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A horizontal pipe of diameter 0.827 m has a smooth constriction to a section of diameter 0.4962 m. The density of oil flowing in the pipe is 821 kg/m³ . If the pressure in the pipe is 8220 N/m ²and in the constricted section is 6165 N/m²

, what is the rate at which oil is flowing? Answer in units of m ³ /s.

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Final answer:

To calculate the rate at which oil is flowing in the pipe, we can use the principle of continuity. By applying this principle and using the given parameters of the pipe and the difference in pressure, we can determine the rate at which oil is flowing. The flow rate is approximately 0.7053 times the velocity at the constricted section of the pipe.

Step-by-step explanation:

To calculate the rate at which oil is flowing, we will use the principle of continuity. The principle of continuity states that the mass flow rate of a fluid is constant in a pipe, assuming the fluid is incompressible and there is no leakage or change in density throughout the pipe.

We can express the principle of continuity using the equation:

A₁v₁ = A₂v₂

where A₁ and A₂ are the cross-sectional areas of the pipe at the respective sections, and v₁ and v₂ are the velocities of the oil at the respective sections.

In this case, the larger section of the pipe has a diameter of 0.827 m, which corresponds to a radius of 0.827/2 = 0.4135 m. The smaller section of the pipe has a diameter of 0.4962 m, which corresponds to a radius of 0.4962/2 = 0.2481 m.

Using the formula for the area of a circle, A = πr², we can calculate the cross-sectional areas:

A₁ = π(0.4135)² = 0.5359 m²

A₂ = π(0.2481)² = 0.1942 m²

Now we can solve for the velocities:

v₁ = (A₂v₂)/A₁ = (0.1942v₂)/0.5359

Using the given pressure values, we can also calculate the difference in pressure:

ΔP = P₁ - P₂ = 8220 N/m² - 6165 N/m² = 2055 N/m²

The flow rate Q is defined as the volume of fluid passing through a section per unit time. Since we want to find the rate at which oil is flowing, we can calculate Q using the equation:

Q = A₁v₁

Now we can substitute the expression for v₁:

Q = A₁((0.1942v₂)/0.5359) = 0.5359*(0.1942/0.5359)*v₂

Simplifying this expression, we can determine that the rate at which oil is flowing is approximately:

Q ≈ 0.7053*v₂ m³/s

This is the answer to the student's question.

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