Jump to content

alpha2cen

Senior Members
  • Posts

    1100
  • Joined

  • Last visited

Posts posted by alpha2cen

  1. Im still confused as to why the rate isnt constant. Can someone explain this in simple terms please.

    Water pressure

    P=(density). g h

     

    But, air is compressible fluid.

    Air density is not constant. It is a function of pressure, P.

     

    And, other factors, i.e., temperature, humidity, etc. give an influence on pressure. They can vary with altitude.

  2. Without an existance of the background back side something, transforming from energy to particle can happen? At the BigBang point, the high energy seperate away the background back side particles too far. But, current low collision energy does not have a capacity to push away the background back side particles so far. So, antimatters are background back side particles which exist not so far away ?

  3. The elevation at 0 meters has an air pressure of 1000mb,

    3000 meters has an air pressure of 700mb,

    6000 meters has 500mb.

    9000 meters has an air pressure of 330mb.

     

    Why is the rate of decrease of air pressure not constant with increasing altitude?

     

     

    On the ground, Z=0

     

    [latex]\Delta P=-\rho \ g \ \Delta Z[/latex]

     

    From the ideal gas equation.

    [latex]\rho =\frac{PM}{RT}=K \ P[/latex]

    If T is constant, K=constant

     

    [latex]\Delta P=-C \ P \ \Delta Z[/latex]

    C; constant

    [latex]\Delta Z \rightarrow 0[/latex]

    [latex]d P=-C \ P \ d Z[/latex]

     

    Pressure equation

    Let [latex]Z_{1}=P_{1}, \ Z_{2}=P_{2}[/latex]

     

    [latex]\frac{P_{2}}{P_{1}}=exp (-C \ (Z_{2}-Z_{1}))[/latex]

    [latex] C =\frac{gM}{RT}[/latex]

  4. Also, it's important to note that one generally obtains the correct equation (both dimensionally sound and physically correct) through means of experimentation (i.e. empirically) not simply by using mathematics - of course, the theoretical derivations can help you to achieve the correct equation for some quantity, however at some point you are going to have to do an experiment(s) in order to obtain a meaningful result which is physically correct.

     

    This goes for differential equations too, as you may be able to find the general solution of a differential equation through mathematics - however to obtain the complete solution you will need boundary/initial conditions which are obtained through experiment.

    In the case of solving mathematical equations, using dimensionless forms of the equations is more reliable.

  5.  

     

     

    If you are saying that a stronger spring is needed at top because this is the main compressor and a better seal is needed to prevent leakage back through the valve. That would make more sense.

    In general, the piston type compressor is used at the high pressure compression. We can control the compression ratio by reducing the empty space. However, without good sealing, the compressor can not generate high pressure gas output.

  6. I agree with you but the problem is the net angular momentum of the water from the Coriolis effect in the sink is less than

    than the net angular momentum of the water leaving a small circular drain in the center of the sink which can easily

    be confirmed from a simple visual inspection. I think this should be more thoroughly investigated by the scientific

    community to find out whats going on.

    This is a Figure about the coriolis effect.

    The first Figure is about the height effect, and then, the second Figure is about the latitude effect.

    coriolis.jpg

    coriolis2.jpg

  7. Okay so I believe this is a fairly straight forward enthalpy problem, but for what ever reason it says i'm wrong.

     

    Did I miss something completely or is there an error in my conversions?

    Before doing this experiment, have you done any experiment about the reference reagents by using the same apparatus?

    Quantitative experiment always should be exact, even the experiment is in the elementary course.

  8. "Nobody said it was a two stage pump."

     

    OK, my mistake. I thought that was what you were implying. Because the valve on the lower left of the bottom part of the cylinder is also a (fresh air) inlet valve.

     

    "It may look that way to you, but it doesn't to me."

     

    The main compression is occuring at the upper part of the machine. Botton part sealing state is not so good.

    To maintain a good sealing state , suitable spring pressure is required.

    sealing.jpg

    Sealing is one of the important factor in the compressor design.

  9. How about thinking about these conditions again?

    1) Iron particle size. The particle size is too large?

    2) Charcoal property.

    Charcoal is an important material for the oxygen supplier. To occur that oxidation reaction, sufficient oxygen must be supplied to the iron surface.

    And, always watch out, dust explosion!!!

  10. In conclusion the reason is the expansion of the universe right?

    Using modified Beer Lambert Law, I tested the Steady State Universe theory.

    But I could not fit the obtained redshift data to that model anyway.

    http://www.scienceforums.net/topic/71791-light-interaction;-calculated-results-about-the-steady-state-universe/#entry720407

    The blue line is a redshift data line, and the red line is a model data line.

    steady-model2.jpg

     

     

    steady-state1.jpg

  11. In the case of incompressible fluid and the same height.


    [latex]\frac{\Delta P}{\rho } +\frac{\alpha \Delta v^{2}}{2} + f=0[/latex]

    When there is no fraction loss, the fluid velocity difference between two points appeares to their pressure difference.

    And, pressure difference also appeares to their velocity difference.

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.