SOLUTION OF DIFFERENT YEAR PROBLEMS 1. A reservoir with a storage capacity of 300 million cubic meters is able to irrigate 40,000 hectares with 2 fillings each year. The crop season is 120 days. What is the duty? Solution: ∆ = V/A = 300×106/40,000×104 = 0.75 m B = 120 days We know, ∆ = 8.64 B/D D = 8.64 B/∆ = 8.64×120/0.75 = 1382.4 hectares/cumec (ans) 2. Find out the capacity of a reservoir from the following data. The culturable command area is 80,000 hectares. Crop
Base in days
Rice Wheat Sugarcane
120 120 320
Duty in hectares/cumec 1800 2000 2500
Irrigation Intensity (%) 25 30 20
Assume the canal and reservoir losses as 5% and 10% respectively Solution: Rice: Area for Rice = C.C.A × Intensity of irrigation = 80,000 × (25/100) = 20,000 hectares ∆ = 8.64 B/D = 8.64×120/1800 = 0.576 m Volume of water required for rice, V = A×∆ = 20,000 × 0.576 = 11520 hec-m Wheat: Volume of water required for wheat, V = A×∆ = [80,000 × (30/100)]× (8.64×120/2000) = 12,441.6 hec-m Sugarcane: Volume of water required for sugarcane, V = A×∆ = [80,000 × (20/100)] × (8.64×320/2500) = 17760 hec-m Total volume of water = (11520+12,441.6+17760) hec-m = 41721.6 hec-m = 41721.6×104 m3 After considering canal losses (5%) and reservoir loss (10%), Volume of water = 1.15 × (41721.6×104) m3 = 479.8×106 m3 = 479.8 M m3
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3. Determine the head discharge of a canal from the following data. The value of time factor may be assumed as 0.75. Crop Rice Wheat Sugarcane
Base Period in days 120 120 310
Area in hectares 4000 3500 3000
Duty in hectares/cumec 1500 2000 1200
Solution: Discharge of canal required (a) For rice = 4000/1500 = 2.667 cumec (Kharif) (b) For wheat = 3500/2000 = 1.75 cumec (Rabi) (c) For sugarcane = 3000/1200 = 2.5 cumec (Perennial) As, the base period of sugarcane is 310 days, it will require water both in Kharif and Rabi seasons Now, actual discharge required in Kharif season = 2.667 + 2.5 = 5.167 cumec Actual discharge required in Rabi season = 1.75 + 2.5 = 4.25 cumec So, the maximum discharge in Kharif season (i.e. 5.167 cumec) should be taken into consideration as it will be able to serve both the seasons. Time factor = 0.75 = Actual discharge/Design discharge = 5.167/ Design discharge discharge Design discharge = 5.167/0.75 = 6.889 cumec Therefore, the required head discharge of the canal is 6.889 cumec. 4. Determine the required flow to cover a strip of land of 0.02 hectares in area from a tube-well with a time of 45 minutes. The infiltration capacity of the soil may be taken as 3 cm/h and the average depth of flow on the field as 12 cm. (4) Solution: Here, A = 0.04 hectares = 0.02 × 104 m2 = 200 m2 Q y f = 3 cm/hr = (3/100) m/hr = 0.03 m/hr We know, t 2.3 log f y = 12 cm = 0.12 m Q fA (tf/2.3y) t = 45 minutes = 0.75 hr Required flow, Q = 10 fA/0.21 (0.750.03/2.30.12) = 10 0.03200/0.21 = 34.47 m3/hr = 0.01 cumec 5. A channel is to be designed for irrigating 5000 hectares in Kharif crop and 4000 hectares in Rabi crop. The water requirement for Kharif and Rabi are 60 cm and 20 cm, respectively. The base period for Kharif is 3 weeks and for Rabi is 4 weeks. Determine the discharge of the channel for which it is to be designed. Solution: Kharif Crop: Duty, D = 864 B/∆ = 864(3×7)/60 = 302.4 hectares/cumec Discharge required for Kharif = Area/Duty = (5000/302.4) cumec = 16.53 cumec Rabi Crop: Duty, D = 864 B/∆ = 864(4×7)/20 = 1209.6 hectares/cumec Discharge required for Rabi = Area/Duty = (4000/1209.6) cumec = 3.31 cumec The required discharge is maximum of the two, i.e. 16.53 cumec (Ans)
