A horizontal pipe (Venturi Tube) 10.0 cm in diameter has a smooth reduction to a pipe 5.00 cm in diameter. If the pressure of the water in the larger pipe is 8.0 x 105 Pa and the pressure in the smaller pipe is 3.0 x 104 Pa, at what rate does water flow through the pipes. Hint: Use the continuity equation to express one speed in terms of the another.
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To determine the rate at which water flows through the pipes, we can use the principle of continuity equation, which states that the mass flow rate of a fluid is constant at any given point in an incompressible flow.
The continuity equation can be expressed as:
A1v1 = A2v2
Where A1 and A2 are the cross-sectional areas of the two pipes, and v1 and v2 are the velocities of the water in the larger and smaller pipes, respectively.
Given that the larger pipe has a diameter of 10.0 cm (radius of 5.0 cm) and the smaller pipe has a diameter of 5.00 cm (radius of 2.50 cm), we can calculate the cross-sectional areas:
A1 = πr1^2 = π(0.05 m)^2
A2 = πr2^2 = π(0.025 m)^2
We are also given the pressure in the larger pipe (P1) as 8.0 x 10^5 Pa and the pressure in the smaller pipe (P2) as 3.0 x 10^4 Pa.
To find the velocities in the pipes, we can use Bernoulli's equation:
P1 + 0.5ρv1^2 = P2 + 0.5ρv2^2
Where ρ is the density of water, which is approximately 1000 kg/m^3 for water at room temperature.
Now, let's solve for v1:
P1 + 0.5ρv1^2 = P2 + 0.5ρv2^2
(8.0 x 10^5 Pa) + (0.5 x 1000 kg/m^3 x v1^2) = (3.0 x 10^4 Pa) + (0.5 x 1000 kg/m^3 x v2^2)
Simplifying the equation, we have:
(v1^2 - v2^2) = (2(P2 - P1))/ρ
v1^2 - v2^2 = (2(3.0 x 10^4 Pa - 8.0 x 10^5 Pa))/(1000 kg/m^3)
v1^2 - v2^2 = -770 m^2/s^2
Now, let's use the continuity equation to express v1 in terms of v2:
A1v1 = A2v2
(π(0.05 m)^2)v1 = (π(0.025 m)^2)v2
v1 = (r2^2/r1^2)v2
Substituting the values:
v1 = (0.025 m)^2/(0.05 m)^2) x v2
v1 = 0.25 x v2
Now, plug this into the earlier equation:
(0.25v2)^2 - v2^2 = -770 m^2/s^2
0.0625v2^2 - v2^2 = -770 m^2/s^2
-0.9375v2^2 = -770 m^2/s^2
v2^2 = (770 m^2/s^2)/0.9375
v2 = sqrt((770 m^2/s^2)/0.9375)
v2 ≈ 26.9 m/s
Finally, we can find the rate at which water flows through the pipes by multiplying the velocity v2 by the cross-sectional area A2:
Rate of flow = A2 x v2
Rate of flow = (π(0.025 m)^2) x (26.9 m/s)