CE 6403 APPLIED HYDRAULIC ENGINEERING UNIT - II GRADUALLY VARIED FLOW Dynamic equations of gradually varied and spatially varied flows - Water surface flow profile classifications: Hydraulic Slope, Hydraulic Curve - Profile determination by Numerical method: Direct step method and Standard step method, Graphical method - Applications. BY. MR.KUMAR.G M.E. STRUCTURAL ENGINEERING ASSISTANT PROFESSOR DEPARTMENT OF CIVIL ENGINEERING SRI VENKATESWARA COLLEGE OF ENGINEERING SRIPERUMBUDUR 60 7
Dynamic Equation of Gradually Varied Flow Objective: get the relationship between the water surface slope and other characteristics of flow. The following assumptions are made in the derivation of the equation. The flow is steady.. The streamlines are practically parallel (true when the variation in depth along the direction of flow is very gradual). Thus the hydrostatic distribution of pressure is assumed over the section. 3. The loss of head at any section, due to friction, is equal to that in the corresponding uniform flow with the same depth and flow characteristics. (Manning s formula may be used to calculate the slope of the energy line) 4. The slope of the channel is small. 5. The channel is prismatic. 6. The velocity distribution across the section is fixed.. 7. The roughness coefficient is constant in the reach.
Consider the profile of a gradually varied flow in a small length dx of an open channel the channel as shown in the figure below. The total head (H) section is given by: at any H = Z + y + V g Taking x-axis along the bed of the channel and differentiating the equation with respect to x: dh dx dz dy = + + dx dx d dx V g
dh/dx = the slope of the energy line (S f ). dz/dx = the bed slope (S 0 ). Therefore, dy Sf = S0 + + dx d V dx g Multiplying the velocity term by dy/dy and transposing, we get dy dy d V + = S S dy d V 0 f + dx dy dx g or dx dy g = S 0 S f dy dx = S 0 d + dy S f V g This Equation is known as the dynamic equation of gradually varied flow. It gives the variation of depth (y) with respect to the distance along the bottom of the channel (x).
The dynamic equation can be expressed in terms of the discharge Q: dy dx = S 0 S f Q T g A 3 The dynamic equation also can be expressed in terms of the specific energy E : dy dx = de / dx Q T g A 3
Depending upon the type of flow, dy/dx may take the values: (a) dy dx = 0 The slope of the water surface is equal to the bottom slope. (the water surface is parallel to the channel bed) or the flow is uniform. (b) dy dx = positive The slope of the water surface is less than the bottom slope (S 0 ). (The water surface rises in the direction of flow) or the profile obtained is called the backwater curve. (c) dy dx = negative The slope of the water surface is greater than the bottom slope. (The water surface falls in direction of flow) or the profile obtained is called the draw-down curve.
Notice that the slope of water surface with respect to horizontal (S w ) is different from the slope of water surface with respect to the bottom of the channel (dy/dx). A relationship between the two slopes can be obtained: Consider a small length dx of the open channel. The line ab shows the free surface, The line ad is drawn parallel to the bottom at a slope of S 0 with the horizontal. The line ac is horizontal. The water surface slope (Sw) is given by S w = = bc cd bd ab = sinφ ab Let q be the angle which the bottom makes with the horizontal. Thus cd cd S0 = sinθ = ad ab
The slope of the water surface with respect to the channel bottom is given by dy dx bd = ad bd ab S w = S 0 dy dx This equation can be used to calculate the water surface slope with respect to horizontal. dy dx = S 0 S w
Water Profile Computations (Gradually Varied Flow) Engineers often require to know the distance up to which a surface profile of a gradually varied flow will extend. To accomplish this we have to integrate the dynamic equation of gradually varied flow, so to obtain the values of y at different locations of x along the channel bed. The figure below gives a sketch of calculating the M curve over a given weir.
Direct Step Method One of the most important method used to compute the water profiles is the direct step method. In this method, the channel is divided into short intervals and the computation of surface profiles is carried out step by step from one section to another. For prismatic channels: Consider a short length of channel, dx, as shown in the figure. dx
Applying Bernoulli s equation between section and, we write: S dx V V + y + = y + + g g 0 S dx f or S dx + E = E + S dx 0 f or dx = E S E 0 S f where E and E are the specific energies at section and, respectively. This equation will be used to compute the water profile curves.
