Answer:
From the Bernoulli energy principle,
ΔP + 1/2ρΔv² = 0 -------------------------- eqn 1
where
ΔP = pressure drop = P2 - P1 = (1 - 0.25)x10⁵ N/m =7.5 x 10⁴N/m
Δv²= velocity change = v₂² - v₁²
ρ = water density = 1kg/m3
Recall volumetric flow rate, Q=A v = constant
A = cross sectional area = πr²=πd²/4
d=pipe diameter at point 2 = 2.5cm = 0.025m and Q =0.20m³/min = 0.00333m³/s
So A= 0.000491m²
we can get v2 = Q/A = 6.79m/s
From eqn 1, v₁² = 2(P2 - P1)/ρ + v₂²
v₁² = (2 x 7.5 x 10⁴)/1000 + 6.79²
v₁² = 196
v₁ = 14m/s
we can now get the area of the constriction point 2, A₁ = Q/v₁
A₁ = 0.000238m² and the diameter now will be d₁
d₁² = 4 x A₁ / π
= 4 x 0.000238/3.14 = 0.000303m²
d₁ = √0.000303 = 0.0174m
Therefore, the diameter of a constriction in the pipe at the new pressure = 0.0174m = 1.74cm
