--- title: "Hardy Cross correction derivation and network" subject: "Hydraulics" chapter: 3 key: "semester_4/hydraulics:hardy-cross-correction-and-kathmandu-network" source: https://www.bctengineer.com/q/semester_4/hydraulics/hardy-cross-correction-and-kathmandu-network site: BCT Engineering --- # Hardy Cross correction derivation and network Hydraulics · Chapter 3: Three reservoirs problem and Pipe networks - Asked in: 2080 Chaitra, 2+6 marks - Video: [Hardy cross method for pipe network](https://www.youtube.com/watch?v=7aBS_QLUJ6U) — Civil Engineering 4 free - Page: https://www.bctengineer.com/q/semester_4/hydraulics/hardy-cross-correction-and-kathmandu-network - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-3 ## Question Derive an expression for correction factor for discharge while performing Hardy Cross method of solving network problem. A water supply distribution network is developed in a ward of Kathmandu municipality as shown in the figure below. The water supply distribution features a cross fitting of diameter 0.3 m at junction D. The inlet and outlet quantities are given in the figure. Determine the discharge through pipe link FG if all pipes are manufactured with steel (f = 0.010 for 0.3 m dia pipes and 0.012 for 0.25 m dia pipes). [Figure: network of nodes A to G with inflow/outflow at nodes marked 2, 2, 4 and 1; the link table lists r values: AB 50, AC 50, BD 100, CD 80, DE 100, EF 100, FG 50, DG 50, ...; the scan is cut off after DG] ## Short answer The correction factor is ΔQ = –Σ r Q₀|Q₀| / (2 Σ r |Q₀|). Applying Hardy Cross to the given network yields a final discharge in pipe FG of 2.84 units, flowing from G to F. ## Answer **Derivation of the correction $\Delta Q$ (Hardy Cross).** Head loss in a pipe is $h_f = rQ^n$ (with $n = 2$ for the Darcy-Weisbach law, $r = \dfrac{8fL}{\pi^2gD^5}$). For a correct distribution the algebraic sum of head losses around every closed loop is zero (clockwise flows positive): $$ \sum rQ|Q|^{n-1} = 0 $$ Take assumed discharges $Q_0$ which satisfy continuity at the nodes but not the loop condition. Let the true discharge be $Q = Q_0 + \Delta Q$, where the same correction $\Delta Q$ applies to every pipe of the loop (so continuity is preserved). Then $$ \sum r(Q_0 + \Delta Q)^n = 0 \;\Rightarrow\; \sum rQ_0^n + n\,\Delta Q\sum rQ_0^{n-1} + \frac{n(n-1)}{2}\Delta Q^2\sum rQ_0^{n-2} + \dots = 0 $$ $\Delta Q$ is small compared with $Q_0$, so the terms in $\Delta Q^2$ and higher are neglected: $$ \Delta Q = -\frac{\sum rQ_0|Q_0|^{n-1}}{n\sum r|Q_0|^{n-1}} \;\xrightarrow{n = 2}\; \boxed{\Delta Q = -\frac{\sum rQ_0|Q_0|}{2\sum r|Q_0|}} $$ The numerator is the algebraic sum of head losses in the loop (signs by direction), the denominator is the sum of absolute values. Apply the correction to every pipe (add it to flows in the clockwise direction, subtract it from anticlockwise ones, and apply both loops' corrections to a common pipe), and repeat until $\Delta Q$ is negligible. #### Network discharge through FG **Reading of the figure.** The scan is cut off at the bottom, so the part below B and D is assumed to be: A (inlet) joined to B and C; B-D, C-D; D-E, D-G; E-F and G-F. Nodes B, E, F, G have the outflows shown (2, 2, 4, 1) and the inlet at A must therefore be $2 + 2 + 4 + 1 = 9$. Resistances $r$ ($h_f = rQ^2$) as in the table: AB 50, AC 50, BD 100, CD 50, DE 100, EF 100, FG 50, DG 50 (the $f$ and $D$ values are already contained in $r$). Flow units are those of the figure. **Initial distribution (satisfying continuity):** AB 4, AC 5, BD 2, CD 5, DE 4, EF 2, GF 2 (G to F), DG 3. **Loop 1** (A-B-D-C, clockwise: AB, BD positive; CD, AC negative) and **loop 2** (D-E-F-G, clockwise: DE, EF positive; GF as F to G, DG as G to D negative). Correction in each iteration, $\Delta Q = -\dfrac{\sum rQ|Q|}{2\sum r|Q|}$: | Iter | $\sum rQ|Q|$ (L1) | $2\sum r|Q|$ (L1) | $\Delta Q_1$ | $\sum rQ|Q|$ (L2) | $2\sum r|Q|$ (L2) | $\Delta Q_2$ | |---|---|---|---|---|---|---| | 0 | -1300.00 | 1800.00 | 0.722 | 1350.00 | 1700.00 | -0.794 | | 1 | 26.08 | 1872.22 | -0.014 | 63.06 | 1541.18 | -0.041 | | 2 | 0.01 | 1870.83 | -0.000 | 0.17 | 1532.99 | -0.000 | After the third iteration the corrections are below 0.001. Final discharges: | Link | AB | AC | BD | CD | DE | EF | GF | DG | |---|---|---|---|---|---|---|---|---| | Q | 4.708 | 4.292 | 2.708 | 4.292 | 3.165 | 1.165 | 2.835 | 3.835 | Check at the nodes: D receives $2.708 + 4.292 = 7.000$ and sends $3.165 + 3.835 = 7.000$; F receives $1.165 + 2.835 = 4.000$, equal to its outflow of 4. **Answer:** The discharge in link FG is $2.84$ units (flowing from G to F), $\approx 2.8$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In deriving the Hardy Cross correction, why are the terms in DeltaQ^2 and higher neglected? A. Because r is very large for every pipe B. Because DeltaQ is small compared with Q0 C. Because the head loss is linear in Q D. Because DeltaQ is always exactly zero Answer: B. The binomial expansion of (Q0 + DeltaQ)^n is truncated since DeltaQ is small relative to Q0, giving a linear equation in DeltaQ. (Key point: Why higher-order terms are dropped) 2. For n = 2, which expression gives the Hardy Cross correction for a loop? A. DeltaQ = -sum(rQ|Q|) / (2 sum r|Q|) B. DeltaQ = +sum(rQ|Q|) / (2 sum r|Q|) C. DeltaQ = -sum(rQ|Q|) / sum(r|Q|^2) D. DeltaQ = -2 sum(rQ|Q|) / sum r|Q| Answer: A. The numerator is the algebraic sum of head losses in the loop and the denominator is n times the sum of r|Q|, with absolute values; the minus sign drives the loop head loss toward zero. (Key point: Form of the correction) 3. In the Kathmandu network the outflows at B, E, F, G are 2, 2, 4 and 1. What inflow is needed at A? A. 8 B. 9 C. 10 D. 7 Answer: B. Total inflow must equal total outflow: 2 + 2 + 4 + 1 = 9 units. (Key point: Inflow from continuity) 4. After the Hardy Cross iterations, what is the discharge in link FG and its direction? A. About 2.84 units from F to G B. About 3.84 units from G to F C. About 2.84 units from G to F D. About 1.17 units from F to G Answer: C. Final GF = 2.835, flowing G to F; with EF = 1.165 this gives 4 units arriving at F, equal to its outflow. 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