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Find: Omax [cfs] Given Data 29,000 33,000 37,000 41,000 inflow outflow

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Presentation on theme: "Find: Omax [cfs] Given Data 29,000 33,000 37,000 41,000 inflow outflow"— Presentation transcript:

1 Find: Omax [cfs] Given Data 29,000 33,000 37,000 41,000 inflow outflow
time time elevation elevation Find the maximum outflow from the reservoir, O max, in cubic feet per second. [pause] Given Data outflow area

2 Find: Omax [cfs] Given Data 29,000 33,000 37,000 41,000 inflow outflow
time time elevation elevation This problem provides the data the generate the plots shown here, which includes --- Given Data outflow area

3 Find: Omax [cfs] Given Data 29,000 33,000 37,000 41,000 inflow outflow
time time elevation elevation inflow hydrograph, Given Data outflow area

4 Find: Omax [cfs] Given Data 29,000 33,000 37,000 41,000 inflow outflow
time time elevation elevation the reservoir outflow, for the first time period, --- Given Data outflow area

5 Find: Omax [cfs] Given Data 29,000 33,000 37,000 41,000 inflow outflow
time time elevation elevation the outflow, as a function of water surface elevation, often called the rating curve, and --- Given Data outflow area

6 Find: Omax [cfs] Given Data 29,000 33,000 37,000 41,000 inflow outflow
time time elevation elevation the area of the reservoir as a function of elevation, used to define the reservoir geometry. Given Data outflow area

7 Find: Omax [cfs] Given Data 29,000 33,000 37,000 41,000 inflow outflow
time time elevation elevation Let’s begin by taking a look at the inflow hydrograph. Given Data outflow area

8 Find: Omax [cfs] inflow time
The problem asks to find the maximum flowrate out of the reservoir, so if we re-label the vertical axis to be --- time

9 Find: Omax [cfs] flowrate inflow time
flowrate, and add in an approximate --- time

10 Find: Omax [cfs] flowrate inflow outflow time
outflow hydrograph in blue, the problem is asking we find --- outflow time

11 ? Find: Omax [cfs] flowrate inflow outflow time
the maximum outflow. [pause] Also note that we’ve been given the --- outflow time

12 ? Find: Omax [cfs] flowrate inflow outflow time O1=4,000 [cfs]
initial outflow flowrate, which is 4,000 cubic feet per second. [pause] outflow time O1=4,000 [cfs]

13 ? Find: Omax [cfs] flowrate inflow outflow time O0-12=4,000 [cfs] Time
[hours] [cfs] 0-12 5,000 flowrate inflow 12-24 14,600 24-36 35,800 36-48 45,000 48-60 33,400 60-72 17,000 72-84 14,500 ? The inflow hydrograph is in 12 hour time steps, with inflows ranging from --- 84-96 11,000 outflow time O0-12=4,000 [cfs]

14 ? Find: Omax [cfs] flowrate inflow outflow time O0-12=4,000 [cfs] Time
[hours] [cfs] 0-12 5,000 flowrate inflow 12-24 14,600 24-36 35,800 36-48 45,000 48-60 33,400 60-72 17,000 72-84 14,500 ? 5,000, to 45,000, cubic feet per second. As defined, the inflow hydrograph would look more like a --- 84-96 11,000 outflow time O0-12=4,000 [cfs]

15 Find: Omax [cfs] flowrate inflow time Time inflow [hours] [cfs] 0-12
5,000 flowrate inflow 12-24 14,600 24-36 35,800 36-48 45,000 48-60 33,400 60-72 17,000 72-84 14,500 bar graph, with discrete changes in flowrate ever 12 hours, 84-96 11,000 time

16 Find: Omax [cfs] flowrate inflow outflow time Time inflow [hours]
0-12 5,000 flowrate inflow 12-24 14,600 24-36 35,800 36-48 45,000 outflow 48-60 33,400 60-72 17,000 72-84 14,500 as would the outflow hydrograph. For this problem we’ll think of all flowrates, in terms of these 12 hour time periods. 84-96 11,000 time

17 ? Find: Omax [cfs] flowrate inflow outflow time Time inflow [hours]
0-12 5,000 flowrate inflow 12-24 14,600 24-36 35,800 36-48 45,000 outflow 48-60 33,400 60-72 17,000 72-84 14,500 ? In order to determine the maximum flowrate, out of the reservoir, 84-96 11,000 time

18 ? Find: Omax [cfs] + = +Oj+1 +Oj 2*Sj+1 2*Sj (Ij+Ij+1) Δt Δt time
We’ll use the general level-pool routing equation, which relates the --- time