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6. Water is released at the rate of 5 cumec at the head sluice. If the duty at the field is 100 ha/cumec and the loss of water in transit is 30 %, find the area of the land that can be irrigated. Solution: Here,
Q = 5 m3/s, D = 100 ha/cumec, Loss of water = 30 %
Remaining water after loss = 5 – 30% of 5 = 5 – 1.5 = 3.5 m3/s 1 cumec water can be irrigated for area = 100 hectares 3.5
”
”
”
”
= 100×3.5 = 350 hectares
350 hectares of land can be irrigated (ans) 7. The discharge available from a tubewell is 136 m3/hr. Assuming 3000 hours of working for this tubewell in a year, estimate the culturable area that this tubewell can command. The intensity of irrigation is 50 % and the average water depth required for the Rabi and Kharif crops is 51 cm. Solution: System-1: Volume of water, V = Q×t = 136×3000 = 408×103 m3 Area can be irrigated, A = V/h = 408×103 /0.51 = 800×103 m2 CCA = Area can be irrigated/Intensity of irrigation = 800×103/0.50 = 160×104 m2 = 160 hectares System-2: Area can be irrigated, A = Duty×Q = [864×(3000/24)/51] ×(136/3600) = 80 hectares CCA = Area can be irrigated/Intensity of irrigation = 80/0.50 = 160 hectares 3. Compute the water application efficiency of an irrigation system based on the following data: 1.8 m3/min of water is diverted to the farm each day of 24 hrs. Each day 0.8 hectare of maize and 1 hectare of wheat are irrigated. The irrigation requirement of maize is 8 cm and that of wheat is 12 cm. Solution: Volume of water, V = Q×t = 1.8×60×24 = 2592 m3 Volume of water required to irrigate maize = Ah = 0.8×104 × (8/100) = 640 m3 Volume of water required to irrigate wheat = Ah = 1×104 × (12/100) = 1200 m3 Total water use by maize and wheat = 640 + 1200 = 1840 m3 Water application efficiency, E a = (1840/2592)×100 = 70.99% say 71% 71% of the water applied is beneficially utilized while 29% is lost as deep percolation, runoff, evaporation and spillage.
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4. The culturable commanded area for a distributary is 20,000 hectares. The intensity of irrigation during Kharif rice and jute are 50% and 30% respectively. During Rabi intensities of wheat and cotton are 60% and 35% respectively. If the total water requirements of rice, jute, wheat and cotton are 45 cm, 50 cm, 25 cm and 30 cm and their periods of growth are 160 days, 150 days, 170 days and 140 days respectively; determine the outlet discharge from average demand considerations. Solution: Kharif Crop: (i) Area to be irrigated for rice = C.C.A × Intensity of irrigation = 20,000 × (50/100) = 10,000 hectares Discharge required for rice = Area/Duty = [10,000/(864×160/45)] cumec = 3.26 cumec (ii) Area to be irrigated for jute = C.C.A × Intensity of irrigation = 20,000 × (30/100) = 6000 hectares Discharge required for jute = Area/Duty = [6000/(864×150/50)] cumec = 2.31 cumec Maximum discharge for Kharif crops = (3.26 + 2.31) cumec = 5.57 cumec Rabi Crop: (i) Area to be irrigated for wheat = C.C.A × Intensity of irrigation = 20,000 × (60/100) = 12,000 hectares Discharge required for wheat = Area/Duty = [12,000/(864×170/25)] cumec = 2.04 cumec (ii) Area to be irrigated for cotton = C.C.A × Intensity of irrigation = 20,000 × (35/100) = 7000 hectares Discharge required for cotton = Area/Duty = [7000/(864×140/30)] cumec = 1.74 cumec Maximum discharge for Rabi crops = (2.04 + 1.74) cumec = 3.78 cumec The design discharge is maximum of the two crops (Kharif & Rabi), i.e. 5.57 cumec (Ans)
