First slide

Fluid dynamics

Question

Air is blown through a pipe AB at a rate of 15 L/min. The cross-sectional area of the wide portion of the pipe AB is 2 cm² and that of the narrow portion is 0.5 cm². The difference in water level h is: $(ρ_{a i r} = 1.3 k g / m^{3})$

Moderate

A

16 mm
B

1.55 mm
C

10 mm
D

3.2 mm

Solution

$V = A_{1} V_{1} = A_{2} V_{2} V = A_{1} V_{1} \frac{15 \times 10^{- 3}}{60} = 2 \times 10^{- 4} V_{1} v_{1} = 1.25 m / s - - - - (1) V = A_{2} V_{2} \frac{15 \times 10^{- 3}}{60} = 0.5 \times 10^{- 4} V_{2} v_{2} = 5 m / s - - - - (2) P_{1} - P_{2} = \frac{1}{2} ρ_{a i r} ({v_{2}}^{2} - {v_{1}}^{2}) h (ρ_{w}) g = \frac{1}{2} \times 1.3 ({v_{2}}^{2} - {v_{1}}^{2}) h (998.7) 10 = \frac{1}{2} \times 1.3 (5^{2} - {1.25}^{2}) h = 1.55 m m$

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