How Do You Calculate Air Flow Through a Loop Defined by Specific Points?

[joker_900]

Homework Statement

Air is flowing with a speed of 0.4m/s in the direction of the vector (-1, -1, 1). Calculate the volume of air flowing per second through the loop which consists of straight lines joining, in turn, the following (1,1,0), (1,0,0), (0,0,0), (0,1,1), (1,1,1) and (1,1,0).

Homework Equations

Possibly flux = integral A.n dS if projection is required?

The Attempt at a Solution

I've tried doing this by dot producting he velocity vector with the surface area vector of three planes, a square of area 1 in the xy plane, the same in the xz plane and a triangle in the zy plane of area 0.5. However I don't think this is right. Please help!

 

[anathema]

Hello,

To calculate the volume of air flowing per second through the loop, we can use the formula:

Volume = Area x Velocity x Time

First, we need to calculate the area of the loop. We can do this by finding the area of each face and adding them together.

The first face has an area of 1 (1x1), the second face has an area of 1 (1x1), the third face has an area of 0.5 (1x0.5), and the fourth face has an area of 0.5 (1x0.5). The total area of the loop is 3 square meters.

Next, we need to calculate the velocity of the air flow in the direction of the loop. Since the air is flowing in the direction of the vector (-1, -1, 1), we can use the Pythagorean theorem to find the magnitude of this vector:

|v| = √((-1)^2 + (-1)^2 + 1^2) = √3

Therefore, the velocity of the air flow in the direction of the loop is √3 m/s.

Lastly, we need to determine the time interval. Since we are calculating the volume per second, the time interval is 1 second.

Now, we can plug in our values into the formula:

Volume = 3 x √3 x 1 = 3√3 cubic meters per second.

Therefore, the volume of air flowing per second through the loop is approximately 5.2 cubic meters per second.

I hope this helps! Let me know if you have any further questions.

 

[Moetasim]

I would approach this problem by first understanding the physical principles involved. In this case, we are dealing with the flow of air, which can be described by fluid dynamics. The given vector (-1, -1, 1) represents the direction and speed of the air flow.

To calculate the volume of air flowing per second, we can use the equation Q = Av, where Q is the volumetric flow rate, A is the cross-sectional area, and v is the velocity. In this case, we are given the velocity, but we need to determine the cross-sectional area.

To do this, we can construct a loop using the given points and calculate the area of the loop. This will give us the cross-sectional area through which the air is flowing. Once we have the cross-sectional area, we can plug it into the equation Q = Av to calculate the volume of air flowing per second.

To calculate the area of the loop, we can use the formula for the area of a polygon. In this case, the loop consists of a square and a triangle. We can calculate the area of each shape and then add them together to get the total area of the loop.

Once we have the total area, we can plug it into the equation Q = Av and solve for Q. This will give us the volume of air flowing per second through the given loop.

In summary, to solve this problem, we need to use the principles of fluid dynamics and the formula for the area of a polygon. By understanding the physical principles involved and using the appropriate equations, we can accurately calculate the volume of air flowing per second through the given loop.