Alternate Required discharge for rice, Q r = C.C.A × Intensity of irrigation /Duty = 0.5×20,000/{864×(160/45)}= 3.26 cumec Required discharge for jute, Q f = 0.30×20,000/{864×(150/50)}= 2.31 cumec Maximum discharge for Rabi crops = (3.26 + 2.31) cumec = 5.57 cumec Similarly, Required discharge for wheat, Q w = 0.6×20,000/{864×(170/25)}= 2.04 cumec Required discharge for cotton, Q C = 0.35×20,000/{864×(140/30)}= 1.74 cumec Maximum discharge for Rabi crops = (2.04 + 1.74) cumec = 3.78 cumec The design discharge is maximum of the two crops (Kharif & Rabi), i.e. 5.57 cumec (Ans)
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5. A loam soil has field capacity of 22 % and wilting coefficient of 10 %. The dry unit weight of soil is 1.5 gm/cc. If the root zone depth is 70 cm, determine the storage capacity of the soil. Irrigation water is applied when moisture content falls to 14 %. If the water application efficiency is 75 %, determine the water depth required to be applied in the field. Solution: Available moisture = Field Capacity – Permanent wilting point = 22 – 12 = 10 % Optimum moisture = 22 – 5 = 17 % Maximum storage capacity =
d d [FC – OMC] w
= 1.5 × (70/100) × [0.22 – 0.10]/1 = 0.126 m = 12.6 cm (ans) Since the moisture is allowed to vary between 22% and 14%, The efficiency created in this fall = 1.5 × (70/100) × [0.22 – 0.14]/1 = 0.084 m = 8.4 cm Hence, 8.4 cm depth of water is the net irrigation requirement (NIR) Quantity of water required to be applied in the field (FIR) = NIR/ η a = 8.4/0.75 = 11.2 cm (cm) 6. Compute the depth and frequency of irrigation required for a certain crop with data given below: Root zone depth Field capacity Wilting Point Specific gravity of soil Consumptive use Efficiency of irrigation
100 cm 22 % 12 % 1.5 25 mm/day 50 %
Assume 5% depletion on available moisture before application of irrigation water at field capacity. Solution: Available moisture = Field Capacity – Permanent wilting point = 22 – 12 = 10 % Optimum moisture = 22 – 5 = 17 % Depth of water stored in root zone between these two limits =
d d [FC – OMC] w
= 1.5 × (100/100) × [0.22 – 0.17]/1 = 0.075 m = 7.5 cm Here, consumptive use = 25 mm/day = 2.5 cm/day 2.5 cm of water is utilized by the plant in 1 day 7.5 cm of water will be utilized by the plant in = 1×7.5/2.5 days = 3 days Frequency of irrigation is 3 days
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7. A canal of length 5 km and of discharge capacity 3.5 m3/s is proposed to be lined with boulder lining. The total cost of lining is estimated as 4 lakhs. The life of lining is considered as 60 years. Justify the lining in the canal from the following data: Rate of interest Seepage loss Revenue for irrigation water Maintenance cost per km for lined canal Maintenance cost per km for unlined canal Base period of crop Additional benefit/km
8% 2% Tk. 75 per hec-m Tk. 1000 Tk. 2500 120 days Tk. 1000
