tmpst Posted November 11, 2013 Share Posted November 11, 2013 (edited) Consider a flow in a one layer, two dimensional tunnel/bay. Assume edges in the y-direction across the bay, but no edges in the x-direction of the flow. If we assume the velocity across the bay to be 0, and the situation to be stationary ([math]\dfrac{\partial}{\partial t} = 0 [/math]), the the left side simplifies to a one dimensional directional derivative [math]u_x \dfrac{\partial u_x}{\partial x}[/math]. Assuming constant pressure, then the right side simplifies to [math]\nu {\nabla}^2 u_x[/math]. We only have friction from the edges, so the viscosity term further reduces to [math]\nu \dfrac{\partial}{\partial y}(\dfrac{\partial u_x}{\partial y})[/math], or [math]\dfrac{1}{\rho}(\dfrac{\partial \tau_y}{\partial y})[/math], depending on how you want to write it. The question is: How can these be in balance, when on the left side there is no [math]\dfrac{\partial u_x}{\partial y}[/math] term, but on the right there is only a term dependant on [math]\dfrac{\partial u_x}{\partial y}[/math]? To make things clearer I left out the continuity equation, which means that we might have a sing at the "end" of the bay so we are not assuming [math]\dfrac{\partial u_x}{\partial x}=0[/math]. If we want [math]\nabla \cdot U = 0[/math], we can reconstruct the scenario by adding depth. Then the directional derivative will still not have any [math]\dfrac{\partial u_x}{\partial y}[/math] term, but the the viscosity term will. EDIT: This might fit better under Earth Scienses or Engineering. Sorry if it ended up in the wrong place. Edited November 11, 2013 by tmpst Link to comment Share on other sites More sharing options...

studiot Posted November 11, 2013 Share Posted November 11, 2013 (edited) A picture is worth 1000 words. I think you mean [math]\frac{{\partial p}}{{\partial y}} = \frac{{\partial p}}{{\partial z}} = 0[/math] as conditions for [math]\frac{{\partial u}}{{\partial t}} = \frac{{ - 1}}{\rho }\frac{{\partial p}}{{\partial x}} + \nu \frac{{{\partial ^2}u}}{{\partial {y^2}}}[/math] The first equations tell us that p is a function of x and t only. The second (NS) equation tells us that [math]\frac{{\partial p}}{{\partial x}}[/math] is equal to the difference between two terms that are both independent of x. Thus [math]\frac{{\partial p}}{{\partial x}}[/math] must be a function of t alone. since we have steady state [math]\frac{{\partial p}}{{\partial x}}[/math] must therefore be zero. Am I reading you right?#Edit My apologies I see a serious clanger crept in during the copy/pasting. I said [math]\frac{{\partial p}}{{\partial t}}[/math] must be zero when I really meant [math]\frac{{\partial p}}{{\partial x}}[/math] I have corrected the above. Edited November 12, 2013 by swansont fix math typo by request Link to comment Share on other sites More sharing options...

tmpst Posted November 11, 2013 Author Share Posted November 11, 2013 (edited) I am assuming [math]\dfrac{\partial}{\partial t}=0[/math], so [math]\dfrac{{\partial u}}{{\partial t}}=0[/math], and [math]\frac{{\partial p}}{{\partial t}}=0[/math], What I am not assuming is, that [math](u \cdot \nabla)u = u_x \dfrac{\partial u_x}{\partial x}=0[/math]. Assuming one layer (i.e. no z-direction), my NS simplifies to: [math]u_x \dfrac{\partial u_x}{\partial x} = \nu \dfrac{\partial^2 u_x}{\partial y^2}[/math] The question is, what is the connection between those terms? Is this equation even possible if I assume no [math]\nabla p = 0[/math]? Of course if we add the z-direction we have to add the hydrostatic pressure [math]\frac{{\partial p}}{{\partial z}}[/math] right? I will also need to account for the bottom friction and possible wind stress by adding a term [math]\nu \frac{{{\partial ^2}u_x}}{{\partial {z^2}}}[/math]. The question still stands: On the left I have no other terms then one describing the change in the x-direction [math]\frac{{\partial}}{{\partial x}}[/math]. What is the interpretation to get these balanced out with the terms consisting only of changes in the y- [math]\frac{{\partial}}{{\partial y}}[/math] and z- direction [math]\frac{{\partial}}{{\partial z}}[/math]? Edited November 11, 2013 by tmpst Link to comment Share on other sites More sharing options...

