--- title: "MOC numerical: discharge and head at P (30 cm pipe)" subject: "Computational Techniques in Civil Engineering" chapter: 6 key: "semester_8/computational_techniques_in_civil_engineering:moc-pipe-discharge-head-pc-30cm" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/moc-pipe-discharge-head-pc-30cm site: BCT Engineering --- # MOC numerical: discharge and head at P (30 cm pipe) Computational Techniques in Civil Engineering · Chapter 6: Method of Characteristics - Asked in: 2078 Kartik, 6 marks - Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/moc-pipe-discharge-head-pc-30cm - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-6 ## Question The following data are given at two points A and B along a pipe of diameter 30 cm carrying water as shown in the figure: $Q_A = 0.4$ m$^3$/s, $Q_B = 0.45$ m$^3$/s, $H_A = 26.5$ m, $H_B = 27.5$ m, $\Delta x = 500$ m, $\Delta t = 0.4$ sec, $f = 0.02$, $a$ (or $c$) $= 1200$ m/s, elevation difference between A and P $= 1$ m. Using the finite difference form of the characteristics equations, compute discharge and head at point P. [Figure: x-t grid with points A, C, B at time $t$ spaced $\Delta x$ apart and P at time $t + \Delta t$ above C, with $C^+$ from A and $C^-$ from B] ## Short answer Use C+ from A and C- from B with Bp=1730.53 s/m^2 and Rp=340.03 s^2/m^5. K1=664.31 and K2=-682.38 give Q_P=(K1-K2)/(2Bp)=0.3891 m^3/s and H_P=(K1+K2)/2=-9.038 m. ## Answer Use the finite difference form of the characteristic equations on the rectangular grid (A and B at time $t$, P at $t+\Delta t$): $$ C^+:\ H_P=H_A-B_p(Q_P-Q_A)-R_pQ_A|Q_A|,\qquad C^-:\ H_P=H_B+B_p(Q_P-Q_B)+R_pQ_B|Q_B| $$ $$ B_p=\frac{a}{gA_p},\qquad R_p=\frac{f\,\Delta x}{2gDA_p^2} $$ $H$ is the piezometric head (it already contains the elevation), so the 1 m elevation difference between A and P is not needed to find $H_P$ and $Q_P$. #### Step 1: Constants $$ A_p=\frac{\pi}{4}(0.3)^2=0.07069\ \text{m}^2 $$ $$ B_p=\frac{1200}{9.81\times0.07069}=1730.53\ \text{s/m}^2,\qquad R_p=\frac{0.02\times500}{2\times9.81\times0.3\times0.07069^2}=340.03\ \text{s}^2/\text{m}^5 $$ (Here $a\Delta t=1200\times0.4=480$ m is close to $\Delta x=500$ m, so the characteristics are taken through A and B as in the figure.) #### Step 2: Known quantities $$ K_1=H_A+B_pQ_A-R_pQ_A|Q_A|=26.5+1730.53(0.4)-340.03(0.16)=664.31 $$ $$ K_2=H_B-B_pQ_B+R_pQ_B|Q_B|=27.5-1730.53(0.45)+340.03(0.2025)=-682.38 $$ #### Step 3: Solve for P $$ Q_P=\frac{K_1-K_2}{2B_p}=\frac{664.31-(-682.38)}{2\times1730.53}=0.3891\ \text{m}^3/\text{s} $$ $$ H_P=\frac{K_1+K_2}{2}=-9.038\ \text{m} $$ **Answer:** $Q_P\approx0.3891$ m$^3$/s and $H_P\approx-9.038$ m. The negative head means the data give a pressure below the vapour pressure at P (column separation would occur), because the large impedance $B_p$ makes the $Q$ difference between A and B produce a big head change. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the 1 m elevation difference between A and P not needed when computing Q_P and H_P? A. Because H is the piezometric head, which already contains the elevation B. Because the pipe is assumed horizontal in MOC C. Because elevation effects are cancelled by friction D. Because elevation only matters for pressure, never for head Answer: A. H is piezometric head (pressure head plus elevation), so the elevation is already inside H_A and H_B. (Key point: Piezometric head includes elevation) 2. Which expression gives the characteristic impedance coefficient B_p used in the C+ and C- equations? A. B_p = g A_p / a B. B_p = a / (g A_p) C. B_p = f dx / (2 g D A_p^2) D. B_p = a dt / dx Answer: B. B_p = a/(g A_p) (here 1730.53 s/m^2); the friction coefficient R_p is the f dx/(2 g D A_p^2) term. (Key point: Constant B_p) 3. After finding K1 (from C+) and K2 (from C-), how are Q_P and H_P obtained? A. Q_P = (K1 + K2)/(2B_p), H_P = (K1 - K2)/2 B. Q_P = K1 - K2, H_P = K1 + K2 C. Q_P = (K1 - K2)/(2B_p), H_P = (K1 + K2)/2 D. Q_P = (K1 - K2)/B_p, H_P = K1 - K2 Answer: C. Equating the two characteristic equations gives Q_P = (K1-K2)/(2B_p) and H_P as the average (K1+K2)/2. (Key point: Solving for Q_P and H_P) 4. The computed H_P is about -9.04 m. What does this indicate? A. The computation is correct and means the flow reverses B. Pressure below vapour pressure, so column separation could occur C. The pipe is flowing at full capacity D. The Courant condition has been satisfied exactly Answer: B. A negative head means pressure falls below vapour pressure, which would cause column separation. 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