Pitot Stagnation Pitot Stagnation Tubes Under Tubes Under

Laminar Laminar ConditionsConditions

*P7.24*P7.24

Jace BenoitJace Benoit

February 15, 2007February 15, 2007

Problem StatementProblem Statement

Air at 20°C and 1 atm flows past Air at 20°C and 1 atm flows past the flat plate in the flat plate in Fig. P7.24 Fig. P7.24 under under laminar conditions. There are two laminar conditions. There are two equally spaced pitot stagnation equally spaced pitot stagnation tubes, each placed 2 mm from the tubes, each placed 2 mm from the wall. The manometer fluid is water wall. The manometer fluid is water at 20°C. If at 20°C. If UU=15m/s and =15m/s and LL=50cm, =50cm, determine the values of the determine the values of the monometer readings monometer readings hh11 and and hh22. . Determine these values if the Determine these values if the manometer fluid is replaced with manometer fluid is replaced with mercury.mercury.

SketchSketch

Fig P7.24Fig P7.24

u1 u2

AssumptionsAssumptions

Laminar Flat-Plate Flow Laminar Flat-Plate Flow Incompressible Flow Incompressible Flow Constant TemperatureConstant Temperature No Mixing at Liquid-Air InterfaceNo Mixing at Liquid-Air Interface Frictionless FlowFrictionless Flow Steady FlowSteady Flow

SolutionSolutionThe following gives a summary of the densities and kinematic velocity from tables A.1, A.2, and A.3.

AirAir WaterWater MercuryMercury

Density (Density (ρρ)) 1.20 kg/m1.20 kg/m33 998 kg/m998 kg/m33 13,550 13,550 kg/mkg/m33

Kinematic Kinematic Velocity (Velocity (vv))

1.50 E-5 1.50 E-5 m2/sm2/s

Not Not RequiredRequired

Not Not RequiredRequired

SolutionSolution

2/1

vx

Uy

828.2)5.0)(/1050.1(

/15002.0

2/1

251

msm

smm

Since this is laminar flow along a boundary layer, the Blasius equation (7.21) can be used to determine η.

000.2)0.1)(/1050.1(

/15002.0

2/1

252

msm

smm

SolutionSolution

Uu

84605.0

828.20.3

81152.084605.0

8.20.3

Now that η is known, Table 7.1 on page 462 can be used to approximate u/U. For η = 2.8, u/U = 0.81152 and for η = 3.0, u/U = 0.84605, so

for η1 = 2.828:

8164.01 U

u

Fortunately, the u/U value corresponding to η = 2.0 can be read directly from Table 7.1 to be 0.62997.

SolutionSolution

smsmu /25.12)8164.0()/15(1

Plugging U = 15 m/s into the u/U expression yields the velocities at the entrances of each manometer.

smsmu /447.9)6298.0()/15(2

SolutionSolutionWith the inlet velocities known, Bernoulli’s equation (3.77) can be used to solve for the inlet pressure assuming atmospheric pressure at the outlet of the manometer.

02

112

21

22

zzguu

P

SolutionSolution

2

2uPP outletinlet

Pasm

mkgPaPi 1012352

)/25.12()/20.1(101325

22

1,

Rearranging this equations with the appropriate cancellations yields the following:

Pasm

mkgPaPi 1012712

)/447.9()/20.1(101325

22

2,

SolutionSolution

owateri PghP

g

PPh

water

io

Now that the inlet and outlet pressures are known, this problem can now be solved as a manometer problem.

SolutionSolution

mmsmmkg

Pah 2.9

)/81.9)(/998(

)101235101325(231

mmsmmkg

Pah 5.5

)/81.9)(/998(

)101271101325(232

Finally, the manometer heights for water are as follows:

SolutionSolution

mmsmmkg

Pah 68.0

)/81.9)(/550,13(

)101235101325(231

All that needs to be done to determine the height if mercury is the manometer fluid is changed the density in the previous equation to 13,550 kg/m3.

mmsmmkg

Pah 41.0

)/81.9)(/550,13(

)101271101325(232

Application to BMEApplication to BMEThe manometer with one end being submerged in water and the other in another medium can mimic a catheter which is used to measure various biological effects such as cardiac output and blood pressure. The catheter is filled with a saline solution similar to how this manometer is filled with water and mercury, and pressure is exerted at each ends of the catheter.

ReferenceReference

White, Frank M. White, Frank M. Fluid MechanicsFluid Mechanics.. 5th 5th Ed. McGraw-Hill Companies, Inc.: Ed. McGraw-Hill Companies, Inc.: New York. 2003.New York. 2003.