studiot Posted November 11, 2013 Share Posted November 11, 2013 Perhaps the correction to my last post will make more sense as an answer. Link to comment Share on other sites More sharing options...

tmpst Posted November 12, 2013 Author Share Posted November 12, 2013 You are still missing the [math](u \cdot \nabla)u = u_x \dfrac{\partial u_x}{\partial x}[/math] term on the left, which is not zero. Assuming stationary conditions, the term on the left in your equation should be zero i.e. [math]\dfrac{{\partial u}}{{\partial t}}=0[/math]. Then the left term is not independent of x, and I guess we might get a [math]\dfrac{\partial p}{\partial x}[/math] term . My question is, how does the convective derivate (describing only change in the x-direction) relate to the viscosity term (describing only change (or actually acceleration of the speed) in the y-direction). I can understand that the closer we are the edge of the bay, the more the friciton affects the flow, and the smaller the change in the speed in the x-direction. What i dont "see", is how that follows from the balance in the NS-equation, and I'm hoping someone could nudge me in the right direction. Link to comment Share on other sites More sharing options...

Bignose Posted November 12, 2013 Share Posted November 12, 2013 Consider a flow in a one layer, two dimensional tunnel/bay. Assume edges in the y-direction across the bay, but no edges in the x-direction of the flow. If we assume the velocity across the bay to be 0, and the situation to be stationary ([math]\dfrac{\partial}{\partial t} = 0 [/math]), the the left side simplifies to a one dimensional directional derivative [math]u_x \dfrac{\partial u_x}{\partial x}[/math]. Assuming constant pressure, So, let me make sure I understand correctly. You want to assume a no influx of flux ("velocity across the bay to be 0"), because if you just assume the velocity is zero, you only have a trivial solution. So, it is looking something like a driven cavity. But then you assume that the pressure is zero everywhere, too. With no flux of mass or momentum, and no pressure difference, you will get a trivial solution. You have nothing that drives flow, and the equations -- continuity + N-S -- essentially are telling you that. Link to comment Share on other sites More sharing options...

tmpst Posted November 12, 2013 Author Share Posted November 12, 2013 I have come to see that I have to allow for difference in pressure, so [math]\dfrac{\partial p}{\partial x}[/math] is not assumed to be trivial. I can obviously assume a wind forcing [math]\tau_w[/math] within the viscosity term in the x-direction instead, right? It all comes down to, that I can't seem to wrap my head around how the term [math]\nu \nabla^2 u_x[/math], which reduces to just a [math]\dfrac{\partial}{\partial y}[/math] term, affects the [math]u_x\dfrac{\partial u_x}{\partial x}[/math]-term. If we have no friction, then e.g. it makes perfect sense to me, that a reduce in pressure (so an increase in the [math]-\dfrac{\partial p}{\partial x}[/math]) would be balanced out by an increase in the velocity. In our case that increase in in velocity will be thwarted by the viscosity term in some proportion to the distance to the edge. But I don't "see" the connection between the derivate of the velocity in the x-direction, and the second derivate of the velocity in the y-direction. How do those cancel out? Should I be thinking of the viscosity term as being a "profile" of the the magnitude of the friction near the edge? I really appreciate everyones help this far. Link to comment Share on other sites More sharing options...

