1. Channel RoutingChannel Routing Simulate the movement of water through a channel Used to predict the magnitudes, volumes, and temporal patterns of the flow (often a flood wave) as it translates down a channel. 2 types of routing : hydrologic and hydraulic. both of these methods use some form of the continuity equation. Continuity equation Hydrologic Routing Hydraulic Routing Momentum Equation

2. Continuity EquationContinuity Equation The change in storage (dS) equals the difference between inflow (I) and outflow (O) or : O-I= dt dS For open channel flow, the continuity equation is also often written as : q= x Q + t A A = the cross-sectional area, Q = channel flow, and q = lateral inflow Continuity equationContinuity equation Hydrologic Routing Hydraulic Routing Momentum Equation 3. Hydrologic RoutingHydrologic Routing Methods combine the continuity equation with some relationship between storage, outflow, and possibly inflow. These relationships are usually assumed, empirical, or analytical in nature. An of example of such a relationship might be a stage-discharge relationship. Continuity equation Hydrologic RoutingHydrologic Routing Hydraulic Routing Momentum Equation 4. Use of Manning EquationUse of Manning Equation Stage is also related to the outflow via a relationship such as Manning's equation 2 1 3 249.1 fh SAR n Q = Continuity equation Hydrologic RoutingHydrologic Routing Hydraulic Routing Momentum Equation 5. Hydraulic RoutingHydraulic Routing Hydraulic routing methods combine the continuity equation with some more physical relationship describing the actual physics of the movement of the water. The momentum equation is the common relationship employed. In hydraulic routing analysis, it is intended that the dynamics of the water or flood wave movement be more accurately described Continuity equation Hydrologic Routing Hydraulic RoutingHydraulic Routing Momentum Equation 6. Momentum EquationMomentum Equation Expressed by considering the external forces acting on a control section of water as it moves down a channel )S-Sg(= A vg + 2x A)y( A g + x v V+ t v fo Henderson (1966) expressed the momentum equation as : t v g 1 - x v g v - x y -S=S of Continuity equation Hydrologic Routing Hydraulic Routing Momentum EquationMomentum Equation 7. Combinations of EquationsCombinations of Equations Simplified Versions : t v g 1 - x v g v - x y -S=S of x v g v - x y -S=S of x y -S=S of Sf = So Unsteady -Nonuniform Steady - Nonuniform Diffusion or noninertial Kinematic Continuity equation Hydrologic Routing Hydraulic Routing Momentum Equation 8. Routing MethodsRouting Methods Modified Puls Kinematic Wave Muskingum Muskingum-Cunge Dynamic Modified Puls Kinematic Wave Muskingum Muskingum-Cunge Dynamic Modeling Notes 9. Modified PulsModified Puls The modified puls routing method is probably most often applied to reservoir routing The method may also be applied to river routing for certain channel situations. The modified puls method is also referred to as the storage-indication method. The heart of the modified puls equation is found by considering the finite difference form of the continuity equation. Modified PulsModified Puls Kinematic Wave Muskingum Muskingum-Cunge Dynamic Modeling Notes 10. Modified PulsModified Puls t S-S = 2 O+O( - 2 I+I 122121 Continuity Equation O+ t S2 =O- t S2 +I+I 2 2 1 1 21 Rewritten The solution to the modified puls method is accomplished