nerg Conept Components of the energ euation z is the elevation head is the pressure head-potential head V /g is the dnami head-kineti head H z + + V g V g S f x V g GL HGL S o x x
SPCIFIC NRGY CONCPT Speifi nerg,, is the energ referred as the hannel bed as datum Taking the datum z0 as the bottom of the hannel, the speifi energ is the sum of the depth of flow and the veloit head. V Q + + g g[ A ( )] AA()
SPCIFIC NRGY CONCPT V + g V + g V V + + + (Sf So) x g g + ( S f S o ) x If hannel bottom is horizontal and no head loss For uniform flow S f S o then
Speifi-nerg Curve For a given Q, is onl a funtion of, i.e: () V Q + + g g A [ ( )] Q [ ] g ( ) * A( ) onstant A Qons. 45 o B
Speifi-nerg Curve For a given Q, is onl a funtion of, i.e: () V Q + + g g A ( ) * A( ) onstant [ The plot of vs is alled speifi-energ urve. Above euation has two asmptotes: (-)0 and 0 ( )] Q [ ] g in fat one setion of the urve falls within the 45 o angle between these two asmptotes in the first uadrant. There is another setion of the urve shown as broken line, but this is of no pratial interest as it ield negative values for.
A Qons. V g Q [ ] g ( ) A( ) onstant 45 o V g B
If we regard this urve as a means of solving.: V Q + + g g A ( )] for, given and Q, the three solutions of ubi are learl shown b drawing a vertial line oresponding to the given value of. Onl two of them are phsiall real, so for given values of and Q, there are two possible depths of flow, unless the vertial line referred misses the urve altogether. These two possible flow depths, for a given and Q, are referred as alternate depths. One ma sa that the urve represents two possible regimes of flowslow and deep on the upper limb, fast and shallow on the lower limbmeeting at the rest of the urve, C. Other urves might be drawn for other values of Q; sine, for a given value of, inreases with Q, urves having higher values of Q will our inside and to the right of those having lower values of Q. [
45 o V g Q>Q>Q Q>Q Q C V g
Speifi nerg for retangular hannels For retangular hannels, one an define unit disharge,, as: Q V ( b) V b b V Speifi nerg,, for retangular hannels is defined in terms of unit disharge as: + V g + g and speifi- energ urve is drawn for a given unit disharge.
45 o V g C V g > > >
Speifi nerg Plot vs for onstant Q as to see breakdown of into pressure () & dnami (V /g) heads as 0 for large reahes a minimum 4 What is the phsial interpretation of loal minimum? V /(g) 0 45 0 4
Minimum Speifi nerg Reall Froude Number Fr V Q A Ë gd gd Q A Fr gd Q gda Q A g T Fr A Q T ga F r Q T ga
d d d d + Q + g Minimum Speifi nerg d da Q ga A For a given Q, when speifi energ is minimum? F r Q T ga Q T ga or sine da d Q g d d F d r A 0 da d Af() d d F r, i.e: flow is ritial T 0 dat*d da d T
A Qons. V g F r < 45 o F r C V g ( ) [ A( ) ] onstant F r > B Q g
SPCIFIC NRGY CONCPT For a given Q, () onl. The plot of vs. gives Speifi nerg Curve. V Q + + g ga As Q inrease, the urves will move towards right
Charateristis Of The Speifi nerg Curve min Critial depth Curve has asmptotes: & 0 lines Curve has limbs AC and BC Limb AC approahes the horizontal axis as 0 Limb BC approahes to line as On this urve Q is onstant At an point P on this urve, the ordinate represents the depth and the absissa represents the speifi energ. For a given speifi energ, there are possible flow depths: and
Charateristis Of The Speifi nerg Curve At point C, the speifi energ is minimum Minimum speifi energ orresponds to ritial state of flow, i.e. Fr. At the ritial state, the two alternate depths beome one, whih is known as ritial depth,. min If <, V > V Fr > the limb AC orresponds to superritial flow and is superritial depth V Q T Fr Fr gd ga If >, V < V Fr < the limb BC orresponds to subritial flow and is subritial depth The depths and are alled alternate depths.
