Water flows in a horizontal pipe of non-uniform area of cross-section at a pressure difference of 1.6 cm of mercury. If the velocity of water at the larger cross-section of pipe is 50 cm/s, find the velocity of water at the other end?

Question

Question

This question was previously asked in

AAI ATC Junior Executive 25 March 2021 Official Paper (Shift 3)

Answer 1

CONCEPT:

Applying Bernoulli's theorem,

\(P + \frac{1}{2}\rho v^2+ Z = C\)

CALCULATION:

Given: v₁ = 50 cm/s = 0.5 m/s

P₁ - P₂ = 0.016 × 13600 × 9.81 = 2134.65 Pa

Z₁ = Z₂ Horizontal pipe, v₂ = ?

\(\Delta P = \frac{1}{2}\rho(v_2^2 - v_1^2)\)

\(⇒ 2134.65 =\frac{1}{2}\times 1000 (v_2^2 - 0.5^2)\)

\(⇒ \frac{2134.65\times 2}{1000} + 0.5^2 = v_2^2\)

⇒ v₂ = \({\sqrt {4.51} }\)

Hence the correct option is option 3

Additional Information

\(\frac{P_1}{\rho g}+ z_1 + \frac{v_1^2}{2g}= \frac{P_2}{\rho g} +z_2 + \frac{v_2^2}{2g}\)

Assumptions of Bernoulli's theorem: