QUESTIONS

  1. What is the difference between hydrologic and hydraulic methods of catchment routing?

    Hydrologic methods of catchment routing are based on the storage concept and are spatially lumped to provide a runoff hydrograph at the catchment outlet. Examples of hydrologic catchment routing methods are the time-area method and the cascade of linear reservoirs.

    Hydraulic methods use kinematic or diffusion waves to simulate surface runoff within a catchment in a distributed context. Unlike hydrologic methods, hydraulic methods can provide a runoff hydrograph inside the catchment.

  2. What are catchment isochrones? How are they determined?

    Catchment isochrones (or isochrone lines) are the loci of points of equal travel time to the catchment outlet.

    Catchment isochrones are determined by estimating the time of concentration at several points inside the catchment, and interpolating to find the contours of equal travel time.

  3. How is time of concentration defined when using hydrographs generated by the time-area method?

    In the absence of runoff diffusion, as would be the case of a hydrograph calculated by the time-area method, the concentration time can be defined as the difference between hydrograph time base and effective rainfall duration. Alternatively, assuming that the point of inflection of the receding limb of a measured hydrograph occurs at the same time as the end of the hydrograph time base, the concentration time can be defined as the difference between the time to point of inflection and the effective rainfall duration.

  4. When can the Clark unit hydrograph be considered synthetic? Explain.

    The Clark unit hydrograph can be considered synthetic when the linear reservoir storage constant, required to provide the necessary runoff diffusion, is obtained by means of a regionally derived formula.

  5. What is the difference between translation and diffusion? How are the time-area and rational methods related?

    Translation (or concentration) can be interpreted as the movement of water in a direction parallel to the channel bottom. Diffusion (runoff diffusion, or channel storage) can be interpreted as the movement of water in a direction perpendicular to the channel bottom. Translation is a first-order process; diffusion is a second-order process.

    The time-area method is essentially an extension of the runoff concentration principle used in the rational method. Unlike the rational method, however, the time-area can account for the temporal variation of rainfall intensity. The translation phase of the time-area method is similar to the rational method in the absence of runoff diffusion. The storage phase of the time-area method can take into account the effect of runoff diffusion.

  6. How can the linear reservoir storage coefficient be determined for runoff data?

    To determine the linear reservoir routing coefficient from runoff data, the differential equation of storage can be evaluated at the tail of a measured (i.e., outflow) hydrograph, past the end of the translated-only (i.e., inflow) hydrograph. Since at this point the inflow is equal to zero:

                 dS
    - O  =  _____
                 dt

    and since S = KO, the following expression for K is obtained:

                    O
    K  =  - ________
                   dO
                 _____
                   dt

    and K can be evaluated from the outflow at a point past the end of the translated-only hydrograph and its time rate of change.

    Alternatively, K can be evaluated at the point of inflection of the receding limb of a measured outflow hydrograph.

  7. What is the principle behind the method of cascade of linear reservoirs used in catchment routing?

    The principle behind the method of cascade of linear reservoirs is that several reservoirs in series can provide enough diffusion so that translation (runoff concentration) and storage (runoff diffusion) can be simulated. In this way, a one-parameter routing method is extended to a two-parameter routing method, while remaining within the computational framework of the linear reservoir.

  8. Why is it necessary to use two schemes in catchment routing using first-order kinematic wave techniques?

    First-order kinematic wave schemes are conditionally stable, i.e., they are stable only for a certain range of Courant numbers. For instance, the forward-in-time, backward-in-space scheme is stable for Courant numbers C < 1. Conversely, the forward-in-space, backward-in-time scheme is stable for Courant numbers C > 1. To provide numerical stability for all Courant numbers, it is necessary to alternate between the two schemes.

  9. Why is a kinematic wave solution using numerical techniques usually grid-dependent? Why is the diffusion wave solution grid-independent?

    A kinematic wave solution using numerical techniques is usually grid dependent because of the creation of numerical diffusion which is not specifically related to the chosen grid size. Therefore, the solution varies with the grid size, i.e., it is grid dependent.