by developing a graph (or table) of O -vs- [2S/t + O]. In order to do this, a stage-discharge-storage relationship must be known, assumed, or derived. Modified PulsModified Puls Kinematic Wave Muskingum Muskingum-Cunge Dynamic Modeling Notes 11. Modified Puls ExampleModified Puls Example Given the following hydrograph and the 2S/t + O curve, find the outflow hydrograph for the reservoir assuming it to be completely full at the beginning of the storm. The following hydrograph is given: Hydrograph For Modified Puls Example 0 30 60 90 120 150 180 0 2 4 6 8 10 Time (hr) Discharge(cfs) 12. Modified Puls ExampleModified Puls Example 2S/t + O curve for Modified Puls example 0 200 400 600 800 1000 1200 0 10 20 30 40 50 60 70 Outflow (cfs) 2S/t+O(cfs) The following 2S/t + O curve is also given: 13. Modified Puls ExampleModified Puls Example A table may be created as follows: Time In In+In+1 2Sn/t - On 2Sn/t + On+1 On+1 (hr) (cfs) (cfs) (cfs) (cfs) (cfs) 0 1 2 3 4 5 6 7 8 9 10 11 12 14. Modified Puls ExampleModified Puls Example Hydrograph For Modified Puls Example 0 30 60 90 120 150 180 0 2 4 6 8 10 Time (hr) Discharge(cfs) Next, using the hydrograph and interpolation, insert the Inflow (discharge) values. For example at 1 hour, the inflow is 30 cfs. Time In In+In+1 2Sn/t - On 2Sn/t + On+1 On+1 (hr) (cfs) (cfs) (cfs) (cfs) (cfs) 0 0 1 30 2 60 3 90 4 120 5 150 6 180 7 135 8 90 9 45 10 0 11 0 12 0 15. Modified Puls ExampleModified Puls Example Time In In+In+1 2Sn/t - On 2Sn/t + On+1 On+1 (hr) (cfs) (cfs) (cfs) (cfs) (cfs) 0 0 30 1 30 2 60 3 90 4 120 5 150 6 180 7 135 8 90 9 45 10 0 11 0 12 0 The next step is to add the inflow to the inflow in the next time step. For the first blank the inflow at 0 is added to the inflow at 1 hour to obtain a value of 30. 16. Modified Puls ExampleModified Puls Example Time In In+In+1 2Sn/t - On 2Sn/t + On+1 On+1 (hr) (cfs) (cfs) (cfs) (cfs) (cfs) 0 0 30 1 30 90 2 60 150 3 90 210 4 120 270 5 150 330 6 180 315 7 135 225 8 90 135 9 45 45 10 0 0 11 0 0 12 0 0 This is then repeated for the rest of the values in the column. 17. Modified Puls ExampleModified Puls Example Time In In+In+1 2Sn/t - On 2Sn/t + On+1 On+1 (hr) (cfs) (cfs) (cfs) (cfs) (cfs) 0 0 30 0 0 1 30 90 30 2 60 150 3 90 210 4 120 270 5 150 330 6 180 315 7 135 225 8 90 135 9 45 45 10 0 0 11 0 0 12 0 0 O+ t S2 =O- t S2 +I+I 2 2 1 1 21 The 2Sn/t + On+1 column can then be calculated using the following equation: Note that 2Sn/t - On and On+1 are set to zero. 30 + 0 = 2Sn/t + On+1 18. Modified Puls ExampleModified Puls Example 2S/ t + O curve for Modified Puls example 0 200 400 600 800 1000 1200 0 10 20 30 40 50 60 70 Outflow (cfs) 2S/t+O(cfs) Time In In+In+1 2Sn/t - On 2Sn/t + On+1 On+1 (hr) (cfs) (cfs) (cfs) (cfs) (cfs) 0 0 30 0 0 1 30 90 30 5 2 60 150 3 90 210 4 120 270 5 150 330 6 180 315 7 135 225 8 90 135 9 45 45 10 0 0 11 0 0 12 0 0 Then using the curve provided outflow can be determined. In this case, since 2Sn/t + On+1 = 30, outflow = 5 based on the graph provided. 