each type of channel, the flow depth may occur at any 1 of the 3 different regions:
Region 1: Space above the top most line. Above than any flow depth around that free surface region M1 or C1 or S1 ...
Region 2: Space between the top line and the next lower (second) line. Region between normal and critical depth M2 or S2 ...
Region 3: Space between the second line and the bed. Lower than any flow depth near the channels bottom region M3 or C3 or S3 ...
Region 1
Region 2
Region 3
Water Surface ProfilesGradually Varied Flow (GVF) based on the slope ‘s0’ type of the channel are lettered as (M, C, S, H, A) the water depths ‘y’ location within the possible regions numbered
and named as (1, 2, 3).
Region1: y > yn and y > ycr. The water depth is above both ofM1; C1 ; S1. these depths
Region 2: yn > y > ycr or ycr > y > yn. The water depth is betweenM2; S2; H2; A2. these depths.
Region 3: y < yn and y < ycr . The water depth is below both of M3; C3; S3; H3; A3. these depths.
Region 1
Region 3
Region 2
NDL
CDLNDLCDL
NDLCDL
Region 1 Region 2 Region 3
NDL
CDL
NDL
CDL
NDL
CDL
Region 1 Region 2 Region 3
NDL
CDLNDL
CDL
NDL
CDL
Region 1 Region 2 Region 3
Region 1 Region 2 Region 3
Region 1 Region 2 Region 3
Occurence of Non-uniform flow zones
due to differentsuccessive reachesand the occured
water surface profiles
WATER SURFACE PROFILED DUE TO DIFFERENT LONGITUDUNAL CHANNAL CROSS-SECTIONS(GRADUALLY VARIED FLOWS)
1- MILD SLOPE TO MILDER SLOPE ‘M1’ SLOPE [y1< y2]
M1
MILD MILDER
2- MILD SLOPE TO LESS MILD SPOPE ‘M2’ SLOPE [y1> y2]
Ykr MILD
LESS MILD
GVF Curve OCCURS within the UPPER slope
Sub-critical (Fr < 1) Sub-critical (Fr < 1)
Sub-critical (Fr < 1)
Sub-critical (Fr < 1)
Question 8.4:A rectangular channel of bottom width B = 9.65 m carries an
average discharge of Q = 10.845 m3/s. There are two succesive long reaches. The first reach has s01=0.0022 and n1=0.018; and the second reach has s02=0.0019 and n2=0.021a) Classify the slopes,b) Draw the longitudunal flow profile by showing the flow depths andc) Name the occurance of non-uniform portion.
3- STEEP SLOPE TO STEEPER SLOPE ‘S2’ SLOPE [y1> y2]
STEEP
STEEPER
4- STEEP SLOPE TO LESS STEEP SLOPE ‘S3’ SLOPE [y1 < y2]
S3
STEEP LESS STEEP Super critical (Fr >1)
Super critical (Fr >1)
Super critical (Fr >1)
Super critical (Fr >1)
Question 8.5:A rectangular channel of bottom width B = 7.55 m carries an
average discharge of Q = 8.965 m3/s. There are two succesive long reaches. The first reach has s01=0.018 and n1=0.018; and the second reach has s02=0.013 and n2=0.021. a) Classify the slopes,b) Draw the longitudunal flow profile by showing the flow depths andc) Name the occurance of non-uniform portion.
7.55 m
0.0018
s01= 0.018n1= 0.018
s02= 0.013n2 = 0.021
Answer: ycr= 0.523 m
yn1= 0.343 m yn2= 0.418 mscr1= 0.0047 scr2= 0.0064s01: STEEP s02: LESS STEEP
S3 SLOPE
(Hydraulic drop)
GVF Curve OCCURS ON BOTH slopes
‘Hydraulic Drop’M2- S2 curves
Super critical (Fr >1)
Sub-critical (Fr < 1)
Example 8.6:A rectangular channel of bottom width B = 6.55 m carries
an average discharge of Q = 8.965 m3/s. There are two succesivelong reaches where first reach s01=0.0032 and n1=0.018; and thesecond reach s02=0.013 and n2=0.021.1- Classify the slopes,2- Draw the longitudunal flow profile and show flow depth and namethe occurance of non-uniform portion.3- Estimate the Length of each GVF curve (LM2, LS2).
Answer1- sb1: Mild (scr1 = 0.0047)
sb2 : Steep (scr2 = 0.0064);2- y1: 0.653 m, y2: 0.462 m , ycr:0.576 m
M2 - S2 . 3. LM2 =16.91 m, LS2 =14.05 m.
s01=0.0032n1= 0.018
s01=0.013n1= 0.021
6- STEEP SLOPE TO MILD SLOPE ‘M3 with Hydraulic jump or SLOPES [y1 < y2]
Hydraulic jump with S1’
Sub-critical (Fr < 1)
Sub-critical (Fr < 1)
Super critical (Fr >1)
Super critical (Fr >1)
Question 8.7T:A rectangular channel of bottom width B = 6.55 m carries an
average discharge of Q = 8.965 m3/s. There are two succesive long reaches. The first reach has s01=0.0032 and n1=0.018; and the second reach has s02=0.0013 and n2=0.021. a) Classify the slopes,b) Draw the longitudunal flow profile by showing the flow depths,c) Name the occurance of non-uniform portion.
s01=0.0032n1= 0.018
s02=0.0013 n2= 0.021
Question 8.8T:A common rectangular channel cross-section of bottom width
B = 9.65 m carries an average discharge of Q =10.845 m3/s. There are three succesive long reaches as detailed with relevant characteristics. a) Classify the slopes,b) Draw the longitudunal flow profile by showing the flow depths,c) Name the occurance of non-uniform portions.
Answer: ycr= 0.505 m
yn1= 0.732 m yn2= 0.6342 m yn3= 0.427 m scr1= 0.0062 scr2= 0.0045 scr3= 0.00316s01: MILD s02: MILD s03: STEEP