    In the diffusion wave technique the physical and numerical diffusivities are matched. Therefore, the solution does not vary appreciably with grid size, i.e., it is grid independent.

PROBLEMS

  1. A 45-km2 catchment has a 6-h time fo concentration with isochrones at 2-h intervals, resulting in the following time-area histogram:

    Time (h) 0-2 2-4 4-6
    Area (km2) 9 21 15

    Use the time-area method to calculate the outflow hydrograph from the following effective storm pattern:

    Time (h) 0-2 2-4 4-6 6-8 8-10 10-12
    Effective rainfall (cm/h) 0.5 1.0 2.0 3.0 1.0 0.5

    Use a spreadsheet. Verify your results with ONLINE ROUTING06.

    The accumulated effective rainfall depth is:

    (0.5 + 1.0 + 2.0 + 3.0 + 1.0 + 0.5) cm/h × 2 h = 16 cm.

    The total runoff volume is: 45 km2 × 16 cm = 720 km2-cm = 7.2 hm3.

    The calculations are shown in the following table.

    Time
    (h)    		 Flows
    (km2-cm/h)    		 Outflow
    (m3/s)
    Effective rainfall (cm/h)
    0.5 1.0 2.0 3.0 1.0 0.5 Outflow
    0 0 0 0
    2 4.5 0 4.5 12.50
    4 10.5 9.0 0 19.5 54.17
    6 7.5 21.0 18.0 0 46.5 129.17
    8 0 15.0 42.0 27.0 0 84.0 233.33
    10 0 30.0 63.0 9.0 0 102.0 283.33
    12 0 45.0 21.0 4.5 70.5 195.83
    14 0 15.0 10.5 25.5 70.83
    16 0 7.5 7.5 20.83
    18 0 0 0
    Sum 360.0

    The sum of outflow hydrograph ordinates is: 360 km2-cm/h.

    The total runoff volume is: 360 km2-cm/h × 2 h = 720 km2-cm = 7.2 hm3.

    The outflow hydrograph (in m3/s) is shown in the last column. ANSWER.

    The results are confirmed by running ONLINE ROUTING06 as shown in the table below.

  2. A 68-km2 catchment has a 4-h time of concentration with isochrones at 1-h intervals resulting in the following time-area histogram:

    Time (h) 0-1 1-2 2-3 3-4
    Area (km2) 12 19 26 11

    Use the time-area method to calculate the outflow hydrograph from the following storm:

    Time (h) 0-1 1-2 2-3 3-4 4-5 5-6
    Total rainfall (cm/h) 1 2 4 3 2 1

    The runoff curve number is CN = 75. Use a spreadsheet. Verify your results with ONLINE ROUTING06.

    The total rainfall depth is: P = 13 cm. With CN = 75, and R = 2.54 cm/in, the effective rainfall depth (i.e., runoff) (Eq. 5-9) is: Q = 6.465 cm.

    Assuming φ between 1 and 2 cm/h:

    [(2 - φ) × 1 + (4 - φ) × 1 + (3 - φ) × 1 + (2 - φ) × 1] = 6.465

    From which: φ = 1.13 cm/h. The total and effective rainfall pattern is:

    Time (h) 0-1 1-2 2-3 3-4 4-5 5-6
    Total rainfall (cm/h) 1.00 2.00 4.00 3.00 2.00 1.00
    Effective rainfall (cm/h) 0 0.87 2.87 1.87 0.87 0.00

    The accumulated effective rainfall depth is: 6.48 cm.

    The total runoff volume is:

    68 km2 × 6.48 cm = 440.64 km2-cm = 4.41 hm3.

    The calculations are shown in the following table.

