Transcript Answer: 10 kg

Conditions/Assumptions for Bernoulli
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Five conditions
A)
B)
C)
D)
E)
Solution for Bernoulli Assumptions
A.
B.
C.
D.
E.
Steady state
No friction
No work
Same streamlines
Newtonian fluid
Boat Pitot Tube Problem
• The speed of a boat is measured by a Pitot
tube, as shown below. When traveling in
sea water, the tube measures a pressure of
2.5 lbf/in2. Given the density of sea water
is 64 lbm/in3. Derive the mathematical
relationship to show how the velocity of
the boat relates to “L”. What is the velocity
of the boat in this problem (ft/sec)?
L
d
Solution – Boat Pitot Tube
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𝑢12
2
𝑃1
+ 𝑔𝑧1 +
𝜌
• 𝑢1 =
• 𝑢1 =
2𝑔𝐿
𝑓𝑡
19
𝑠
=
𝑢32
2
+ 𝑔𝑧3 +
𝑃3
𝜌
Velocity of fluid on a moving sheet
• A polymer solid sheet is being drawn in a
horizontal direction at a constant velocity. The
sheet is coated with a liquid film. The liquid
film has a constant thickness (𝛿). The liquid is
incompressible. Solve for Vx of the fluid.
𝛿
y
x
u
Solution for Moving Sheet Problem
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𝜕𝜌
𝜕𝑡
+
𝜕(𝜌𝑉𝑥 )
𝜕𝑥
+
𝜕(𝜌𝑉𝑦 )
𝜕𝑦
𝜕(𝜌𝑉𝑧 )
+
𝜕𝑧
=0
• 𝐵𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 1: 𝑦 = 0, 𝑉𝑥 =
𝑢 𝑛𝑜 𝑠𝑙𝑖𝑝 𝑏𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛
• 𝐵𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 2: 𝑦 = 𝛿, 𝑉𝑥 = 0
• 𝑉𝑥 =
𝑦 2 𝜕𝑃
2𝜇 𝜕𝑥
−
𝛿 𝜕𝑃
2𝜇 𝜕𝑥
+
𝑢
𝛿
𝑦−𝑢
Navier Stoke Solution Method
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Make Good Assumptions
Write Down Navier Stokes and simplify
Integrate the equations
Find the boundary conditions
Easy Boundary Conditions
• The fluid at the surface whether it’s a pipe or
plate will move at the velocity of the surface.
No Slip Condition (usually 0 because surfaces
are generally stationary)
• The fluid in the middle of the pipe moves the
fastest thus dv=0 because it’s the maximum
Navier Stoke Assumptions
• Steady State
• Constant Density and Viscosity
• Entrance effects are neglected thus only 1D
motion
• No slip condition
Velocity Problem
• A tank containing acetone at 20°C has a drain
pipe connected vertically at the bottom. The
pipe is made of galvanized iron, 0.5 in NPS
Schedule 40 with length L= 2m. The acetone
depth in the tank is H= 5 m. Calculate the
velocity and flow rate of acetone at the exit.
Remember there is a pipe entrance as a
“fitting” in this problem
Solution
• Find Viscosity using eq 1.17 because of a non
standard temperature
• Assumptions
• P1=P2 thus deltaP=0
• Z1=5+2 and z2=0
• Standard energy equation simplifies to
• Gz1=u2^2/2+F
• F=2*ff*um^2*L/D+2*ff*um^2(25)
• 25 value comes from pipe fitting
Solution Cont
• Assume a Reynolds in order to calculate ff
using charts
• Find ff values to find velocity
• Use velocity for new Reynolds.
• Iterate
• Answer: 4.53m/s
Momentum Balance
Solution
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Law of Continuity Q1=Q2 thus A1=A2
Sum of Forces=Momentum
Y direction first to find the mass
Fy-W=change in momentum
W=98 so m=10
Answer: 10 kg
One more Momentum
Solutions
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Fx=-25.8kN
Fz=-3522N
Momentum Balance
Fx+PinAin+PoutAout*cos(45)=pAoutUout^2*cos45-pAinUin^2
Fz-Welbow-Wh20+PoutAout*sin45=pAoutUout^2*sin(45)
Laminar Problem
• The distribution of velocity, u, in metres/sec
with radius r in metres in a smooth bore tube
of 0.025 m bore follows the law, u = 2.5 - kr2.
Where k is a constant. The flow is laminar and
the velocity at the pipe surface is zero. The
fluid has a coefficient of viscosity of 0.00027
kg/m s. Determine (a) the rate of flow in m3/s
(b) the shearing force between the fluid and
the pipe wall per metre length of pipe.
Answers
• [6.14x10-4 m3/s, 8.49x10-3 N]
• Steps
• Find K through boundary equation at r=.0125
velocity is 0 solve for k
• Dq=v*da
• DQ=(2.5-16000r^2)*2pi*rdr from 0 to .0125
Shear Force Calc
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Newton’s law of viscosity
F=tau(2pir)
Tau=viscosity*du/dr
Du/dr=-32000r
F=-.00027*32000*.0125*2pir
F=8.48E-3 N