- #211
erobz
Homework Helper
Gold Member
- 3,658
- 1,538
This is the problem fully worked. as well as some other surrounding questions.
But it is equal and opposite to the drag on the vane, as usually defined in aerodynamics: component of the aerodynamic force parallel to the incoming relative flow.erobz said:##F_{nc}## is not a drag force in the traditional sense.
Doesn't matter. Look at the definition: drag is the component of the aerodynamic force parallel to the incoming relative flow. There is nothing in there about inviscid vs viscid.erobz said:There is no drag on the vane. The flow is inviscid.
You seem to be arguing about nothing to avoid getting to anything of substance? You are setting up strawman arguments to knock down. It's really annoying. Can we be grown ups now?A.T. said:Doesn't matter. Look at the definition: drag is the component of the aerodynamic force parallel to the incoming relative flow. There is nothing in there about inviscid vs viscid.
I'm clarifying the terms so we can communicate.erobz said:You seem to be arguing about nothing to avoid getting to anything of substance?
There is absolutely zero need to talk about drag forces and lift forces here. Force ##F_x## and ##F_y## are more than sufficient.A.T. said:I'm clarifying the terms so we can communicate.
If we assume that in the frame of the cart: outflow speed = inflow speed = ##v_j -v_c## (idealized case), then you can transform the vectors into the ground frame:erobz said:@A.T. What are the incoming/outgoing flow velocities w.r.t a frame fixed to the ground.
Something isn't sitting right with me.A.T. said:If we assume that in the frame of the cart: outflow speed = inflow speed = ##v_j -v_c## (idealized case), then you can transform the vectors into the ground frame:
$$Vin_{ground} = \begin{pmatrix} v_j \\ 0 \end{pmatrix}$$
$$Vout_{ground} = \begin{pmatrix} cos \theta (v_j - v_c) - v_c \\ sin \theta (v_j - v_c) \end{pmatrix}$$
erobz said:In an attempt to find out when exactly opinions diverge:
Lets say the cart below is moving at a constant velocity ##v_c## with application of ##F_{nc}##. With respect to an inertial frame fixed to the ground the jet has velocity ##v_j##
Assumptions:
Constant flow properties across the jet
Steady Flow
Inviscid Flow
The diagram is indicating the rate of momentum entering\exiting the cart w.r.t the inertial frame per the assumptions.
View attachment 322941
What is the force ##F_{nc}## acting on the cart?
That's clear. You have mentioned all in a citation above. It is a steady inviscid flow. Flow is transferred momentum into control volume. That's all.erobz said:The blue arrows aren't forces. That’s partially my fault. I have hybridized what is traditionally two separate diagrams ( momentum diagram, and force diagram) for brevity without stating that.
Well, I would like to get this ironed out in the straight forward case first. Do you have any input on what I'm asking in post #220?Gleb1964 said:But I hope we will come to the relevant configuration.
I think, if it should be ##+v_c## for x component for back transformation.A.T. said:If we assume that in the frame of the cart: outflow speed = inflow speed = ##v_j -v_c## (idealized case), then you can transform the vectors into the ground frame:
$$Vin_{ground} = \begin{pmatrix} v_j \\ 0 \end{pmatrix}$$
$$Vout_{ground} = \begin{pmatrix} cos \theta (v_j - v_c) - v_c \\ sin \theta (v_j - v_c) \end{pmatrix}$$
I like that:Gleb1964 said:I think, if it should be ##+v_c## for x component for back transformation.
Yes, fixed it.Gleb1964 said:I think, if it should be ##+v_c## for x component for back transformation.
Note that in the ground frame the air slows down and loses kinetic energy. So if for example ##F_{nc}## is a frictional force, that results in heat dissipation, this is where the energy for it comes from.erobz said:I like that:
$$\vec{v_{o/O} } = \vec{v_{j/c}}+\vec{v_{c/O}}$$
We’ll, I just need to be careful to have an inertial frame of reference for the governing equations to be valid. The cart frame is the obvious choice when the control volume is not accelerating, but soon I hope it will be and I needed to be careful about referencing all momentum correctly from the fixed ground frame. I got this wrong in some of those earlier analysis and hope to correct that moving forward.A.T. said:Note that in the ground frame the air slows down and loses kinetic energy. So if for example ##F_{nc}## is a frictional force, that results in heat dissipation, this is where the energy for it comes from.
