Answer :
Answer:
The flow speed of the gas in the pipeline is 31.57 m/s.
Explanation:
Given;
diameter of the pipeline, d = 0.25 m
radius of the pipeline, r = d/2 = 0.125 m
volumetric flow rate of the gas, Q = 1.55 m³/s
The cross sectional area of the pipeline is given as;
A = πr²
A = π(0.125²)
A = 0.0491 m²
Volumetric flow rate is given by;
Q = Av
Where;
v is the speed of the fluid in the pipeline
v = Q / A
[tex]v = \frac{1.55 \ \frac{m^3}{s} }{0.0491 \ m^2} \\\\v = 31.57 \ m/s\\[/tex]
Therefore, the flow speed of the gas in the pipeline is 31.57 m/s.
Final answer:
The speed is approximately 31.57 m/s. This is calculated using the equation for flow rate (Q = AV), with the given flow rate of 1.55 cubic meters per second, and the pipe's cross-sectional area (calculated from its diameter).
Explanation:
The question is asking about the flow speed of the gas moving through the pipeline, also known as the velocity of the gas. To find this, we apply the principle of incompressibility of the fluid (in this case the gas), which implies that the flow rate is consistent throughout the pipeline. This is often represented by the equation Q = AV, where Q is the flow rate, A is the cross-sectional area of the pipe, and V is the velocity.
The cross-sectional area of the pipeline can be calculated using the formula A = π(r^2), where r is the radius of the pipe. In this case, the diameter is given as 0.250m, so the radius r will be 0.250/2 = 0.125m. Thus, the area A is π*(0.125)^2 = 0.0491 m².
The flow rate, Q, is given as 1.55 cubic meters per second. By substituting these values into the equation Q = AV, we get 1.55m³/s = 0.0491 m² * V. Isolate V by dividing both sides of the equation by 0.0491, we get V = 1.55m³/s / 0.0491 m² = 31.57 m/s. So, the flow speed of the gas in the pipeline is about 31.57 meters per second.
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