    Time
    (h)    		 Flows
    (km2-cm/h)    		 Outflow
    (m3/s)
    Effective rainfall (cm/h)
    0.00 0.87 2.87 1.87 0.87 0.00 Outflow
    0 0 0 0
    1 0 0 0 0
    2 0 10.44 0 10.44 29.00
    3 0 16.53 34.44 0 50.97 141.58
    4 0 22.62 54.53 22.44 0 99.59 276.64
    5 0 9.57 74.62 35.53 10.44 0 130.16 361.56
    6 0 31.57 48.62 16.53 0 96.72 268.67
    7 0 20.57 22.62 0 43.19 119.97
    8 0 9.57 0 9.57 26.58
    9 0 0 0 0
    Sum 440.64

    The sum of outflow hydrograph ordinates is 440.64 km2-cm/h.

    The integration of the outflow hydrograph results in:

    440.64 km2-cm/h × 1 h = 440.64 km2-cm ≅ 4.41 hm3 , which is the same as the total runoff volume.  ANSWER.

    The results are confirmed by running ONLINE ROUTING06 as shown in the table below.

  3. An 82-km2 catchment has the following characteristics:

    Time (h) 0-1 1-2 2-3 3-4 4-5 5-6
    Area (km2) 14 22 29 17 0 0
    Effective rainfall (cm/h) 0.8 1.2 1.3 2.1 0.7 0.5

    Using a spreadsheet, calculate the outflow hydrograph by the time-area method. Verify your results with ONLINE ROUTING06.

    The accumulated effective rainfall depth is: 6.6 cm.

    The total runoff volume is:

    82 km2 × 6.6 cm = 541.2 km2-cm = 5.412 hm3.

    The results of the routing by the time-area method are shown in the following table.

    Time
    (h)    		 Outflow
    (m3/s)
    0 0
    1 31.111
    2 95.556
    3 188.333
    4 295.556
    5 316.944
    6 292.778
    7 186.111
    8 73.333
    9 23.611
    10 0
    Sum 1503.333

    The sum of outflow hydrograph ordinates is 1503.333 m3/s.

    The integration of the outflow hydrograph results in:

    1503.333 m3/s × 1 h = 1503.333 (m3/s)-h

    1503.333 (m3/s)-h × 3600 s/h × 10-6 hm3/m3 = 5.412 hm3 , which is the same as the total runoff volume.  ANSWER.

    The results are confirmed by running ONLINE ROUTING06 as shown in the table below.

  4. Use the Clark method with a spreadsheet to derive a 3-h unit hydrograph for a catchment with the following time-area diagram:

    Time (h) 0-3 3-6 6-9 9-12
    Area (km2) 57 72 39 15

    Use Δt = 3 h and K = 3 h. Verify your results with ONLINE ROUTING07.

    Since Δt = 3 h, and K = 3 h: Δt /K = 1, and from Table 8-1, the linear reservoir-routing coefficients are the following: C0 = C1 = C2 = 1/3.

    The unit runoff volume is:

    (57 + 72 + 39 + 15) km2 × 1 cm = 183 km2-cm = 1.83 hm3.

    Since the duration of the SI unit hydrograph (1 cm of runoff) is 3 h, the rainfall intensity is 1/3 cm/h. The calculations are shown in the following table.

    Time
    (h)    		 Flows
    (km2-cm/h)    		 Outflow
    (m3/s)
    Effective rainfall (cm/h)
    Ai × 1/3 cm/h C0 I2 C1 I1 C2 O1 Outflow
    0 0 - - - 0 0
    3 19 6.333 0 0 6.333 17.593
    6 24 8.000 6.333 2.111 16.444 45.679
    9 13 4.333 8.000 5.481 17.815 49.486
    12 5 1.667 4.333 5.938 11.938 33.162
    15 0 0 1.667 3.979 5.646 15.684
    18 0 0 0 1.882 1.882 5.228
    21 0 0 0 0.627 0.627 1.743
    24 0 0 0 0.209 0.209 0.581
    27 0 0 0 0.070 0.070 0.194
    30 0 0 0 0.023 0.023 0.065
    33 0 0 0 0.007 0.007 0.022
    36 0 0 0 0.003 0.003 0.007
    Sum 61 60.996

    The unit runoff volume is obtained by integrating the unit hydrograph:

    60.996 km2-cm/h × 3 h = 182.998 km2-cm = 1.83 hm3.

