An open-circuit wind tunnel draws in air from the atmosphere through a well-contoured nozzle. In the test section, where the flow is straight and nearly uniform, a static pressure tap is drilled into the tunnel wall. A manometer connected to the tap shows that static pressure within the tunnel is 45 mm of water below atmospheric. Assume that the air is incompressible, and at 25 C, 100 kPa absolute. Calculate the air speed in the wind-tunnel test section

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batolisis batolisis · Answer 1 · 2020-12-03T14:56:54+01:00

Answer:

Air speed in the wind-tunnel [tex]v_{2}[/tex] = 27.5 m/s

Explanation:

Given data :

Manometer reading ; p1 - p2 = 45 mm of water

Pressure at section ( I ) p1 = 100 kPa ( abs )

temperature ( T1 ) = 25°C

Pw ( density of water ) = 999 kg/m3

g = 9.81 m/s^2

next we apply Bernoulli equation at section 1 and section 2

p1 - p2 = [tex]\frac{PairV^{2} _{2} }{2}[/tex] ---------- ( 1 )

considering ideal gas equation

Pair ( density of air ) = [tex]\frac{P}{RT}[/tex] ------------------- ( 2 )

R ( constant ) = 287 NM/kg.k

T = 25 + 273.15 = 298.15 k

P1 = 100 kN/m^2 = 100 * 10^3 or N/m^2

substitute values into equation ( 2 )

= 100 * 10^3 / (287 * 298.15)

= 1.17 kg/m^3

Also note ; p1 - p2 = PwgΔh ------- ( 3 )

finally calculate the Air speed in the wind-tunnel test section by equating equation ( 1 ) and ( 3 )

[tex]\frac{PairV^{2} _{2} }{2}[/tex] = PwgΔh

[tex]V^{2} _{2}[/tex] = [tex]\frac{2*999* 9.81* 0.045}{1.17}[/tex] = 753.86

[tex]v_{2}[/tex] = 27.5 m/s