Solution: Annual Benefits: (a) Saved seepage water by lining: Amount of water, V = (3.5 × 120 × 24 × 60 × 60) m3 = 36288000 m3 = 36288000/104 = 3628.8 hec-m Saving in seasonal seepage loss per km, m = 2% of V = (0.02 × 3628.8) = 72.58 hec-m Money saved by lining = mR 1 = Tk. (72.58 × 75) = Tk.5443.5 (b) Saving in maintenance cost: Saving in maintenance cost, R 2 = Tk. (2500 – 1000) = Tk.1500 Assume, p = 40% of this is saved in lined channel Annual saving in maintenance cost = pR 2 = Tk. (0.4 × 1500) = Tk.600 Total annual benefit/km = Tk. (5443.5 + 600 + 1000) = Tk.7043.5 Annual Cost: Cost of lining per km, C = Tk. (4,00,000/5) = Tk.80,000 Annual cost of lining, C/Y = Tk. (80,000/60) = Tk.1333.33 Total annual cost = C/Y + C/2(r/100) = Tk. {1333.33 + 80,000/2(8/100)} = Tk. 4533.33
Benefit cost ratio =
mR1 pR2 C C r Y 2 100
= 7043.5/4533.33 = 1.55 > 1 Lining is justified
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8. Use Khosla’s curves to calculate the percentage uplift pressure at various key points for a barrage foundation profile shown in figure below applying necessary corrections. Assume the thickness of the floor is 0.8 m. Also find exit gradient considering upstream pond level at 103 m. 100 m E1
98 m
1:4
C1
E2
C2
E3
C3
D 1 (94 m) D 2 (92 m)
D 3 (92 m)
8m 30 m
90 m
Solution: For upstream pile line no. (1) b = 120 m d = 100 – 94 = 6 m φ E = 100 % φ C1 = 100 – 20 = 80 % φ D1 = 100 – 14 = 86 %
α = b/d = 120/6 = 20 1/α = 1/20 = 0.05
(i) Correction for φ C1 Correction = 19
= 19×
Dd D b' b
D = Depth of Pile no.2 = 97.2 – 92 = 5.2 m d = Depth of Pile no.1 = 99.2 – 94 = 5.2 m b′ = 30 m b = 120 m
5.2 5.2 5.2 30 120
= 0.69 % (+ve) (ii) Floor thickness C 1 = (86 – 80)/(100 – 94)×0.8 = 0.8 % (+ve) Corrected (φ C1 ) = (80 + 0.69 + 0.8) % = 81.49 %
For upstream pile line no. (2) b = 120 m α = b/d = 120/6 = 20 d = 100 – 94 = 6 m 1– b 1 /b = 1 – 0.25 b 1 /b = 30/120 = 0.25 φ E2 = 100 – 30 = 70 % φ C2 = 63 % (For a base ratio 0.25 and α = 20) φ D2 = 100 – 34 = 66 %
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(i) Correction for φ E2 Correction = 19
= 19×
Dd D b' b
3.2 5.2 3.2 30 120
= 0.43 % (–ve) (ii) Floor thickness = (70 – 66)/(98 – 92)×0.8 = 0.8 % (+ve) Corrected = (80 + 0.69 + 0.8) % = 81.49 % (iii) Slope thickness Correction factor for 4:1 slope from table 11.4 = 3.3 Horizontal length of the slope = 8 m Distance between two pile lines between which the slopping floor is located = 30 m Actual correction = (3.3×(3/30) = 0.88 % (+ve) Corrected (φ E2 ) = (70 – 0.43 – 0.533 + 0.88) % = 69.92 % (i) Correction for φ C2 Correction = 19
= 19×
Dd D b' b
D = Depth of Pile no.3 = 97.2 – 92 = 5.2 m d = Depth of Pile no.2 = 97.2 – 92 = 5.2 m b′ = 30 m b = 120 m
5.2 5.2 3.2 30 120
= 0.69 % (–ve) (ii) Floor thickness = (70 – 66)/(98 – 92)×0.8 = 0.533% (+ve) Corrected (φ C2 ) = (63 + 0.69 + 0.533) % = 64.223 %
Exit gradient H = 103 – 98 = 5 m d = 98 – 92 = 6 m b = 120 m GE =
α = b/d = 120/6 = 20 For α = 20 1/(π√λ) = 0.098 (From chart)
5 H 1 = × 0.098 = 0.082 (ans) d 6
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9. Using Khosla’s curves, determine the following for the apron shown below: (a) Uplift pressure at points E, D, C, E 1 and D 1 (b) Exit gradient Neglect the effect of floor thickness
6m
Impervious Floor
E C
E1 6m
8m
D D1 6m
20 m
Solution: b = 26 m α = b/d = 26/6 = 4.33 d=6m 1– b 1 /b = 1 – 0.231 = 0.77 b 1 /b = 6/26 = 0.231 φ E = 100 – 18.9 = 81.1 % φ C = 50.1 % (For a base ratio 0.231 and α = 4.33) φ D = 100 – 36 = 64 % (i) Correction for φ E Correction = 19
