1. Department of Civil Engineering, I.T.U
Reservoirs
Tarkan Erdik, PhD
Associate Professor,
Department of Civil Engineering, I.T.U

2. Primary function is to store
A reservoirs has two categories:
1- Storage (conservation) [i.e., Atatürk dam]
2- Distribution (service) [for emergencies & fire fighting]
Physical Characteristics of Reservoirs
Primary function is to store
Most important characteristic:“storage capacity”

3. Reservoir: Collects water behind a dam or barrier
Reservoirs are constructed for:
Drinking water,
Irrigation,
Hydropower,
Flood mitigation
During a specified time interval;
S (supply) < D (demand)
Need for “water storage”

4. Reservoir
Spillway crest
Upstream
Spillway
Dam body
Downstream

5. Spillways are structures constructed to provide safe release of flood waters from a dam to a downstream area.

8. Elevation-Area-Volume Curves
Area-elevation curve:
is obtained by measuring the area enclosed within each contour in the reservoir site using a planimeter.
Usually a 1/5000 scaled topographic map
Elevation-storage curve:
is the integration of an area-elevation curve.
The storage between any two elevations can be obtained by the product of average surface area at two elevations multiplied by the difference in elevation.

9. Elevation-Area-Volume Curves
A planimeter is a device used to calculate the area of an arbitrary two-dimensional shape.

11. h (m)
A (Km2 )
0.00
1.00
1.62
2.00
2.16
3.00
2.64
4.00
3.09
5.00
3.51
6.00
3.90
7.00

12. The Q stands for volume.
The exact answer is16.08 whereas the approximate value is 15.96

13. Recep YURTAL
Ç.Ü., İnş.Müh.Böl.

14. Recep YURTAL
Ç.Ü., İnş.Müh.Böl.

15. Recep YURTAL
Ç.Ü., İnş.Müh.Böl.

16. Elevation-Area-Volume Curves
In the selection of a suitable reservoir site, storage ability of the topography must be searched carefully.
An ideal reservoir site must have sufficiently impervous and sound formations both at the bed and sides to store huge quantities of water without excessive seepage and slope stability problems.
Deep and narrow reservoirs are usually preferable to shallow and wide reservoirs due to less evaporation and lower costs of expropriation.
To determine reservoir volume with given location & dam height
Zero Pool
Minimum Operating Level
Spillway Crest
Max. Operating Level
VOLUME (106 m3)
ELEVATION ABOVE MEAN SEA LEVEL (m)
Area
Volume
Typical reservoir elevation-area-volume curves

17. Total reservoir storage components:
Normal pool level
Minimum pool level
Active storage
Dead storage
Flood control storage
Surcharge storage

18. Storage zones in a reservoir
Dead storage
Sluiceway
Spillway crest
Active storage or Useful storage
Minimum pool level
Normal pool level
Flow
Flood control storage
Maximum pool level
Surcharge storage
Retarding pool level
Sediment accumulation

19. Normal pool level is the maximum elevation to which the reservoir surface will rise for ordinary reservoir operations.
Minimum pool level is the lowest allowable elevation to which the reservoir surface level can fall.
Dead storage is located below minimum pool level. The top elevation is dictated by amount of sediment accumulation at the end of the life time of reservoir.
Surcharge storage: This is required as a reserve between Full Reservoir Level and the Maximum Water level to contain the peaks of floods that might occur when there is insufficient storage capacity for them below Full Reservoir Level.

20. Therefore, the elevation of the lowest sluiceway must be located at least at minimum pool level. Water stored below this level is not available for any use.
The storage between minimum and normal pool levels is named as useful or active storage.
The flood control storage occupies between the retarding and normal pool levels.
The surcharge storage stays between retarding and maximum pool level.