    It is confirmed that this value is the same as that obtained from the catchment area.  ANSWER.

    The results are confirmed by running ONLINE ROUTING07 as shown in the table below.

  5. Use the Clark method with a spreadsheet to derive a 1-h unit hydrograph for a catchment with the following time-area diagram:

    Time (h) 0-1 1-2 2-3 3-4 4-5 5-6
    Area (km2) 12 20 42 66 30 16

    Use Δt = 1 h and K = 2 h. Verify your results with ONLINE ROUTING07.

    Since Δt = 1 h, and K = 2 h: Δt /K = 1/2, and from Table 8-1, the linear reservoir-routing coefficient are the following:

    C0 = 1/5; C1 = 1/5; C2 = 3/5

    The unit runoff volume is:

    (12 + 20 + 42 + 66 + 30 + 16) km2 × 1 cm = 186 km2-cm = 1.86 hm3

    Since the duration of the SI unit hydrograph (1 cm of runoff) is 1 h, the rainfall intensity is 1 cm/h. The calculations are shown in the following table.

    Time
    (h)    		 Flows
    (km2-cm/h)    		 Outflow
    (m3/s)
    Effective rainfall (cm/h)
    Ai × 1 cm/h C0 I2 C1 I1 C2 O1 Outflow
    0 0 - - - 0 0
    1 12 2.40 0.00 0.00 2.40 6.67
    2 20 4.00 2.40 1.44 7.84 21.78
    3 42 8.40 4.00 4.70 17.10 47.51
    4 66 13.20 8.40 10.26 31.86 88.51
    5 30 6.00 13.20 19.11 38.32 106.44
    6 16 3.20 6.00 22.99 32.19 89.42
    7 0 0 3.20 19.31 22.51 62.54
    8 0 0 0 13.50 13.51 37.52
    9 0 0 0 8.10 8.10 22.51
    10 0 0 0 4.86 4.86 13.51
    11 0 0 0 2.92 2.92 8.11
    12 0 0 0 1.75 1.75 4.86
    13 0 0 0 1.05 1.05 2.92
    14 0 0 0 0.63 0.63 1.75
    15 0 0 0 0.38 0.38 1.05
    16 0 0 0 0.23 0.23 0.63
    17 0 0 0 0.14 0.14 0.38
    18 0 0 0 0.08 0.08 0.23
    19 0 0 0 0.05 0.05 0.14
    20 0 0 0 0.03 0.03 0.08
    21 0 0 0 0.02 0.02 0.05
    22 0 0 0 0.01 0.01 0.03
    Sum 186 185.96

    The unit runoff volume is obtained by integrating the unit hydrograph:

    185.96 km2-cm/h × 1 h = 185.96 km2-cm ≅ 1.86 hm3.

    It is confirmed that this value is the same as that obtained from the catchment area. ANSWER.

    The results are confirmed by running ONLINE ROUTING07 as shown in the table below.

  6. Use the Clark method with a spreadsheet to derive a 2-h unit hydrograph for a catchment with the following time-area diagram:

    Time (h) 0-1 1-2 2-3 3-4 4-5 5-6
    Area (km2) 10 20 30 20 12 8

    Use Δt = 1 h and K = 4 h. Verify your results with ONLINE ROUTING07.

