A fan causes a velocity field that falls off
inversely with the distance from the fan. For
one fan, the air speed at 1.3 m is 6.3 m/s. If
two of these fans are placed 3 m apart, pointed
at each other, what will be the air speed 1 m
away from one of the fans?
v = k/r^2
solve for k
Now use k and find v for r=1 and r=2.
The difference of those (remember v is a vector) will be your answer.
To find the air speed 1 meter away from one of the fans, we can use the principle of superposition. The velocity field created by two fans pointed at each other can be obtained by summing the individual velocity fields created by each fan at that point.
Given that the air speed at a distance of 1.3 meters from one fan is 6.3 m/s, we can use this information to determine the value of the velocity field created by a single fan at 1.3 meters.
Let's assume the velocity field created by a single fan at a distance of 1 meter is V1. We want to find this value.
According to the problem, we know that the air speed at a distance of 1.3 meters from the fan is 6.3 m/s. Therefore, we can set up the following equation:
V1 / 6.3 = (1.3 / 1)^(-1)
Let's solve this equation:
V1 / 6.3 = (1 / 1.3)
Cross-multiplying:
V1 = 6.3 / 1.3
V1 ≈ 4.846 m/s
Therefore, the velocity field created by a single fan at a distance of 1 meter is approximately 4.846 m/s.
Since the two fans are placed 3 meters apart and pointed at each other, we can consider them as two sources of the velocity field. So, at a distance of 1 meter from one of the fans, we need to consider the contribution of both fans.
Using the principle of superposition, we can sum the velocity fields created by each fan:
Velocity field at a distance of 1 meter from one fan = V1 + V1
Velocity field at a distance of 1 meter from one fan = 4.846 m/s + 4.846 m/s
Velocity field at a distance of 1 meter from one fan ≈ 9.692 m/s
Therefore, the air speed 1 meter away from one of the fans, when two of these fans are placed 3 meters apart and pointed at each other, will be approximately 9.692 m/s.