A fluid of density 820 kg/m³ has a mass flow of 10 kg/s. What diameter of pipe will give the flow a velocity of 2 m/s? (1000 pts, answer in cm)?
The correct answer and explanation is:
Given:
- Fluid density, ρ=820 kg/m3\rho = 820 \, \text{kg/m}^3 (though this is not needed directly here)
- Mass flow rate, m˙=10 kg/s\dot{m} = 10 \, \text{kg/s}
- Desired velocity, v=2 m/sv = 2 \, \text{m/s}
We need to find the diameter dd of the pipe such that the fluid flows at velocity vv.
Step 1: Relate mass flow rate to velocity and cross-sectional area
The mass flow rate m˙\dot{m} relates to velocity vv, density ρ\rho, and cross-sectional area AA by the equation: m˙=ρAv\dot{m} = \rho A v
Solving for AA: A=m˙ρvA = \frac{\dot{m}}{\rho v}
Step 2: Calculate cross-sectional area AA
A=10 kg/s820 kg/m3×2 m/s=101640=0.00609756 m2A = \frac{10 \, \text{kg/s}}{820 \, \text{kg/m}^3 \times 2 \, \text{m/s}} = \frac{10}{1640} = 0.00609756 \, \text{m}^2
Step 3: Find diameter from area
The cross-sectional area of a circular pipe is: A=πd24A = \frac{\pi d^2}{4}
Solve for dd: d=4Aπ=4×0.006097563.1416=0.007761=0.0881 md = \sqrt{\frac{4A}{\pi}} = \sqrt{\frac{4 \times 0.00609756}{3.1416}} = \sqrt{0.007761} = 0.0881 \, \text{m}
Step 4: Convert to centimeters
d=0.0881 m×100=8.81 cmd = 0.0881 \, \text{m} \times 100 = 8.81 \, \text{cm}
Final answer:
The pipe diameter should be approximately 8.81 cm to achieve the velocity of 2 m/s with the given flow.
Explanation:
The problem involves a fluid moving through a pipe, with known density and mass flow rate. The goal is to find the pipe diameter so that the velocity of the fluid is exactly 2 m/s.
Mass flow rate is the mass of fluid passing through the pipe per unit time. This depends on the fluid density, the velocity, and the pipe’s cross-sectional area. First, by dividing the mass flow rate by the product of density and velocity, the pipe’s cross-sectional area is determined. The cross-sectional area is then used to calculate the diameter since the pipe has a circular cross-section.
It is important to note that density plays a role in converting between volumetric flow rate and mass flow rate, which is why it is necessary to include it in the equation. Velocity relates to how fast the fluid moves through the pipe, and the diameter controls the area through which the fluid flows. By adjusting the diameter, the velocity can be controlled for a given mass flow.
This approach assumes steady, incompressible flow and neglects friction or pressure losses. The solution gives the pipe size needed to maintain the specified velocity with the given fluid and flow rate conditions.