Jump to content


  • Posts

  • Joined

  • Last visited

Everything posted by arkie87

  1. This question pertains to "viscous dissipation" term in the energy equation in flows.I am a bit confused about viscous dissipation. It seems to me that viscous dissipation is essentially the same thing as pumping power, such that if I pump 1 m3/s of air at 1 kPa, it takes 1 kW. To get 1 kPa pressure drop, I need a lot of friction, and that friction will cause viscous dissipation, which will result in 1 kW of heating i.e. ~0.8 K temperature rise in the air. For instance, for the simple case of steady state, fully developed laminar flow between parallel plates, one case show that the viscous dissipation term exactly equals the pumping power. However, for an ideal gas, enthalpy is only a function of temperature. If I expand the air through a valve instead, thermodynamics tells me the temperature wont have changed if the process is isenthalpic.How do i reconcile these two? Why wont the air going through the valve heat up if it is dissipating the same amount of power?
  2. Ok, thanks! Sounds like something a Kerbal would do.
  3. Apologies in advance if this question is not clear; I will revise based on feedback to make it more clear. I have a thought experiment: (1) consider a centrifugal fan with the inlet in the central hub (at radius=0) and the outlet at the outer radius. If the fan blades spin, the centrifugal/centripetal forces causes a radial pressure gradient. The pressure at the outer radius is higher than the inner radius. Since the outlet is venting to ambient pressure, flow is created. (2) Now imagine you weld the fan blades to the housing. If you spun the housing at the same RPM, you would still get flow, since the outlet pressure is still above ambient. (3) Now imagine you welded a duct connecting the outlet to the inlet hub. The system is now fully contained. If you spun the device, would you get flow? My inclination is no, since in the return duct, there is the same centripetal acceleration that causes the same pressure gradient in the radial direction. Thus, i cant think of a reason the flow would go one way vs. the other. Thoughts? Clarifications? TIAA
  4. My wife is not a happy flier, and particularly dislikes the feeling of turbulence. I've told her that she should probably book seats towards the center of the plane, since this should reduce the g-forces from turbulence. I was wondering: (a) Is this true, and if so, is the reduction significant? (b) Is the best place in the center, or the center of lift/drag? Is that likely to be in a specific location (center of wing? edge of wing? center of mass? etc... TIA
  5. Well, dont assume that shape. Assume a cylindrical shape... regardless, even for a cylinder water rushes in from directly below but also from the sides. But it is nearly impossible to know the relative balance of each without doing some sort of CFD. Agreed, that is my opinion.
  6. @Nightside Thanks again for your reply. Very often, just knowing what something is called provides a huge breakthrough in being able to learn about it i.e. knowing it is called "added mass" or "virtual mass" allowed me to find much more references out there. And interestingly, @Scotius, the navy has studied it way before ballistic missiles, to be able to predict acceleration of a submarine underwater, for instance (concept of added mass applies even without buoyancy). See these references below: https://core.ac.uk/download/pdf/36709212.pdf So, for a sphere, added mass is 1/2 of the mass of the displaced fluid, but for more streamlined bodies, added mass can approach 0, as shown here: https://www.researchgate.net/publication/233581794_Translational_Added_Mass_of_Axisymmetric_Underwater_Vehicles_with_Forward_Speed_Using_Computational_Fluid_Dynamics
  7. Not quite. The form drag force is zero (presumably) when the object isnt moving. I am asking about the acceleration obtained at the start. The quora answer @Nightside gave was pretty useful, and explains the answer to my question. TL;DR: there is a concept of "added/virtual mass" (it is basically the mass of fluid that gets dragged along with the rocket) that limits its acceleration at the start (when velocity and all forms of drag are zero). See @Nightside's answer, which directly conflicts with yours. maximum acceleration is 2g, even when velocity is zero (where form drag would be zero).
  8. There is form drag and viscous drag/skin friction. Viscous drag/skin friction can be zero (which is what i meant by zero drag), but form drag cannot i.e. the thought experiment would be ill-posed or self-contradictory. zero velocity... so skin friction is zero. yes, acceleration, and yes from standstill ( i never said not from a standstill)?
  9. lol, my bad in this instance. replies like that, i usually do ignore. its hard to ignore when they begin attacking your assumptions even though its a thought experiment...
  10. I agree, my question could have been more clear. But in no way did the original response actually respond to the approximately 2/3 of my question either... And while you might be upset at people who get annoyed at people who try to help them for free with their problems, it is also equally annoying to be on the other end of it, when you ask for help with a specific problem and someone answers a different question altogether without fully reading your question. Or worse-- they start a whole argument about one of your assumptions when its supposed to be a thought experiment... etc.... if i had a penny for every time that has happened on this forum (and others).... i'd have like 10 pennies (at least)
  11. I asked for maximum acceleration, which occurs the instant the object is let go, where drag is zero. Yeah, i didnt specifically state it, but i went ahead and discussed a lot of other items, which the responder didnt address. Just told me stuff i already knew. It can be very frustrating when you ask a specific and highly technical question, and someone responds with basic physics and doesnt address your question. @mikegarrison for instance, this response above... not sure what do with it...
  12. yeah, i've run CFD simulations and i have a phd in mechanical engineering in the fields of fluid flow & heat transfer, but because of the accelerating frame of reference, typical navier stokes equations are not applicable..