studiot Posted November 12, 2013 Share Posted November 12, 2013 (edited) I did ask for a sketch to confirm your flow. Bignose was also unsure. I think you mean what is called plane parallel shear flow given by the flow field. u = u(y,x,z,t) = (u(y,t), 0, 0) Note line 6 of post#2 should read The first equations tell us that p is a function of x and t only. Further since u is independent of x, [math]\nabla .u = 0[/math] Edited November 12, 2013 by studiot Link to comment Share on other sites More sharing options...

tmpst Posted November 13, 2013 Author Share Posted November 13, 2013 Sorry about tha absence of a picture. I now have a picture of a scenario know. Please don't mind if this contradicts something I have said previously. I constructed this scenario to isolate the the terms I'm having problem with. So we have a constant flow that enters a bay. Because there is no driving force, the fluid will slow down due to the friction of the walls and the bottom. The fluid is incompressible, and we have no flow in the y-direction. The continuity equation is [math]\dfrac{\partial u_x}{\partial x} + \dfrac{\partial u_z}{\partial z}=0[/math]. We have no sink, so a change in velocity must mean a change in pressure, right? if the change of the velocity is negative, the pressure must rise. The change in velocity is due to the friction. In the x-direction the NS-simplifies to: [math]u_x \dfrac{\partial u_x}{\partial x} = -\dfrac{1}{\rho}\dfrac{\partial p}{\partial x} + \nu \nabla^2 u_x[/math] and in the z-direction to: [math]u_z \dfrac{\partial u_z}{\partial z} = -\dfrac{1}{\rho}\dfrac{\partial p}{\partial z} + \nu \nabla^2 u_z[/math] Now if we concentrate on the x-direction, how should I understand it? I'm pretty sure I'm not able to ask the right questions, but what exactly is the "pressure" here? Is it the dynamic pressure, the static pressure or the total pressure? How would a change in air pressuare be accounted for? My guts tell me that an increase in air pressure would result in an increase in velocity, so do you have to switch the sign when accounting for outer pressure? The viscosity term is reducing the velocity, but the reduction in velocity must then again increase the pressure. In my mind I would like to write then in two separate equations. The friction reduces the velocity [math]u_x \dfrac{\partial u_x}{\partial x} = \nu \nabla^2 u_x[/math], and the reduction in velocity increases the pressure [math]u_x \dfrac{\partial u_x}{\partial x} = -\dfrac{1}{\rho}\dfrac{\partial p}{\partial x}[/math]. When written in the same equation, the friction term is actually smaller then the reduction in velocity, which doesn't make any senso to me. Am I understanding the pressure wrong, or am I switching some sign somewhere? The second part that I find confusing is the value of the viscosity term. At the edges the shear stress is just the friction, but how do I get the shear stress [math]\tau_y[/math], say, 1m from the edge? By definition it follows from knowing the gradient of [math]u_x[/math] as a function of [math]y[/math], but that is exactly what I am trying to calculate. Is there some law or rule that says what kind of relationship there is between the shear stress and the distance from the wall? It can't be linear, because then the shear stress would be constant in the whole bay due the symmetrical situation. It feels that I am making some wrong assumption, or misunderstanding something. Then I'm just going thrue the same steps over and over, but they don't add up, and I can't really put my finger on the problem. PS. The increase in pressure could manifest itself as a rise in water level, right? Would the whole term then be accounted for by change in the static pressure, or should I always use the total pressure (unless I decide to use an approximation). If no rise in water level is possible, would the change in pressure then reduce to a change only in the dynamic pressure? Link to comment Share on other sites More sharing options...

studiot Posted November 13, 2013 Share Posted November 13, 2013 How would a change in air pressuare be accounted for? My guts tell me that an increase in air pressure would result in an increase in velocity, so do you have to switch the sign when accounting for outer pressure? Let's start with this one since it introduces a complication that is unhelpful until you are happy with the basic flow equations. I am assuming you we are discussing flow of water in an open channel with sides and a bed, and that your diagram is a plan view with water depth measured in the z direction ? As such there will be a particular pressure at any given depth due to the sum of the imposed atmospheric pressure and the pressure due to the water depth. If the atmosphere is constant in the x direction we may subtract the constant atmospheric pressure and deal in what is known a gauge pressure. If the bed is horizontal the pressure is the same everywhere and there water is still. There is no reason for it to flow. Now incline the bed so the left hand end is higher than the right hand end. This difference in elevation will provide the energy to cause flow along the lines you have indicated, except that it will develop a curved profile due to the friction with the sides and bottom. You should look up Poiselle's equation. Link to comment Share on other sites More sharing options...

