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A horizontal pipe of diameter 0.842 m has a smooth constriction to a section of diameter 0.5052 m . The density of oil flowing in the pipe is 821 kg/m3. If the pressure in the pipe is 7370 N/m2 and in the constricted section is 5527.5 N/m2, what is the rate at which oil is flowing?

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We can use Bernoulli's equation to solve this problem, which states that the total pressure at any point in a fluid flow system is the sum of the static pressure and the dynamic pressure. The equation can be written as:

P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2

where P is the pressure, ρ is the density, and v is the velocity of the fluid at two different points in the flow.

We can assume that the fluid is incompressible, so the mass flow rate (m_dot) is constant throughout the pipe. The mass flow rate is given by:

m_dot = ρA1v1 = ρA2v2

where A is the cross-sectional area of the pipe at two different points in the flow.

We can use the above equations to solve for the rate at which oil is flowing:

From Bernoulli's equation:

P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2

Substituting the given values:

7370 N/m2 + (1/2)821 kg/m3v1^2 = 5527.5 N/m2 + (1/2)821 kg/m3v2^2

From the continuity equation:

A1v1 = A2v2

Substituting the given values:

(π/4)(0.842 m)^2v1 = (π/4)(0.5052 m)^2v2

Simplifying, we get:

v2 = (0.842/0.5052)^2v1 = 2.628v1

Substituting v2 into Bernoulli's equation and simplifying, we get:

v1 = 6.08 m/s

Substituting v1 into the continuity equation and simplifying, we get:

m_dot = ρA1v1 = 177.4 kg/s

Therefore, the rate at which oil is flowing is 177.4 kg/s.

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