0 60 Dd D =0 = 19× ' 6 26 b b
Corrected (φ E ) = 81.1 %
D=0 d=6m b′ = 6 m b = 26 m
(i) Correction for φ C Correction = 19
= 19×
Dd D b' b
D = 8m d=6m b′ = 20 m b = 26 m
8 6 8 20 26
= 6.47 % (+ve) Corrected (φ C ) = (50.1 + 6.47) % = 56.57 % Exit gradient H=6m d=8m b = 26 m GE =
α = b/d = 26/6 = 4.33 For α = 4.33 1/(π√λ) = 0.20 (From chart)
6 H 1 = × 0.20 = 0.15 (ans) d 8 9
10. Using Khosla’s curves, determine the following for the apron shown below: (a) Uplift pressure at points C, E 1 and D 1 (b) Exit gradient Assume floor thickness = 1 m
6m Impervious Floor E C
E1 6m
10 m
D D1 25 m
6m
Solution: b = 31 m α = b/d = 31/7 = 4.33 d=7m 1– b 1 /b = 1 – 0.194 = 0.806 b 1 /b = 6/31 = 0.194 φ E = 100 – 15.5 = 84.5 % φ C = 56.5 % (For a base ratio 0.194 and α = 4.33) φ D = 100 – 34 = 66 % (i) Correction for φ C Correction = 19
= 19×
Dd D b' b
D = 10 m d=6m b′ = 25 m b = 31 m
10 10 6 25 31
= 6.20 % (+ve) (ii) Floor thickness = (84.5 – 66)/(7)×1 = 2.64 % (+ve) Corrected (φ C ) = (56.5 + 6.20 + 2.64) % = 65.34 % Exit gradient H=6m d = 11 m b = 31 m GE =
α = b/d = 31/11 = 2.82 For α = 2.82 1/(π√λ) = 0.26 (From chart)
6 H 1 = × 0.26 = 0.142 (ans) d 11 10
11. Using the Khosla’s curves, determine the following for the apron shown below: (a) If percentage of pressure at C 2 is 56%, what will be the percentage of pressure at this point after corrections due to pile interference and slope (b) Find exit gradient where, corrections factor for slope, 3:1 = 4.5, Assume floor thickness = 1 m 158 m
155 m 3:1 154 m
C2
152 m
3m 40 m 147 m 141.7 m 57 m
Solution: φ C2 = 56 % (i) Correction for φ C Correction = 19
= 19×
Dd D b' b
D = Depth of pile no. (3) = 153 – 141.7 = 11.3 m d = 153 – 147 = 6 m b′ = 40 m b = 57 m
11.3 6 11.3 40 57
= 3.07 % (+ve) (ii) Slope thickness Correction factor for 3:1 slope from table 11.4 = 4.5 Horizontal length of the slope = 3 m Distance between two pile lines between which the sloping floor is located = 40 m Actual correction = 4.5×(3/40) = 0.34 % (-ve) Corrected (φ C2 ) = (56 + 3.07 – 0.34) % = 58.73 % Exit gradient H = (158 – 152) m = 6 m d = (152 – 141.7) m = 10.3 m b = 57 m GE =
α = b/d = 57/10.3 = 5.53 For α = 5.53 1/(π√λ) = 0.18 (From chart)
6 H 1 = × 0.18 = 0.105 (ans) d 10.3 11
12. Design and sketch the shape of an Ogee type spillway for the following data using the empirical equation developed by US Army Corps Engineers, Upstream Head, H = 20 m Shape of the upstream face = 1: 1.5 (H:V) Values of K and n are 1.393 and 1.81 respectively Solution: r 1 = 0.5H = 0.5×20 = 10 m r 2 = 0.21H = 0.21×20 = 4.2 m
X (m)
Y (m)
a = 0.139H = 0.139×20 = 2.78 m
2
1.59
b = 0.237H = 0.237×20 = 4.74 m
4
5.60
6
11.67
8
19.64
10
29.42
We know,
Xn = YkHn–1
21.81 = Y×1.939×(20)1.81 – 1 Y = 1.59 m
13. A centrifugal pump is required to lift water at the rate of 100 lit/sec. Calculate the horse power of the engine from the following data when the water is directly supplies to the field channel. (a) Suction head = 6 m (b) Coefficient of friction = 0.01 (c) Efficiency of pump = 65% (d) Water is directly supplied to the field channel (e) Diameter of pipe = 15 cm Solution:
Q = 100 lit/sec = (100/1000) m3/s = 0.1 m3/s
