hidraulika vezba 8

8
Stefan Spasojević 190/08 str. 1 Zadatak 8. Transformacija talasa u akumulaciji α= 9.1 β= 8.9 Tabela 7.1 Kriva površine jezera Z 18.9 m [ha] [mnm] 0.49 13 104.55 1.89 m 10 95.45 19.1 m n= 0.014 Tabela 7.2 Ulazni hidrogram t Q [h] 0 0 100 mnm 1 200 82 mnm 3 0 81 mnm A jez B prel = C p = D T,isp = L T,isp = m -1/3 s [m 3 /s] Z KP = Z ul,T,isp = Z izl,T,isp = 0 0.5 1 1.5 2 2.5 3 0 50 100 150 200 Ulazni hidrogram vreme [h] protok [m3/s] 94 95 96 97 98 99 100 101 102 103 104 105 106 10 10.5 11 11.5 12 12.5 13 f(x) = 0.3296703297x - 21.467032967 Kriva površine jezera Z [mnm] Aak [ha]

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Page 1: hidraulika vezba 8

Stefan Spasojević 190/08 str. 1

Zadatak 8. Transformacija talasa u akumulaciji

α= 9.1β= 8.9

Tabela 7.1 Kriva površine jezera

Z 18.9 m

[ha] [mnm] 0.49

13 104.55 1.89 m

10 95.45 19.1 m

n= 0.014

Tabela 7.2 Ulazni hidrogramt Q

[h]

0 0 100 mnm

1 200 82 mnm

3 0 81 mnm

Ajez Bprel=

Cp=

DT,isp=

LT,isp=

m-1/3s

[m3/s]

ZKP=

Zul,T,isp=

Zizl,T,isp=

0 0.5 1 1.5 2 2.5 30

50

100

150

200

Ulazni hidrogram

vreme [h]

protok [m3/s]

94 95 96 97 98 99 100 101 102 103 104 105 10610

10.5

11

11.5

12

12.5

13f(x) = 0.3296703297x - 21.467032967

Kriva površine jezera

Z [mnm]

Aak [ha]

Page 2: hidraulika vezba 8

Stefan Spasojević 190/08 str. 2

a) Određivanje najvećeg proticaja preko preliva

Δt= 10 min

Tabela 7.3 Proračun najvećeg proticaja metodom 'prediktor-korektor'

t Z Z*

[min] [mnm] [ha] [mnm] [ha]

0 0 100 11.50 0.00 100 11.50 0.00 ### ###10 33.33 100.09 11.53 1.05 ### 11.59 5.28 ### ###20 66.67 100.33 11.61 7.77 ### 11.72 20.72 ### ###30 100 100.69 11.73 23.26 ### 11.87 45.88 ### ###40 133.33 101.10 11.87 47.48 ### 12.02 78.10 ### ###50 166.67 101.54 12.02 78.42 ### 12.17 114.36 ### ###60 200 101.97 12.17 ### ### 12.31 152.32 ### ###70 183.33 102.26 12.26 ### ### 12.34 159.75 ### ###80 166.67 102.38 12.31 ### ### 12.33 158.35 ### ###90 150 102.40 12.31 ### ### 12.31 151.47 ### ###

100 133.33 102.35 12.29 ### ### 12.27 141.25 ### ###110 116.67 102.26 12.26 ### ### 12.23 129.02 ### ###120 100 102.13 12.22 ### ### 12.17 115.58 ### ###130 83.33 101.98 12.17 ### ### 12.12 101.47 ### ###140 66.67 101.82 12.12 ### ### 12.06 87.05 ### ###150 50 101.64 12.06 86.42 ### 11.99 72.53 ### ###160 33.33 101.45 11.99 71.98 ### 11.93 58.11 ### ###170 16.67 101.25 11.92 57.59 ### 11.86 44.01 ### ###180 0 101.04 11.85 43.48 ### 11.78 30.43 ### ###

152.75

Qul Aak Qprel Aak* Qprel* fn fn+1*

[m3/s] [m3/s] [m3/s]

Qprelmax= m3/s

0 0.5 1 1.5 2 2.5 30

50

100

150

200

Ulazni hidrogram

vreme [h]

protok [m3/s]

0 20 40 60 80 100 120 140 160 180 2000

50

100

150

200

250

Hidrogram

Qul Qprel

vreme [min]

protok [m3/s]

