--- title: "Hardy Cross network with valve on diagonal" subject: "Hydraulics" chapter: 3 key: "semester_4/hydraulics:hardy-cross-network-six-links-resistance" source: https://www.bctengineer.com/q/semester_4/hydraulics/hardy-cross-network-six-links-resistance site: BCT Engineering --- # Hardy Cross network with valve on diagonal Hydraulics · Chapter 3: Three reservoirs problem and Pipe networks - Asked in: 2076 Bhadra, 10 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/hardy-cross-network-six-links-resistance - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-3 ## Question Determine discharge distribution in the pipes shown in the figure using Hardy Cross method. | Pipes | AB | BC | CD | AD | DB | Valve | |---|---|---|---|---|---|---| | Resistance coefficient (r) | 1 | 2.5 | 1.5 | 2 | 3 | 1.5 | [Figure: nodes A, B, C, D with a diagonal BD carrying a valve; 70 lit/min enters at A; 30 lit/min leaves at B; 40 lit/min leaves at C] ## Short answer After iterative head-balance corrections, the final discharges are AB 41.6, AD 28.4, BC 16.7, CD 23.3, and BD 5.1 l/min (D to B). ## Answer **Data.** Resistance $r$ in $h_f = rQ^2$ (discharges in l/min): AB 1, BC 2.5, CD 1.5, AD 2, DB 3. The valve (r = 1.5) sits in the diagonal, in series with DB, so the diagonal has $r_{BD} = 3 + 1.5 = 4.5$. Inflow 70 at A; outflows 30 at B and 40 at C. Sign rule: clockwise flow positive. #### Initial distribution (continuity satisfied) A to B 40, A to D 30; at B (out 30): B to C 10, B to D 0; at C (out 40): the 10 from B plus 30 from D (D to C 30). **Loop I** = A-B-D-A, **Loop II** = B-C-D-B. #### Hardy-Cross iterations $\Delta Q = -\dfrac{\sum rQ|Q|}{2\sum r|Q|}$ for each loop, applied to all its pipes (the diagonal BD gets both corrections). | Iter | $\sum rQ|Q|$ (I) | $2\sum r|Q|$ (I) | $\Delta Q_I$ | $\sum rQ|Q|$ (II) | $2\sum r|Q|$ (II) | $\Delta Q_{II}$ | |---|---|---|---|---|---|---| | 1 | -200.00 | 200.00 | 1.000 | -1100.00 | 140.00 | 7.857 | | 2 | -212.59 | 259.71 | 0.819 | 273.33 | 217.43 | -1.257 | | 3 | 57.52 | 239.40 | -0.240 | -29.55 | 196.23 | 0.151 | | 4 | -7.23 | 243.39 | 0.030 | 11.05 | 200.05 | -0.055 | After about 10 iterations the corrections fall below 0.001. #### Final distribution (l/min) | Pipe | Discharge | Direction | |---|---|---| | AB | 41.60 | A to B | | AD | 28.40 | A to D | | BC | 16.70 | B to C | | CD | 23.30 | D to C | | BD (with valve) | 5.10 | D to B | Checks: node A: $41.60 + 28.40 = 70.00$; node B: $41.60 + 5.10 = 30 + 16.70$ (46.70 = 46.70); node C: $16.70 + 23.30 = 40.00$. **Answer:** AB = 41.6, AD = 28.4, BC = 16.7, CD = 23.3 (D to C) and the diagonal BD = 5.1 l/min (flowing D to B). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the valve in the diagonal BD handled? A. Treated as a separate loop B. Added in series: r_BD = 3 + 1.5 = 4.5 C. Added in parallel to DB D. Ignored since it has no head loss Answer: B. The valve sits in series with DB, so the resistances add. (Key point: Valve treated as series resistance) 2. Which condition must the assumed initial discharges satisfy? A. Continuity at every node B. Zero head loss around every loop C. Equal flow in all pipes D. Equal head loss in all pipes Answer: A. Initial flows satisfy continuity; the loop head-loss condition is then achieved through corrections. (Key point: Initial distribution) 3. The diagonal BD belongs to both loops. How is it corrected? A. It receives only the correction of loop I B. It is not corrected C. It receives the average of the two corrections D. It receives both loops' corrections Answer: D. A shared pipe gets the correction of its own loop and the opposite-signed correction of the neighbouring loop. (Key point: Common pipe in two loops) 4. What is the final flow in the diagonal BD (with valve)? A. 0 l/min B. 10 l/min from D to B C. 5.1 l/min from D to B D. 5.1 l/min from B to D Answer: C. The converged result shows BD = 5.10 l/min flowing D to B; node B check: 41.60 + 5.10 = 30 + 16.70. 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