tmpst Posted November 13, 2013 Author Share Posted November 13, 2013 I am assuming you we are discussing flow of water in an open channel with sides and a bed, and that your diagram is a plan view with water depth measured in the z direction ? As such there will be a particular pressure at any given depth due to the sum of the imposed atmospheric pressure and the pressure due to the water depth. If the atmosphere is constant in the x direction we may subtract the constant atmospheric pressure and deal in what is known a gauge pressure. Yes and yes. Constant atmospheric pressure will of course not show up in any partial derivates. But if the atmospheric pressure would drop at some point x, would that show as an decrease in the [math]\dfrac{\partial p}{\partial x}[/math] term, and thus inducing an increase in the [math]\dfrac{\partial u_x}{\partial x}[/math] term? This would mean a decrease in atmospheric pressure on the left side of the channel would induce a flow away from that area. I find this counterintuitive, as one would think that "pressing down" on the left side would cause the water to flow away from the area, not the other way around? In addition to the gauge pressure we have the dynamic pressure if we have any flow present, right? If the bed is horizontal the pressure is the same everywhere and there water is still. There is no reason for it to flow. This is true, but here we have a flow entering the channel. This flow is due to some external force (e.g. wind, elevation) outside the channel. The external forces do not exist inside the channel. So we have a constant flow entering the channel, and we want to know how it is affected by the friction. Now incline the bed so the left hand end is higher than the right hand end. This difference in elevation will provide the energy to cause flow along the lines you have indicated, except that it will develop a curved profile due to the friction with the sides and bottom. You should look up Poiselle's equation. Then I have to add the gravitational term to the NS-equation, right? This is interesting, but this case is about the attenuation of the flow due to the friction. Adding a gravitational term will not solve the question concerning the balance between the change in velocity and pressure, and the friction. After a quick glance, the Poiselle's equation is not applicable for my exact scenario, because it does not allow for acceleration of the fluid (an I assume a negative acceleration due to friction). That said, it is closely related and seems very interesting. Will definitely read up on it. Thank you. Link to comment Share on other sites More sharing options...

studiot Posted November 13, 2013 Share Posted November 13, 2013 You have described 'flow entering the channel' from the uncharted area (Here be dragons ) to the left. Why would flow enter the channel? I assume that the ocean to the left has a common water surface level at the interface with the channel, even if there is an abrupt and enormous change of bed level. In a real world water approaching such a sharp profile would create vortices in the lee of both right angles and laminar flow euqations would no longer be applicable. Water can only enter the channel if water moves along the channel and something cause that movement. Bignose has already alluded to that. What ever causes the movement imparts changes of pressure etc to the water. Yes, Poiselle's equation is the simplest profile. Real world hydraulic profiles in open channel flow are associated with the names of Chezy, Manning and others. The water surface profile is complicated and full of features and heavily influenced by changes of bed slope (and obstructions). In addition to uniform flow with a surface parallel to the bed (note not horizontal) the flow depth can be gradually varied (also called draw down) or gradually increased to pass over an obstacle (called a backwater curve|) or it can show a step change at an abrupt reduction of gradient. This is called a hydraulic jump and can be quite spectacular. In addition to the NS equations you should study the momentum balance equations. These are the ones that describe the forces that act between the layers of water and the boundaries. I suspect these will supply the information you are lacking. In order to study viscoscity/ friction you need to study momentum transport between the layers. Link to comment Share on other sites More sharing options...

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