--- title: "Correct trial network flows and pressure at C" subject: "Hydraulics" chapter: 3 key: "semester_4/hydraulics:network-distribution-trial-pressure-at-c" source: https://www.bctengineer.com/q/semester_4/hydraulics/network-distribution-trial-pressure-at-c site: BCT Engineering --- # Correct trial network flows and pressure at C Hydraulics · Chapter 3: Three reservoirs problem and Pipe networks - Asked in: 2068 Magh, 8+2 marks - Video: [Hardy cross method for pipe network](https://www.youtube.com/watch?v=7aBS_QLUJ6U) — Civil Engineering 4 free - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/network-distribution-trial-pressure-at-c - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-3 ## Question For a pipe network shown in figure below, trial discharge distribution is shown, n = 2 for all the pipes. Obtain the correct distribution. Find also the available pressure at C, if the supply pressure at A is provided by 6 m high water tank. [Figure: network of nodes A, B, C, D, E with inflow 5 at A and outflows 2 at B, 2 at D and 1 at E; trial discharges and resistances: BC 1 (r = 20), CD 1 (r = 20), AB 1 (r = 40), AC 2 (r = 10), AE 2 (r = 30), DE 1 (r = 40)] ## Short answer Applying Hardy‑Cross corrections until convergence gives final discharges: AB 1.23 A→B, CB 0.77 C→B, AC 2.19 A→C, CD 1.42 C→D, ED 0.58 E→D, AE 1.58 A→E. Available pressure head at C is supply head 6 m minus head loss in AC, yielding –42.1 m (negative, indicating the given supply is insufficient if units are metres). ## Answer **Data.** Inflow 5 at A; outflows 2 at B, 2 at D, 1 at E. Trial flows and resistances ($h_f = rQ^2$, $n = 2$): AB 1 ($r$ = 40), AC 2 ($r$ = 10), AE 2 ($r$ = 30), BC 1 ($r$ = 20, flowing C to B), CD 1 ($r$ = 20), DE 1 ($r$ = 40, flowing E to D). Continuity check: A: $1 + 2 + 2 = 5$; B: $1 + 1 = 2$; C: $2 = 1 + 1$; D: $1 + 1 = 2$; E: $2 = 1 + 1$. **Loop I** = A-B-C-A (AB +, CB $-$, CA $-$). **Loop II** = A-C-D-E-A (AC +, CD +, ED $-$, EA $-$). #### Hardy-Cross iterations | 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 | -20.00 | 160.00 | 0.125 | -100.00 | 280.00 | 0.357 | | 2 | -14.51 | 169.64 | 0.086 | -10.84 | 248.93 | 0.044 | | 3 | -1.82 | 172.22 | 0.011 | -3.90 | 243.73 | 0.016 | The corrections fall below 0.001 after 14 iterations. #### Correct distribution | Pipe | Discharge | Direction | |---|---|---| | AB | 1.226 | A to B | | CB | 0.774 | C to B | | AC | 2.194 | A to C | | CD | 1.419 | C to D | | ED | 0.581 | E to D | | AE | 1.581 | A to E | Checks: A: $1.226 + 2.194 + 1.581 = 5.000$; B: $1.226 + 0.774 = 2.000$; D: $1.419 + 0.581 = 2.000$. #### Pressure at C The supply head at A from the 6 m tank is $H_A = 6$ m. The loss in AC is $h_{AC} = rQ^2 = 10(2.194)^2 = 48.12$ (in the units of the data). Taking A and C at the same level, the pressure head at C is $$ \frac{p_C}{\gamma} = H_A - h_{AC} = 6 - 48.12 = -42.12 $$ With $r$ and $Q$ used exactly as given, the loss exceeds the 6 m of supply head, so the available pressure at C comes out negative, which is not physically possible. Either the 6 m tank is not enough for these resistances, or $r$ is on a scale in which $rQ^2$ is not in metres. The method is: **available pressure head at C = 6 m $-$ $h_{AC}$ (in metres), and the pressure is $\gamma$ times this.** **Answer:** Correct discharges: AB = 1.23, CB = 0.77, AC = 2.19, CD = 1.42, ED = 0.58, AE = 1.58; $p_C/\gamma = 6 - h_{AC}$ (negative, $-42.1$, if the data are taken literally). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Loop I is A-B-C-A taken clockwise. How is the trial flow in pipe CB (flowing C to B) treated? A. Excluded from the loop B. Taken as zero C. Negative, as it opposes the clockwise direction D. Positive, as it is a trial flow Answer: C. In A-B-C-A, clockwise means B to C, so a flow from C to B is negative. (Key point: Signs in loop I) 2. Which check confirms the trial flows are acceptable starting values? A. Equal r in all pipes B. Continuity at each node C. Equal flows in both loops D. Zero pressure at all nodes Answer: B. At A: 1 + 2 + 2 = 5, at B: 1 + 1 = 2, etc. (Key point: Continuity of trial flows) 3. How is the available pressure head at C found with the supply head from the 6 m tank at A? A. p_C/gamma = 6 - h_AC B. p_C/gamma = 6 + h_AC C. p_C/gamma = h_AC - 6 D. p_C/gamma = 6 x h_AC Answer: A. Pressure head at C equals the head at A minus the loss along AC (h = rQ^2); with the given data the loss 48.12 exceeds 6, so the result is negative. 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