--- title: "Oil through two parallel pipes" subject: "Hydraulics" chapter: 2 key: "semester_4/hydraulics:oil-two-parallel-pipes-pump-power" source: https://www.bctengineer.com/q/semester_4/hydraulics/oil-two-parallel-pipes-pump-power site: BCT Engineering --- # Oil through two parallel pipes Hydraulics · Chapter 2: Simple pipe flow problems and solution - Asked in: 2071 Bhadra, 8 marks - Video: [BYU ChEn 285 Fluids Lab: Parallel Pipe Flow](https://www.youtube.com/watch?v=I2SrSkTVC20) — ignite.byu.edu - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/oil-two-parallel-pipes-pump-power - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-2 ## Question A total 12 liters per sec of oil is pumped through 2 pipes in parallel, one 12 cm in diameter and the other 10 cm in diameter, both pipes 1000 m long. The specific gravity of oil is 0.85, average roughness height is 0.26 mm for both pipes and kinematic viscosity is 9 cm²/sec. Calculate the flow rate through each pipe, and power generated by pump. ## Short answer Assuming laminar flow, head loss equality gives Q ∝ D⁴. Flow rates: Q1=8.10 L/s (12 cm pipe), Q2=3.90 L/s (10 cm pipe). Pump head 145.9 m of oil; power = 14.6 kW. ## Answer **Data.** $Q = 12$ l/s $= 0.012$ m³/s, $D_1 = 0.12$ m, $D_2 = 0.10$ m, $L = 1000$ m each, s.g. 0.85 ($\rho = 850$ kg/m³), $k = 0.26$ mm, $\nu = 9$ cm²/s $= 9\times10^{-4}$ m²/s. The kinematic viscosity is very high, so the Reynolds numbers are small. Assume laminar flow in both pipes and verify afterwards. #### Flow in each pipe In parallel, both pipes have the same head loss. For laminar flow $h_f = \dfrac{128\nu LQ}{\pi gD^4}$, so $Q \propto D^4$: $$ \frac{Q_1}{Q_2} = \left(\frac{0.12}{0.10}\right)^4 = 2.0736 $$ $$ Q_1 = 0.012\left(\frac{2.0736}{1 + 2.0736}\right) = 0.008096\ \text{m}^3/\text{s} = 8.096\ \text{l/s},\qquad Q_2 = 0.003904\ \text{m}^3/\text{s} = 3.904\ \text{l/s} $$ #### Check for laminar flow $$ V_1 = \frac{Q_1}{\pi(0.12)^2/4} = 0.7158\ \text{m/s},\ Re_1 = \frac{V_1D_1}{\nu} = 95.4 \qquad V_2 = 0.4971\ \text{m/s},\ Re_2 = 55.2 $$ Both are well below 2000, so the flow is laminar and the roughness (0.26 mm) has no effect. #### Head loss and pump power $$ h_f = \frac{128\nu LQ_1}{\pi gD_1^4} = \frac{128(9\times10^{-4})(1000)(0.008096)}{\pi(9.81)(0.12)^4} = 145.94\ \text{m of oil} $$ The same value is obtained from pipe 2 ($145.94$ m). Pump head $= 145.94$ m (equal to the friction head, with no static lift). Power delivered to the oil: $$ P = \rho g Q h_f = 850(9.81)(0.012)(145.94) = 14603\ \text{W} \approx 14.60\ \text{kW} $$ **Answer:** $Q_1 = 8.10$ l/s (12 cm pipe), $Q_2 = 3.90$ l/s (10 cm pipe); pump power $\approx 14.6$ kW (head 145.9 m of oil). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is laminar flow assumed for the oil pipes? A. The pipes are smooth B. The kinematic viscosity is very high, so Re is small C. The roughness is large D. The flow rate is large Answer: B. nu = 9 cm^2/s is very large, giving low Reynolds numbers; this is then verified (Re about 95 and 55). (Key point: Laminar assumption) 2. In laminar flow with equal head loss and length, how does Q vary with D? A. Q proportional to D^2 B. Q proportional to D^4 C. Q proportional to D^(5/2) D. Q proportional to D Answer: B. From h_f = 128 nu L Q/(pi g D^4), equal h_f gives Q proportional to D^4. (Key point: Flow split in parallel laminar) 3. What is the effect of the 0.26 mm roughness on the results? A. It sets f through Colebrook B. It increases the head loss by 10% C. None, since laminar friction does not depend on roughness D. It makes the flow turbulent Answer: C. In laminar flow f = 64/Re, independent of roughness. (Key point: Role of roughness) 4. How is pump power obtained? A. P = gamma_water Q h_f B. P = rho g Q h_f with rho = 850 kg/m^3 and total Q C. P = rho g Q1 h_f using one pipe only D. P = Q x h_f Answer: B. Power to the oil uses its density (s.g. 0.85) and the total discharge 12 l/s with the common head loss, giving about 14.6 kW. 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