Question

Need help solving this problem for Applied Fluids Mechanics

11. The Figure shown below is a parallel pipeline system with two branches used to supply lubricating water to the bearings. The main line and two branches use the same size of pipes. The pressures at section 1 and section 2 are pi = 398.1kPa, p2 = 300kPa, respectively. The resistance coefficents for two bearings are Ky 5, K2 12. The cross section areas of two branch pipes are Aa Ap = 1.5 x 10 m2. The engergy loss caused by friction can be ignored. The engergy loss cuased by one bend is h3 0.2m. 1) Write down the general engergy enquation for section 1 and 2 and simplify it. 2) Calculate the total engery loss htot between section 1 and 2. 3) Find the engery losses h,a and hybfor branch a and b. 4) Calculate the energy loss caused by the two bearings hy and h2, respectively. 5) Calculate the velocities Va and v, 6) Calculate the volume flow rates Qa and Qb P +7 Va + 1) KT K2 Figure 1 Parallel pipeline system

Answer 1

[1] Write the general energy equation for section 1 and 2 р. . P2 V + Pg 2g Pg 2g V 2 (1) Р. и? Z2+2h3K K 2g + +21 = Pg 2g + + 2g Pg 2g Write the continuity equation for the flow - DV 4 V. AVAV 4 Also, -DVb 4 2 2 a same size (D D2 Da D D). Therefore having Pipes are &

Write the continuity equation for the flow. )k-;o*}>-(;a}e 2 2 2 4. 4 a 4 4 VVV V = V 2 V 2V having at same level. Therefore, Also, pipes are = Substitute values in equation [1] V 2 -K2 2g 2 Z+ -+Z = 2g 2g Pg Pg 2g lal-las PA-P2- 2hK + K2) 8g Activate Go to DC Pg

[2] Calculate the total energy loss between section 1 and 2 (398.1-300)x103 (1000)(9.81) h, 2hKK,) 8g =10 m _ Pg [3] Calculate the velocity of flow in main line P-P2-2hKi+ K2)| 8g pg (398.1-300)x103 = 2 (0.2)+(5+12) 8(9.81) (1000) (9.81) (10-0.4)x8(9.81) (17) - 6.657 m/s (6.657) V 3.329 m/s 2 Va-V 2 _ Activate Go to PC set

[4] Calculate the energy loss caused by first bearing. (3.329) (5) 2(9.81) 2 h 3 к, -2.824 m 2g Calculate the energy loss caused by second bearing. (3.329) (12) 2 (9.81) h, 3D к, = 6.778 m 2g [6] Calculate the volume flow rate a AV-1.5x 10(3.329) 5x10 m3/s _ Q A(15x10)(3.329) = 5x10 -4 '/s