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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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
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