Unit Hydrograph

ENGINEERING HYDROLOGY Prof. Rajesh Bhagat Asst. Professor Civil Engineering Department Yeshwantrao Chavan College Of Engineering Nagpur

B. E. (Civil Engg.) GCOE, Amravati

M. Tech. (Enviro. Engg.) VNIT, Nagpur

Mobile No.:- 8483003474 / 8483002277 Email ID:- [email protected] Website:- www.rajeysh7bhagat.wordpress.com

Unit-III 1) Runoff: Runoff, sources and component, classification of streams, factors

affecting runoff, Estimation Methods. Measurement of discharge of a stream by Areaslope and Area-velocity methods.

2) Hydrograph: Flood hydrographs and its components, Base flow & Base flow separation, S-Curve technique, unit hydrograph, synthetic hydrograph. Instantaneous Unit hydrograph.

2

HYDROGRAPH:1) A plot of the discharge in stream against time chronologically. 2) Depending upon unit of time involved:

1) Annual hydrograph 2) Monthly hydrograph 3) Seasonal hydrograph

4) Flood hydrograph or storm hydrograph or hydrograph: it shows stream flow due to storm over catchment. It is used flooding characteristics of stream. 

Above Hydrograph 1,2,3 are called long term hydrograph and are used for longed term studies like calculating the surface potential of stream, reservoir studies, drought studies. 3

HYDROGRAPH storm of Duration D Precipitation

P

tl tp

peak flow

Discharge Q

baseflow new baseflow

w/o rainfall 6

DTEL

6

Watershed Urbanization

Factors affecting Hydrograph: 1) 2) 3) 4) 5) 6) 7) 8)

Size Shape Slope Drainage density Land use or vegetation Rainfall intensity Rainfall duration Direction of storm movement

9

Hydrograph: 1) Hydrograph is a graphical variation of discharge against time. 2) It is a response of a given catchment to a rainfall input. 3) The discharge noted in hydrograph is the combined effect of surface runoff, interflow & base flow. 4) If two storms occurs in a catchment such that the 2nd one doesn’t start before the direct runoff due to 1st one has ceased, we get a singled peaked hydrograph.

10

Hydrograph: 1) If however, the second storm start before the direct runoff due to 1st storm has ceased, (complex storm) then multipeak hydrograph are obtained.

11

1) 2) 3) 4) 5) 6) 7) 8)

A1ABCDEE1 is called hydrograph due to isolated storm I1. AB is rising limb or concentration curve. BCD is crest curve. DE is falling curve or recession curve. C is point of crest or peak. E is end of direct runoff. EA’ is the hydrograph in the period of ground water recession. A’ is beginning of direct runoff due to 2nd storm.

12

1) 2) 3) 4) 5) 6) 7)

T is base period of 1st storm hydrograph. A1AEE1 is the base flow contribution to total discharge. ABCDE direct runoff contribution to total discharge. G1 is the centre of mass of rainfall. G2 is the centre of mass of hydrograph. TL = lag time. tpk = time of peak from starting point A

13

Hydrograph Separation: 1) In hydrological analysis it is necessary to obtain Direct Runoff Hydrograph (DRH) from Total Storm Hydrograph (TSH). 2) To separate DRH from TSH, various methods are available.

14

1) The flood data and base flow in a storm are estimated for a storm in a catchment area of 600 km2. calculate the effective rainfall. Time in Days

0

1

2

3

4

5

6

7

8

9

Discharge (m3/s)

20

63

151

133

90

63

44

29

20

20

Base flow (m3/s)

20

22

25

28

28

26

23

21

20

20

15

Ordinates of DRH after the separation of the base flow are: Time in Days

0

1

2

3

4

5

6

7

8

9

Discharge (m3/s)

20

63

151

133

90

63

44

29

20

20

Base flow (m3/s)

20

22

25

28

28

26

23

21

20

20

Ordinates of DRH after the separation of the base flow

0

41

126

105

62

37

21

8

0

0

Plot the DRH for given Ordinate. Volume of DRH = rainfall excess x catchment area Rainfall excess = (Volume of DRH / Catchment Area)

16

Ordinates of DRH after the separation of the base flow are: Time in Days

0

1

2

3

4

5

6

7

8

9

Discharge (m3/s)

20

63

151

133

90

63

44

29

20

20

Base flow (m3/s)

