AL Fluid P.44. V~ 1m x 0.4 m x 0.2m ~ 0.01 m 3 Density  = M/V ~ 100kg / 0.01 = 10000 kg m -3.

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Presentation transcript:

AL Fluid P.44

V~ 1m x 0.4 m x 0.2m ~ 0.01 m 3 Density  = M/V ~ 100kg / 0.01 = kg m -3

P.44  =M/V, V=  (d/2) 2 L d=  (4M)/(  L) %d=(1/2)[ %M + %  + %L] %d=(1/2)[ 4% + 0% + 2%] = 3%

P.47

P.48 A 1 v 1 = A 2 v 2 By principle of continuity

P.48 Assume the fluid is incompressible and viscosity is neglected (P 1 – P 2 ) A 1 v 1 t = (1/2)  tA 1 v 1 (v v 1 2 ) +  gA 1 v 1 t(h 2 – h 1 ) Bernoulli’s Equation (P 1 – P 2 ) = (1/2)  (v v 1 2 ) +  g (h 2 – h 1 ) P 1 + (1/2)  v  gh 1 = P 2 + (1/2)  v  gh 2

P.49 P o + (1/2)  (0) 2 +  gh = P o + (1/2)  v 2 +  g(0) gh = (1/2) v 2 v =  2gh v =  2(10)(0.1) = (m/s)

P.49 Flow rate are the same v large A large = v small A small (1.5) (2.4) = v small (20)(2x10 -2 ) v small = 9 (m/s)

P.50 P = P s + (1/2)  v x 10 4 = 4.3 x (1/2) (1000)v 2 v =2.828 (m/s) dV/dt = A v =(20 x ) (2.828) = x (m 3 /s)

P.50 P a + (1/2)  v a 2 +  gh a = P b + (1/2)  v b 2 +  gh b A a v a = A b v b dm/dt =  dV b /dt =  A b v b =180 (850)(0.2) v b =180 v b = v a = P b - P a = (1/2)  (v a 2 -v b 2 ) +  g(h a – h b ) P b - P a = (Pa) P b - P a = (1/2) (850)( – ) + (850)(10)(2)

P.50 P u + (1/2)  v u 2 +  gh u = P b + (1/2)  v b 2 +  gh b P u + (1/2)  v u 2 = P b + (1/2)  v b 2 P u - P b = F / A = (1/2)  (v b 2 - v u 2 ) F / A = (1/2)  v b 2 v b =  2F/(  A)

P.50 (1/2)  air (v Y 2 - v X 2 ) =  oil g(h X – h Y ) (1/2) (1.2)(v Y 2 ) = (800)(10)(0.01) v Y = (m/s)

P.50 dV/dt = A dh/dt = – v(0.5 x )  gh = (1/2)  v 2 v 2 =2gh Set dV/dt = 0 => = v(0.5 x ) v = 2 (2) 2 =2(10)h h =0.2 (m)

P.50  oil gh 1 +  water gh2 = (1/2)  water v 2 (800)(10)(0.5) + (1000)(10)(3.5) = (1/2) (1000)v 2 v= (m/s)

P.51

P.52

P.54

P u + (1/2)  v u 2 = P b + (1/2)  v b 2 P b - P u = 1000 = (1/2) (1.2) ((v b +16) 2 – (v b ) 2 ) 2000 = (1.2) ((v b +16)+ (v b ) ) ((v b +16)- (v b ) ) 2000 = (1.2) (2v b +16) (16) v b = (m/s)

P.54 P a + (1/2)  v a 2 = P b + (1/2)  v b 2 P a - P b = (1/2)  (v b 2 - v a 2 ) A a v a = A b v b (20)(2) = (5)v b v b = 8 P a - P b = (1/2)(800)( )=24000 P a - P b should be There is energy loss. Energy loss =PV = (30000 – 24000) (10 x ) = 6 x (J)

P.55 (1/2)  (v b 2 - v a 2 ) =  g(h a – h b ) (1/2) (v b 2 - v a 2 ) = g(H) A a v a = A b v b v a = (0.25)v b (1/2) ((4v a ) 2 - v a 2 ) = g(H) (1/2) (15v a 2 ) = (10)(0.2)=2 v a = (m/s) v b = (m/s)

P.55 P d = P e +  water gh 2 = P e P f = P b + (1/2)  water v b 2 c d e P c = P f +  water gh 1 = P f f P e = P a + (1/2)  water v a 2 P c = P d +  Hg gh 3 = P d v a A a = v b A b v a (0.02) = v b (0.08) v a = 4 v b

P.55 Total pressure P = static pressure + (1/2)  v 2 1 x 10 4 = 0.6 x (1/2) (1.2 x 10 3 ) v 2 v = Volume flow rate = A v = (2.5 x )(2.582) = 6.5 x 10 -3

P.56 (1) For a steady and incompressible water flow, the flow rate is constant. Y Y N (2) Flow rate is constant. v X A X = v Y A Y, hence v X = v Y (3) P + (1/2)  v 2 +  gh = constant (1/2) (v Y 2 - v X 2 ) = g(h X – h Y ) Faster speed at point Y, hence h Y should be lower

P.56 (1) Terminal vel. related to the air resistance and the gravitational force only. Y N N (2) Spinning ball has pressure difference between two sides, thus a force is formed to change the direction of ball. (3) Due to the fluid is incompressible, not viscous and steady flow, Bernoulli’s equation can explain it.

P.56