Calculate Drag Force & Power for Long Cylindrical Antenna

AI Thread Summary
The discussion focuses on calculating the drag force and power required for a long cylindrical antenna mounted on a car moving at 100 km/h. The drag force was initially calculated using a drag coefficient (CD) of 1.14, resulting in a power requirement of 121.4 W, but adjustments to CD based on a more accurate reading brought the force down to 3.83 N and the power to 106 W. The standard air density used in calculations was confirmed to be 1.2 kg/m3. Participants noted the challenges of accurately determining CD and the effects of airflow around the car on the antenna's performance. Overall, the calculations were deemed acceptable, with a consensus that using a CD of 1 is reasonable for this scenario.
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Homework Statement


A long cylindrical antenna of 9 mm (0.009 m) diameter is mounted vertically on a car moving at 100 km/h (27.778 m/s). Antenna is 920 mm long (0.92 m). Calculate the power required to move antena through the air with standard kinematic viscositiy ν = 1.46 ⋅ 10-5 m2/s.

The given result is P = 104 W.


2. Homework Equations

F = 0.5 ⋅ ρ ⋅ v2 ⋅ CD ⋅ A

where:
F - drag force
ρ = 1.2 kg/m3 - density of the air
v = 27.778 m/s - speed
CD - drag coefficient
A - relevant area

P = F ⋅ v

where:
P - reguired power


3. The Attempt at a Solution


First, I tried to find drag coefficient. To do that, I must find Reynolds number:

Re = (v ⋅ d) / ν
Re = (27.778 ⋅ 0.009) / (1.46 ⋅ 10-5)
Re = 17123.4

According to the diagram, for infinitely long cylinders, CD ≈1

b010a828f330f2c43cc493c60e7c70b5.jpg


Also, relevant area is:
A = diameter ⋅ height
A = 0.009 ⋅ 0.92
A = 8.28 ⋅ 10-3 m2

Now, the drag force is:
F = 0.5 ⋅ 1.2 ⋅ 27.7782 ⋅ 1 ⋅ 8.28 ⋅ 10-3
F = 3.83 N

Required power is equal to:
P = F ⋅ v
P = 4.37 ⋅ 27.778
P = 106.4 W

Is this the correct way to calculate this? Am I getting 2 watts more just because of "not so accurate" reading from the diagram above?

Thanks.



 
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Why did the force change from 3.83 N to 4.37 N?
Where does the value for the density of air come from?
2% deviation is certainly fine if you take CD from the diagram.
 
Thanks for reply.

When I calculated the force for the first time, I read CD = 1.14 from the diagram and I got F = 4.37 N and P = 121.4 W, which was too much.
Then I realized that the CD might be a little lower, so in second reading from the diagram, I read CD = 1. Then I got F = 3.83 N and P = 106 W.
Since I typed that in a hurry, I missed to correct the value for the force in all equations. Sorry for that.

In the book where I found this example under the text it is said that the air has standard characteristics, so the density is 1.2 kg/m3.

So, this procedure is fine?

Is there any other way to calculate CD in this case, because the "reading from the diagram" method is not so accurate?

Thanks.
 
Last edited:
The assumption that the antenna is in a free air flow is not so accurate either. Usually car antennas are attached to cars, and those influence the air flow. Taking a value of 1 should be fine, you won't get a 1% accuracy anyway.
 
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