Delivery head, H d = 0 (As water is directly supplied to field) H=5m f = 0.01 d = 15 cm = 0.15 m The length of the pipe where frictional effect may occur is taken equal to the suction head, so l = 6 m Hf =
f l Q 2 0.01 6 0.12 = = 2.63 m 5 3d 5 3 0.15
So, total head, H = H s + H d + H f = 6 + 0 + 2.63 = 8.63 m Efficiency, η = 65% = 0.65 Horse Power (H.P) =
w Q H 1000 0.1 8.63 = = 17.70 ≈ 18 75 0.65 75 12
14. An area of 300 hectares is to be irrigated from a minor channel with one outlet; C.C.A is 80% of total area. The intensity of irrigation is 50% for Rabi and 30% for Kharif crop. Taking loss in conveyance system as 5% of outlet discharge, determine the design discharge of the channel. Take outlet discharge factor for wheat season as 1500 ha/m3/sec and for rice season 1000 ha/m3/sec. Solution: For Rabi: C.C.A = 300×0.8 = 240 Area under irrigation = 240×0.5 = 120 hectares Outlet discharge = 120/1500 = 0.08 m3/s Outlet discharge (considering conveyance loss) = (0.08 + 0.08×0.05) = 0.084 m3/s For Kharif: C.C.A = 300×0.8 = 240 Area under irrigation = 240×0.3 = 72 hectares Outlet discharge = 72/1000 = 0.072 m3/s Outlet discharge (considering conveyance loss) = (0.072 + 0.072×0.05) = 0.0756 m3/s Design discharge = 0.084 m3/s (maximum of two) 15. A clayey soil has field capacity of 22 % and wilting coefficient of 10 %. The dry unit weight of soil is 1.3 gm/cc. If the root zone depth is 70 cm, determine the storage capacity of the soil. Irrigation water is applied when moisture content falls to 14 %. If the water application efficiency is 75 %, determine the water depth required to be applied in the field. Solution: Available moisture = Field Capacity – Permanent wilting point = 22 – 12 = 10 % Optimum moisture = 22 – 5 = 17 % Maximum storage capacity =
d d [FC – OMC] w
= 1.5 × (70/100) × [0.22 – 0.10]/1 = 0.126 m = 12.6 cm (ans) Since the moisture is allowed to vary between 22% and 14%, The efficiency created in this fall = 1.5 × (70/100) × [0.22 – 0.14]/1 = 0.084 m = 8.4 cm Hence, 8.4 cm depth of water is the net irrigation requirement (NIR) Quantity of water required to be applied in the field (FIR) = NIR/ η a = 8.4/0.75 = 11.2 cm (cm) 16. A crop having 4 months crop period is to be grown in a place. Average monthly temperature is 180C, effective rainfall is 5 cm and average monthly percent of annual day light that occur during the period is 7.5. Determine consumptive irrigation requirement of that crop in cm using Blaney-Criddle equation and a crop factor equal to 0.7. Solution: C u = (k×p)/40 [1.8t + 32] = [(0.7×7.5)/40]× [1.8×18 + 32] = 8.45 cm Consumptive irrigation requirement, C.I.R = C u – R e = 8.45 – 5 = 3.45 cm
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17. Determine the possible delivery head for a residential building if you purchase a centrifugal pump from the market with the following specifications: Brake Horse Power = 35 Suction Head =5m Coefficient of friction = 0.01 Pump efficiency = 80 % Pipe diameter = 15 cm Required delivery flow rate = 150 l/s Solution:
Q = 150 lit/sec = (150/1000) m3/s = 0.15 m3/s
The length of the pipe where frictional effect may occur is taken equal to the suction head, so l = 5 m f l Q 2 0.01 5 0.152 Hf = = = 4.94 m 5 3d 5 3 0.15 Efficiency, η = 80% = 0.80
wQ H = 35 75 1000 0.15 H = 35 75 0.80
Brake Horse Power (B.H.P) =
H = 13.44 m We know, H = H s + H d + H f 13.44 = 5 + H d + 4.94 H d = 3.5 m
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