21. General guidelines for a reservoir location:
Cost of the dam
Cost of real estate (expropriation costs).
Topographic conditions to store water
Possibility of deep reservoir
Quality of stored water
Reliable hill-slopes

22. Reservoir Yield
Yield: Amount of water that reservoir can deliver in a prescribed interval of time.
The yield is based on
inflow
capacity
Firm (safe) yield: Amount of water that can be supplied during a critical period. Can be never determined by certainty

23. Target yield: specified for a reservoir based on the estimated demands in most cases.
Secondary yield: Water available in excess of safe yield during high flow periods

24. Selection of Capacity of a Storage Reservoir
Designing the capacity of a storage reservoir involves with determination of the critical period during Inflow < Demand

25. There are 4 approaches to determine the capacity
Mass curve (Ripple diagram) method;
Sequent-peak algorithm;
Operation study;
Optimization analysis

26. For the determination of reservoir capacity, the critical period must be determined first. A long period of observed flow is required. When short period of observed flows or no observations area available stochastic methods are used to generate synthetic flows that has the same statistical properties such as mean, variance, correlations etc.

27. 1) Mass curve (Ripple diagram) method, developed by Ripple in 1883
Features of Mass Curve
Cumulative plotting of net reservoir inflow.
Slope of mass curve gives the value of inflow (S) at that time.
Slope of demand curve gives the demand rate (D) or yield.

28. Mass curve (Ripple diagram) method
The difference between the lines (a+b) tangent to the demand line (∑D) drawn at the highest and lowest points (A and B, respectively) of mass curve (∑S) gives the rate of withdrawal from reservoir during that critical period.
The maximum cumulative value between tangents is the required storage capacity (active storage).

29. Mass curve method (Chin, 2013)B
A critical period always begins at the end of a preceding high-flow period that leaves the reservoir full. A critical period ends when the reservoir has refilled after the drought period. E
The critical drawdown period begins when the reservoir is full and ends when the reservoir is empty.
Assumptions:
1-The Ripple method is applicable when the release is constant.
2-The total release over the time interval of analysis does not exceed the total reservoir inflows.

30. Points B and C define the beginning of low-flow periods and The point E define the end of the low-flow periods. A line representing the constant reservoir release is placed tangent to the beginning of each low-flow period, i.e., B and C. Then the vertical distance is the required active storage.

31. Example:Monthly Water Flows in the Little Weiser River Near Indian Valley, Idaho, 1966-1970
Year
Month
Flow (AF)
∑ Flow (AF) 1
1965 10
742 31 4
6720
129105 2 11
1060
1802 32 5
13,290
142395 3 12
1000
2802 33 6
9290
151685
1966
1500
3302 34 7
1540
153225
1080
4382 35 8
915
154140
6460
10,842 36 9
506
154646
10,000
20,842 37
886
155532
13,080
33,922 38
3040
158572
4910
38,832 39
2990
161562
981
39,813 40
1969
8170
169732
283
40,096 41
2800
172532
322
40,398 42
4590
177122 13
404
40,822 43
21,960
199082 14
787
41,609 44
30,790
229872 15
2100
43,709 45
14,320
244192 16
1967
4410
48,119 46
2370
246562 17
2750
50,869 47
709
247271 18
3370
54,239 48
528
247799 19
5170
59,409 49
859
248658 20
19,680
79,089 50
779
249437 21
19,630
98,719 51
* 12
1250
250687 22
3590
102,309 52
1970
11,750
262437 23
710
103,019 53
5410
267847 24
518
103,537 54
5560
273407 25
924
104,461 55
5610
279017 26
1020
105,481 56
24,330
303347 27
874
106,355 57
32,870
336217 28
1968
107,375 58
7280
343497 29
8640
116,015 59
1150
344647 30
6370
122,385 60
916
345563

32. A reservoir should be designed for the constant reservoir release of 2000 ac-ft per month
Points A1, A2, A3, and A4 define the beginning of low-flow periods whereas points B1, B2, B3, and B4 define the end of the low-flow periods.