    Since Δt = 1 h, and K = 4 h: Δt /K = 1/4, and from Table 8-1, the linear reservoir-routing coefficient are the following:

    C0 = 1/9; C1 = 1/9; C2 = 7/9

    The unit runoff volume is:

    (10 + 20 + 30 + 20 + 12 + 8) km2 × 1 cm = 100 km2-cm = 1.0 hm3

    Since the duration of the SI unit hydrograph (1 cm of runoff) is 2 h, the rainfall intensity is 0.5 cm/h

    Time
    (h)    		 Flows
    (km2-cm/h)    		 Outflow
    (m3/s)
    Effective rainfall (cm/h)
    Ai × 0.5 cm/h Ai × 0.5 cm/h Sum C0 I2 C1 I1 C2 O1 Outflow
    0 0 0 0 - - - 0 0
    1 5 0 5 0.556 0.000 0.000 0.556 1.543
    2 10 5 15 1.667 0.555 0.432 2.654 7.373
    3 15 10 25 2.778 1.667 2.064 6.509 18.08
    4 10 15 25 2.778 2.778 5.063 10.618 29.495
    5 6 10 16 1.778 2.778 8.259 12.814 35.595
    6 4 6 10 1.111 1.778 9.967 12.855 35.709
    7 0 4 4 0.444 1.111 9.999 11.554 32.095
    8 0 0 0 0 0.444 8.986 9.431 26.197
    9 0 0 0 0 0 7.335 7.335 20.376
    10 0 0 0 0 0 5.705 5.705 15.848
    11 0 0 0 0 0 4.437 4.437 12.326
    12 0 0 0 0 0 3.451 3.451 9.587
    13 0 0 0 0 0 2.683 2.683 7.453
    14 0 0 0 0 0 2.088 2.088 5.799
    15 0 0 0 0 0 1.624 1.624 4.511
    16 0 0 0 0 0 1.263 1.263 3.508
    17 0 0 0 0 0 0.982 0.982 2.729
    18 0 0 0 0 0 0.764 0.764 2.122
    19 0 0 0 0 0 0.594 0.594 1.651
    20 0 0 0 0 0 0.462 0.462 1.284
    21 0 0 0 0 0 0.359 0.359 0.999
    22 0 0 0 0 0 0.28 0.28 0.777
    23 0 0 0 0 0 0.217 0.217 0.604
    24 0 0 0 0 0 0.169 0.169 0.47
    25 0 0 0 0 0 0.132 0.132 0.365
    26 0 0 0 0 0 0.102 0.102 0.284
    27 0 0 0 0 0 0.08 0.08 0.221
    28 0 0 0 0 0 0.062 0.062 0.172
    29 0 0 0 0 0 0.048 0.048 0.134
    30 0 0 0 0 0 0 0 0
    Sum 100 99.842

    The results are confirmed by running ONLINE ROUTING07 as shown in the table below.

  7. The 2-h unit hydrograph for a 92-km2 catchment is the following:

    Time (h) 0 2 4 6 8 10 12 14
    Flow (m3/s) 0 2.778 8.611 14.333 19.433 21.938 20.942 15.897
    Time (h) 16 18 20 22 24 26 28 30
    Flow (m3/s) 9.538 5.722 3.433 2.061 1.236 0.742 0.444 0.267

    Given the following time-area diagram, what is the linear reservoir storage constant in the Clark method?

    Time (h) 0-2 2-4 4-6 6-8 8-10 10-12
    Area (km2) 10 15 18 21 16 12

    The time of concentration is: tc = 12 h.

    The effective rainfall duration is:  tr = 2 h.

    The time base of the translated-only hydrograph (Eq. 10-1) is:

    Tb = tc + tr = 14 h.

    Therefore, the evaluation of K can be based on Eq. 10-4, using two consecutive hydrograph ordinates after 14 h. Using the ordinates at 14 and 16 h:

    K = - [Ō/ (dO/dt)]

    K = - [(15.897 + 9.538)/2] / [(9.538 - 15.897)/ 2 h] = 4 h. ANSWER.

  8. A 1-h unit hydrograph derived from measured data has the following ordinates:

    Time (h) 0 1 2 3 4 5 6 7
    Flow (m3/s) 0 7 22 48 60 90 74 47
    Time (h) 8 9 10 11 12 13 14
    Flow (m3/s) 28 17 10 6 4 3 2

    Assuming a time of concentration tc = 6 h, calculate the linear reservoir storage constant in the Clark method.

