Understanding Drag Forces and Coefficients: Thermofluids Lab Help

[Edward Trail]

<< Mentor Note: Thread moved from the technical forums, so no Template is used >>

hey guys

As the title suggests i need a little help with my results from my thermofluids lab, i am trying to come to a conclusion about two of my graphs which are attached.

The lab is looking at measuring the drag forces involved with a model car using pitot tubes to measure the pressure drops.

my understanding is that because Fd(total drag force) is proportional to V^2 then the drag coefficient is not affected due to the equation cd= Fd/ 1/2 ρ v^2 A . as the fd and v^2 cancel each other out? am i correct in this ? am i right in say " the graph shows that an increase in the velocity does not affect the drag coefficient" ? and the same with Reynolds number?

If i am completely wrong let me know

 

[collinsmark]

<< Mentor Note: Thread moved from the technical forums, so no Template is used >>

hey guys

As the title suggests i need a little help with my results from my thermofluids lab, i am trying to come to a conclusion about two of my graphs which are attached.

The lab is looking at measuring the drag forces involved with a model car using pitot tubes to measure the pressure drops.

my understanding is that because Fd(total drag force) is proportional to V^2 then the drag coefficient is not affected due to the equation cd= Fd/ 1/2 ρ v^2 A . as the fd and v^2 cancel each other out? am i correct in this ? am i right in say " the graph shows that an increase in the velocity does not affect the drag coefficient" ? and the same with Reynolds number?

If i am completely wrong let me know

That sounds about right to me.

Be careful how you word it though. It's not that $F_d$ and $V^2$ "cancel each other out." Instead, I would phrase it as a the measured aerodynamic forces and $\frac{1}{2} \rho V^2 A$ are both proportional to the the velocity squared, and thus both proportional to each other.

Given that, what can you say about the flow in terms of it being laminar flow vs. turbulent flow?

By the way, in one of your excel graphs, you have velocity, U, having units of "m/s^2." Those are units of acceleration, not velocity.

 

[Edward Trail]

hi mark

Thank you for the response, I am still a little unsure to be honest. is it not just fd and v^2 that are proportional therefore the velocity has no affect on drag coefficient? and the same with Reynolds as my understanding is because Reynolds is a function of velocity and we know velocity does not affect the coefficient of drag neither does the Reynolds number is this correct?

im being asked to explain each graph what why and how, I am struggling with the why and how.

In regards to the flow it seams to be very turbulent due to the Reynolds number being greater than 4000.

Ahhh thanks for that, not sure why I put that to be honest.

 

[collinsmark]

hi mark

Thank you for the response, I am still a little unsure to be honest. is it not just fd and v^2 that are proportional therefore the velocity has no affect on drag coefficient? and the same with Reynolds as my understanding is because Reynolds is a function of velocity and we know velocity does not affect the coefficient of drag neither does the Reynolds number is this correct?

im being asked to explain each graph what why and how, I am struggling with the why and how.

In regards to the flow it seams to be very turbulent due to the Reynolds number being greater than 4000.

Ahhh thanks for that, not sure why I put that to be honest.

Yes, your wording looks correct to me.

Assuming turbulent flow, the drag coefficient is the ratio of $F_d$ and $\frac{1}{2} \rho V^2 A.$ But since both $F_d$ and $\frac{1}{2} \rho V^2 A$ both scale (approximately*) with $V^2$, the ratio -- and thus the drag coefficient -- remain (approximately*) constant, and not a function of velocity.

And, as you said (and as you put it in your own words), the same goes for drag coefficient vs Reynolds number.

*(For reasons that are probably outside the scope of your coursework, the drag coefficient can vary slightly with velocity due to different "boundary layer" properties [for example, if the model car went supersonic]. But but your data does not have extreme enough data points, such that this matters. So you can just treat the drag coefficient as a constant over the range of velocities which you measured.)