19 ? Find: Omax [cfs] + = +Oj+1 +Oj outflow 2*Sj+1 2*Sj (Ij+Ij+1) Δt Δt
storage inflow ? the storage volume of water, currently in the reservoir, with the out flow and inflow. [pause] This equation also time

20 ? j+1 j Find: Omax [cfs] + = +Oj+1 +Oj 2*Sj+1 2*Sj (Ij+Ij+1) Δt Δt
relates two consecutive time steps, j, and j + 1. [pause] time

21 Find: Omax [cfs] Given Data inflow outflow time time elevation
So we return to the given data, and look at the bottom 2 plots, --- Given Data outflow area

22 Find: Omax [cfs] Given Data inflow outflow time time elevation
which relate the outflow and area, as a function of the water surface elevation. Given Data outflow area

23 Find: Omax [cfs] elevation outflow elevation area Elev. Outflow Area
[ft] [cfs] [acre] 2,015 4,360 2,017 1,718 4,630 2,019 4,860 4,940 outflow 2,021 8,920 5,260 2,023 13,750 5,620 elevation By knowing the area and elevation, an approximation for the storage volume can be computed for each elevation by --- 2,025 19,210 6,000 2,027 25,250 6,390 2,029 31,820 6,790 2,031 38,880 7,210 area

24 Find: Omax [cfs] Δ elevation [ft] x storage [acres-ft]
Outflow Area Storage [ft] [cfs] [acre] [acre-ft] 2,015 4,360 2,017 1,718 4,630 2,019 4,860 4,940 2,021 8,920 5,260 Δ elevation [ft] 2,023 13,750 5,620 multiplying the change in elevation by the average area, at the given elevation and the previous elevation. x ave area [acres] 2,025 19,210 6,000 2,027 25,250 6,390 storage [acres-ft] 2,029 31,820 6,790 2,031 38,880 7,210

25 Find: Omax [cfs] Δ elevation [ft] x storage [acres-ft]
Outflow Area Storage [ft] [cfs] [acre] [acre-ft] 2,015 4,360 2,017 1,718 4,630 2,019 4,860 4,940 2,021 8,920 5,260 Δ elevation [ft] 2,023 13,750 5,620 The storage at 2,015 feet is 0 acre-feet. We’ll assume this is the bottom of the reservoir. [pause] The storage at 2,017 feet of elevation equals --- x ave area [acres] 2,025 19,210 6,000 2,027 25,250 6,390 storage [acres-ft] 2,029 31,820 6,790 2,031 38,880 7,210

26 Find: Omax [cfs] Δ elevation [ft] x storage [acres-ft]
Outflow Area Storage [ft] [cfs] [acre] [acre-ft] 2,015 4,360 2,017 1,718 4,630 2 [ft] 2,019 4,860 4,940 4,495 [acre] 2,021 8,920 5,260 Δ elevation [ft] 2,023 13,750 5,620 2 feet of elevation change, times the average of the first two areas, or, 4,495 acres, which computes to x ave area [acres] 2,025 19,210 6,000 2,027 25,250 6,390 storage [acres-ft] 2,029 31,820 6,790 2,031 38,880 7,210

27 Find: Omax [cfs] Δ elevation [ft] x storage [acres-ft]
Outflow Area Storage [ft] [cfs] [acre] [acre-ft] 2,015 4,360 2,017 1,718 4,630 2 [ft] 2,019 4,860 4,940 4,495 [acre] 2,021 8,920 5,260 Δ elevation [ft] 2,023 13,750 5,620 8,990 acre-feet of storage, in the first 2 feet of depth in the reservoir. x ave area [acres] 2,025 19,210 6,000 2,027 25,250 6,390 storage [acres-ft] 2,029 31,820 6,790 2,031 38,880 7,210 8,990 [acres-ft]

28 Find: Omax [cfs] Δ elevation [ft] x storage [acres-ft]
Outflow Area Storage [ft] [cfs] [acre] [acre-ft] 2,015 4,360 2,017 1,718 4,630 8,990 2,019 4,860 4,940 2,021 8,920 5,260 Δ elevation [ft] 2,023 13,750 5,620 The next storage depth continues this calculation, adding the storage value from the lower entry. x ave area [acres] 2,025 19,210 6,000 2,027 25,250 6,390 storage [acres-ft] 2,029 31,820 6,790 2,031 38,880 7,210

