Water moves through a constricted pipe in steady, ideal flow. At the lower point shown in the figure below, the pressure is 1.65 ✕ 105 Pa and the pipe radius is 2.50 cm. At the higher point located at y = 2.50 m, the pressure is 1.22 ✕ 105 Pa and the pipe radius is 1.20 cm
(a) Find the speed of flow in the lower section.
(b) Find the speed of flow in the upper section.
(c) Find the volume flow rate through the pipe.
Bernoulli’s equation:
p1 + ½ρ(v1)² = p2 + ½ρ(v2)² + ρgy
(a) The flow rate at both ends must be the same, or
Q = v1*A1 = v1*π(r1)² = v2*π(r2)² = v2*A2, so
v2 = v1(r1/r2)². So
p1 + ½ρ(v1)² = p2 + ½ρ(v1)²(r1/r2)⁴ + ρgy
Plugging in values:
1.65e5Pa + ½(1000kg/m³)(v1)² = 1.22e5Pa + ½(1000kg/m³)(v1)²(2.5/1.2)⁴ + 1000kg/m³*9.8m/s²*2.5m
18500Pa = (500kg/m³)(v1)²((2.5/1.2)⁴ - 1) = 8919kg/m³ * (v1)²
(v1)² = 2.07 m²/s²
v1 = 1.44 m/s
(b) v2 = 1.44m/s * (2.5/1.2)² = 6.25 m/s
(c) Q = 6.25m/s * π(0.012m)² = 0.00283 m³/s
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