--- title: "Maximum length of pipe portion AB" subject: "Hydraulics" chapter: 2 key: "semester_4/hydraulics:pipe-abc-max-length-ab" source: https://www.bctengineer.com/q/semester_4/hydraulics/pipe-abc-max-length-ab site: BCT Engineering --- # Maximum length of pipe portion AB Hydraulics · Chapter 2: Simple pipe flow problems and solution - Asked in: 2079 Chaitra, 8 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/pipe-abc-max-length-ab - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-2 ## Question Two reservoirs whose difference of level is 15 m are connected by a pipe ABC whose highest point B is 2 m below the level in the upper reservoir A. The portion AB has a diameter of 200 mm, and the portion BC has a diameter of 150 mm, the friction coefficient being the same for both portions. Total length of pipe ABC is 3 km. Find the maximum allowable length of the portion AB if the pressure head at B is not to be more than 2 m below atmospheric pressure. Neglect the secondary losses and take f = 0.012. [Figure: free surface of reservoir 1; pipe rising from A to the summit B (2 m below the upper water level) then falling to C in reservoir 2; $D_{AB}$ = 200 mm, $D_{BC}$ = 150 mm; level difference 15 m] ## Short answer Maximum allowable length of portion AB is 1806 m, with BC about 1194 m, satisfying the limiting pressure head of -2 m gauge at summit B. ## Answer **Data.** $H = 15$ m, $D_1 = D_{AB} = 0.2$ m, $D_2 = D_{BC} = 0.15$ m, $L_1 + L_2 = 3000$ m, $f = 0.012$. Summit B is 2 m below the upper water level; pressure head at B $\ge -2$ m (gauge). Secondary (minor) losses are neglected. Let $h_1 = \dfrac{V_1^2}{2g}$. By continuity $V_2 = V_1\left(\dfrac{D_1}{D_2}\right)^2 = 1.7778V_1$, so $\dfrac{V_2^2}{2g} = \left(\dfrac{D_1}{D_2}\right)^4h_1 = 3.1605\,h_1$. #### Condition at B (limiting pressure) Energy equation from the upper reservoir surface (datum, $p = 0$) to B, which is 2 m below the surface ($z_B = -2$), with $\dfrac{p_B}{\gamma} = -2$ m: $$ 0 = -2 + (-2) + h_1\left(1 + f\frac{L_1}{D_1}\right) \;\Rightarrow\; h_1\left(1 + 0.060L_1\right) = 4 \quad\cdots(1) $$ (here $f/D_1 = 0.012/0.2 = 0.060$ per metre). #### Condition for the total head Level difference = friction in AB + friction in BC (no minor losses): $$ 15 = h_1\,(0.060)L_1 + 3.1605\,h_1\left(\frac{0.012}{0.15}\right)(3000 - L_1) $$ $$ 15 = h_1\left[0.060L_1 + 0.080(3.1605)(3000 - L_1)\right] \quad\cdots(2) $$ #### Solve Substitute $h_1 = \dfrac{4}{1 + 0.060L_1}$ from (1) into (2): $$ 15\left(1 + 0.060L_1\right) = 4\left[0.060L_1 + 0.080(3.1605)(3000 - L_1)\right] $$ $$ L_1 = 1806\ \text{m} $$ Check: $h_1 = \dfrac{4}{1 + 0.06(1806)} = 0.03657$ m, so $V_1 = 0.847$ m/s, $V_2 = 1.506$ m/s, $Q = 0.0266$ m³/s; friction in AB $= 3.963$ m and in BC $= 11.037$ m, which add to 15 m. **Answer:** The maximum allowable length of portion AB $\approx 1806$ m (so BC is about 1194 m). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the limiting pressure at B used to get the first equation? A. Energy equation from the upper reservoir surface to B with p_B/gamma = -2 m and z_B = -2 m, including velocity head and friction in AB B. Energy equation between the two reservoir surfaces only C. Set the pressure at B equal to atmospheric pressure D. Apply continuity at B only Answer: A. The pressure limit at B is imposed by writing Bernoulli from reservoir A to B, with the velocity head and friction loss in AB. (Key point: Condition at summit B) 2. If D_AB = 200 mm and D_BC = 150 mm, how is V2^2/2g in BC related to h1 = V1^2/2g? A. V2^2/2g = (D1/D2)^2 h1 B. V2^2/2g = (D1/D2)^4 h1 = 3.16 h1 C. V2^2/2g = (D2/D1)^4 h1 D. V2^2/2g = h1 Answer: B. Continuity gives V2 = V1 (D1/D2)^2, so the velocity head scales with the fourth power of the diameter ratio. (Key point: Velocity relation) 3. With minor losses neglected, what does the 15 m level difference equal? A. Friction loss in AB only B. Friction loss in BC only C. Sum of friction losses in AB and BC D. Sum of friction losses and the velocity head at B Answer: C. The whole level difference is consumed by friction in the two series pipes when minor losses are ignored. (Key point: Total head equation) 4. How is the maximum length of AB found? A. Substitute h1 = 4/(1 + 0.06 L1) from the B-condition into the total-head equation and solve for L1 B. Divide 3000 m in proportion to the diameters C. Take L1 as 2 m divided by the hydraulic gradient D. Assume equal velocities in both portions Answer: A. The two equations in h1 and L1 are combined; eliminating h1 gives L1 of about 1806 m. 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