33. A line representing 2000 ac-ft per month is placed tangent to the beginning of each low-flow period, i.e., A1 , A2, A3, and A4. Then the locations in each low-flow period with the largest vertical distance between the mass curve at points B1 , B2, B3, and B4 are identified. The largest of the four vertical distances B1 is 7200 ac-ft, which is the required active storage.

34. 2) Sequent-Peak Analysis (SPA)
The mass curve approach is easy to use when short periods of data are to be analyzed.
SPA is a modification of the Mass Curve analysis for lengthy time series and particularly suited to computer coding.

35. The steps of sequent-peak analysis are as follows:
Plot ∑ (Inflow-Withdrawal) : in symbolized fashion ∑(S-D)
Locate the initial peak and the next peak
Compute the storage required which is the difference between the initial peak and the lowest trough in the interval,
Repeat the process for all sequent peaks,
Determine the largest value of storages as “STORAGE CAPACITY

36. Illustration of the sequent –peak algorithm

37. Example by Erkek and Ağıralioğlu, 1995.
2.Sequent-peak
1.Sequent-peak
Maximum storage
Time
∑(S-D)
Illustration of the sequent –peak algorithm 2

38. Illustration of the sequent –peak algorithm 3
Maximum storage
Illustration of the sequent –peak algorithm 3

39. Example (http://www. slideserve
2.Sequent-peak
1.Sequent-peak
Maximum storage
Illustration of the sequent –peak algorithm 4

40. Analytical solution to SPA is good for computer coding
Equations below are used:
Vt = Dt – St + Vt if positive
Vt = 0 otherwise
Vt :required storage capacity at the end of period t
Vt-1 : required storage capacity at the end of previous period t
Dt : release during period t
St : inflow during period t

41. 3) OPERATION STUDY
It is presumed that the reservoir is adequate if the reservoir can supply all types of demands under possible losses such as seepage and evaporation.
The operation study is based on the solution of the continuity equation.
Where dV is differential storage during time dt
I and Q are the instantaneous total inflow outflow, respectively.

42. Operation study is used to
a) Determine the required capacity,
b) Define the optimum rules for operation,
c) Select the installed capacity for powerhouses,

43. Operation is carried out
only for an extremely low flow period and presents the required capacity to overcome the selected drought;
for the entire period and presents the power production for each year.

44. 4) OPTIMIZATION ANALYSIS & STOCHASTIC MODELS
Reliability of Reservoir Yield

45. 2.8 Reservoir Sedimentation
Sediments  eventually fill all reservoirs
determine the useful life of reservoirs
important factor in planning
■ River carry some suspended sediment and move
bed load (larger solids along the bed).
■ Large suspended particles + bed loads  deposited at the head of the reservoir & form delta.
■ Small particles  suspend in the reservoir or flow over the dam.

47. ☻ Unfortunately, the total rate of sediment transport in Turkey > 18 times that in the whole Europe
(500x106 tons/year)

48. Sedimentation stored behind a dam

49. “RESERVOIR SEDIMENTATION RATE”
based on survey of existing reservoirs, containing
* Specific weight of the settled sediments
* % of entering sediment which is deposited
“TRAP EFFICIENCY”:
% of inflowing sediment retained in the reservoir, which is a function of the ratio of reservoir capacity to total inflow.

50. Sediment trap efficiency and capacity to inflow ratio (Brune, 1953), where S=reservoir capacity (106 m3). I=Mean annual inflow (106 m3/year).

51. ► “Prediction of sediment accumulation”
IMPORTANT NOTES:
► “Prediction of sediment accumulation”
-- Difficult due to high range of variability in sediment discharge
► SOLUTION: “Continuous hydrologic simulation models”
-- used for prediction purposes
< But, at least, 2-3 years daily data are needed for calibration of the model. >

52. In practice sediment load measurements are carried out in streams and a calibration curve showing the relationship between suspended load Qs and stream discharge Q is obtained.
Relationship between discharge and sediment load (Sediment rating curve)

53. The United States Bureau of Reclamation (USBR) recommends a sampling period of at least 5 years to cover the full range in water discharges.
In the absence of suspended sediment data, the total sediment transport of a stream may be estimated from adjacent similar watersheds whose sediment rates have been recorded previously.