29 Find: Omax [cfs] Elev. Outflow Area Storage [ft] [cfs] [acre]
[acre-ft] 2,015 4,360 2,017 1,718 4,630 8,990 2,019 4,860 4,940 18,560 2,021 8,920 5,260 28,760 2,023 13,750 5,620 39,640 Once the storage volumes has been calculated, --- 2,025 19,210 6,000 51,260 2,027 25,250 6,390 63,650 2,029 31,820 6,790 76,830 2,031 38,880 7,210 90,830

30 Find: Omax [cfs] Elev. Outflow Area Storage [ft] [cfs] [acre]
[acre-ft] 2,015 4,360 2,017 1,718 4,630 8,990 2,019 4,860 4,940 18,560 2,021 8,920 5,260 28,760 2,023 13,750 5,620 39,640 we can remove the elevation values, to make room for --- 2,025 19,210 6,000 51,260 2,027 25,250 6,390 63,650 2,029 31,820 6,790 76,830 2,031 38,880 7,210 90,830

31 Find: Omax [cfs] Outflow Area Storage [cfs] [acre] [acre-ft] 4,360
4,360 1,718 4,630 8,990 4,860 4,940 18,560 8,920 5,260 28,760 13,750 5,620 39,640 two of the terms we saw in the routing equation, 19,210 6,000 51,260 25,250 6,390 63,650 31,820 6,790 76,830 38,880 7,210 90,830

32 Find: Omax [cfs] 2*Sj Δt Δt +Oj+1 -Oj Outflow Area Storage 2*Sj+1
[acre] [acre-ft] Δt Δt 4,360 1,718 4,630 8,990 4,860 4,940 18,560 8,920 5,260 28,760 13,750 5,620 39,640 Where the left term contains the storage and outflow terms for the next time period, --- 19,210 6,000 51,260 25,250 6,390 63,650 31,820 6,790 76,830 38,880 7,210 90,830

33 Find: Omax [cfs] 2*Sj Δt Δt +Oj+1 -Oj Outflow Area Storage 2*Sj+1
[acre] [acre-ft] Δt Δt 4,360 1,718 4,630 8,990 4,860 4,940 18,560 8,920 5,260 28,760 13,750 5,620 39,640 and the right term contains the storage and outflow for the current time period. 19,210 6,000 51,260 25,250 6,390 63,650 31,820 6,790 76,830 38,880 7,210 90,830

34 Find: Omax [cfs] 2*Sj Δt Δt +Oj+1 -Oj Outflow Area Storage 2*Sj+1
[acre] [acre-ft] Δt Δt 4,360 1,718 4,630 8,990 4,860 4,940 18,560 8,920 5,260 28,760 13,750 5,620 39,640 Plugging in the next storage and next outflow into the equation. 19,210 6,000 51,260 25,250 6,390 63,650 31,820 6,790 76,830 38,880 7,210 90,830

35 Find: Omax [cfs] + 2*Sj Δt Δt 2*8,990 [acre-ft] ft2 ft3 1,718 s 43,560
Outflow Area Storage 2*Sj+1 2*Sj +Oj+1 -Oj [cfs] [acre] [acre-ft] Δt Δt 4,360 1,718 4,630 8,990 4,860 4,940 18,560 8,920 5,260 28,760 13,750 5,620 39,640 and converting acre feet per hour into cubic feet per second, this left term equates to --- 2*8,990 [acre-ft] 19,210 6,000 51,260 ft2 ft3 + 1,718 s 43,560 25,250 6,390 63,650 * 12 [hr] * 3,600 acre s 31,820 6,790 76,830 hr 38,880 7,210 90,830

36 Find: Omax [cfs] + 2*Sj Δt Δt 2*8,990 [acre-ft] ft2 ft3 1,718 43,560 s
Outflow Area Storage 2*Sj+1 2*Sj +Oj+1 -Oj [cfs] [acre] [acre-ft] Δt Δt 4,360 1,718 4,630 8,990 4,860 4,940 18,560 8,920 5,260 28,760 13,750 5,620 39,640 19,848 cubic feet per second. 19,210 2*8,990 [acre-ft] 6,000 51,260 ft2 ft3 + 1,718 43,560 25,250 6,390 63,650 s * 12 [hr] * 3,600 acre s 31,820 6,790 76,830 hr ft3 38,880 7,210 90,830 19,848 s

37 Find: Omax [cfs] 2*Sj Δt Δt +Oj+1 -Oj Outflow Area Storage 2*Sj+1
[acre] [acre-ft] Δt Δt 4,360 19,848 1,718 4,630 8,990 4,860 4,940 18,560 8,920 5,260 28,760 13,750 5,620 39,640 The remaining values in both columns are calculated likewise. 19,210 6,000 51,260 25,250 6,390 63,650 31,820 6,790 76,830 38,880 7,210 90,830