The following steps summarize the direct step method:. Calculate the specific energy at section where depth is known. For example at section -, find E, where the depth is known (y ). This section is usually a control section.. Assume an appropriate value of the depth y at the other end of the small reach. Note that: y > y y < y if the profile is a rising curve and, if the profile is a falling curve. 3. Calculate the specific energy (E ) at section - for the assumed depth (y ). 4. Calculate the slope of the energy line (S f ) at sections - and - using Manning s formula V n R / = 3 S f and And the average slope in reach is calculated V n R / = 3 S f S fm = S + S f f
5. Compute the length of the curve between section - and - L, = dx = E S E 0 S fm or L, = S 0 E S f E + S f 6. Now, we know the depth at section -, assume the depth at the next section, say 3-3. Then repeat the procedure to find the length L,3. 7. Repeating the procedure, the total length of the curve may be obtained. Thus L = L,+ L,3+... + L n, n where (n-) is the number of intervals into which the channel is divided.
. A river 90 m wide and 3 m deep has stable bed and vertical banks with a bed slope of in 500. Estimate the length of back water curve produced by an afflux of m. Assume Manning s in N = 0.035. Given data Wide (b) = 90 m Deep (d) =3 m Bed slope ( i b ) = / 500 = 4 x 0-4 Afflux (h - h ) = m Manning s N = 0.035 Find out Length of back water curve ( L ) =? Solution h - h = m Area (A ) = b x d = 90 x 3 = 70 m² Wetted Perimeter (P ) = b + h = 90 + ( x 3 ) = 96 m Hydraulic mean depth (m ) = A / P = 70 m² / 96 m =.83 m Manning s formula V N 3 3 = m i V = m i b N
V = N m 3 i b V 3-4 = (.83) (4 x 0 ) 0.035 V = ( / 0.035 ) (.83 ) 0.667 ( 4 x 0-4 ) 0.5 V =.39 m / sec E = h + V g E = 3 + [ (.39 ) / ( x 9.8 ) ] E = 3.066 m To Find Velocity ( V ) Continuity Equation A x V = A x V h = + h = + 3 = 5 m A = b x h = 90 x 5 = 450 m Wetted Perimeter ( P ) = b + h = 90 + ( x 5 ) = 00 m Hydraulic mean depth (m ) = A / P = 450 m² / 00 m = 4.500 m
A x V = A x V 70 x.39 = 450 x V V = (307.53 / 450 ) = 0.683 m / sec E = h + V g E = 5 + [ ( 0.683 ) / ( x 9.8 ) ] E = 5.04 m To Find Average Velocity ( V av ) First Average Depth ( h av ) = [ ( h + h ) / ] = [ ( 3 + 5 ) / ] ( h av ) = 4 m Continuity Equation A av x V av = A x V A av = b x h av = 90 x 4 = 360 m V av = [ ( A x V ) / A av ] = [ ( 450 x 0.683 ) / 360 ] V av = 0.854 m / sec Average Hydraulic Mean Depth ( m av ) = [ (m + m ) / ] = [ (.83 + 4.500 ) / ] ( m av ) = 3.657 m
To Find Energy Line Slope ( i e ) Vav = N m av 3 i e 0.854 = 0.035 (3.657) 3 (i e ) 0.854 = ( / 0.035 ) ( 3.657 ) 0.667 ( i e ) 0.5 0.854 = 67.848 ( i e ) 0.5 ( i e ) 0.5 = ( 0.854 / 67.848 ) ( i e ) = ( 0.854 / 67.848 ) ( i e ) =.584 x 0-4 Length of back water curve L E E i b i e = 5.04 3.066 L= 4 4 0.584 0 ( L) = 804.305 m = 8.04 Km Answer * Length of back water curve ( L ) = 804.305 m = 8.04 Km 4