54. By examining the data of the drainage basins of 16 different dams in Turkey, Göğüş and Yalçınkaya (1992) proposed the following formula (R2=0.89).
Y= mean sediment yield in m3
A= drainage area in km2

55. ► “To control amount of entering sediment”:
IMPORTANT NOTES:
► “To control amount of entering sediment”:
(a) Upstream sedimentation basins,
(b) Vegetative screens,
(c) Soil conservation methods (i.e., terraces),
(d) Implementing sluice gates at various levels.
(e) Dredging of settled materials, but not economical!

56. Q1: Design reservoir capacity by applying S-D differences (Erkek and Ağıralioğlu, 1995)
Jan.
Feb.
Mar.
Apr.
May
Jun.
Jul.
Aug.
Sep.
Oct.
Nov.
Dec.
Volume influx
(106 m3) 21 20 40 45 86 15 9 7 12 27 25
Volume outflux 24
Differences

57. Jan.
Feb.
Mar.
Apr.
May
Jun.
Jul.
Aug.
Sep.
Oct.
Nov.
Dec.
Volume influx
(106 m3) 21 20 40 45 86 15 9 7 12 27 25
Volume outflux 24
Differences -3 -4 16 62 -9
-15
-17
-12 3 1
V1=(3+4) ×106=7×106 m3
F1=( ) ×106=99×106 m3
V2=( ) ×106=57×106 m3
F2=(3+1) ×106=4×106 m3
∑F=103×106 m3
∑V=64×106 m3
Since ∑F>∑V max V should be selected, which is 57×106 m3. Operation study should be performed to test if reservoir volume is sufficient. The reservoir should be assumed full.

58. Volume in the reservoir
Differences
(106 m3)
Volume in the reservoir
Spilled water -3 54 -4 50 16 57 9 21 62 53 -9 44
-15 29
-17 12
-12 3 1 4
∑= 92
Reservoir is sufficient enough
For consecutive reasons reservoir volume might be selected as 60×106 m3 (please pay attention to the difference in January. Please pay attention to the red numbers (4-3-4=-3)). The reservoir capacity should not go below zero level.

59. Q2 Design reservoir capacity by applying S-D differences (Erkek and Ağıralioğlu, 1995)
F1=200×106 m3
V1=50×106 m3
F2=75×106 m3
V2=90×106 m3
F3=30×106 m3
V3=60×106 m3
F4=75×106 m3
V4=80×106 m3
∑F=380×106 m3
∑V=280×106 m3
Since ∑F>∑V max V should be selected, which is 90×106 m3. Operation study should be performed to test if reservoir volume is sufficient.

60. 90×106 m3 is not sufficient enough because reservoir ends with -35
90×106 m3 is not sufficient enough because reservoir ends with -35. New reservoir capacity is, therefore, 90+35=125 ×106 m3.

62. Question 3 Design reservoir capacity by using SPA (Erkek and Ağıralioğlu,1995)
Jan.
Feb.
Mar.
Apr.
May
Jun.
Jul.
Aug.
Sep.
Oct.
Nov.
Dec.
Volume influx (S) 21 20 40 45 86 15 9 7 12 27 25
Volume outflux (D) 24
Differences (S-D) -3 -4 16 62 -9
-15
-17
-12 3 1

63. (92-35)=57×106 m3 Jan. Feb. Mar. Apr. May Jun. Jul. Aug. Sep. Oct.
Nov.
Dec.
Volume influx (S) 21 20 40 45 86 15 9 7 12 27 25
Volume outflux (D) 24
Differences (S-D) -3 -4 16 62 -9
-15
-17
-12 3 1
∑(S-D) -7 30 92 88 79 64 47 35 38 39
(92-35)=57×106 m3

