Air at 110 kPa and 50°C flows upward through a 6- cm-diameter inclined duct at a rate of 45 L/s. The duct diameter is then reduced to 4 cm through a reducer. The pressure change across the reducer is measured by a water manometer. The elevation difference between the two points on the pipe where the two arms of the manometer are attached is 0.20 m. Determine the differential height between the water levels of the two arms of the manometer.

Question

contestada

Respuesta :

Xema8 Xema8 · Answer 1 · 2019-10-12T10:00:23+02:00

Answer:

6.45 cm

Explanation:

Volume of air passing through duct per second

v = cross sectional area x velocity

45 x 1000 = 3.14 x 3 x 3 x v

v₁ = 15.92 m /s ,

As air is in-compressible

a₁ x v₁ = a₂ x v₂

v₂ = ( a₁ / a₂ )x 15.92

= (9 / 4) x 15.92

= 35.82 m /s

To find pressure difference at two points of a flowing water , we apply Bernauli''s theorem

P₁ + ρ gh₁ + 1/2 ρ v₁² = P₂ + ρ gh₂ + 1/2 ρ v₁²

P₁ + 1/2 ρ v₁² = P₂ + ρ g( h₂-h₁) + 1/2 ρ v₁²

110000+ .5 x 1.109 x 15.92² = P₂ + 1.109 x 9.8 x .2 + .5 x 1.109 x 35.82 x 35.82

110000+ 140.53 = P₂ + 2.17 + 771.46

P₂ = 109367

Pressure difference = 633 Pa

If h be the height in difference of water level in manometer

h x 1000 x 9.8 = 633

h = 6.45 cm .