19. Modified Puls ExampleModified Puls Example Time In In+In+1 2Sn/t - On 2Sn/t + On+1 On+1 (hr) (cfs) (cfs) (cfs) (cfs) (cfs) 0 0 30 0 0 1 30 90 20 30 5 2 60 150 3 90 210 4 120 270 5 150 330 6 180 315 7 135 225 8 90 135 9 45 45 10 0 0 11 0 0 12 0 0 To obtain the final column, 2Sn/t - On, two times the outflow is subtracted from 2Sn/t + On+1. In this example 30 - 2*5 = 20 20. Modified Puls ExampleModified Puls Example Time In In+In+1 2Sn/t - On 2Sn/t + On+1 On+1 (hr) (cfs) (cfs) (cfs) (cfs) (cfs) 0 0 30 0 0 1 30 90 20 30 5 2 60 150 74 110 18 3 90 210 4 120 270 5 150 330 6 180 315 7 135 225 8 90 135 9 45 45 10 0 0 11 0 0 12 0 0 The same steps are repeated for the next line. First 90 + 20 = 110. From the graph, 110 equals an outflow value of 18. Finally 110 - 2*18 = 74 21. Modified Puls ExampleModified Puls Example Time In In+In+1 2Sn/t - On 2Sn/t + On+1 On+1 (hr) (cfs) (cfs) (cfs) (cfs) (cfs) 0 0 30 0 0 1 30 90 20 30 5 2 60 150 74 110 18 3 90 210 160 224 32 4 120 270 284 370 43 5 150 330 450 554 52 6 180 315 664 780 58 7 135 225 853 979 63 8 90 135 948 1078 65 9 45 45 953 1085 65 10 0 0 870 998 64 11 0 0 746 870 62 12 0 0 630 746 58 This process can then be repeated for the rest of the columns. Now a list of the outflow values have been calculated and the problem is complete. 22. MuskingumMuskingum MethodMethod Sp = K O Sw = K(I - O)X Prism Storage Wedge Storage CombinedS = K[XI + (1-X)O] Modified Puls Kinematic Wave MuskingumMuskingum Muskingum-Cunge Dynamic Modeling Notes 23. Muskingum,Muskingum, cont...cont... O2 = C0 I2 + C1 I1 + C2 O1 Substitute storage equation, S into the S in the continuity equation yields : S = K[XI + (1-X)O] O-I= dt dS t0.5+Kx-K t0.5-Kx -=C0 t0.5+Kx-K t0.5+Kx =C1 t0.5+Kx-K t0.5-Kx-K =C2 Modified Puls Kinematic Wave MuskingumMuskingum Muskingum-Cunge Dynamic Modeling Notes 24. Muskingum Notes :Muskingum Notes : The method assumes a single stage-discharge relationship. In other words, for any given discharge, Q, there can be only one stage height. This assumption may not be entirely valid for certain flow situations. For instance, the friction slope on the rising side of a hydrograph for a given flow, Q, may be quite different than for the recession side of the hydrograph for the same given flow, Q. This causes an effect known as hysteresis, which can introduce errors into the storage assumptions of this method. Modified Puls Kinematic Wave MuskingumMuskingum Muskingum-Cunge Dynamic Modeling Notes 25. Estimating KEstimating K K is estimated to be the travel time through the reach. This may pose somewhat of a difficulty, as the travel time will obviously change with flow. The question may arise as to whether the travel time should be estimated using the average flow, the peak flow, or some other flow. The travel time may be estimated using the kinematic travel time or a travel time based on Manning's equation. Modified Puls Kinematic Wave MuskingumMuskingum Muskingum-Cunge Dynamic Modeling Notes 26. Estimating XEstimating X The value of X must be between 0.0 and 0.5. The parameter X may be thought of as a weighting coefficient for inflow and outflow. As inflow becomes less important, the value of X decreases. The lower limit of X is 0.0 and this would be indicative of a situation where inflow, I, has little or no effect on the storage. A reservoir is an example of this situation and it should be noted that attenuation would be the dominant process compared to translation. Values of X = 0.2 to 0.3 are the most common for natural streams; however, values of 0.4 to 0.5 may be calibrated for streams with little or no flood plains or storage effects. A value of X = 0.5 would represent equal weighting between inflow and outflow and would produce translation with little or no attenuation. Modified Puls Kinematic Wave MuskingumMuskingum Muskingum-Cunge Dynamic Modeling Notes 27. Did you know? Lag and K is a special case of Muskingum -> X=0! 