Water flows at 0.860m/s from a hot water heater, through a 450-kPa pressure regulator. The pressure in the pipe supplying an upstairs bathtub 3.70 m above the heater is 414-kPa.
What's the flow speed in this pipe?
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To find the flow speed in the pipe, we can make use of the principle of conservation of energy. The energy at the inlet of the pipe will be equal to the energy at the outlet of the pipe. We can calculate the flow speed using the Bernoulli's equation:
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
Where:
P1 and P2 are the pressures at the inlet and the outlet respectively,
ρ is the density of the fluid (water in this case),
v1 and v2 are the flow speeds at the inlet and the outlet respectively,
g is the acceleration due to gravity,
h1 and h2 are the heights at the inlet and the outlet respectively.
In this problem, we are given:
v1 = 0.860 m/s (flow speed at the inlet),
P1 = 450 kPa (pressure at the inlet),
P2 = 414 kPa (pressure at the outlet),
h1 = 0 (height at the inlet, since it is not specified),
h2 = 3.70 m (height at the outlet).
Since the height at the inlet is not specified, we can assume that it is at the same level as the hot water heater, which is usually located on the ground floor. Therefore, h1 = 0.
Substituting the given values into the equation, we have:
450 kPa + (1/2)ρ(0.860 m/s)^2 + ρgh1 = 414 kPa + (1/2)ρv2^2 + ρgh2
Since h1 = 0, the term ρgh1 cancels on both sides, simplifying the equation to:
450 kPa + (1/2)ρ(0.860 m/s)^2 = 414 kPa + (1/2)ρv2^2 + ρgh2
Simplifying further, we have:
450 kPa + (1/2)ρ(0.860 m/s)^2 = 414 kPa + (1/2)ρv2^2 + ρgh2
Rearranging the terms to isolate v2, we get:
v2^2 = ((450 kPa - 414 kPa) / (1/2)ρ) + (1/2)(0.860 m/s)^2 + 2gh
Substituting the values of the known variables, ρ (density of water is approximately 1000 kg/m^3), and g (acceleration due to gravity is approximately 9.8 m/s^2), we can calculate the flow speed in the pipe.