Question

water is flowing with a speed of 2m/s in a horizontal pipe with cross section are goes on decreasing from 0.02 sq.m to 0.01 sq.m at pressure 40000 pa. what will be the pressure at smaller cross section?

Answer 1

Given,
A₁ = 0.02 cm²
A₂ = 0.01 cm²
V₁ = 2m/s
V₂ = ?
Use equation of continuity ,
e.g., A₁V₁ = A₂V₂
0.02 × 2 = 0.01 × V₂
V₂ = 4 m/s

Now, apply Bernoulli's theorem,
$\bold{P +\frac{\rho v^2}{2}+\rho gh}=\bold{\text{constant}}$
Here, h = 0 so, formula will be
P₁ + ρV₁²/2 = P₂ + ρV₂²/2

Here, P₁ = 40000 Pa
V₁ = 2m/s , V₂ = 4m/s and ρ = 1000 kg/m³ [ density of water ]
now, 40000 + 1000×2²/2 = P₂ + 1000× 4²/2
42000 - 8000 = P₂
P₂ = 34000 Pa

Hence, pressure of small cross section area = 34000 Pa

Answer 2

Answer:

$3 .4\times \ {10 }^{4}$ pascal