38 Find: Omax [cfs] 2*Sj Δt Δt +Oj+1 -Oj Outflow Area Storage 2*Sj+1
[acre] [acre-ft] Δt Δt 4,360 19,848 1,718 4,630 8,990 42,289 16,412 4,860 4,940 18,560 66,919 32,569 8,920 5,260 28,760 93,691 49,079 13,750 5,620 39,640 122,584 66,191 We will return to this table later in the problem. Let’s call this table, 19,210 6,000 51,260 153,611 84,164 25,250 6,390 63,650 186,761 103,111 31,820 6,790 76,830 222,054 123,121 38,880 7,210 90,830 - 144,294

39 Find: Omax [cfs] Table 1: Lookup 2*Sj Δt Δt +Oj+1 -Oj Outflow Area
Storage 2*Sj+1 2*Sj +Oj+1 -Oj [cfs] [acre] [acre-ft] Δt Δt 4,360 19,848 1,718 4,630 8,990 42,289 16,412 4,860 4,940 18,560 66,919 32,569 8,920 5,260 28,760 93,691 49,079 13,750 5,620 39,640 122,584 66,191 Table 1, the lookup table. [pause] Now it’s time to route the flows through the reservoir. 19,210 6,000 51,260 153,611 84,164 25,250 6,390 63,650 186,761 103,111 31,820 6,790 76,830 222,054 123,121 38,880 7,210 90,830 - 144,294

40 Find: Omax [cfs] Given Data inflow outflow time time elevation
We’ll used the data from the top 2 graphs, --- Given Data outflow area

41 Find: Omax [cfs] Given Data inflow outflow time time elevation
the inflow hydrograph and the beginning outflow data point. Given Data outflow area

42 Find: Omax [cfs] 2*Sj 2*Sj Δt Δt outflow +Oj -Oj [cfs] Time inflow
[hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 In this table, each row represents a 12 hour time period --- 60-72 17,000 166,248 110,739 27,775 72-84 14,500 142,239 96,167 23,036 84-96 11,000 121,667 83,593 19,037

43 Find: Omax [cfs] 2*Sj 2*Sj Δt Δt outflow +Oj -Oj [cfs] Time inflow
[hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 with the inflow values, 60-72 17,000 166,248 110,739 27,775 72-84 14,500 142,239 96,167 23,036 84-96 11,000 121,667 83,593 19,037

44 Find: Omax [cfs] 2*Sj 2*Sj Δt Δt outflow +Oj -Oj [cfs] Time inflow
[hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 the initial outflow value, --- 60-72 17,000 166,248 110,739 27,775 72-84 14,500 142,239 96,167 23,036 84-96 11,000 121,667 83,593 19,037

45 Find: Omax [cfs] 2*Sj 2*Sj Δt Δt outflow +Oj -Oj [cfs] Time inflow
[hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 and these two intermediate terms, which we just computed in Table 1. [pause] We’ll call this table, --- 60-72 17,000 166,248 110,739 27,775 72-84 14,500 142,239 96,167 23,036 84-96 11,000 121,667 83,593 19,037

46 Find: Omax [cfs] Table 2: Results 2*Sj 2*Sj Δt Δt outflow +Oj -Oj
Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 Table 2, the results table. [pause] We’ll begin by determining the quantity --- 60-72 17,000 166,248 110,739 27,775 72-84 14,500 142,239 96,167 23,036 84-96 11,000 121,667 83,593 19,037

47 Find: Omax [cfs] Table 2: Results 2*Sj 2*Sj Δt Δt outflow +Oj -Oj
Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 ? 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 2 times the storage, divided by the time period, minus the outflow, for the first time period. To do so, --- 60-72 17,000 166,248 110,739 27,775 72-84 14,500 142,239 96,167 23,036 84-96 11,000 121,667 83,593 19,037

48 Find: Omax [cfs] Table 2: Results 2*Sj 2*Sj Δt Δt outflow +Oj -Oj
Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 ? 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 all we need is the initial outflow quantity, 4,000 cubic feet per second, and Table 1, --- 60-72 17,000 166,248 110,739 27,775 72-84 14,500 142,239 96,167 23,036 84-96 11,000 121,667 83,593 19,037

49 Find: Omax [cfs] Table 1: Lookup 2*Sj Δt Δt +Oj+1 -Oj Outflow Area
Storage 2*Sj+1 2*Sj +Oj+1 -Oj [cfs] [acre] [acre-ft] Δt Δt 4,360 19,848 1,718 4,630 8,990 42,289 16,412 4,860 4,940 18,560 66,919 32,569 8,920 5,260 28,760 93,691 49,079 13,750 5,620 39,640 122,584 66,191 the lookup table. Here we notice, 4,000 cubic feet per second of outflow --- 19,210 6,000 51,260 153,611 84,164 25,250 6,390 63,650 186,761 103,111 31,820 6,790 76,830 222,054 123,121 38,880 7,210 90,830 - 144,294

50 Find: Omax [cfs] Table 1: Lookup 2*Sj Δt Δt +Oj+1 -Oj Outflow Area
Storage 2*Sj+1 2*Sj +Oj+1 -Oj [cfs] [acre] [acre-ft] Δt Δt 4,360 19,848 1,718 4,630 8,990 42,289 16,412 4,000 4,860 4,940 18,560 66,919 32,569 8,920 5,260 28,760 93,691 49,079 13,750 5,620 39,640 122,584 66,191 is between 1,718 cubic feet per second and 4,860 cubic feet per second. Therefore we can we can interpolate to find our desired value. 19,210 6,000 51,260 153,611 84,164 25,250 6,390 63,650 186,761 103,111 31,820 6,790 76,830 222,054 123,121 38,880 7,210 90,830 - 144,294

51 Find: Omax [cfs] Table 1: Lookup 2*Sj Δt Δt +Oj+1 -Oj Outflow Area
Storage 2*Sj+1 2*Sj +Oj+1 -Oj [cfs] [acre] [acre-ft] Δt Δt 4,360 19,848 1,718 4,630 8,990 42,289 16,412 4,000 4,860 4,940 18,560 66,919 32,569 8,920 5,260 28,760 93,691 49,079 13,750 5,620 39,640 122,584 66,191 We know it’s a number between 16,000 and 32,000, but closer to 32,000. 19,210 6,000 51,260 153,611 84,164 25,250 6,390 63,650 186,761 103,111 31,820 6,790 76,830 222,054 123,121 38,880 7,210 90,830 - 144,294

52 Find: Omax [cfs] Table 1: Lookup 2*Sj Δt Δt 2*Sj Δt +Oj+1 -Oj
Outflow Area Storage 2*Sj+1 2*Sj +Oj+1 -Oj [cfs] [acre] [acre-ft] Δt Δt 4,360 19,848 1,718 4,630 8,990 42,289 16,412 4,000 4,860 4,940 18,560 66,919 32,569 8,920 5,260 28,760 93,691 49,079 13,750 5,620 39,640 122,584 66,191 After performing the interpolation, --- 19,210 6,000 51,260 153,611 84,164 2*Sj (4,000-1,718) -Oj=16,412+(32,569-16,412) * Δt (4,860-1,718)

53 Find: Omax [cfs] Table 1: Lookup 2*Sj Δt Δt 2*Sj Δt =28,147 [cfs]
Outflow Area Storage 2*Sj+1 2*Sj +Oj+1 -Oj [cfs] [acre] [acre-ft] Δt Δt 4,360 19,848 1,718 4,630 8,990 42,289 16,412 4,000 4,860 4,940 18,560 66,919 32,569 8,920 5,260 28,760 93,691 49,079 13,750 5,620 39,640 122,584 66,191 the value equals, 28,147 cubic feet per second. [pause] Back to Table 2, the results table, --- 19,210 6,000 51,260 153,611 84,164 2*Sj (4,000-1,718) -Oj=16,412+(32,569-16,412) * Δt (4,860-1,718) =28,147 [cfs]

54 Find: Omax [cfs] Table 2: Results 2*Sj 2*Sj Δt Δt outflow +Oj -Oj
Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 ? 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 we plug the value in. 60-72 17,000 166,248 110,739 27,775 72-84 14,500 142,239 96,167 23,036 84-96 11,000 121,667 83,593 19,037

55 Find: Omax [cfs] Table 2: Results 2*Sj 2*Sj Δt Δt outflow +Oj -Oj
Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 Next we’ll use the routing equation from before, --- 60-72 17,000 166,248 110,739 27,775 72-84 14,500 142,239 96,167 23,036 84-96 11,000 121,667 83,593 19,037

56 Find: Omax [cfs] + = +Oj+1 +Oj Table 2: Results 2*Sj 2*Sj Δt Δt 2*Sj+1
Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 And solve for the left hand side, for the second time period, by knowing --- 2*Sj+1 2*Sj + +Oj+1 = +Oj (Ij+Ij+1) Δt Δt

57 Find: Omax [cfs] + = +Oj+1 +Oj Table 2: Results 2*Sj 2*Sj Δt Δt 2*Sj+1
Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 our value for twice the storage of the time period minus the outflow, plus--- 2*Sj+1 2*Sj + +Oj+1 = +Oj (Ij+Ij+1) Δt Δt

58 Find: Omax [cfs] + = +Oj+1 +Oj Table 2: Results 2*Sj 2*Sj Δt Δt 2*Sj+1
Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 the inflow from the current and previous time periods. 2*Sj+1 2*Sj + +Oj+1 = +Oj (Ij+Ij+1) Δt Δt

59 Find: Omax [cfs] + = 28,147+5,000+14,600 = +Oj+1 +Oj Table 2: Results
Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 Adding these three terms together we get, --- 2*Sj+1 2*Sj + +Oj+1 = +Oj (Ij+Ij+1) Δt Δt = 28,147+5,000+14,600

60 Find: Omax [cfs] + = 28,147+5,000+14,600 = 47,747 [cfs] = +Oj+1 +Oj
Table 2: Results Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 47,747 cubic feet per second. 2*Sj+1 2*Sj + +Oj+1 = +Oj (Ij+Ij+1) Δt Δt = 28,147+5,000+14,600 = 47,747 [cfs]

61 Find: Omax [cfs] + = 28,147+5,000+14,600 = 47,747 [cfs] = +Oj+1 +Oj
Table 2: Results Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 This value is added to the table. From here, we’ll take our value of ---- 2*Sj+1 2*Sj + +Oj+1 = +Oj (Ij+Ij+1) Δt Δt = 28,147+5,000+14,600 = 47,747 [cfs]

62 Find: Omax [cfs] + = 28,147+5,000+14,600 = 47,747 [cfs] = +Oj+1 +Oj
Table 2: Results Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 47,747, and lookup the values for outflow and --- 2*Sj+1 2*Sj + +Oj+1 = +Oj (Ij+Ij+1) Δt Δt = 28,147+5,000+14,600 = 47,747 [cfs]

63 Find: Omax [cfs] + = 28,147+5,000+14,600 = 47,747 [cfs] = +Oj+1 +Oj
Table 2: Results Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 ? 5,760 ? 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 twice the storage over delta t minus O. 2*Sj+1 2*Sj + +Oj+1 = +Oj (Ij+Ij+1) Δt Δt = 28,147+5,000+14,600 = 47,747 [cfs]

64 Find: Omax [cfs] + = 28,147+5,000+14,600 = 47,747 [cfs] = +Oj+1 +Oj
Table 2: Results Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 ? 5,760 ? 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 We’re interested in the outflow because, the question asks to find the maximum outflow, --- 2*Sj+1 2*Sj + +Oj+1 = +Oj (Ij+Ij+1) Δt Δt = 28,147+5,000+14,600 = 47,747 [cfs]

65 Find: Omax [cfs] + = 28,147+5,000+14,600 = 47,747 [cfs] = +Oj+1 +Oj
Table 2: Results Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 ? 5,760 ? 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 and we’re interested in twice the storage over delta t minus O because we need it to find subsequent outflow values. 2*Sj+1 2*Sj + +Oj+1 = +Oj (Ij+Ij+1) Δt Δt = 28,147+5,000+14,600 = 47,747 [cfs]

66 Find: Omax [cfs] + = 28,147+5,000+14,600 = 47,747 [cfs] = +Oj+1 +Oj
Table 2: Results Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 ? 5,760 ? 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 Taking 47,747 over to the lookup table, --- 2*Sj+1 2*Sj + +Oj+1 = +Oj (Ij+Ij+1) Δt Δt = 28,147+5,000+14,600 = 47,747 [cfs]

67 Find: Omax [cfs] Table 1: Lookup 2*Sj Δt Δt +Oj+1 -Oj Outflow Area
Storage 2*Sj+1 2*Sj +Oj+1 -Oj [cfs] [acre] [acre-ft] Δt Δt 4,360 19,848 1,718 4,630 8,990 42,289 16,412 47,747 4,860 4,940 18,560 66,919 32,569 8,920 5,260 28,760 93,691 49,079 13,750 5,620 39,640 122,584 66,191 it falls between 42,289 and 66,919. At this point, we should adjust this table, --- 19,210 6,000 51,260 153,611 84,164 25,250 6,390 63,650 186,761 103,111 31,820 6,790 76,830 222,054 123,121 38,880 7,210 90,830 - 144,294

68 Find: Omax [cfs] Table 1: Lookup 2*Sj Δt Δt +Oj+1 -Oj Outflow Area
Storage 2*Sj+1 2*Sj +Oj+1 -Oj [cfs] [acre] [acre-ft] Δt Δt 4,360 19,848 1,718 4,630 8,990 42,289 16,412 47,747 4,860 4,940 18,560 66,919 32,569 8,920 5,260 28,760 93,691 49,079 13,750 5,620 39,640 122,584 66,191 to account for the differences in time periods, j+1 and j, --- 19,210 6,000 51,260 153,611 84,164 25,250 6,390 63,650 186,761 103,111 31,820 6,790 76,830 222,054 123,121 38,880 7,210 90,830 - 144,294