    The time of concentration is:  tc = 6 h.

    The effective rainfall duration is:  tr = 1 h.

    The time base of the translated-only hydrograph (Eq. 10-4) is:

    Tb = tc + tr = 7 h.

    Therefore, the evaluation of K can be based on Eq. 10-4, using two consecutive hydrograph ordinates after 7 h. Alternatively, the end of the translated-only hydrograph can be assumed to occur at the point of inflection on the receding limb of the measured hydrograph, which can be readily identified at 7 h.

    Using Eq. 10-4 and the ordinates at 7 and 8 h:

    K = - [(47 + 28)/2] / [(28 - 47)/ 1 h] = 1.97 h.

    Since this is a unit hydrograph derived from measured data, it may be appropriate to average several values of K computed at the tail of the hydrograph.

    Accordingly, using the ordinates at 8 and 9 h:  K = 2.04 h.

    Using the ordinates at 9 and 10 h:  K = 1.92 h.

    Using the ordinates at 10 and 11 h:  K = 2.0 h. The average of four calculations is: K = 1.98 h, say K = 2 h. ANSWER.

  9. Using ONLINE ROUTING08, route the effective rainfall hyetograph of Example 10-3 (with time interval Δt = 6 h using: (a) K = 12 h and N = 4; and (b) K = 18 h and N = 3.

    The calculations are shown in the following table.

    Time
    (h)    		 Outflow
    (m3/s)
    K = 12 h; N = 4 K = 18 h; N = 3
    0 0 0
    6 1.778 3.239
    12 18.489 29.617
    18 79.644 107.098
    24 206.080 233.935
    30 379.193 374.072
    36 549.125 489.028
    42 670.129 560.611
    48 723.116 589.121
    54 713.299 582.541
    60 657.812 550.891
    66 575.947 503.530
    72 483.973 448.089
    78 393.396 390.243
    84 311.134 333.888
    90 240.497 281.460
    96 182.314 234.293
    102 135.916 192.929
    108 99.866 157.381
    114 72.452 127.329
    120 51.976 102.268

    The results are confirmed by running ONLINE ROUTING08 as shown in the table below.

    The results are confirmed by running ONLINE ROUTING08 as shown in the table below.

  10. Use ONLINE ROUTING08 to route the following storm hyetograph through a 535-km2 catchment.

    Time (h) 0-6 6-12 12-18 18-24
    Effective rainfall depth (cm) 1.0 1.5 2.5 1.2
    Effective rainfall depth (cm/hr) 0.1667 0.2500 0.4167 0.2000

    Set K = 12 h and N = 3. Report peak outflow and time-to-peak.

  11. Using the method of cascade of linear reservoirs, derive a 3-h unit hydrograph for a 432-km2 basin. Assume Δt = 3 h, K = 9 h and two reservoirs (N = 2). Verify the results with ONLINE ROUTING08.

    The basin area is 432 km2. The calculations follow Example 10-3, with Δt= 3 h and K = 9 h, leading to C = Δt/K = 1/3, and to the routing coefficients (Eqs. 10-7 and 10-8): C1 = 1/7; and C2 = 5/7.

    The routing equation is Eq. 10-10. The results of the routing with 2 linear reservoirs (N = 2) are summarized in the following table.

    t
    (h)    		Outflow
    (m3/s)
    00.000
    316.327
    639.650
    948.313
    1248.789
    1545.049
    1839.464
    2133.392
    2427.569
    2722.347
    3017.859
    3314.111
    3611.047
    398.582
    426.623
    455.084
    483.883
    512.954
    542.238
    571.690
    601.273
    630.956
    660.716
    690.536
    720.400
    750.298
    780.221
    810.164
    840.122
    870.090
    900.067
    930.049
    960.036
    990.027
    1020.020
    1050.015
    1080.011
    1110.008
    1140.006
    1170.004
    1200.003
    1230.002
    1260.002
    1290.001
    1320.001
    Sum399.999

    The sum of ouflow hydrograph ordinates is 399.999, or 400 m3/s.