20

22

25

28

28

26

23

21

20

20

Ordinates of DRH after the separation of the base flow

0

41

126

105

62

37

21

8

0

0

Volume of DRH = rainfall excess x catchment area Rainfall excess = (Volume of DRH / Catchment Area)

Volume of DRH = direct runoff = (41 + 126 + 105 + 62 + 37 + 21 + 8) x 1 = 400 m3/s = 34560000 m3/day = 34560000 m3 Rainfall excess = (Volume of DRH / Catchment Area)

Rainfall excess = (34560000/ (600 x 106)) = 0.0576 m = 5.76 cm

17

Excess Rainfall & Effective Rainfall:1) Excess rainfall: if the initial losses and infiltration subtracted from the total rainfall, the remaining portion of rainfall is called rainfall excess. Surface

runoff occurs only when there is rainfall excess. Rainfall excess = Total rainfall – Φ.t

1) Effective rainfall: it is that portion of rainfall which cause direct runoff. As direct runoff includes both surface runoff and interflow, the effective rainfall is slightly greater than rainfall excess. Effective rainfall = (direct runoff volume / area of catchment)

Interflow is small, so direct runoff is equal to surface runoff & therefore they are used synonymously.

18

Effective Rainfall Hyetograph:1) When initial losses and filtration losses are subtracted from the rainfall hyetograph, we get Effective Rainfall Hyetograph (ERH).

2) It is also known as Hyetograph of rainfall excess. 3) Direct Runoff Hydrograph (DRH) is the result of Effective Rainfall Hyetograph (ERH).

4) Area under ERH x Catchment area = Runoff Volume = Area under direct DRH

19

2) A storm over catchment of area 5 km2 had a duration of 14 hours. If the Φ index for the catchment is 0.4 cm/hr, determine the effective rainfall hyetograph and the volume of direct runoff from the catchment due to the

storm. The mass curve of rainfall of the storm are as below. Time from start of storm, Hr

0

2

4

6

8

10

12

14

Accumulated rainfall, cm

0

0.6

2.8

5.2

6.7

7.5

9.2

9.6

20

Hour Accumulated rainfall, cm

Time interval, hour

Depth of rainfall, cm

Φ x (time interval)

ER, cm

Intensity of ER, cm/hr

0

0

-

-

-

-

-

2

0.6

2

0.6

0.8

0

0

4

2.8

2

2.2

0.8

1.4

0.7

6

5.2

2

2.4

0.8

1.6

0.8

8

6.7

2

1.5

0.8

0.7

0.35

10

7.5

2

0.8

0.8

0

0

12

9.2

2

1.7

0.8

0.9

0.45

14

9.6

2

0.4

0.8

0

0

Plot the hyetograph for above Intensity of ER against time. Area under ERH x Catchment area = Direct Runoff Volume Total Effective Rainfall = (0.7 + 0.8 + 0.35 + 0.45) x 2 = 4.6 cm Direct runoff volume = (4.6 / 100 ) x 5 x 1000000) = 230000

m3

21

Unit Hydrograph: 1) The Unit Hydrograph of the catchment is defined as hydrograph of direct runoff (DRH) results from 1cm depth of effective rainfall occurring uniformly over the catchment at a uniform rate during a specified period of time (D-hr). 2) Thus we can have 6-Hr Unit Hydrograph, 12-Hr Unit Hydrograph, etc. 3) 6-Hr unit hydrograph will have an effective rainfall intensity of 1/6 cm/hr.

22

Unit Hydrograph: 1) The D-hr Unit Hydrograph, D should not be more than any of the following: 1) Time of concentration 2) Lag time 3) Period of rise

2) Volume of water contained inside the unit hydrograph (ie area of unit of hydrograph) is equal to (1cm x catchment area)

23

Unit Hydrograph: Assume that a 6-hour unit hydrograph(UH) of a catchment has been derived, whose ordinates are given in the following table and a corresponding graphical representation is shown in Figure. Time, Hr

0

6

12

18

24

30

36

42

48

54

60

66

72

78

84

Discharge, m3/s

0

5

15

50

120

201

173

130

97

66

40

21

9

3.5

2

24

Unit Hydrograph: 2) Assume further that the effective rainfall hyetograph(ERH) for a given storm on the region has been given as in the following table. Time, Hrs

0

6

12

18

Effective rainfall, cm

0

2

4

3

3) This means that in in the first 6 hours, 2cm excess rainfall has been recorded, 4cm in the next 6 hour & 3cm in the next. 4) Direct runoff hydrograph can then be calculated by the three separate hydrograph for three excess rainfalls by multiplying the ordinates of the 6hrunit hydrograph by corresponding rainfall amounts.

