A wind tunnel is a large apparatus used for scale model testing of various objec

Question

A wind tunnel is a large apparatus used for scale model testing of various objects through a fluid flow field. The largest in the world is the National Full-Scale Aerodynamics Complex at NASA Ames Research Center in Moffitt Field, California. Suppose you are testing a scale model in an open-circuit tunnel at NASA Ames. Atmospheric pressure is 98.7 kPa (abs) and the temperature is 27 degrees Celsius. Determine the pressure at the stagnation point on the nose of the airplane. Why is this answer significant? Determine the manometer reading, h, for the manometer attached to the static pressure tap within the test section of the wind tunnel if the air velocity within the test section is 50 m/s

Explanation / Answer

Part 1)

a wind tunnel is an application of the Bernoulli principle, the expression for the conservation of energy in fluid

P + ½ v2+ g y = cte

We will determine two points'll call entry point 1 and point smaller diameter will call 2

P1 + ½ v12+ g y1 = P2 + ½ v22+ g y2

we can assume that the system is horizontally so y1 =y2

P1 + ½ v12= P2 + ½ v22

P2 = P1 + ½ (v12- v22) (1)

inlet pressure

P1 = 98.7 Kpa = 98.7 103 Pa

we must know the air speed, for this we use the continuity equation

Q = A1 v1 = A2 v2

where Q is called flow

A1 and A2 are known areas because construction diameters are known

A1= R2 = D2 /4

A1/A2 = (D1/D2)2

v1/v2 = A2/A1 = (D2/D1)2

I'm not quite sure, but when analyzing your graphic believe that the flow is a known value

substituting 1

P2 = P1 + ½ (v12- (v1 (d1/d2)2)2)

P2 = P1 + ½ v12 (1 – (d1/d2)4 )

with this expression we can calculate the pressure at point 2, if the diameters of the tubes and the speed of air inlet or outlet is known, I think this speed value is not known

Let's write this equation in terms of inflow

Q = A1 v1

Q = ( d12 /4) v1

v1 = 4 Q / ( d12 )

P2 = P1 + ½ (4 Q / ( d12 ) )2 (1 – (d1/d2)4 )

P2 = P1 + 8/ Q2/d14 ( d24 – d14 )/d24

evaluate

d14 ( d24 – d14 )/d24 = 1/d14 – 1/d24

replace it in the above expression

P2 = P1 + 8/ Q2(1/d14– 1/d24)

This expression allows us to calculate the pressure at the point 2

P1 = 98.7 103 Pa

Part 2)

Consider the pressure in the venturi tube

Pa = g h + P2

P2 = Pa - g h (2)

Pa = 98.7 103Pa

= 1000 Kg/m³

They give us v2, so change slightly the expression obtained from Bernoulli's equation 1

P2 = P1 + ½ (v12- v22)

v1 = (A2/A1) v2

P2 = P1 + ½ v22( (d2/d1)2-1)

= 1.2 Kg / m³at T =20C

v2 = 50 m/s

P1 = 98.7 103 Pa

to know the pressure we must know the diameters of the two sections, unfortunately I do not see that this is a fact to complete the calculation

To make an explicit calculation I will assume that d2 d1 = 5, if you have another value you have to just change

d2/d1 = 1/5

P2 = P1 + ½ v22( (d2/d1)2-1)

P2 = 98.7 103 + ½ 1.2 502 (1/52 -1)

P2 = 98.7 103 + 1.44 103

P2 = 97.26 103 Pa

calculate of 2

P2 = Pa - g h

h = (Pa-P2)/ g

h = ( 98.7 – 97.26) 103 / (1.2 9.8)

h = 0.1224 m = 12.24 cm

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