.The normal depth of flow of water, in a rectangular channel.5 m wide, is one metre. The bed slope of the channel is 0.0006 and Manning s roughness coefficient N = 0.0. Find the critical depth. At a certain section of the same channel the depth is 0.9 m while at a second section the depth is 0.86 m. Find the distance between the two sections. Also Find whether the section is located downstream or upstream with respect to the first section. Given data Wide ( b ) =.5 m Normal Depth ( d n ) = m Bed slope ( i b ) = 0.0006 = 6 x 0-4 Manning s N = 0.0 h = 0.9 m h = 0.86 m Find out Length of back water curve ( L ) =? Solution Area (A) = b x d n =.5 x =.5 m² Wetted Perimeter (P ) = b + d n =.5 + ( x ) = 3.5 m Hydraulic mean depth (m ) = A / P =.5 m² / 3.5 m = 0.49 m
V = N m 3 i b V = 0.0 (0.49) 3 (6 x 0 V = ( / 0.0 ) ( 0.49 ) 0.667 ( 6 x 0-4 ) 0.5 V =.6 m / sec Q = A x V Q =.5 x.6 =.74 m³ / sec q = Q / b =.74 /.5 =.6 m² / sec Critical Depth ( y c ) y c = ( 0.37 ) 0.333 Critical Depth ( y c ) = 0.56 m -4 ) 3. 6 3 y c = y c = q g 9. 8 E = h + V g V = Q / A =.74 / b x h =.74 /.5 x 0.9 V =.6 m / sec
E = h + V g E = 0.9 + [ (.6 ) / ( x 9.8 ) ] E =.00 m V = Q / A =.74 / b x h =.74 /.5 x 0.86 V =.350 m / sec E = h + V g E = 0.86 + [ (.350 ) / ( x 9.8 ) ] E = 0.953 m Average Height ( h av ) = [ ( h + h ) / ] = [ ( 0.9 + 0.86 ) / ] ( h av ) = 0.890 m Average Hydraulic Mean Depth ( m av ) = [ A / P ] = [ ( b x h av ) / ( b + h av ) = [ (.5 x 0.890 ) / (.5 + x 0.890 ) ] ( m av ) = 0.407 m
To Find Energy Line Slope ( i e ) Q = A N m av 3 i e.74 =.5 x 0.890 0.0 (0.407) 3 (i e ).74 = (.335 / 0.0 ) ( 0.407 ) 0.667 ( i e ) 0.5.74 = 6.080 ( i e ) 0.5 ( i e ) 0.5 = (.74 / 6.080 ) ( i e ) = (.74 / 6.080 ) ( i e ) = 8.34 x 0-4 Distance between the two sections L E E = x = i b i e L = x = 6 0.953.00 4 0 8.34 0 4 L = x = 4. 930 m
Water Surface Slope = b i e Average Velocity ( V av ) = [ ( V + V ) / ] = [ (.6 +.350 ) / ] ( V av ) =.306 m / sec Fe = (V av / g h av ) Fe = [(.306 ) /( 9.8 x 0.890 )] = 0.95 (dy / dx) = -.65 x 0-4 = av av 4 4 = 6 0 8.34 0. Since (dy / dx) is negative, the second section is downstream.
3. A rectangular channel with a bottom width of 4 m and a bottom slope of 0.0008 has a discharge of.5 m³ / sec. In a gradually varied flow in this channel, the depth at a certain location is found to be 0.3 m. Assuming N = 0.06, determine the type of gradually varied flow profile. Given data Width (b) = 4 m Depth (y) = 0.3 m Bed slope ( i b ) = 0.0008 = 8 x 0-4 Discharge (Q) =.5 m³ / sec Manning s N = 0.06 Solution Calculating ( y n ) using Manning s Formula Q = A N m 3 i b Area (A) = b x y n = 4 x y n = 4y n Wetted Perimeter (P) = b + y n = 4 + ( y n ) Hydraulic mean depth (m) = A / P = 4y n / 4 + ( y n )
Substituting all values Q = A N m 3 i b 4yn 4yn.5= 0.06 4+ y n 3 (8x 0-4 ).5 x 0.06 0.08 = 5 [ 4y n]3 [ 4 + y n]3 0. 857 = [ 4y ] 5 3 n [ 4 + y n]3 3 [ 4+ y ] [ n 4y ]3 0.857 = n 5
[ 4y ] 5 3 0.857[ 4+ y ] 3 0 n n = The above equation will be solved by Hit and Trail methid. (i) Assume y n = 0.400 m, then L.H.S. of equation (i) will be as L. H. S = [.6] 5 3 0.857[ 4.8]3 L. H. S =.89.439 = 0.50 (ii) Assume y n = 0.500 m, then L.H.S. of equation (i) will be as L. H. S = [ ] 5 3 0.857 [ 5]3 (i) L. H. S = 3.75.506 = 0.669 where y n = 0.400 m, L.H.S. is ve. But when y n = 0.500 m, L.H.S. is + ve. Hence value of y n lies between 0.400 m and 0.500 m.