69 Find: Omax [cfs] Table 1: Lookup 2*Sj Δt Δt +Oj -Oj Outflow Area
Storage 2*Sj 2*Sj +Oj -Oj [cfs] [acre] [acre-ft] Δt Δt 4,360 1,718 4,630 8,990 19,848 16,412 4,860 4,940 18,560 42,289 32,569 47,747 8,920 5,260 28,760 66,919 49,079 13,750 5,620 39,640 93,691 66,191 to make interpolation easier. Except this time we’ll interpolate twice, --- 19,210 6,000 51,260 122,584 84,164 25,250 6,390 63,650 153,611 103,111 31,820 6,790 76,830 186,761 123,121 38,880 7,210 90,830 222,054 144,294

70 Find: Omax [cfs] Table 1: Lookup 2*Sj Δt Δt +Oj -Oj Outflow Area
Storage 2*Sj 2*Sj +Oj -Oj [cfs] [acre] [acre-ft] Δt Δt 4,360 1,718 4,630 8,990 19,848 16,412 4,860 4,940 18,560 42,289 32,569 47,747 8,920 5,260 28,760 66,919 49,079 13,750 5,620 39,640 93,691 66,191 once to find the outflow, and once to find twice the storage over delta t minus O. 19,210 6,000 51,260 122,584 84,164 25,250 6,390 63,650 153,611 103,111 31,820 6,790 76,830 186,761 123,121 38,880 7,210 90,830 222,054 144,294

71 Find: Omax [cfs] Table 1: Lookup 2*Sj Δt Δt Oj=5,760 [cfs] +Oj -Oj
Outflow Area Storage 2*Sj 2*Sj +Oj -Oj [cfs] [acre] [acre-ft] Δt Δt 4,360 1,718 4,630 8,990 19,848 16,412 4,860 4,940 18,560 42,289 32,569 47,747 8,920 5,260 28,760 66,919 49,079 13,750 5,620 39,640 93,691 66,191 The outflow equals 5,760 cubic feet per second, --- 19,210 6,000 51,260 122,584 84,164 25,250 6,390 63,650 153,611 103,111 Oj=5,760 [cfs]

72 Find: Omax [cfs] Table 1: Lookup 2*Sj Δt Δt 2*Sj Oj=5,760 [cfs]
Outflow Area Storage 2*Sj 2*Sj +Oj -Oj [cfs] [acre] [acre-ft] Δt Δt 4,360 1,718 4,630 8,990 19,848 16,412 4,860 4,940 18,560 42,289 32,569 47,747 8,920 5,260 28,760 66,919 49,079 13,750 5,620 39,640 93,691 66,191 and twice the storage over delta t minus O equals 36,228 cubic feet per second. Returning to the results table, --- 19,210 6,000 51,260 122,584 84,164 25,250 6,390 63,650 153,611 103,111 2*Sj Oj=5,760 [cfs] =36,228 [cfs] -Oj Δt

73 Find: Omax [cfs] Table 2: Results 2*Sj 2*Sj Δt Δt 2*Sj Oj=5,760 [cfs]
Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 the values are plugged in--- 2*Sj Oj=5,760 [cfs] =36,228 [cfs] -Oj Δt

74 Find: Omax [cfs] Table 2: Results 2*Sj 2*Sj Δt Δt 2*Sj Oj=5,760 [cfs]
Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 [pause] Then, the process repeats for all time steps we have inflow data. 2*Sj Oj=5,760 [cfs] =36,228 [cfs] -Oj Δt

75 Find: Omax [cfs] Table 2: Results 2*Sj 2*Sj Δt Δt outflow +Oj -Oj
Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 ? 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 We find our next value for twice the storage over delta t plus the outflow, for the next time period, ---

76 Find: Omax [cfs] + = +Oj+1 +Oj Table 2: Results 2*Sj 2*Sj Δt Δt 2*Sj+1
Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 ? 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 by plugging in the known values. 2*Sj+1 2*Sj + +Oj+1 = +Oj (Ij+Ij+1) Δt Δt

77 Find: Omax [cfs] + = 86,628 [cfs] = +Oj+1 +Oj Table 2: Results 2*Sj
Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 adding the value to the table, returning to the lookup table, --- 2*Sj+1 2*Sj + +Oj+1 = +Oj (Ij+Ij+1) Δt Δt = 86,628 [cfs]