25

26

27

28

Sample calculation for the example solved graphically is given table Time, Hrs

UH Ordinates, m3/s

Direct runoff due to 2cm excess rainfall in first 6hrs

Direct runoff due to 4cm excess rainfall in second 6hrs

Direct runoff due Direct runoff to 3cm excess hydrograph, rainfall in third m3/s 6hrs

0

0

0

0

0

0

6

5

10

0

0

10

12

15

30

20

0

50

18

50

100

60

15

175

24

120

240

200

45

485

30

201

402

480

150

1032

36

173

346

804

360

1510

42

130

260

692

603

1555

48

97

194

520

519

1233

54

66

132

388

390

910

60

40

80

264

291

635

66

21

42

160

198

400

72

9

18

84

120

222

78

3.5

7

36

63

106

84

2

4

14

27

45

90

0

8

10.5

18.5

96

0

0

6

6

29

3) The ordinates of 6 hr unit hydrograph of a catchment is given below: Time, Hr

0

3

6

9

12

15

18

24

30

36

42

48

54

60

69

Ordinates of 6 hr UH

0

25

50

85

125

160

185

160

110

60

36

25

16

8

0

Derive the flood hydrograph in the catchment due to the storm given below: Time from start of storm (hr)

0

6

12

18

Accumulated Rainfall

0

3.5

11

16.5

The storm loss rate for the catchment is estimated 0.25 cm/hr. The base flow can be assumed to be 15 m3/s at the beginning and increasing by 2.0 m3/s for every 12 hours till the end of the direct runoff hydrograph.

30

Time interval of storm (hr)

6

12

18

Accumulated Rainfall

3.5

11

16.5

Rainfall

3.5

7.5

5.5

Loss @ 0.25cm/Hr for 6 Hrs

1.5

1.5

1.5

Effective Rainfall, cm

2

6

4

Due to unequal time interval of UH ordinates, a few entries have to be interpolated to complete the table.

31

Time, Hr

Ordinates of UH

DRH due to 2cm ER

DRH due to 6cm ER

DRH due to 4cm ER

Ordinates of Final DRH

Base Flow, m3/s

Ordinates of Flood Hydrograph, m3/s

A

B

C =(B x 2)

D = (B x 6)

E = (Bx4)

F=(C+D+E)

G

H=(G+F)

0 3 6 9 12 15 18 21 24 27 30 36 42 48 54 60 66 69 72 75 78 81

0 25 50 85 125 160 185 172.5 160 135 110 60 36 25 16 8 0

0 50 100 170 250 320 370 345 320 270 220 120 72 50 32 16 0

0 0 0 150 300 510 750 960 1110 1035 960 660 360 216 150 96 48 0

0 0 0 0 0 100 200 340 500 640 740 640 440 240 144 100 64 32 0

0 50 100 320 550 930 1320 1645 1930 1945 1920 1420 872 506 326 212 112 32 0

15 15 15 15 17 17 17 17 19 19 19 21 21 23 23 25 25 27 27

15 65 115 335 567 947 1337 1662 1949 1694 1939 1441 893 529 349 237 137 59 27

32

Derivation Unit Hydrograph from Flood Hydrograph of Isolated Storm: 4) The following are the ordinates of the flood hydrograph from a catchment area of 780 km2 due to 6 hr storm. Derive the 6 hr unit hydrograph of the basin. Time, Hr

6

12

18

24

30

36

42

48

54

60

66

72

78

Discharge, m3/s

40

64

215

360

405

350

270

205

145

100

70

50

40

Assume the base flow of 40 m3/s.

Direct Surface runoff = (64-40) + (215-40) + (360-40) + (405-40) + (350-40) + (270-40) + (205-40) + (145-40) + (100-40) + (70-40) + (50-40) Direct Surface runoff = 1794 m3/s

DRH in depth = ((1794 x 6 x60 x 60) / (780 x 106)) x 100 = 4.968 cm (Rain Excess)

33

Derivation Unit Hydrograph from Flood Hydrograph of Isolated Storm: 4) The following are the ordinates of the flood hydrograph from a catchment area of 780 km2 due to 6 hr storm. Derive the 6 hr unit hydrograph of the basin. Time, Hr

6

12

18

24

30

36

42

48

54

60

66

72

78

Discharge, m3/s

40

64

215

360

405

350

270

205

145

100

70

50

40

Therefore the ordinates of UH are obtained by dividing the ordinates of DRH

hydrograph by rain excess 4.968 cm to get ordinates of UH. Time, Hr

6

12

18

24

30

36

42

48

54

60

66

72

78

Direct Runoff

0

24

175

320

365

310

230

165

105

60

30

10

0

(Ordinates of UH) Discharge, m3/s

0

4.83

35.22

64.42

74.47

62.4

46.29

33.21

21.13

12.077

6.04

2.01

340

5) The peak of flood hydrograph due to a 3 Hr duration isolated storm in a

catchment is 270 m3/s. The total depth of rainfall is 5.9cm. Assuming an average infiltration losses of 0.3 cm/hr and a constant base flow of 20 m3/s estimates the peak of the 3 hr unit hydrograph of this catchment. If the area of catchment is 567 km2 determine the base width of 3 hr unit hydrograph by assuming it to be triangular in shape. Duration of Rainfall Excess = 3 Hr Total Depth of Rainfall = 5.9 cm Loss @ 0.3 cm/Hr for 3 hours = 0.3 x 3 = 0.9 cm Rainfall Excess = 5 cm Peak Flood Hydrograph = 270 m3/s Base flow = 20 m3/s Peak of DRH = 270 – 20 = 250 m3/s

35

Peak of 3 Hr Unit Hydrograph = (peak of DRH / Rainfall Excess) = 250 / 5 = 50 m3/s

5) The peak of flood hydrograph due to a 3 Hr duration isolated storm in a

catchment is 270 m3/s. The total depth of rainfall is 5.9cm. Assuming an average infiltration losses of 0.3 cm/hr and a constant base flow of 20 m3/s estimates the peak of the 3 hr unit hydrograph of this catchment. If the area of catchment is 567 km2 determine the base width of 3 hr unit hydrograph by assuming it to be triangular in shape. Let B = base width of 3 hr Unit Hydrograph Volume represented by the area of UH = Volume of 1 cm depth over the catchment Area of UH = Area of catchment x 1 cm Peak of 3 Hr Unit Hydrograph = 50 m3/s Area of catchment = 567 km2 (1/2) B x 50 x 60 x 60 = 567 x 106 x (1/100) B = 63 Hours

36

6) Determine the ordinates of flood hydrograph of 3 successive storms of 4 hr duration, each producing rainfall of 3 cm, 4 cm and 2 cm respectively. Φ-index = 0.25 cm/hr and base flow is 10 m3/sec. Time (T) (Hrs.)

Ordinates of 4 I II III DRH hr. UH storm storm storm O R=2 cm

R=3 cm

Base flow

Ordinates Flood hydrograph

R=1 cm

0

0

0

0

10

10

2

16

32

32

10

42

4

22

44

0

44

10

54

6

43

86

48

134

10

144

8

64

128

66

0

194

10

204

10

49

98

129

16

243

10

253

12

32

64

192

22

278

10

288

14

20

40

147

43

230

10

240

16

0

0

96

64

160

10

170

18

60

49

109

10

119

20

0

32

32

10

42

22

20

20

10

30

24

0

0

10

10

6) A storm produces rainfall intensities of 0.75, 2.25 and 1.25 cm/hr on a drainage area in 3 successive time period of 4 hr. Φ-index = 0.25 cm/hr and base flow is 10 m3/sec. Time (T) Ordinates of 4 (Hrs.) hr. UH A B

I Storm C

II storm D

III storm E

R=2 cm

R=8 cm

R=4 cm

DRHO F=B+C +D+E

Base flow

Ordinates Flood hydrograph

G

H=F+G

0

0

0

0

10

10

2

12.52

25.04

25.04

10

35.04

4

21.31

42.62

0

42.62

10

52.62

6

23.54

47.08

100.16

147.24

10

157.24

8

14.79

29.58

170.48

200.06

10

210.06

10

12.18

24.36

188.32 50.08

262.76

10

272.76

12

0

0

118.32

85.24

203.56

10

213.56

14

97.44

94.16

191.6

10

201.6

16

0

59.16

59.16

10

69.16

18

48.72

48.72

10

58.72

20

0

0

10

10

0

7) Determine the ordinates of unit hydrograph from flood hydrograph. Neglect base flow. Area= 405 hectare. Time (T) (Hrs.)

Ordinates of Flood hydrograph, m3/s

0

0

2

0.3

4

1.7

6

2.6

8

5.4

10

4

12

2.6

14

1.1

16

0.6

18

0

7) Determine the ordinates of unit hydrograph from flood hydrograph. Neglect base flow. Area= 405 hectare. Excess rainfall x Catchment area = runoff volume = Area of Hydrograph Excess rainfall = (runoff volume) / Catchment area Excess rainfall = (131760) / 4050000 = 0.0325m = 3.25 cm

7) Determine the ordinates of unit hydrograph from flood hydrograph. Neglect base flow. Area= 405 hectare. Time (T) (Hrs.) A

Ordinates of Flood hydrograph, m3/s B

Ordinates of 2 hr. UH C = B /3.25

0

0

0.000

2

0.3

0.092

4

1.7

0.523

6

2.6

0.800

8

5.4

1.662

10

4

1.231

12

2.6

0.800

14

1.1

0.338

16

0.6

0.185

18

0

0.000

8) Determine the ordinates of flood hydrograph of 3 hr rainfall resulting into total rainfall of 15 cm. initial loss is 0.5 cm and Φ-index = 1 cm/hr. Sol: Excess rainfall = 15 – 0.5 – (1 x 3) = 11.5 cm Time (T) (Hrs.) A

Ordinates of 3 hr UH B

Ordinates of Flood hydrograph C = B x 11.5

0

0

0.00

6

3

34.50

12

5

57.50

18

9

103.50

24

11

126.50

30

7

80.50

36

5

57.50

42

4

46.00

48

2

23.00

54

1

11.50

60

0

0.00

S-CURVE METHOD S-curve or the summation curve is the hydrograph of direct surface discharge that would result from a continuous succession of unit storms producing 1 cm in time (T) hrs.

Time (T) (Hrs.)

Ordinates of 4 hr UH

Ordinates of 4 hr UH lagged by 4hr

S-Curve ordinate

S-Curve lagged by 12 hr

Difference

Ordinates of 12 hr UH

A

B

C

D=B+C

E

F=D-E

G=(4/12)*(F)

0

0

0

0

0.00

4

20

0

20

20

6.67

8

80

20

100

100

33.33

12

130

100

230

0

230

76.67

16

150

230

380

20

360

120.00

20

130

380

510

100

410

136.67

24

90

510

600

230

370

123.33

28

52

600

652

380

272

90.67

32

27

652

679

510

169

56.33

36

15

679

694

600

94

31.33

40

5

694

699

652

47

15.67

44

0

699

699

679

20

6.67

Q.9 The ordinates of 4-hr unit hydrograph are given below. Determine the ordinates of 3-hr UH using SCurve technique. Ordinates of Ordinates Ordinates of 4-hr U.H. of 4-hr Time 3 3 4-hr U.H. (m /s) U.H. (m /s) (Hours) 3 (m /s) lagged by 4- lagged by hr 8-hr C

D

S-Curve ordinates 3 (m /s)

S-Curve ordinates 3 (m /s) lagged by 3-hr

Difference

Ordinates of 33 hr U.H. (m /s)

E=B+C+D

F

G=E-F

H=(4/3)*(G)