(iii) Assume y n = 0.40 m, then L.H.S. of equation (i) will be as L. H. S = [.64 ] 5 3 0.857 [ 4.8 ]3 L. H. S =.8.445= 0.64 (iv) Assume y n = 0.430 m, then L.H.S. of equation (i) will be as L. H. S = [.7 ] 5 3 0.857 [ 4.86 ]3 L. H. S =.469.459 = 0.00 The value of L.H.S. =0.00 is negligible in comparison to the value of 0.669. Therefore value of y n = 0.430 m. q = Q / b =.5 / 4 = 0.375 m² / sec Critical Depth ( y c ) Critical Depth ( y c ) = 0.43 m 3 0. 375 3 y c = q g y c = 9. 8
Type of profile : since y n > y c, the channel is a mild slope channel. Also given y = 0.3 m such that y n > y > y c. Hence it is a M type curve. CLASSIFICATION OF CHANNELS S.No Channel Category Symbol Characteristics Condition Remarks. Mild slope M y n > y c Subcritical flow at normal depth. Steep slope s y c > y n Supercritical flow at normal depth 3. Critical slope C y c = y n Critical flow at normal depth 4. Horizontal bed H i b = 0 Cannot sustain uniform flow 5. Adverse slope A i b < 0 Cannot sustain uniform flow
TYPES OF GRADUALLY VARIED FLOW PROFILES S.No Channel Region Condition Type. y > y n > y c M Mild slope. y n > y > y c M 3. 3 y n > y c > y M 3 4. y > y c > y n S Steep slope 5. y c > y > y n S 6. 3 y c > y n > y S 3
TYPES OF GRADUALLY VARIED FLOW PROFILES S.No Channel Region Condition Type 7. Critical slope y > y n = y c C 8. 3 y < y n = y c C 3 9. y > y c H Horizontal 0. bed 3 y < y c H 3. Adverse slope y > y c A. 3 y < y c A 3
Classification of Channel-Bed Slopes The slope of the channel bed is very important in determining the characteristics of the flow. Let S 0 : the slope of the channel bed, S c : the critical slope or the slope of the channel that sustains a given discharge (Q) as uniform flow at the critical depth (y c ). y n is is the normal depth when the discharge Q flows as uniform flow on slope S 0.
The slope of the channel bed can be classified as: ) Critical Slope C : the bottom slope of the channel is equal to the critical slope. S0 = Sc or yn = yc ) Mild Slope M : the bottom slope of the channel is less than the critical slope. S0 < Sc or yn > yc 3) Steep Slope S : the bottom slope of the channel is greater than the critical slope. S 0 > S c or yn < yc 4) Horizontal Slope H : the bottom slope of the channel is equal to zero. S 0 = 0. 0 5) Adverse Slope A : the bottom slope of the channel rises in the direction of the flow (slope is opposite to direction of flow). S 0 = negative
Classification of Flow Profiles (water surface profiles) The surface curves of water are called flow profiles (or water surface profiles). The shape of water surface profiles is mainly determined by the slope of the channel bed S o. For a given discharge, the normal depth y n and the critical depth y c may be calculated. Then the following steps are followed to classify the flow profiles: - A line parallel to the channel bottom with a height of y n is drawn and is designated as the normal depth line (N.D.L.) - A line parallel to the channel bottom with a height of y c is drawn and is designated as the critical depth line (C.D.L.) 3- The vertical space in a longitudinal section is divided into 3 zones using the two lines drawn in steps & (see the next figure)
4- Depending upon the zone and the slope of the bed, the water profiles are classified into 3 types as follows: (a) Mild slope curves M, M, M 3. (b) Steep slope curves S, S, S 3. (c) Critical slope curves C, C, C 3. (d) Horizontal slope curves H, H 3. (e) Averse slope curves A, A 3. In all these curves, the letter indicates the slope type and the subscript indicates the zone. For example S curve occurs in the zone of the steep slope.
