Answer:
(a) the flow speed of the second section is 11 m/s
(b) the pressure of the second section is 6.33 x 10⁴ Pa
Explanation:
Given;
flow rate in the first section, Q₁ = 2750 cm³/sec
area of the first cross section, A₁ = 10 cm²
pressure in the first cross section, P₁ = 1.2 x 10⁵ Pa
area of the second section, A₂ = 2.5 cm²
(a) the flow speed of the second section (V₂)
Apply continuity equation;
Q₁ = Q₂
Q₁ = A₂V₂
V₂ = Q₁ / A₂
V₂ = (2750) / (2.5)
V₂ = 1100 cm/s = 11 m/s
(b) the pressure of the second section (P₂)
Apply Bernoulli's equation;
P₁ + ¹/₂ρV₁² = P₂ + ¹/₂ρV₂²
where;
ρ is density of water = 1000 kg/m³
V₁ is the speed of water in the first section;
Q₁ = A₁V₁
V₁ = Q₁ / A₁
V₁ = (2750) / (10)
V₁ = 275 cm/s = 2.75 m/s
P₂ = P₁ + ¹/₂ρV₁² - ¹/₂ρV₂²
P₂ = P₁ + ¹/₂ρ(V₁² - V₂²)
P₂ = 1.2 x 10⁵ Pa + ¹/₂ x 1000 (2.75² - 11²)
P₂ = 1.2 x 10⁵ Pa + 500(-113.438)
P₂ = 1.2 x 10⁵ Pa - 0.567 x 10⁵ Pa
P₂ = 0.633 x 10⁵ Pa
P₂ = 6.33 x 10⁴ Pa