28. More Notes - MuskingumMore Notes - Muskingum The Handbook of Hydrology (Maidment, 1992) includes additional cautions or limitations in the Muskingum method. The method may produce negative flows in the initial portion of the hydrograph. Additionally, it is recommended that the method be limited to moderate to slow rising hydrographs being routed through mild to steep sloping channels. The method is not applicable to steeply rising hydrographs such as dam breaks. Finally, this method also neglects variable backwater effects such as downstream dams, constrictions, bridges, and tidal influences. Modified Puls Kinematic Wave MuskingumMuskingum Muskingum-Cunge Dynamic Modeling Notes 29. Muskingum Example ProblemMuskingum Example Problem Time Inflow C0I2 C1I1 C2O1 Outflow 0 3 3 1 5 2 10 3 8 4 6 5 5 A portion of the inflow hydrograph to a reach of channel is given below. If the travel time is K=1 unit and the weighting factor is X=0.30, then find the outflow from the reach for the period shown below: 30. Muskingum Example ProblemMuskingum Example Problem The first step is to determine the coefficients in this problem. The calculations for each of the coefficients is given below: t0.5+Kx-K t0.5-Kx -=C0 C0= - ((1*0.30) - (0.5*1)) / ((1-(1*0.30) + (0.5*1)) = 0.167 t0.5+Kx-K t0.5+Kx =C1 C1= ((1*0.30) + (0.5*1)) / ((1-(1*0.30) + (0.5*1)) = 0.667 31. Muskingum Example ProblemMuskingum Example Problem C2= (1- (1*0.30) - (0.5*1)) / ((1-(1*0.30) + (0.5*1)) = 0.167 t0.5+Kx-K t0.5-Kx-K =C2 Therefore the coefficients in this problem are: C0 = 0.167 C1 = 0.667 C2 = 0.167 32. Muskingum Example ProblemMuskingum Example Problem Time Inflow C0I2 C1I1 C2O1 Outflow 0 3 0.835 2.00 0.501 3 1 5 2 10 3 8 4 6 5 5 The three columns now can be calculated. C0I2 = 0.167 * 5 = 0.835 C1I1 = 0.667 * 3 = 2.00 C2O1 = 0.167 * 3 = 0.501 33. Muskingum Example ProblemMuskingum Example Problem Next the three columns are added to determine the outflow at time equal 1 hour. 0.835 + 2.00 + 0.501 = 3.34 Time Inflow C0I2 C1I1 C2O1 Outflow 0 3 0.835 2.00 0.501 3 1 5 3.34 2 10 3 8 4 6 5 5 34. Muskingum Example ProblemMuskingum Example Problem This can be repeated until the table is complete and the outflow at each time step is known. Time Inflow C0I2 C1I1 C2O1 Outflow 0 3 0.835 2.00 0.501 3 1 5 1.67 3.34 0.557 3.34 2 10 1.34 6.67 0.93 5.57 3 8 1.00 5.34 1.49 8.94 4 6 0.835 4.00 1.31 7.83 5 5 3.34 1.03 6.14 35. Muskingum Example - Tenkiller Look at R-5 36. Reach at Outlet K = 2 hrs X = 0.25 37. K = 2 & 4 Hrs. X = 0.25 38. K = 2, 4, & 6 Hrs. X = 0.25 39. Lets Alter X K = 2 hrs. K = 2 hrs X = 0.25 K = 2 hrs X = 0.5 40. And Again.... K = 2 hrs X = 0.5 K = 2 hrs X = 0.05 41. Muskingum-CungeMuskingum-Cunge Muskingum-Cunge formulation is similar to the Muskingum type formulation The Muskingum-Cunge derivation begins with the continuity equation and includes the diffusion form of the momentum equation. These equations are combined and linearized, Modified Puls Kinematic Wave Muskingum Muskingum-CungeMuskingum-Cunge Dynamic Modeling Notes 42. Muskingum-CungeMuskingum-Cunge working equationworking equation where : Q = discharge t = time x = distance along channel qx = lateral inflow c = wave celerity m = hydraulic diffusivity Latcq x Q x Q t Q + = + 2 2 Modified Puls Kinematic Wave Muskingum Muskingum-CungeMuskingum-Cunge