  13. arg... when the object is at rest, there is no drag. turbulence isnt an issue either for obvious reasons (Reynolds=0). sheesh. Yes, i didnt call it an air bubble since air bubbles can deform under drag force and also expand as the static pressure changes, as you noted. It's hard to tell a priori what minute details strangers on the internet will latch onto instead of answering the question at hand, so i opted for a volume of air with an infitinitely small, massless, rigid shell instead of a bubble. Yes, thank you for the quora link. Cant say i understand why the limit is 2g though. The answer also appears to indicate that shape would be a factor (since they solve potential flow around a sphere) i.e. 2g is the solution for a sphere, but what about a rocket-shaped object? And no, i'm not considering launching objects from the sea floor. just had a discussion where someone claimed "maximum acceleration due to buoyancy is 1g since 1g acceleration causes buoyancy," but that appears to be incorrect...
  14. Let's limit the fluid to water, let's limit the acceleration due to gravity to 1 g, and let's limit the object to being filled with air (and zero wall thickness). My main question is if the acceleration that you compute from a force balance i.e. (weight of displaced water - weight of object)/mass of object is physical, because the fluid underneath or on the sides of the object must accelerate to follow the object (or else you have cavitation), and this acceleration will "absorb" some of the available pressure drop or pressure gradient.
  15. My main question is *neglecting drag*, what is the maximum acceleration of the body. Theoretically, the limit should be the ratio of the densities of the fluid and the object times g. In reality, drag can be neglected as diameter --> infinity. Unless there is cavitation, there needs to be enough volume flow to fill the void left by the object. I dont know how you can say there isnt? I
  16. sorry to say it, but you didnt really answer my question (or get the point of it).
  17. If you have a one cubic meter streamlined-shape filled with air submerged under water on earth, what is the maximum acceleration possible from the object? A simple calculation using Newton's law shows that the acceleration can be rho_water/rho_air*g, but this fails to take into account the fact that water has to be accelerated under the object to take its place. However, it is unclear to me whether this water rushing in underneath can accelerate faster than g. While the water directly below the rocket must accelerate at 1 g, the surrounding water and above would be accelerating significantly slower than 1 g, and the weight of this fluid might be able to accelerate the fluid following the object to faster than 1 g, since its weight is much larger than the weight of the fluid following the object. A related thought experiment is that since water in a tank of pressurized air can be accelerated through an infinitely small nozzle at a g proportional to the pressure, one can achieve the same pressure with a tall rocket-shaped air-filled object as long as the rocket is sufficiently tall. As long as the rocket diameter is much smaller than the container diameter, the velocity in the container will be negligible. Thoughts? Am i making a mistake somewhere?
  18. If you are allowed downward thrust i.e. thrust perpendicular to your velocity vector, it should be possible. Otherwise, as you approach orbital velocity, your weight drops to zero, and your wheels loose traction
  19. ok, so it only applies to perfect spheres. no other shape?
  20. Could be. 351 J out of 107,559 J is about 0.326% , which is comparable to the oblateness of the spheroid 22 km out of 6371 km (~0.345%). Though, on the other hand, if you used the value of g at the equator and poles, though, you might reduce the error as well...? However, i thought that there was some law that says that the law of gravitation -- GMm/R^2 -- applies to point masses as well as solid bodies, as long as the distance is between centers of mass. Based on that, i'm not sure why an oblate spheroid would have any effect?
  21. Well, i think i have figured it out. Thanks to all for your help. The main mistake we were making is the assumption that gravitational force scales with the inverse square law-- this is only true outside the body; but inside the body, it increases linearly. My solution is as follows: The gravitational acceleration inside the earth is: g=g_0*r/R where g_0 is the gravitational acceleration on the surface at R. The specific energy required to ascend from the center of the earth up to the north pole is: E_p/m = integral(g_0*r/R_p*dr,0,R_p)= 1/2 g_0 *r^2/R_p (@r=0,R_p) = 1/2*g_0*R_p Similarly, the specific energy required to ascend from the center of the earth up to the equator is: E_e/m = integral((g_0*r/R_e-w^2*r)*dr,0,R_e)= 1/2*(g_0/R_e-w^2)*r^2 (@r=0,R_e) = 1/2*g_0*R_e - 1/2*w^2*R_e^2 Plugging in values for g_0 = 9.81 m/s2 R_p = 6356 km R_e = 6378 km w = 2*pi/(60*60*24) = 7.272e-5 rad/s We get E_p/m = 31.17618 MJ/kg E_e/m = 31.17653 MJ/kg Alternatively, we can write: 1/2 (w*R_e)^2 = 1/2*g_0*(R_e-R_p) 1/2 (w*R_e)^2 = 107,559 J/kg 1/2*g_0*(R_e-R_p) = 107,910 J/kg So we have energy balance within 351 J, the error of which is probably due our assumption of g_0. Thus, the integral of the potential energy from the center of the earth contributes the needed factor of 1/2 to make centripetal energy and gravitational potential energy balance ( @m4v @sevenperforce). Accordingly, it appears the gravitational potential energy of an oblate spheroid has nothing to do with it. Sorry again, @Snark
  22. So to summarize so far: (1) Ignoring gravitational potential, the energy required to walk from the equator to the pole (equivalent to pulling oneself to the center of a spinning object) is exactly equal to the kinetic energy due to rotation, and that kinetic energy is half of the potential energy difference due to the difference in height. (2) The other half is supposedly due to different gravitational potential of a perfect ellipsoid vs. a perfect sphere. Does anyone know the formula for the gravitational force of an ellipsoid. A linearization when a~=b would probably be sufficient to predict the difference of potential for 20km/6000km.
  23. What's your point? I am talking about A and you are talking about B and C.
  24. I'm not sure when i said that delta_energy needed to be zero... just that energy needed to be conserved.
  • Create New...