Speifi nerg For retangular Channel A b Q unit disharge b Q AV ( b) V + V g + g
Critial Flow:Retangular hannel Q T ga T b Q b A b T g b b g A g g / Onl for retangular hannels! Given the depth one an find the disharge V g V
Critial Flow Relationships: Retangular Channels / sine V g V g Froude number for ritial flow V g V g V g + V g + veloit head 0.5 * depth
Critial Depth Minimum energ for 4 d d 0 kineti potential! V g Fr 0 0 4 Fr> Superritial Fr< Subritial
Charateristis of Critial Flow Arbitrar Cross Setion Retangular Cross Setion T A g Q Fr g g Fr D D g V + g V For a given Q, min For a given, min For a given speifi energ, For a given speifi energ o, Q Q max o, max
H H Disussion A long retangular hannel arries water with a flow depth of on a horizontal hannel. If there is a rise on the hannel bed: a)what is the relation of total head between setion and, if head loss assumed to be negligible? H H b)what is the relation of speifi energ between setion and? + )How does the water surfae profile hange? z + z + z Side view
CHANNL TRANSITION for retangular hannels Change on the bottom elevation of hannel Change on the width of the hannel Change on the bottom elevation and width of the hannel
Q Upward Step-Constant width VA V ( b) V ( b) H H ) Subritial flow g / + z
Upward Step-Constant width H H g + z /
Downward Step-Constant width H H z + ) Subritial flow / g ) ( ) ( b V b V VA Q
Downward Step-Constant width H H g + z /
Upward & Downward Step-Constant width Subritial flow + z + z + z ()
Channel xpansion (onstant bed elevation) H g H /
Channel Contration (onstant bed elevation) H g H /
Figure 0.7
Speifi nerg: Step Up Additional Consideration z 0 4 0 4 Z z H H g / +. m
CHOKING
CHOKING
xample Water is flowing in a retangular hannel. Find the hange in depth and in absolute water level produed b a smooth downward step of 0.0 m if the upstream veloit and depth are given as. a) V m/s and m. b) V 5 m/s and 0.60 m. abs? V m/s m z0 m Datum () () b
xample Solution part a nerg n between () and () : +Dz +h loss 0 V + +.46m.46 + 0.0 g 9.6 +, V 9m g 9.76 + + 9.6 4.8 / s.76 m There are possible solutions. To determine whih one ours ompute upstream Froude number. V F 0.55 < g 9.8 subritialflow. Therefore, will orrespond to subritial flow.
xample Solution part a.40 m 9 m /s must be greater than m. The root of greater than m an found b trial and error as;.00 m.40 m abs - ( z + ).40 - (.0 + 0.0 ).46 m z0.0 m.76 m abs 0.0 m. abs? V m/s m z0 m Datum () () b
xample Solution part b V 5m/s 60m abs? z0 m b nerg n between () and () : +Dz +h l 0 V 5 z + + h l + 0.60 +.874m g 9.6 0.0 +.874.74 m +, V 5 0.6 m / s g 0.4587.74 m +.74 m + 9.6 There are possible solutions. To determine whih one ours ompute upstream Froude number. V 5 Fr.06 > g 9.8 0.60, superriti al flow
xample Solution part b must be smaller than 0.60 m m /s an be solved b trial and error to obtain 0.58 m. Or 0.5 m. 0.60m 0.5m.874m z0.0m.74m abs ( z + ) - ( 0.0 + 0.60 ) - 0.5 abs 0.7 m. V 5m/s 60m abs? z0 m b
Solution of Speifi nerg uation + g + + g C Subritial Root * C Superritial root * C
Koh Parabola For a given, Q [ga ( )] / Q Q() [() for retangular hannels] The plot of Q vs. [ vs ] gives Koh Parabola. 0 Fr < 0 Fr < Fr Fr Fr > Fr > Q Q Q max max
Variation of speifi energ and (unit) disharge with depth: (a) versus for onstant ; (b) Q versus for onstant.
xample 4 For an approah flow in a retangular hannel with depth of.0 m and veloit of. m/s, determine the depth of flow over a gradual rise in the hannel bottom of z 0.5 m. Repeat the solution for z 0.50 m. GL V. Y.0 v /g z 0.5 datum Assuming no energ loss through the transition : v v + + + z g g whih sas for a onstant total energ (T), there is a speifi energ loss between - of z (meters).
Steps Determine Condition (sub- or super-) Calulate ritial depth, ompare to sub super depth g / Y [(*.) / 9.8] / Y.54 m Thus, the flow is subritial Solve speifi energ euation for orret root for, V Q VA V ( b) V ( b) or V (.*)/ + [(.) / (*9.8)] + {[(.*) / ] }/*9.8+ 0.5 Solve for Water surfae elevation + z
Repeat for a step of 0.50 m Chek value of min Determine value of ( - z) Find <min (CAN T B!!!!) Flow baks-up to allow passage of Y inreases; onditions at step are ritial, with ; min