Read the description of the pipe loss experiment, found in the file \"ME 440 Aer

Question

Read the description of the pipe loss experiment, found in the file "ME 440 Aerolab 7-Boundary Layer Plate" posted on BbLean in the "Lab Related" folder. The information in this file will direct you to read the lab description from the Acrolab manual, so plan on reading that too Succinctly outline the procedure for the experiment. AT LAB. Refer to the Aerolab manual for this experiment and you fluids book, answer the following questions and TURN THESE IN WHEN YOU ARRIVE AT LAB. TURN THISIN WHEN YOU ARRIVE o On the web, find the "NAU Weather Station" Record the barometric pressure and temperature from this website Express the barometric pressure in inches of Hg, mm of Hg, psi and kPa. Show your unit coversions. Compute the air density in kg/m using this information and the ideal gas law Show you work and carry your unit o Write down Bernoulli's equation as applied to a Pitot tube In the equation, identify the terms that correspond to the stagnation pressure, the static pressure and the dynamic pressure ' Simplify Bernoulli's equation to solve for the dynamic pressure " Solve Bemoulli's equation for the air speed measured by the Pitot-tube, assuming the stagnation and static pressure are known. o Assume our wind tunnel operates at a wind speed of 30 m/s. Assume the static pressure is the same as the Barometric pressure from the previous calculation. ' At this speed, what is the value of the dynamic pressure? Express you answer in Pa and inches of H2O What is the value of the stagnation pressure? Express you answer in Pa and inches of H:O The Aerolab manual is fairly terse. Please write at least one question that came to mind while reading the lab description that you would like to have clarified. o

Explanation / Answer

1) From NAU weather station website,

Barometric pressure = 23.52 in

Temperature = 56.6 0F = 286.8167 K

Barometric pressure = 23.52 inches of Hg

                                 = (23.52 x 25.4) mm of Hg = 597.408 mm of Hg

                                 = (23.52 x 0.491) psi = 11.552 psi

                                 = (23.52 x 3.386) kPa = 79.648 kPa

Air Density (in kg/m3) = P/RT

where P = Air pressure (in Pa)

R = Specific gas constant = 287.05 J/kgK

T = Temperature (in K)

So, Air density = 79647.861 / (287.05 x 286.8167) = 0.967 kg/m3

2) For Pitot tube, let 1 is the measurement point and 2 is the stagnation point.

Elevation heads at the two points are equal and velocity of fluid at point 2 is zero.

Applying Bernoulli’s equation between 1 and 2,

(P1/rho x g) + (v12/2g) + z1 = (P2/rho x g) + (v22/2g) + z2

=> (P1/rho x g) + (v12/2g) = (P2/rho x g)

First term in LHS is static pressure head and corresponds to static pressure.

Second term in LHS is velocity head and corresponds to dynamic pressure.

First term in RHS is stagnation pressure head and corresponds to stagnation pressure.

Now, (P1/rho x g) + (v12/2g) = (P2/rho x g)

=> P1 + (rho x v12/2) = P2 [Multiplying rho x g on both sides]

where P1 is static pressure

P2 is stagnation pressure

(1/2) rho v12 is dynamic pressure

P1 + (rho x v12/2) = P2

=> (1/2) rho v12 = P2 - P1

=> v12 = 2 x (P2 - P1) / rho

=> v1 = (2 x (P2 - P1) / rho)0.5

P1, P2 and rho are known

So, Air speed v1 measured by pitot tube = (2 x (P2 - P1) / rho)0.5

3) Wind speed = 30 m/s

Air density = 0.967 kg/m3

Dynamic pressure = 0.5 x Air density x (Wind speed)2 = 0.5 x 0.967 x 302

= 435.15 Pa

= (435.15 x 0.004) in of H2O = 1.7487 in of H2O

Static pressure = Barometric pressure = 79647.861 Pa

Stagnation pressure = Static pressure + Dynamic pressure

                                 = 79647.861 + 435.15 Pa

                                 = 80083.01 Pa

                                 = (80083.01 x 0.004) in of H2O = 321.8253 in of H2O

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