Water moves through a constricted pipe in steady, ideal flow. At the lower point

Question

Water moves through a constricted pipe in steady, ideal flow. At the lower point shown in the figure below, the pressure is P_1 = 1.80 times 10^4 Pa, and the pipe diameter is 5.0 cm. At another point y = 0.40 m higher, the pressure is P_2 = 1.25 times 10^4 Pa and the pipe diameter is 2.50 cm. (a) Find the speed of flow in the lower section. m/s (b) Find the speed of flow in the upper section. Your response differs from the correct answer by more than 10%. Double check your calculations. m/s (c) Find the volume flow rate through the pipe. m^3/s

Explanation / Answer

Bernoulli:
p1 + ½(v1)² = p2 + ½(v2)² + gy

(a) The flow rate at both ends must be the same, or
Q = v1*A1 = v1*(r1)² = v2*(r2)² = v2*A2, so
v2 = v1(r1/r2)². So
p1 + ½(v1)² = p2 + ½(v1)²(r1/r2) + gy
Plug in values:
1.8e4Pa + ½(1000kg/m³)(v1)² = 1.25e4Pa + ½(1000kg/m³)(v1)²(2.5/1.25) + 1000kg/m³*9.81m/s²*0.4m

1576Pa = (500kg/m³)(v1)²((2.5/1.25) - 1) = 7500kg/m³ * (v1)²
(v1)² = 0.2101333 m²/s²
v1 = 0.4584 m/s

(b) v2 = 0.4584m/s * (2.5/1.25)² = 1.8336 m/s

(c) Q = 1.8336m/s * (0.0125m)² = 0.0009 m³/s
Check: Q = 0.4584m³/s * (0.025m)² = 0.0009 m³/s

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