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Answer :
Final answer:
The speed at the second floor is 5.61 m/s. The water pressure is -1700 Pa
Explanation:
To determine the speed and water pressure at the second floor, we can use Bernoulli's equation. First, let's calculate the speed at the second floor. Since the pipe tapers and the water is incompressible, the continuity equation tells us that the volume flow rate is constant. Therefore, we can use the equation A1v1 = A2v2, where A1 and A2 are the cross-sectional areas of the pipe at the input point and second floor respectively, and v1 and v2 are the speeds at the input point and second floor respectively. Given that the inside diameter at the input point is 2.99 cm and the speed at the input point is 0.836 m/s, we can calculate the speed at the second floor:
A1 = (π/4)(2.99 cm)^2 = 7.04 cm²
A2 = (π/4)(1.16 cm)^2 = 1.05 cm²
v2 = (A1v1)/A2 = (7.04 cm²)(0.836 m/s)/(1.05 cm²) = 5.61 m/s
Next, let's calculate the water pressure at the second floor. Again, using Bernoulli's equation, we can relate the pressures at the input point and second floor to the speeds at those points. The equation is P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2, where P1 and P2 are the pressures at the input point and second floor respectively, ρ is the density of water (1000 kg/m³), and v1 and v2 are the speeds at the input point and second floor respectively. Given that the pressure at the input point is 181 kPa, we can calculate the pressure at the second floor:
P2 = P1 + (1/2)ρ(v1^2 - v2^2) = (181 kPa) + (1/2)(1000 kg/m³)((0.836 m/s)^2 - (5.61 m/s)^2) = -1700 Pa
Learn more about Bernoulli's equation here:
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