64. Question 4 Design reservoir capacity by using S-D differences, SPA, Ripple diagram, SPA computational approach t D S 1 40 35 2 50 3 60 4 5 25 6 30 7 20 8 13 9 27 10 55 11 76 12 14 15 16 17 18 19

65. V1=-5×106 m3 F1=30×106 m3 V2=-90×106 m3 F2=66×106 m3 ∑F=126×106 m3T
(Months) Dt
(106 m3) St
St-Dt
∑(S-D) 1 40 35 -5 2 50 10 5 3 60 20 25 4
-15 6 30
-10 7
-20
-25 8 13
-27
-52 9 27
-13
-65 55 15
-50 11 76 36
-14 12 14 -9
-24 16
-44 17
-49 18
-39 19
-19
V1=-5×106 m3
F1=30×106 m3
V2=-90×106 m3
F2=66×106 m3
V3=-50×106 m3
F3=30×106 m3
∑F=126×106 m3
∑V=145×106 m3
Since ∑V>∑F max F should be selected, which is 66×106 m3. Operation study should be performed to test if reservoir volume is sufficient.

66. For operation study T (Months) Dt (106 m3) St St-Dt ∑(S-D) 1 40 35 -52 50 10 5 3 60 20 25 4
-15 6 30
-10 7
-20
-25 8 13
-27
-52 9 27
-13
-65 55 15
-50 11 76 36
-14 12 14 -9
-24 16
-44 17
-49 18
-39 19
-19
Volume in reservoir
(106 m3)
Spilled water 61 66 5 20 46 36 16
-11
-24 -9 27 42 32 17 -3 -8 2 22

67. 66×106 m3 is not sufficient enough because maximum deficient capacity reaches to -24 ×106. New reservoir capacity should be 66+24=90 ×106 m3.

68. By using SPA
90×106 m3

69. By using Ripple diagram

70. By using SPA computational approachDt St
St-Dt
∑(S-D)
Dt-St
Vbegining
Vend
(Months)
(106 m3) 1 40 35 -5 5 2 50 10
-10 3 60 20 25
-20 4
-15 15 6 30 7
-25 8 13
-27
-52 27 77 9
-13
-65 90 55
-50 75 11 76 36
-14
-36 39 12 24 14 -9 34
-24 49 16
-44 69 17
-49 74 18
-39 64 19
-19 44

71. Question 5 by Design reservoir capacity by using SPA (Chin, D. A
Month
Cumulative inflow (×1010 m3)
Cumulative demand (×1010 m3)
Cumulative inflow -demand (×1010 m3) 1
0.39
0.43 2
0.81
0.84 3
1.32 4
1.73
1.81 5
2.21
2.35 6
2.69
2.87 7
3.09
3.4 8
3.5
3.94 9
4.05
4.43 10
4.64
4.91 11
5.15
5.36 12
5.64
5.78 13
6.05
6.21 14
6.39
6.62 15
6.83
7.11 16
7.32
7.6 17
7.84
8.13 18
8.33
8.65 19
8.79
9.19 20
9.28
9.72 21
9.78
10.22 22
10.27
10.7 23
10.75
11.14 24
11.2
11.57