Dynamic Modeling Notes 43. Muskingum-Cunge, cont...Muskingum-Cunge, cont... Method attempts to account for diffusion by taking into account channel and flow characteristics. Hydraulic diffusivity is found to be : O BS Q 2 = The Wave celerity in the x-direction is : dA dQ C = Modified Puls Kinematic Wave Muskingum Muskingum-CungeMuskingum-Cunge Dynamic Modeling Notes 44. Solution of Muskingum-CungeSolution of Muskingum-Cunge Solution of the Method is accomplished by discretizing the equations on an x-t plane. QC+QC+QC+QC= 1+j 1+n Q L4 n 1+j3 1+n j2 n j1 x)-2(1+ k t 2x+ k t =C1 x)-2(1+ k t 2x- k t =C2 x)-2(1+ k t k t -x)-2(1 =C3 x)-2(1+ k t k t 2 =C4 X t Modified Puls Kinematic Wave Muskingum Muskingum-CungeMuskingum-Cunge Dynamic Modeling Notes 45. Calculation of K & XCalculation of K & X c x =k = xcBS Q X O 1 2 1 Estimation of K & X is more physically based and should be able to reflect the changing conditions - better. Modified Puls Kinematic Wave Muskingum Muskingum-CungeMuskingum-Cunge Dynamic Modeling Notes 46. Muskingum-Cunge - NOTESMuskingum-Cunge - NOTES Muskingum-Cunge formulation is actually considered an approximate solution of the convective diffusion equation. As such it may account for wave attenuation, but not for reverse flow and backwater effects and not for fast rising hydrographs. Properly applied, the method is non-linear in that the flow properties and routing coefficients are re-calculated at each time and distance step Often, an iterative 4-point scheme is used for the solution. Care should be taken when choosing the computation interval, as the computation interval may be longer than the time it takes for the wave to travel the reach distance. Internal computational times are used to account for the possibility of this occurring. Modified Puls Kinematic Wave Muskingum Muskingum-CungeMuskingum-Cunge Dynamic Modeling Notes 47. Muskingum-Cunge ExampleMuskingum-Cunge Example The hydrograph at the upstream end of a river is given in the following table. The reach of interest is 18 km long. Using a subreach length x of 6 km, determine the hydrograph at the end of the reach using the Muskingum-Cunge method. Assume c = 2m/s, B = 25.3 m, So = 0.001m and no lateral flow. Time (hr) Flow (m3 /s) 0 10 1 12 2 18 3 28.5 4 50 5 78 6 107 7 134.5 8 147 9 150 10 146 11 129 12 105 13 78 14 59 15 45 16 33 17 24 18 17 19 12 20 10 48. Muskingum-Cunge ExampleMuskingum-Cunge Example First, K must be determined. K is equal to : c x K = seconds3000 /2 /10006 = = sm kmmkm K x = 6 km, while c = 2 m/s Is c a constant? 49. Muskingum-Cunge ExampleMuskingum-Cunge Example The next step is to determine x. = xcBS Q x O 1 2 1 All the variables are known, with B = 25.3 m, So = 0.001 and x =6000 m, and the peak Q taken from the table. 253.0 /)6000)(2)(001.0)(3.25( /150 1 2 1 3 3 = = sm sm x Time (hr) Flow (m3 /s) 0 10 1 12 2 18 3 28.5 4 50 5 78 6 107 7 134.5 8 147 9 150 10 146 11 129 12 105 13 78 14 59 15 45 16 33 17 24 18 17 19 12 20 10 50. Muskingum-Cunge ExampleMuskingum-Cunge Example The coefficients of the Muskingum-Cunge method can now be determined. )1(2 2 1 x K t x K t C + + = 7466.0 )253.01(2 3000 7200 )253.0(2 3000 7200 1 = + + =C Using 7200 seconds or 2 hrs for timestep - this may need to be changed 51. Muskingum-Cunge ExampleMuskingum-Cunge Example The coefficients of the Muskingum-Cunge method can now be determined. )1(2 2 2 x K t x K t C + = 4863.0 )253.01(2 