78 Find: Omax [cfs] Table 1: Lookup 2*Sj Δt Δt +Oj -Oj Outflow Area
Storage 2*Sj 2*Sj +Oj -Oj [cfs] [acre] [acre-ft] Δt Δt 4,360 1,718 4,630 8,990 19,848 16,412 4,860 4,940 18,560 42,289 32,569 8,920 5,260 28,760 66,919 49,079 86,628 13,750 5,620 39,640 93,691 66,191 interpolating to find the next outflow value, --- 19,210 6,000 51,260 122,584 84,164 25,250 6,390 63,650 153,611 103,111 31,820 6,790 76,830 186,761 123,121 38,880 7,210 90,830 222,054 144,294

79 Find: Omax [cfs] Table 2: Results 2*Sj 2*Sj Δt Δt outflow +Oj -Oj
Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 adding that value to the results table, --- 60-72 17,000 166,248 110,739 27,775 72-84 14,500 142,239 96,167 23,036 84-96 11,000 121,667 83,593 19,037

80 Find: Omax [cfs] Table 2: Results 2*Sj 2*Sj Δt Δt outflow +Oj -Oj
Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 and so on, and so forth. 60-72 17,000 166,248 110,739 27,775 72-84 14,500 142,239 96,167 23,036 84-96 11,000 121,667 83,593 19,037

81 Find: Omax [cfs] Table 2: Results 2*Sj 2*Sj Δt Δt outflow +Oj -Oj
Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 If we continue this process to for all 4 days, ---- 60-72 17,000 166,248 110,739 27,775 72-84 14,500 142,239 96,167 23,036 84-96 11,000 121,667 83,593 19,037

82 Find: Omax [cfs] Table 2: Results 2*Sj 2*Sj Δt Δt outflow +Oj -Oj
Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 we will have developed the outflow hydrograph, --- 60-72 17,000 166,248 110,739 27,775 72-84 14,500 142,239 96,167 23,036 84-96 11,000 121,667 83,593 19,037

83 Find: Omax [cfs] Table 2: Results 2*Sj 2*Sj Δt Δt outflow +Oj -Oj
Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 as shown in the far right column. 60-72 17,000 166,248 110,739 27,775 72-84 14,500 142,239 96,167 23,036 84-96 11,000 121,667 83,593 19,037

84 Find: Omax [cfs] Table 2: Results 2*Sj 2*Sj Δt Δt outflow +Oj -Oj
Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 Since the problem asks to find the maximum outflow value, --- 60-72 17,000 166,248 110,739 27,775 72-84 14,500 142,239 96,167 23,036 84-96 11,000 121,667 83,593 19,037

85 Find: Omax [cfs] Table 2: Results 2*Sj 2*Sj Δt Δt outflow +Oj -Oj
Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 We notice it is 29,432, cubic feet per second. 60-72 17,000 166,248 110,739 27,775 72-84 14,500 142,239 96,167 23,036 84-96 11,000 121,667 83,593 19,037

86 Find: Omax [cfs] Table 2: Results 2*Sj 2*Sj Δt Δt 29,000 33,000 37,000
Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 29,000 33,000 37,000 41,000 When reviewing the possible solutions, --- 60-72 17,000 166,248 110,739 27,775 72-84 14,500 142,239 96,167 23,036 84-96 11,000 121,667 83,593 19,037

87 Find: Omax [cfs] Table 2: Results AnswerA 2*Sj 2*Sj Δt Δt 29,000
Time inflow 2*Sj 2*Sj outflow +Oj -Oj [hours] [cfs] Δt Δt [cfs] 0-12 5,000 28,147 4,000 12-24 14,600 47,747 36,228 5,760 24-36 35,800 86,628 61,677 12,476 36-48 45,000 142,477 96,312 23,083 48-60 33,400 174,712 115,848 29,432 29,000 33,000 37,000 41,000 the answer is A. 60-72 17,000 166,248 110,739 27,775 72-84 14,500 142,239 96,167 23,036 84-96 11,000 121,667 83,593 19,037 AnswerA

88 ? Index σ’v = Σ γ d γT=100 [lb/ft3] +γclay dclay 1
Find: σ’v at d = 30 feet (1+wc)*γw wc+(1/SG) σ’v = Σ γ d d Sand 10 ft γT=100 [lb/ft3] 100 [lb/ft3] 10 [ft] 20 ft Clay = γsand dsand +γclay dclay A W S V [ft3] W [lb] 40 ft text wc = 37% ? Δh 20 [ft] (5 [cm])2 * π/4

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