Page 3: hidraulika vezba 8

Stefan Spasojević 190/08 str. 3

t Z Z*

[min] [mnm] [ha] [mnm] [ha]

0 0 95.503 9.99 0.00 95.5 9.99 0.00 ### ###10 33.33 95.60 10.02 0.00 95.80 10.09 0.00 ### ###20 66.67 95.90 10.12 0.00 96.30 10.25 0.00 ### ###30 100 96.39 10.29 0.00 96.97 10.48 0.00 ### ###40 133.33 97.06 10.51 0.00 97.83 10.77 0.00 ### ###50 166.67 97.91 10.80 0.00 98.84 11.11 0.00 ### ###60 200 98.91 11.14 0.00 99.99 11.50 0.00 ### ###70 183.33 99.93 11.48 0.00 ### 11.80 34.31 ### ###80 166.67 100.74 11.75 26.38 ### 11.99 72.45 ### ###90 150 101.30 11.94 60.59 ### 12.09 94.67 ### ###

100 133.33 101.62 12.05 84.40 ### 12.13 104.17 ### ###110 116.67 101.77 12.10 96.63 ### 12.13 104.88 ### ###120 100 101.81 12.11 99.73 ### 12.11 99.84 ### ###130 83.33 101.77 12.10 96.42 ### 12.08 91.16 ### ###140 66.67 101.67 12.07 88.89 ### 12.03 80.24 ### ###150 50 101.54 12.02 78.69 ### 11.97 68.00 ### ###160 33.33 101.39 11.97 66.89 ### 11.91 55.08 ### ###170 16.67 101.20 11.91 54.23 ### 11.84 41.97 ### ###180 0 101.00 11.84 41.24 ### 11.77 29.06 ### ###

99.73 Kota nivoa vode: 95.5025 mnm

c) Određivanje vremena potrebnog da se akumulacija isprazni do kote 95,503

λ= 0.020

2.81 m2Δt= 10 min

Tabela 7.5 Potrebno vreme za pražnjenje akumulacije do kote 95,503

t Z Z*

[min] [mnm] [ha] [mnm] [ha]

0 100 11.50 0 100 11.50 9.34 ### ###10 99.976 11.49 11.16 ### 11.47 9.32 ### ###20 99.922 11.48 11.14 ### 11.46 9.31 ### ###30 99.869 11.46 11.13 ### 11.44 9.30 ### ###40 99.815 11.44 11.11 ### 11.42 9.28 ### ###50 99.762 11.42 11.10 ### 11.40 9.27 ### ###60 99.708 11.40 11.08 ### 11.38 9.26 ### ###70 99.655 11.39 11.06 ### 11.37 9.24 ### ###80 99.601 11.37 11.05 ### 11.35 9.23 ### ###

770 95.892 10.12 9.88 ### 10.10 8.26 ### ###780 95.839 10.10 9.87 ### 10.08 8.24 ### ###790 95.785 10.08 9.85 ### 10.06 8.23 ### ###

b) Određivanje kote tako da najveći proticaj preko preliva bude 100 m3/s

Tabela 7.4 Proračun kote nivoa akumulacije pri protoku 100 m3/sQul Aak Qprel Aak* Qprel* fn fn+1*

[m3/s] [m3/s] [m3/s]

Qprelmax= m3/s

AT=

Aak Qizl Aak* Qizl* fn fn+1*

[m3/s] [m3/s]

Vreme potrebno da se

akumulacija isprazni do kote 95,503 mnm je približno 825

min.

0 20 40 60 80 100 120 140 160 180 2000

50

100

150

200

250

Hidrogram

Qul Qprel

vreme [min]

protok [m3/s]

Page 4: hidraulika vezba 8

Stefan Spasojević 190/08 str. 4

800 95.731 10.06 9.83 ### 10.04 8.21 ### ###810 95.677 10.04 9.81 ### 10.03 8.20 ### ###820 95.623 10.03 9.80 ### 10.01 8.18 ### ###825 95.596 10.02 9.79 ### 10.01 8.19 ### ###

Vreme potrebno da se

akumulacija isprazni do kote 95,503 mnm je približno 825

min.