3000 7200 )253.0(2 3000 7200 2 = + =C 52. Muskingum-Cunge ExampleMuskingum-Cunge Example The coefficients of the Muskingum-Cunge method can now be determined. )1(2 )1(2 3 x K t K t x C + = 232.0 )253.01(2 3000 7200 3000 7200 )253.01(2 3 = + =C 53. Muskingum-Cunge ExampleMuskingum-Cunge Example The coefficients of the Muskingum-Cunge method can now be determined. )1(2 2 4 x K t K t C + = 233.1 )253.01(2 3000 7200 3000 7200 2 4 = + =C 54. Muskingum-Cunge ExampleMuskingum-Cunge Example Then a simplification of the original formula can be made. L n j n j n j n j QCQCQCQCQ 413 1 21 1 1 +++= + ++ + Since there is not lateral flow, QL = 0. The simplified formula is the following: n j n j n j n j QCQCQCQ 13 1 21 1 1 + ++ + ++= 55. Muskingum-Cunge ExampleMuskingum-Cunge Example A table can then be created in 2 hour time steps similar to the one below: Time (hr) 0 km 6 km 12 km 18 km 0 10 2 18 4 50 6 107 8 147 10 146 12 105 14 59 16 33 18 17 20 10 22 10 24 10 26 10 28 10 56. Muskingum-Cunge ExampleMuskingum-Cunge Example It is assumed at time zero, the flow is 10 m3 /s at each distance. Time (hr) 0 km 6 km 12 km 18 km 0 10 10 10 10 2 18 4 50 6 107 8 147 10 146 12 105 14 59 16 33 18 17 20 10 22 10 24 10 26 10 28 10 57. Muskingum-Cunge ExampleMuskingum-Cunge Example Next, zero is substituted into for each letter to solve the equation. n j n j n j n j QCQCQCQ 13 1 21 1 1 + ++ + ++= 0 13 1 02 0 01 1 1 QCQCQCQ ++= 58. Muskingum-Cunge ExampleMuskingum-Cunge Example Using the table, the variables can be determined. 0 13 1 02 0 01 1 1 QCQCQCQ ++= Time (hr) 0 km 6 km 12 km 18 km 0 10 10 10 10 2 18 4 50 6 107 8 147 10 146 12 105 14 59 16 33 18 17 20 10 22 10 24 10 26 10 28 10 = = = 0 1 1 0 0 0 Q Q Q 10 18 10 59. Muskingum-Cunge ExampleMuskingum-Cunge Example Therefore, the equation can be solved. 0 13 1 02 0 01 1 1 QCQCQCQ ++= smQ Q /89.13 )10)(2329.0()18)(4863.0()10)(7466.0( 31 1 1 1 = ++= Time (hr) 0 km 6 km 12 km 18 km 0 10 10 10 10 2 18 13.89 4 50 6 107 8 147 10 146 12 105 14 59 16 33 18 17 20 10 22 10 24 10 26 10 28 10 60. Muskingum-Cunge ExampleMuskingum-Cunge Example Therefore, the equation can be solved. 1 13 2 02 1 01 2 1 QCQCQCQ ++= smQ Q /51.34 )89.13)(2329.0()50)(4863.0()18)(7466.0( 31 1 1 1 = ++= Time (hr) 0 km 6 km 12 km 18 km 0 10 10 10 10 2 18 13.89 4 50 34.51 6 107 8 147 10 146 12 105 14 59 16 33 18 17 20 10 22 10 24 10 26 10 28 10 61. Muskingum-Cunge ExampleMuskingum-Cunge Example This is repeated for the rest of the columns and the subsequent columns to produce the following table. Note that when you change rows, n changes. When you change columns, j changes. Time (hr) 0 km 6 km 12 km 18 km 0 10 10 10 10 2 18 13.89 11.89 10.92 4 50 34.51 24.38 18.19 6 107 81.32 59.63 42.96 8 147 132.44 111.23 88.60 10 146 149.91 145.88 133.35 12 105 125.16 138.82 145.37 14 59 77.93 99.01 117.94 16 33 41.94 55.52 73.45 18 17 23.14 29.63 38.75 20 10 12.17 16.29 21.02 22 10 9.49 9.91 12.09 24 10 10.12 9.70 9.30 26 10 9.97 10.15 10.01 28 10 10.01 9.95 10.08 62. Look again at the Discretization X t 63. Things to Consider As the wave comes into a reach: the flow and depth go up and then down the wave speed and ALL coefficients would then change... What effect would this have? 64. Full Dynamic Wave EquationsFull Dynamic Wave Equations The solution of the St. Venant equations is known as dynamic routing. Dynamic routing is generally the standard to which other methods are measured or