But what about the cart frame, where the air doesn't slow down, but the same heat is generated? Here the ground is moving and can provide energy. The reaction frictional force ##-F_{nc}## by the cart on the ground is doing negative work on the ground in the cart frame.
Another interesting inertial frame is the rest frame of the incoming air, where the air initially doesn't have any kinetic energy, but ends up with some. Here again the energy comes from the ground and goes into heat and air's KE.
You could keep the cart frame inertial, and show that the force ##F_{nc}## needed to keep the cart at constant speed points backwards. This proves that acceleration is possible.erobz said:The cart frame is the obvious choice when the control volume is not accelerating, but soon I hope it will be and I needed to be careful about referencing all momentum correctly from the fixed ground frame.
I was thinking about this approach too. I need to show that the magnitude of the force ##F_{nc}## is non-zero at ##v_x = w##.A.T. said:You could keep the cart frame inertial, and show that the force ##F_{nc}## needed to keep the cart at constant speed points backwards. This proves that acceleration is possible.
Gleb1964 said:I think we are talking that projection of force ##F_{nc}## on a speed vector ##v## is non-zero.
https://www.physicsforums.com/attachments/322992
What you are saying is that you don't agree with what the analysis is going to show. That was the whole point of agreeing on the analysis in the video! This is a complete waste of my time!Gleb1964 said:We are at the frame moving with speed ##v_x=w##, no wind.
Because ##v_x=w##, ##w-v_x = 0##
But the boat speed ##v## has ##v_x## and ##v_y## components. Because is ##v_y## component, the "control volume" is moving up with ##v_y = v sin( \theta )## smashing the air flow upwards and back.
What you are proposing is not parallel to the agreed upon analysis technique shown in the video? Suddenly you are telling me the jet will rebound off of the vane.Gleb1964 said:I just showed my analysis. Have you seen I have mention that agree or disagree with something?
The cv is the reference frame. All the same assumptions about flow characteristics apply. We can do this so long as the cv in not accelerating, it is not accelerating because it being acted on by an external force opposite the direction of motion...just as it was in the video. If said force is non-zero at ##v_x= w## you've won the debate.Gleb1964 said:Because there is outcoming flow, there should be incoming flow.
May be need to go to the reference frame of control volume? I remind, the reference volume is not fixed, its moving up. That mean I have wrong illustration of incoming flux. It is non-zero.
Stop with the apparent jet! Was there an "apparent jet" in the analysis? You are adding it in because you believe you can fool me. I'm not amused...Gleb1964 said:I should been looking on a diagram of apparent wind to not confuse myself.
Now is corrected drawing. The reference frame is fixed on the cart, control volume is fixed.
There is no wind in ##x## -direction, ##v_x=w##, but there is a wind in ##-y## -direction.
From the point of control volume there is incoming flow from top with speed ##w'## and outcoming flow with the same speed, showed as out-flow on the picture.
View attachment 322994
Here is your jet, as you wish. One in the ground frame and another in the cart frame. I have to split it in the volumes to illustrate somehow the transformation.erobz said:Stop with the apparent jet! Was there an "apparent jet" in the analysis? You are adding it in because you believe you can fool me. I'm not amused...
It is apparent you are desperately grasping for straws now because you already realize what the "agreed upon" analysis is going to say...just as I do!
The jet CANNOT push the cart in the direction of the jet FASTER than the jet is going in the direction of the jet! No matter what you do to the sail orientation, or what angle you chose to go at relative to the incoming jet. Making anything you want variable... Momentum CANNOT enter ( nor exit) the cart at ##v_x = w##. The mass flow rate ##\dot m ## is a true "mathematical factor" in the sense that it is multiplied by anything you can come up with for the velocities- any configuration your heart desires. Short of DeVine intervention by the hand of GOD no matter what you do to the sail to maximize the acceleration of the cart there is a hard limit.Gleb1964 said:Here is your jet, as you wish. One in the ground frame and another in the cart frame. I have to split it in the volumes to illustrate somehow the transformation.
View attachment 322996
It certainly can. Just like the stick in the video below can push the cart in the direction of the stick faster than the stick is going in the direction of the stick.erobz said:The jet CANNOT push the cart in the direction of the jet FASTER than the jet is going in the direction of the jet!
erobz said:Stop with the apparent jet! Was there an "apparent jet" in the analysis?