Question

PLEASE WRITE/PRINT CLEARLY AND SOLVE USING THE FINITE ELEMENT METHOD UNCLUDING MATRICIES

2. The flow rate at node 1 is 0.16 liter/s (0.16 *10-2 m/s). The pressure at node 4 is 0 Pa (g). For the given conditions, the flow is laminar throughout the system. Using hand calculation determine i. The pressure in each node. ii. Flow in each element iii. Verify your results (25 + 10 + 5 = 40) (1) L=10 m D=20 mm u = 8 x 10-4 N-s/m2 p=995 kg/m3 L= 7.5 m D= 15 mm [1] L= 7.5 m [4) D=10 mm (2) [3] L= 10 m D= 12.5 mm

Answer 1

AP12 + AP13 APBA Between node [18 and no de [a], [2]-[4] and [1] - [3 paralled. So, + 4P 24 Prenure drop in parallel lines are equal) A830.72 Q, + 13351 285 = 2037.084 Q 2 + 2 4437. 47 Q2 = > 1% 182.005 Qi 26474.5 5 4 62 -> Q1 = 1.456 Q2 at Q2= O. 16 22 1.456Q2 + Qa = 0.16 0.06515 literss = 0.09485 litera 458.2 Pa Flow in each element ii) 12 66.245 Pa 4912 AP24 AP13 APBA 1599.012 132.41 Pa 158 f. 44 Pa 1592.017 Pressure at node 4, Pat o Po (9) i) The pressure est each node P2 = P + AP 24 1266.415Pa (9) Pi P2 + AP 12 = 17 24.675Pa (9) P3 PA + AP 34 1588.144 Pa (9) Verification via [2]- [4 Presure drop between 1 & 4 AP13 + AP34 = 1724.727 Pa Pressure drop between 1 & 4 via [1]-[3] ARI2 + 4P24 1724.615 Pa close match so above As equation ① & 2 give very calculation is OK. 10