72. Cumulative inflow (×1010 m3) Cumulative demand (×1010 m3)
Month
Cumulative inflow (×1010 m3)
Cumulative demand (×1010 m3)
Cumulative inflow -demand (×1010 m3) 1
0.39
0.43
-0.04 2
0.81
0.84
-0.03 3
1.32 4
1.73
1.81
-0.08 5
2.21
2.35
-0.14 6
2.69
2.87
-0.18 7
3.09
3.4
-0.32 8
3.5
3.94
-0.44 9
4.05
4.43
-0.38 10
4.64
4.91
-0.27 11
5.15
5.36
-0.2 12
5.64
5.78
-0.15 13
6.05
6.21
-0.16 14
6.39
6.62
-0.23 15
6.83
7.11 16
7.32
7.6
-0.28 17
7.84
8.13
-0.29 18
8.33
8.65 19
8.79
9.19
-0.4 20
9.28
9.72 21
9.78
10.22 22
10.27
10.7
-0.42 23
10.75
11.14
-0.39 24
11.2
11.57
-0.37

73. 0.44×1010 m3

74. Question 6 by (Chin, D. A., 2013)
Outflow (ac-ft)
Inflow (ac-ft)
Precipitation (ac-ft)
Evaporation (ac-ft)
Volume at the begining of the month (ac-ft)
Volume at the end
of the month
(ac-ft)
Inflow
Precipitation
Evaporation
Volume at the end of the month (ac-ft) 1
2000
742 3
270 31
6720
100 2
1060 5
275 32
13,290
150
1000
280 33
9290 70 4
1500 10
350 34
1540
1080 30
470 35
915 6
6460 50
450 36
506 7
10000
400 37
886 8
13080 38
3040 9
4910
370 39
2990
981
330 40
8170 11
283
300 41
2800 12
322
290 42
4590 13
404 43
21960 14
787 44 15
2100 45
14,320 16
4410 46
2370 17
2750 47
709 18
3370 48
528 19
5170 49
859 20
19,680
779 21
19,630 51
1250 22
3590 52
11,750 23
710 53
5410 24
518 54
5560 25
924 55
5610 26
1020 56
24,330 27
874 57
32,870 28 58
7280 29
8640 59
1150
6370
460 60
916

75. We assume empty reservoir at the begining
Vt=( )-(742+3)+0=1525
Vt=( )-( )+2150=-5550 so it is 0.
Outflow (ac-ft)
Inflow (ac-ft)
Precipitation (ac-ft)
Evaporation (ac-ft)
Volume at the begining of the month (ac-ft)
Volume at the end
of the month
(ac-ft)
Inflow
Precipitation
Evaporation
Volume at the end of the month (ac-ft) 1
2000
742 3
270
1525 31
6720
100
400 2
1060 5
275
2735 32
13,290
150
350
1000
280
4010 33
9290 70
370 4
1500 10
4850 34
1540
330
780
1080 30
470
6210 35
915
300
2163 6
6460 50
450
2150 36
506
290
3944 7
10000 37
886
5325 8
13080 38
3040
4555 9
4910 39
2990
3840
981
1339 40
8170 11
283
3354 41
2800 12
322
5319 42
4590 13
404
7182 43
21960 14
787
8665 44 15
2100
8840 45
14,320 16
4410
6770 46
2370 17
2750 47
709
1589 18
3370
5490 48
528
3348 19
5170
2620 49
859
4756 20
19,680
779
6247 21
19,630 51
1250
7272 22
3590 52
11,750 23
710
1588 53
5410 24
518
3357 54
5560 25
924
4700 55
5610 26
1020
5950 56
24,330 27
874
7351 57
32,870 28
8671 58
7280 29
8640
2471 59
1150
1148
6370
460 60
916
2519

76. Referances
Brune, Gunnar M. "Trap efficiency of reservoirs." Eos, Transactions American Geophysical Union 34.3 (1953):
Chin, David A., Asis Mazumdar, and Pankaj Kumar Roy. Water-resources engineering. Vol. 12. Englewood Cliffs: Prentice Hall, 2000.
Erkek, C., N. Ağıralioğlu, and İ. T. Ü. Su Kaynakları Problemleri. "Yayınları, 1995." KODU/ADI 
Mays, Larry W. Water resources engineering. John Wiley & Sons, 2010.
Yanmaz, A. Melih. Applied water resources engineering. Metu Press, 2006.