--- title: "Water conveyance and application efficiencies" subject: "Irrigation and Drainage Engineering" chapter: 2 key: "semester_6/irrigation_and_drainage_engineering:conveyance-application-efficiency-numerical" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/conveyance-application-efficiency-numerical site: BCT Engineering --- # Water conveyance and application efficiencies Irrigation and Drainage Engineering · Chapter 2: Irrigation Water Requirements - Asked in: 2073 Bhadra, 8 marks - Video: [Irrigation- Irrigation Efficiency / Water conveyance / Application / Storage Eff. (In Hindi)](https://www.youtube.com/watch?v=UM1NPFNOd0g) — Yogendra Ram Meena - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/conveyance-application-efficiency-numerical - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-2 ## Question A stream of 150 liter per second was diverted from canal and 110 liter per second was delivered to the field. An area of 2.2 hectares was irrigated in 8 hrs. Effective depth of root zone was 1.5 m. The runoff loss in the field was 445 m³. The depth of water penetration varied linearly from 1.5 m at the head end of the field to 1.1 m at the tail end. Available moisture holding capacity of the soil is 200 mm per meter depth of soil. Determine the water conveyance efficiency, water application efficiency, water storage efficiency and water distribution efficiency. Irrigation was started at a moisture extraction level of 50%. ## Short answer Conveyance efficiency = Q_field / Q_div = 110/150 = 73.33%. Application efficiency: stored = 3168 - 445 = 2723 m³, applied = 3168 m³, ηa = 85.95%. Storage efficiency: needed = 3300 m³, ηs = 2723/3300 = 82.52%. Distribution efficiency: mean depth d=1.3 m, y=0.1 m, ηd = 92.31%. ## Answer **Given:** $Q_{diverted} = 150$ l/s, $Q_{field} = 110$ l/s, area $= 2.2$ ha $= 22000$ m$^2$, time $= 8$ h, root zone $= 1.5$ m, runoff $= 445$ m$^3$, penetration depth 1.5 m (head) to 1.1 m (tail), available moisture = 200 mm/m, irrigation started at 50% depletion of available moisture. **1. Water conveyance efficiency** $$ \eta_c = \frac{110}{150}\times 100 = 73.33\% $$ **2. Water applied to the field** $$ W_f = 0.110\times 8\times 3600 = 3168\ \text{m}^3 $$ **Water stored in the root zone.** The penetration depth (1.5 m at the head, 1.1 m at the tail) does not exceed the root depth of 1.5 m, so deep percolation is nil (assumed). Hence $$ W_s = W_f - \text{runoff} = 3168 - 445 = 2723\ \text{m}^3 $$ **Water application efficiency** $$ \eta_a = \frac{W_s}{W_f}\times 100 = \frac{2723}{3168}\times 100 = 85.95\% $$ **3. Water storage efficiency.** Water needed to bring the root zone back to field capacity (50% of available moisture was used): $$ W_n = 0.5\times 0.200\times 1.5\times 22000 = 3300\ \text{m}^3 $$ $$ \eta_s = \frac{W_s}{W_n}\times 100 = \frac{2723}{3300}\times 100 = 82.52\% $$ **4. Water distribution efficiency.** Depth varies linearly between 1.5 m and 1.1 m, so mean depth $d = (1.5+1.1)/2 = 1.3$ m and the average numerical deviation $y = 0.2/2 = 0.1$ m. $$ \eta_d = \left(1-\frac{y}{d}\right)\times 100 = \left(1-\frac{0.1}{1.3}\right)\times 100 = 92.31\% $$ **Answer:** $\eta_c = 73.33\%$, $\eta_a = 85.95\%$, $\eta_s = 82.52\%$, $\eta_d = 92.31\%$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With 150 l/s diverted and 110 l/s delivered, eta_c is A. 136% B. 73.33% C. 26.7% D. 85.95% Answer: B. 110/150 = 0.7333. (Key point: Conveyance efficiency) 2. Water applied to the field in 8 h at 110 l/s is A. 880 m3 B. 2723 m3 C. 3168 m3 D. 3300 m3 Answer: C. 0.110 x 8 x 3600 = 3168 m3. (Key point: Water applied) 3. Water stored in the root zone is found by A. adding runoff to water applied B. subtracting water needed C. multiplying by eta_c D. subtracting runoff (445 m3) from water applied, with no deep percolation Answer: D. 3168 - 445 = 2723 m3. (Key point: Water stored) 4. With depth varying linearly 1.5 m to 1.1 m, eta_d is A. 92.31% B. 73.33% C. 84.6% D. 82.52% Answer: A. d = 1.3 m, y = 0.1 m, eta_d = (1 - 0.1/1.3) x 100. 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