I find it very interesting. Kind regards, Mehender The Hague Holland. I wanted to know that, what is the reason of growing of bubble size. Is it because of evaporation at the interface of liquid vapor surface? I also need your help for some material for study please suggest me. I am a under graduate student of final year of mechanical engineering. Please help me …….. Hello kapil, The resulting increase in bubble size is due to the inability for the heated air near the heat source to rise because of lack of or negligible convection in microgravity.
I hope this helps…. Explained well enough. Thanks for making me understand this concept. I have a question :: What is the condition for bubble to be in equilibrium with surrounding? Thanks for your question. Part of the reason we are studying bubbles aboard the space station is to answer questions about phenomena, such as your question on the equilibrium behavior of bubbles in space, which are not currently understood.
My name is Myron. I have just read your report on bubbles and their effects on heat transfer. It was very interesting and enlightening. I was told by my Professor that if no phase change commences such as steam being passed into cold water, condensing and passing off its heat it would make no difference at all in terms of cooling or heating up. All you would get is agitated water. However in this case, we have the opposite situation to the analogy above: the fluid is at a lower temperature than the room temperature and if we follow the 2nd law, heat from the ambient air around it will actually be absorbed by the fluid and if mixed vigorously the heat transfer rate is increased.
However this can only occur if what my Professor says is true, that the colder air bubbles have no effect on cooling, only in agitation. The amount of work, W , done is equal to the area under the PV curve.
Unless you've already got some calculus under your belt, it's a pain and a half to find this kind of area, so we'll just sidestep the issue for now. So how much work is done when we change the pressure of a gas without affecting its volume? It's Picasso time—let's draw a PV graph for this scenario. The area under this PV curve is zero — no work is done here. It's useless. Well, for work. The area under the second PV diagram is much larger, suggesting that a lot more work is required to reach the larger volume under constant pressure.
There's a special name for work under constant pressure like this: isobaric. When pressure remains constant, we have an isobaric process. The process in the first example we discussed, with constant volume while the pressure changes in the gas, is called an isovolumetric or isochoric process. Now give us a second to untie our tongue.
We could also change the volume of a gas while keeping the temperature constant. This is an isothermal process not to be confused with thermal underwear. All of the work done on this type of system is transferred into heat, Q. Calculating the amount of work for this case requires a fancier equation than the others. It is , where n is the number of moles of gas, R is the gas constant, T is temperature, and V is volume.
It left the top hot and monocle at home, though. Lastly, a process can also be adiabatic. Nice, nice. We can then determine the work itself by. Again, this is for adiabatic processes only. Try to use this equation on a diabatic process, and it'll go into diabatic shock. There are two ways to write the first law of thermodynamics. Active 5 years, 10 months ago.
Viewed 10k times. Improve this question. JM97 JM97 3, 6 6 gold badges 29 29 silver badges 57 57 bronze badges. See my answer for more detail. Add a comment. Active Oldest Votes. Improve this answer. Ben Norris Ben Norris 41k 8 8 gold badges silver badges bronze badges. The answer in linked question provides a very thorough explanation. KLuuppo KLuuppo 36 4 4 bronze badges. Maybe it's now more accurate even though not complete.
Feel free to take a tour of the site. Visit the help center to learn more about how it works. Featured on Meta.
0コメント