AL Fluid

P.44

P.44

V~ 1m x 0.4 m x 0.2m ~ 0.01 m3

Density = M/V ~ 100kg / 0.01 = 10000 kg m-3

P.44

=M/V , V=(d/2)2L

d=(4M)/(L)

%d=(1/2)[ %M + % + %L]

%d=(1/2)[ 4% + 0% + 2%] = 3%

P.47

P.47

P.48

A1 v1 = A2 v2

By principle of continuity

P.48

Assume the fluid is incompressible and viscosity is neglected

(P1 – P2 ) A1v1t = (1/2) tA1v1(v22 - v1

2 ) + gA1v1t(h2 – h1)

Bernoulli’s Equation

(P1 – P2 ) = (1/2) (v22 - v1

2 ) + g (h2 – h1)

P1 + (1/2) v12 + gh1 = P2 + (1/2) v2

2 + gh2

P.49

Po + (1/2) (0)2 + gh = Po + (1/2) v2 + g(0)

gh = (1/2) v2

v = 2gh

v = 2(10)(0.1) = 1.414 (m/s)

P.49

Flow rate are the same

vlarge Alarge = vsmall Asmall

(1.5) (2.4) = vsmall (20)(2x10-2)

vsmall = 9 (m/s)

P.50

P = Ps + (1/2) v2

4.7 x 104 = 4.3 x 104 + (1/2) (1000)v2

v =2.828 (m/s)

dV/dt = A v =(20 x 10-4) (2.828) = 5.656 x 10-3 (m3/s)

P.50

Pa + (1/2) va2 + gha = Pb + (1/2) vb

2 + ghb

Aava = Abvb

dm/dt = dVb/dt = Abvb =180

(850)(0.2) vb =180

vb =1.0588

va =0.2647

Pb - Pa= (1/2) (va2 -vb

2 ) + g(ha – hb)

Pb - Pa= 16553 (Pa)

Pb - Pa= (1/2) (850)(0.26472 –1.05882 ) + (850)(10)(2)

P.50

Pu + (1/2) vu2 + ghu = Pb + (1/2) vb

2 + ghb

Pu + (1/2) vu2 = Pb + (1/2) vb

2

Pu - Pb = F / A = (1/2) (vb

2 - vu2 )

F / A = (1/2) vb2

vb = 2F/( A)

P.50

(1/2) air(vY2 - vX

2 ) = oil g(hX – hY)

(1/2) (1.2)(vY2 ) = (800)(10)(0.01)

vY = 11.547 (m/s)

P.50

dV/dt = A dh/dt = 10-3 – v(0.5 x 10-3)

gh = (1/2) v2

v2 =2gh

Set dV/dt = 0 => 10-3 = v(0.5 x 10-3)

v = 2

(2)2 =2(10)h

h =0.2 (m)

P.50

oilgh1 + watergh2 = (1/2) waterv2

(800)(10)(0.5) + (1000)(10)(3.5) = (1/2) (1000)v2

v=8.8317 (m/s)

P.51

P.52

P.52

P.54

P.54

P.54

P.54

Pu + (1/2) vu2 = Pb + (1/2) vb

2

Pb - Pu = 1000 = (1/2) (1.2) ((vb+16)2 – (vb)2 )

2000 = (1.2) ((vb+16)+ (vb) ) ((vb+16)- (vb) )

2000 = (1.2) (2vb+16) (16)

vb= 44.08 (m/s)

P.54

Pa + (1/2) va2 = Pb + (1/2) vb

2

Pa - Pb = (1/2)(vb

2 - va2 ) Aava = Abvb

(20)(2) = (5)vb

vb = 8

Pa - Pb = (1/2)(800)(82 - 22 )=24000

Pa - Pb should be 30000. There is energy loss.

Energy loss =PV = (30000 – 24000) (10 x 10-9) = 6 x 10-5(J)

P.55

(1/2) (vb2 - va

2 ) = g(ha – hb)

(1/2) (vb2 - va

2 ) = g(H) Aava = Abvb

va = (0.25)vb (1/2) ((4va)2 - va

2 ) = g(H)

(1/2) (15va2 ) = (10)(0.2)=2

va = 0.516 (m/s)

vb = 2.066 (m/s)

P.55

Pd = Pe + watergh2 = Pe + 15800

Pf = Pb + (1/2) watervb2

cd

e

Pc = Pf + watergh1 = Pf + 3000 f

Pe = Pa + (1/2) waterva2

Pc = Pd + Hggh3 = Pd + 27200

va Aa = vb Ab

va (0.02) = vb (0.08) va = 4 vb

P.55

Total pressure P = static pressure + (1/2) v2

1 x 104 = 0.6 x 104 + (1/2) (1.2 x 103 ) v2

v = 2.582

Volume flow rate = A v = (2.5 x 10-3)(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. vXAX = vYAY , hence vX = vY

(3) P + (1/2) v2 + gh = constant

(1/2) (vY2 - vX

2 ) = g(hX – hY)

Faster speed at point Y, hence hY 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

P.56

P.56