Page 5: hidraulika vezba 8

Zadatak 8: Transformacija talasa u akumulaciji

α= 6.8 Tabela 7.1 Kriva površine akumulacije

β= 11.2 Z[ha] [mnm]

21.2 m sirina preliva 13 103.4

0.49 koeficijent prelivanja 10 96.6

2.12 m sirina temeljnog ispusta Tabela7.2 Ulazni hidrogram poplavnog talasa

16.8 m duzina temeljnog ispusta t Q

n= 0.014 Maningov koeficijent hrapavosti za temeljni ispust [h]

100 mnm kota krune preliva 0 0

82 mnm kota ulaza u temeljni ispust 1 200

81 mnm kota izlaza iz temeljnog ispusta 3 0

Ajez

Bprel

=

Cp=

DT,isp

=

LT,isp

=

m-1/3s [m3/s]Z

KP=

Zul,T,isp

=

Zizl,T,isp

=

0 0.5 1 1.5 2 2.5 30

20

40

60

80

100

120

140

160

180

200

Kriva površine akumulacije

vreme (h)

protok m3/s

94 95 96 97 98 99 100 101 102 103 104 105 10610

10.5

11

11.5

12

12.5

13

Ulazni hidrogram poplavnog talasa

Z (mnm)

Aak (ha)

Page 6: hidraulika vezba 8

a) Odredjivanje najveceg proticaja preko preliva pomoću metode ˝Prediktor-korektor˝

Δt= 10 minTabela 7.3 Odredjivanje najveceg proticaja preko preliva pomoću metode ˝Prediktor-korektor˝

t Z Z*

[min] [mnm] [ha] [mnm] [ha] [m/s] [m/s]0 0.00 100.00 11.49 0.00 100.00 11.49 0.00 0.00000 0.00029

10 33.33 100.09 11.53 1.18 100.25 11.60 5.90 0.00028 0.0005220 66.67 100.33 11.64 8.64 100.63 11.77 22.84 0.00050 0.0006630 100.00 100.67 11.79 25.46 101.05 11.95 49.75 0.00063 0.0007040 133.33 101.07 11.96 51.18 101.49 12.15 83.32 0.00069 0.0006950 166.67 101.49 12.15 83.30 101.90 12.33 120.23 0.00069 0.0006560 200.00 101.89 12.32 119.11 102.28 12.50 158.33 0.00066 0.0002070 183.33 102.14 12.43 144.28 102.33 12.52 163.73 0.00031 0.0000280 166.67 102.24 12.48 154.63 102.30 12.50 160.65 0.00010 -0.0000990 150.00 102.25 12.48 154.98 102.22 12.47 152.51 -0.00004 -0.00015

100 133.33 102.19 12.46 149.01 102.11 12.42 141.36 -0.00013 -0.00020110 116.67 102.09 12.41 139.17 101.98 12.36 128.46 -0.00018 -0.00023120 100.00 101.97 12.36 127.03 101.84 12.30 114.54 -0.00022 -0.00025130 83.33 101.83 12.30 113.56 101.68 12.23 100.09 -0.00025 -0.00027140 66.67 101.67 12.23 99.35 101.51 12.16 85.40 -0.00027 -0.00029150 50.00 101.50 12.15 84.79 101.33 12.08 70.68 -0.00029 -0.00031160 33.33 101.32 12.07 70.13 101.14 11.99 56.11 -0.00030 -0.00033170 16.67 101.13 11.99 55.58 100.94 11.90 41.90 -0.00032 -0.00035180 0.00 100.93 11.90 41.35 100.72 11.81 28.27 -0.00035 -0.00024

Qul

Aak

Qprel

Aak

* Qprel*

fn

fn+1

*

[m3/s] [m3/s] [m3/s]

0 0.5 1 1.5 2 2.5 30

20

40

60

80

100

120

140

160

180

200

Kriva površine akumulacije

vreme (h)

protok m3/s

94 95 96 97 98 99 100 101 102 103 104 105 10610

10.5

11

11.5

12

12.5

13

Ulazni hidrogram poplavnog talasa

Z (mnm)

Aak (ha)

Page 7: hidraulika vezba 8

t Z Z*

[min] [mnm] [ha] [mnm] [ha]0 0.00 95.96 11.49 0.00 95.96 11.49 0.00 0.00000 0.00029

10 33.33 96.05 11.53 0.00 96.22 11.60 0.00 0.00029 0.0005720 66.67 96.30 11.64 0.00 96.65 11.77 0.00 0.00057 0.0008530 100.00 96.73 11.79 0.00 97.24 11.95 0.00 0.00085 0.0011240 133.33 97.32 11.96 0.00 97.99 12.15 0.00 0.00111 0.0013750 166.67 98.07 12.15 0.00 98.89 12.33 0.00 0.00137 0.0016260 200.00 98.96 12.32 0.00 99.94 12.50 0.00 0.00162 0.0014770 183.33 99.89 12.43 0.00 100.78 12.52 31.46 0.00147 0.0010880 166.67 100.66 12.48 24.54 101.34 12.50 71.45 0.00114 0.0006390 150.00 101.19 12.48 59.57 101.62 12.47 95.10 0.00072 0.00031