compared. The solution of the St. Venant equations is generally accomplished via one of two methods : 1) the method of characteristics and 2) direct methods (implicit and explicit). It may be fair to say that regardless of the method of solution, a computer is absolutely necessary as the solutions are quite time consuming. J. J. Stoker (1953, 1957) is generally credited for initially attempting to solve the St. Venant equations using a high speed computer. Modified Puls Kinematic Wave Muskingum Muskingum-Cunge DynamicDynamic Modeling Notes 65. Dynamic Wave SolutionsDynamic Wave Solutions Characteristics, Explicit, & Implicit The most popular method of applying the implicit technique is to use a four point weighted finite difference scheme. Some computer programs utilize a finite element solution technique; however, these tend to be more complex in nature and thus a finite difference technique is most often employed. It should be noted that most of the models using the finite difference technique are one-dimensional and that two and three-dimensional solution schemes often revert to a finite element solution. Modified Puls Kinematic Wave Muskingum Muskingum-Cunge DynamicDynamic Modeling Notes 66. Dynamic Wave SolutionsDynamic Wave Solutions Dynamic routing allows for a higher degree of accuracy when modeling flood situations because it includes parameters that other methods neglect. Dynamic routing, when compared to other modeling techniques, relies less on previous flood data and more on the physical properties of the storm. This is extremely important when record rainfalls occur or other extreme events. Dynamic routing also provides more hydraulic information about the event, which can be used to determine the transportation of sediment along the waterway. Modified Puls Kinematic Wave Muskingum Muskingum-Cunge DynamicDynamic Modeling Notes 67. Courant Condition?Courant Condition? If the wave or hydrograph can travel through the subreach (of length x) in a time less than the computational interval, t, then computational instabilities may evolve. The condition to satisfy here is known as the Courant condition and is expressed as : c dx dt Modified Puls Kinematic Wave Muskingum Muskingum-Cunge Dynamic Modeling NotesModeling Notes 68. Some DISadvantagesSome DISadvantages Geometric simplification - some models are designed to use very simplistic representations of the cross-sectional geometry. This may be valid for large dam breaks where very large flows are encountered and width to depth ratios are large; however, this may not be applicable to smaller dam breaks where channel geometry would be more critical. Model simulation input requirements - dynamic routing techniques generally require boundary conditions at one or more locations in the domain, such as the upstream and downstream sections. These boundary conditions may in the form of known or constant water surfaces, hydrographs, or assumed stage-discharge relationships. Stability - As previously noted, the very complex nature of these methods often leads to numeric instability. Also, convergence may be a problem in some solution schemes. For these reasons as well as others, there tends to be a stability problem in some programs. Often times it is very difficult to obtain a "clean" model run in a cost efficient manner. Modified Puls Kinematic Wave Muskingum Muskingum-Cunge Dynamic Modeling NotesModeling Notes