Question

glavily UI 0.00! (3.)A 20 in3 sample of oil is compressed in a 2.5 in diameter cylinder until the force applied increased from 1000 lb to 2000 lb. If the bulk modulus equals 300,000 psi, find the change in volume of the oil. 4. For the pipe shown below, calculate: a. The flow rate at point 2 b. The velocity of the fluid at point 1 and point 2 c. The pressure of the fluid at point 2 A2 = 5 in? Z2 = 0 P1 = 200 psi Q1 = 3 gpm A1 = 20 in2 21 = 20 ft

Answer 1

Problem 3 ginen: Valme of Sempre = 20 in 3 Cylinder diameter = 2.5 in. Force applied P = 1000 lb P2 = 2000db Bulla modulin ko 300000 PSI To find: change in volume of ou balution We know that - of de change in change in pressure volume dv a

300 000 here dp = P2P, but P2= 2000 oross Sechonol area a Cylinder [Aces] I x 2-52 Aus = 4 Acco = 4.9087 in 2 P2 = Psi 2 2000 4.9087 = 407,4366 Po ooo 4.9087 203.71992 Psi 10 de des 800,000 dra va 300,000

Page Date : du - 20x [P2-Pi] 300,000 - 20x5 407.4360 - 203.71992] 300,000 dr = 0.01358 in 3

problem 4 given A2 = 5 in2 Z2 = 0 P = 200 psi a = 8 9pm A = 20 in² 2 = 20ft To find da a , qV2 IP2 solution: Formula Required continunite aquahon e, AMI = P₂ AzV2 here l =l2 = because woning fluid at pt 1 and 2 are same

Q = AV de=AV (P = 02] da = A₂ V2 Bemoulies equcahon en 42 - +*+2+ 29 Q = AVI - 19pm - 3.95 in 3x9.95 m20 x (in %) XV () = 0.5175 in = Wk That AVI= AzV2 20x 0.5715 = 5xV2 V2 = 2.31 in sec grauity = 386.09 inches seca a 67 d=0 = 39pm Vi = 0.5775 V2 = 2.31 in in

P I +2 = Pe & V2² +22 eg 29 +22 ( Assuming ancat water is Howding Inside Al lb = 27, 679.90 kg ma in3 * 1000kg - 0.0361 lb m3 m3 *9= 386.084 inches R Itt = 12 inches 2ofa 240 in + (0.5775) 2x 386.085 + (20ft) = 0-0361 X 386.08% to P2 0.0361 +386.088 + [2.310) 2x 386.088 14.3494 +4-3190x wh+240 - 0.07174P2 +0.00691 254.3429 = 0.0717482 P2 = 3545 3432 psi