Question

(a) A horizontal pipe of diameter 10 cm carries crude oil from A to C as shown in Figure 4. At a point B, midway between A and C, oil leaks at a volume flow rate of Q_L . The pressure drop per unit length along AB is 640 Pa/m and along BC is 600 Pa/m. Determine the leakage flow rate Q_L . The density of oil is 800 kg/m³ and its dynamic viscosity is 0.1 N.s/ m² . You may assume that the flow is laminar.

3 (a) A horizontal pipe of diameter 10 cm carries crude oil from A to C as shown in Figure 4. At a point B, midway between A and C, oil leaks at a volume flow rate of Ql. The pressure drop per unit length along AB is 640 Pa/m and along BC is 600 Pa/m. Determine the leakage flow rate QL. The density of oil is 800 kg/m' and its dynamic viscosity is 0.1 N.s/m. You may assume that the flow is laminar. (12 marks) А D = 0.1 m Figure 4 (b) Water enters a parallel pipe system as shown in Figure 5. The pressures at A and B are 40 kPa and 20 kPa respectively. The elevations of A and B are 20 m and 10 m respectively. Determine the volume flow rate in pipes 1 and 2. What is the length of pipe 2? You may make use of the following information: Description Pipe 1 Pipe 2 Diameter (m) 0.06 0.08 | | Length (m) | Friction factor ΣΚΙ 120 0. 02 1 0 ? 10.025 T8 Pipe 1 D = 0.1 m V= 2 m/s V = 2 m/s Pipe 2 Figure 5 (13 marks)

Answer 1

Solution : a foit = 800 kg/m3 . M20. 1NS/m2 > P = 600 Pa I P = Pressure drop at inter a P2 hessie doop at outler ? P1 = 640 Pam D=0.lm. Here, p = 640 Palm and P2 600 Pa/m Usine Darcy - Weisbach equation in terms of pressure drop is given by Pi-Po - Felve t =triction bee 2D where, A - TID² va G = 42 Reynold number Rea PVD for Laminan flow , Re 42100 So, to 64 64 M. Gua PVD - - D (40 to 16 TUDO Then pressure doop P.-P2 = 16 T UD IPLV² lou 20 ~ 40 - 8TM LYZ - 8TML/40 Q (TID - 128 x 0.1X1XQ TX (0.14 > 0 = 9.8 x 104 m3/s. Re 40 = 128 ULO

Pike 1 bi Firstly, We know that: Flow in = Flow out D-0. v =2 m/s = 2 m/s O2 Pipe 2 - + O2 = where Q = Volume flow rate in pipe 1 Q2 = Volume flow rate in pipe 2. Q = AV = 1 Dz x V = I x 01² 42. & Q = 0.01571 m3/s. = 15.71x10 3 m/s 82 = (0, -0.01571) m3/s. Head loss formula using Darcy Weisbach equation , h = f, LV, ² x k h - - 0.02 X 120 X V.? x 10 D, X 2x9.811 h = 24² 1127 - 20.39 ² hz - f2 La Vak 0.025 X L₂ x V x 8 -0.1274 L₂ V₂ 0.08 x 2x 9.81_o But, hi-h₂ = (20- 10) = 10m. → 20.39 4,² - 0.1274 L 2 2 ² = 10 Now , Q = AV, Vi= 0 = 0, da ) V60.0028278 252 to 1/D2 (0.06) 1

Let length of pripe L₂ = 120m = 4. So, equation (1) becomes - 20.39 V, 2 = 10+ 15.288 V 2 TY4(0.08) 0:00SD 2- » Vy - 198.92 82 kar 198.92 (0 - 0.01571) » Y₂ = 198.920,- 3.125 = 2.549 8106 Q2 = 10+ 0.60493 X 106 Q1² Q? (1.94402 x 106 -10 2 Q2 = 5.1438 X1076 » Q, - 2.26 X10-3 and Q2 = 13.45x103 m3/s.
Summary: Both problem is solved using Darcy- Weisbach equation.The length of pipe in question-3b will be 120m.