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Answer :
First, let's calculate the cross-section of the pipeline. The radius of the tube is half the diameter:
[tex]r= \frac{0.250 m}{2}=0.125 m [/tex]
And the cross-sectional area is:
[tex]A= \pi r^2 = \pi (0.125 m)^2 =0.049 m^2[/tex]
The volumetric flow rate is [tex]\rho = 1.55m^3 /s[/tex]. If we divide this value by the cross-sectional area of the pipeline, we get the flow speed of the gas:
[tex]v= \frac{\rho}{A}= \frac{1.55 m^3}{0.049 m^2}=31.6 m/s [/tex]
[tex]r= \frac{0.250 m}{2}=0.125 m [/tex]
And the cross-sectional area is:
[tex]A= \pi r^2 = \pi (0.125 m)^2 =0.049 m^2[/tex]
The volumetric flow rate is [tex]\rho = 1.55m^3 /s[/tex]. If we divide this value by the cross-sectional area of the pipeline, we get the flow speed of the gas:
[tex]v= \frac{\rho}{A}= \frac{1.55 m^3}{0.049 m^2}=31.6 m/s [/tex]
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Final answer:
The gas flow speed in a pipeline with a diameter of 0.250 m delivering 1.55 m³ of gas per second is approximately 31.56 m/s.
Explanation:
The question asks for the flow speed of the gas in a natural-gas pipeline with a specific diameter, which can be calculated through the continuity equation used in fluid dynamics. This equation states that the volume flow rate (Q) is equal to the velocity of the flow (v) times the cross-sectional area of the pipe (A). Therefore, we can rearrange the equation to find the velocity as V = Q/A.
The cross-sectional area A of the pipeline can be calculated using the formula for the area of a circle since the pipe is essentially a circular cylinder: A = πr², where r is the radius of the pipeline. Given the pipeline diameter of 0.250 m, the radius is half this value or 0.125 m. Substituting this into the area formula gives A = π*(0.125 m)² = 0.0491 m².
Finally, we can find the flow speed by dividing the given volume flow rate (1.55 m³/s) by the calculated cross-sectional area: V = Q/A = 1.55 m³/s/0.0491 m² = 31.56 m/s. So, the gas flow speed in the pipeline is approximately 31.56 meters per second.
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