    The integration of the outflow hydrograph leads to:

    400 m3/s × 3 h × 3600 s/h = 4,320,000 m3, or 432 km2-cm, i.e., 1 cm of runoff depth uniformly distributed over the basin area. ANSWER.

    The results are confirmed by runnung ONLINE ROUTING08 as shown in the table below.

  12. Using the method of cascade of linear reservoirs, derive a 6-h unit hydrograph for a 1235-km2 basin. Assume Δt = 6 h, K = 6 h and N = 4. Verify the results with ONLINE ROUTING08.

    With catchment (basin) area = 1235 km2 ; time interval Δt = 6 h; one (1) rainfall increment of 1 cm depth (unit hydrograph); reservoir storage constant K = 6 h; and number of linear reservoirs N = 4. The results are summarized in the following table.

    Time
    (h)    		 Outflow
    (m3/s)
    0 0.000
    6 14.112
    12 61.151
    18 114.463
    24 128.052
    30 103.313
    36 68.585
    42 40.129
    48 21.507
    54 10.808
    60 5.171
    66 2.381
    72 1.062
    78 0.462
    84 0.197
    90 0.082
    96 0.034
    102 0.014
    108 0.005
    114 0.002
    120 0.001
    Sum 571.531

    The integration of the outflow hydrograph leads to:

    571.531 m3/s × 6 h × 3600 s/h = 12,345,069 m3 ≅ 1235 km2-cm, i.e.,

    1 cm or runoff depth uniformly distributed over the basin area. ANSWER.

    The results are confirmed by runnung ONLINE ROUTING08 as shown in the table below.

  13. Derive Eq. 10-16.

    The discretization of Eq. 9-43 in a forward-in-time, backward-in-space linear scheme leads to:

        Q j+1n+1 - Q j+1n               Q j+1n+1 - Q j n
    ___________________   + c   _________________  = cqL
                  Δt                                      Δx

    in which c = βV. Solving for the unknown discharge:

    Q j+1 n+1 = C Q j n + (1- C) Qj+1n + C QL

    in which: C = c ( Δt / Δx ); and QL = qL Δx

    Therefore:

    Q j+1 n+1 = C1 Q j n + C2 Qj+1 n + C3 QL

    in which C1 = C ; C2 = 1 - C; C3 = C. ANSWER.

  14. Derive Eq. 10-17.

    The discretization of Eq. 9-43 in a forward-in-space, backward-in-time linear scheme leads to:

        Q jn+1 - Q jn                     Q j+1n+1 - Q j n+1
    ___________________   + c   _________________  = cqL
                  Δt                                      Δx

    in which c = βV. Solving for the unknown discharge:

    Q j+1 n+1 = [(C- 1)/ C] Q j n+1 + (1/C) Qj n + QL

    in which: C = c ( Δt / Δx ); and QL = qL Δx

    Therefore:

    Q j+1 n+1 = C0 Q j n+1 + C1 Qj n + C3 QL

    in which C0 = [(C - 1)/ C] ; C1 = 1/C; C3 = 1. ANSWER.

  15. Derive Eq. 10-20.

    The discretization of Eq. 9-43 in a way that parallels the Muskingum method, centering the spatial derivative and offcentering the temporal derivative by means of a weighting factor X leads to Eq. 9-61, but with a right-hand side equal to cqL, in which c = βV.

    Multiplying both sides by 2Δt results in Eq. 9-62, with routing coefficients given by Eqs. 9-63 to 9-65, but with the addition of a lateral inflow term on the right-hand side as follows:

                 2Δt cqL  
    _________________________
        2(1- X) + c (Δt / Δx)

    Since C = c (Δt / Δx): c Δt = C Δx, and qLΔx = QL, and using Eq. 9-72, the lateral inflow term is C3QL, in which the routing coefficient C3 is:

                       2C  
    C3 =   _____________ ANSWER.
                  1+ C + D

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