1. AL Fluid P.44

2. V~ 1m x 0.4 m x 0.2m ~ 0.01 m 3 Density  = M/V ~ 100kg / 0.01 = 10000 kg m -3

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%

4. P.47

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

7. 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 2 2 - v 1 2 ) +  gA 1 v 1 t(h 2 – h 1 ) Bernoulli’s Equation (P 1 – P 2 ) = (1/2)  (v 2 2 - v 1 2 ) +  g (h 2 – h 1 ) P 1 + (1/2)  v 1 2 +  gh 1 = P 2 + (1/2)  v 2 2 +  gh 2

8. 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) = 1.414 (m/s)

9. 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)

10. P.50 P = P s + (1/2)  v 2 4.7 x 10 4 = 4.3 x 10 4 + (1/2) (1000)v 2 v =2.828 (m/s) dV/dt = A v =(20 x 10 -4 ) (2.828) = 5.656 x 10 -3 (m 3 /s)

11. 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 =1.0588 v a =0.2647 P b - P a = (1/2)  (v a 2 -v b 2 ) +  g(h a – h b ) P b - P a = 16553 (Pa) P b - P a = (1/2) (850)(0.2647 2 –1.0588 2 ) + (850)(10)(2)

12. 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)

13. 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 = 11.547 (m/s)

14. P.50 dV/dt = A dh/dt = 10 -3 – v(0.5 x 10 -3 )  gh = (1/2)  v 2 v 2 =2gh Set dV/dt = 0 => 10 -3 = v(0.5 x 10 -3 ) v = 2 (2) 2 =2(10)h h =0.2 (m)

15. 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=8.8317 (m/s)

16. P.51

17. P.52

19. P.54

22. 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 = 44.08 (m/s)

23. 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)(8 2 - 2 2 )=24000 P a - P b should be 30000. There is energy loss. Energy loss =PV = (30000 – 24000) (10 x 10 -9 ) = 6 x 10 -5 (J)

24. 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 = 0.516 (m/s) v b = 2.066 (m/s)

25. P.55 P d = P e +  water gh 2 = P e + 15800 P f = P b + (1/2)  water v b 2 c d e P c = P f +  water gh 1 = P f + 3000 f P e = P a + (1/2)  water v a 2 P c = P d +  Hg gh 3 = P d + 27200 v a A a = v b A b v a (0.02) = v b (0.08) v a = 4 v b

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

27. 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

28. 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.

29. P.56