Conditions/Assumptions for Bernoulli

• Five conditions• A)• B)• C)• D)• E)

Solution for Bernoulli Assumptions

A. Steady stateB. No frictionC. No workD. Same streamlinesE. 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

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 (). The liquid is incompressible. Solve for Vx of the fluid.

y

x

u𝛿

Solution for Moving Sheet Problem

Navier Stoke Solution Method

• 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

• 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

• 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

• 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