Flow Profiles in Mild slope Flow Profiles in Steep slope
Flow Profiles in Critical slope Flow Profiles in Horizontal slope Flow Profiles in Adverse slope
4. A river 45 m wide has a normal depth of flow of 3 m and an average bed slope of in 0,000. A weir is built across the river raising the water surface level at the weir site to 5 m above the bottom of the river. Identify the type of gradually varied flow profile considering the river is prismatic. Take the value of N is Manning s Formula as 0.05. Given data Wide ( b ) = 45 m Normal Depth ( y n ) = 3 m Bed slope ( i b ) = / 0,000 = x 0-4 Manning s N = 0.05 y = 5 m Find out Identify the type of gradually varied flow profile? Solution Area (A) = b x y n = 45 x 3 = 35 m² Wetted Perimeter (P ) = b + y n = 45 + ( x 3 ) = 5 m Hydraulic mean depth (m ) = A / P = 35 m² / 5 m =.647 m Q = A N m 3 i b Q = ( 35 / 0.05 ) (.647 ) 0.667 ( x 0-4 ) 0.5 Q = 03.364 m³ / sec
q = Q / b = 03.364 / 45 =.97 m² / sec Critical Depth ( y c ) y c = q g 3 y c =. 97 9. 8 3 Critical Depth ( y c ) = 0.83 m Since y n > y c, the channel is said to be mild slope. Given y = 5 m, Then y > y n > y c Hence this profile is M type profile. 5. A river 00 m wide and 3 m deep has an average bed slope of 0.0005. Identify the type of gradually varied flow profile produced by a low dam which raises the water surface in the upstream by.5 m. Assume N = 0.035. Given data Wide ( b ) = 00 m Normal Depth ( y n ) = 3 m Bed slope ( i b ) = 0.0005= 5 x 0-4 Manning s N = 0.035 Increase in water level / Afflux =.5 m Solution y = y n +.5 = 3 +.5 = 4.5 m
Area (A) = b x y n = 00 x 3 = 300 m² Wetted Perimeter (P ) = b + y n = 00 + ( x 3 ) = 06 m Hydraulic mean depth (m ) = A / P = 300 m² / 06 m =.830 m Q = A N m 3 i b Q = ( 300 / 0.035 ) (.830 ) 0.667 ( 5 x 0-4 ) 0.5 Q = 383.60 m³ / sec q = Q / b = 383.60 / 00 = 3.836 m² / sec Critical Depth ( y c ) Critical Depth ( y c ) =.45 m 3 q 3. 836 3 y c = y = c 9. 8 Since y n > y c, the channel is said to be mild slope channel. 9. 8 Given y = 4.5 m, Then y > y n > y c Hence the gradually varied flow profile is an M type curve.
DETERMINATION OF GRADUALLY VARIED FLOW PROFILES There are three methods. Numerical Method (a) Direct Step method (b) Standard Step method. Graphical Method 3. Direct Integration Method 6. A rectangular channel 0 m wide carries a discharge of 30 cumec. It is laid at the slope of 0.000. If at the section in this channel the depth is.6 m. How for upstream (or) downstream from the section being the depth at m. Take Manning s N = 0.05. Use direct step method. Answer * Length between two section = 047.3 m
ASSIGNMENT II PART A 5 x =0. Write down the dynamic equation of Gradually varied flow?. Define the term Afflux? 3. Write about back water curves? 4. Define transition depth? 5. What are the methods used for finding Gradually varying flow profile? PART B ½ X 6 =40 6)(i) What are the assumptions made in the analysis of Gradually varied flow?(6) (ii) The bed width of a rectangular channel is 4 m and the depth of flow is 6 m. The discharge in the canal is 86 cumecs. The bed slope of the channel is in 4000. Assume Chezy s constant C = 60. Determine the slope of the free water surface? (0) 7)A rectangular channel 8.5 m wide has a uniform depth of flow of.5 m and has a bed slope of in 3000. If due to weir constructed at the downstream end of the channel, water surface at a section is raised by 0.8 m, determine the water surface slope with respect to horizontal at this section. Assume Manning s n = 0.0? (6) 8)Derive an expression for the length of the back water curve? (8)