0

0.00

0.00

6.00 36.00

8.00 48.00

A 0

B 0

1 2

6 36

6 36

3

66

66

0

66.00

88.00

4

91

0

91

6

85.00

113.33

5 6

106 93

6 36

112 129

36 66

76.00 63.00

101.33 84.00

7

79

66

145

91

54.00

72.00

8

68

91

0

159

112

47.00

62.67

9 10

34 27

106 93

6 36

146 156

129 145

17.00 11.00

22.67 14.67

11

13

79

66

158

159

-1.00

-1.33

12

0

68

91

159

146

13.00

17.33

Q.10 The ordinates of surface runoff of 4-hr duration from a catchment area of 357 km2 are measured at 1 hr interval are given below. Determine the ordinates of 6-hr UH using S-Curve technique. Ordinates Ordinates Ordinates Surface of 4-hr of 4-hr of 4-hr Time Ordinates Runoff U.H. U.H. U.H. (Hour of 4-hr 3 3 3 Ordinates 3 (m /s) (m /s) (m /s) s) 3 U.H. (m /s) (m /s) lagged by lagged by lagged by 4-hr 8-hr 12-hr A B C= B/0.816 0.00 0.00 0.00 1.00 15.00 18.36 2.00 25.00 30.61 3.00 36.00 44.07 4.00 38.00 46.52 5.00 48.00 58.77 6.00 69.00 84.48 7.00 91.00 111.41 8.00 113.00 138.34 9.00 101.00 123.65 10.00 88.00 107.74 11.00 71.00 86.92 12.00 54.00 66.11 13.00 31.00 37.95 14.00 21.00 25.71 15.00 9.00 11.02 810

D

0.00 18.36 30.61 44.07 46.52 58.77 84.48 111.41 138.34 123.65 107.74 86.92

E

0.00 18.36 30.61 44.07 46.52 58.77 84.48 111.41

F

0.00 18.36 30.61 44.07

S-Curve ordinates 3 (m /s)

G=C+D+E+F 0.00 18.36 30.61 44.07 46.52 77.13 115.08 155.48 184.87 200.78 222.82 242.41 250.98 238.74 248.53 253.43

S-Curve ordinates 3 Ordinates of 6-hr (m /s) Difference 3 U.H. (m /s) lagged by 6-hr H

0.00 18.36 30.61 44.07 46.52 77.13 115.08 155.48 184.87 200.78

I=H-G J= (4/6)*(I) 0.00 0.00 18.36 12.24 30.61 20.40 44.07 29.38 46.52 31.02 77.13 51.42 115.08 76.72 137.12 91.41 154.26 102.84 156.71 104.47 176.30 117.53 165.28 110.19 135.90 90.60 83.25 55.50 63.66 42.44 52.64 35.10

Q.11. The ordinates of 4-hr UH are given below. Determine the ordinates of 2-hr UH using S-Curve technique and plot the same. 3

3

Time (Hours)

Ordinates of 4-hr U.H. (m /s)

Ordinates of 2-hr U.H. (m /s)

A 0 2 4 6 8 10 12 14 16 18 20 22 24

B 0 12.5 62.5 130 175 180 140 90 50 35 13 3 0

A=B*(4/2) 0 25 125 260 350 360 280 180 100 70 26 6 0

Q.12 The ordinates of S-Curve Hydrograph are given below. Determine the ordinates of 3-hr UH. Effective rainfall is 1 cm/hr.

Ordinates of STime Ordinates of S3 3 Curve (m /s) lagged (Hours) Curve (m /s) by 3-hr A B C 0 0

Difference D

Ordinates of 3-hr U.H. (m3/s) E = D x (1/3)

0

0.00

1

55

55

18.33

2

141

141

47.00

3

251

0

251

83.67

4

344

55

289

96.33

5

413

141

272

90.67

6

463

251

212

70.67

7

501

344

157

52.33

8

523

413

110

36.67

9

538

463

75

25.00

10

546

501

45

15.00

Q.13 The ordinates of 6-hr UH are given below. Determine the ordinates of 4-hr UH using S-Curve technique and plot the same. Time Ordinates of S-Curve S(Hours 6-hr U.H. addition Curve ) (m3/s) ordinat es 3 (m /s) 1 0 6 12 18 24 30 36 42 48 54

2 0 40 90 100 130 80 70 50 30 10

3 0 40 130 230 360 440 510 560 590 600

4=2+3 0 40 130 230 360 440 510 560 590 600 600

Time 5 0 4 8 12 16 20 24 28 32 36 40 44

S-Curve ordinates 3 (m /s) 4hr duration 6=4*( 4/6) 0.00 26.67 86.67 153.33 240.00 293.33 340.00 373.33 393.33 400.00 400.00 0.00

S-Curve Difference Ordinates ordinates of 4-hr 3 (m /s) U.H. 3 lagged by (m /s) 4-hr 7 0.00 26.67 86.67 153.33 240.00 293.33 340.00 373.33 393.33 400.00 400.00

8=6-7 9=8*(6/4) 0 0.00 26.67 40 60.00 90 66.67 100 86.67 130 53.33 80 46.67 70 33.33 50 20.00 30 6.67 10 0.00 0 -400.00 -600