100 133.33 101.50 12.46 84.29 101.73 12.42 105.00 0.00039 0.00009110 116.67 101.64 12.41 96.94 101.74 12.36 105.50 0.00016 -0.00004120 100.00 101.68 12.36 99.99 101.68 12.30 100.00 0.00000 -0.00014130 83.33 101.64 12.30 96.38 101.57 12.23 90.82 -0.00011 -0.00020140 66.67 101.55 12.23 88.45 101.44 12.16 79.44 -0.00018 -0.00024150 50.00 101.42 12.15 77.86 101.28 12.08 66.82 -0.00023 -0.00028160 33.33 101.27 12.07 65.70 101.11 11.99 53.60 -0.00027 -0.00031170 16.67 101.10 11.99 52.73 100.91 11.90 40.25 -0.00030 -0.00034180 0.00 100.90 11.90 39.51 100.70 11.81 27.19 -0.00033 -0.00023

99.99

Dobijena kota nivoa vode je: Z= 95.958 mnm

b) Odredjivanje kote nivoa u akumulaciji tako da najveci proticaj preko preliva bude 100m3/s

Tabela 7.4 Pretpostavljanjem kote nivoa vode u akumulaciji dolazimo do Qprelmax=100 m3/s

Qul

Aak

Qprel

Aak

* Qprel

* fn

fn+1

*

[m3/s] [m3/s] [m3/s] [m/s] [m/s]

Qprel

max= m3/s

Page 8: hidraulika vezba 8

c) Odredjivanje vremena potrebnog da se akumulacija isprazni do kote 95,958 mnm

λ= 0.019

3.53Δt= 10 min

Tabela 7.5 Odredjivanje vremena potrebnog da se akumulacija isprazni do kote 95,958 mnm

t Z Z*

[min] [mnm] [ha] [mnm] [ha]0 100.00 11.49 0.00 100.00 11.49 11.88 0 -0.000103

10 99.97 11.48 14.33 99.89 11.44 11.85 -0.000125 -0.00010420 99.90 11.45 14.31 99.83 11.41 11.83 -0.000125 -0.00010430 99.83 11.42 14.28 99.76 11.38 11.80 -0.000125 -0.00010440 99.76 11.39 14.25 99.69 11.35 11.78 -0.000125 -0.00010450 99.69 11.36 14.23 99.62 11.32 11.76 -0.000125 -0.00010460 99.63 11.32 14.20 99.55 11.29 11.74 -0.000125 -0.00010470 99.56 11.29 14.18 99.48 11.26 11.72 -0.000126 -0.00010480 99.49 11.26 14.15 99.41 11.23 11.70 -0.000126 -0.00010490 99.42 11.23 14.12 99.34 11.20 11.67 -0.000126 -0.000104

100 99.35 11.20 14.10 99.27 11.17 11.65 -0.000126 -0.000104

...

490 96.61 10.00 13.00 96.54 9.96 10.74 -0.00013 -0.000108500 96.54 9.97 12.97 96.46 9.93 10.72 -0.00013 -0.000108510 96.47 9.93 12.94 96.39 9.90 10.69 -0.00013 -0.000108520 96.40 9.90 12.91 96.32 9.87 10.67 -0.00013 -0.000108530 96.33 9.87 12.88 96.25 9.84 10.64 -0.000131 -0.000108540 96.26 9.84 12.85 96.18 9.80 10.62 -0.000131 -0.000108550 96.18 9.81 12.82 96.11 9.77 10.59 -0.000131 -0.000108560 96.11 9.78 12.79 96.03 9.74 10.57 -0.000131 -0.000108570 96.04 9.74 12.76 95.96 9.71 10.54 -0.000131 -0.000109580 95.97 9.71 12.73 95.89 9.68 10.52 -0.000131 -0.000109

Vreme potrebno da se akumulacija isprazni do kote 95,958 mnm iznosi 580 min

AT= m2

Aak

Qizl

Aak

* Qizl* f

nfn+1

*

[m3/s] [m3/s] [m/s] [m/s]