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Water drop evaporating

  1. May 5, 2007 #1
    1. The problem statement, all variables and given/known data

    A spherical drop of water evaporates at a rate proportional to its exposed surface area as it moves.Set up its differential equation of motion.Find the time needed for complete evaporation.What is the velocity at that time?

    2. Relevant equations



    3. The attempt at a solution

    The volume of the sphere is :

    V=(4/3)πr^3, and the area is A=4πr^2

    dV/dt = -k A,the minus sign is to indicate that the volume is decreasing

    dV/dt = (4/3)π(3)r^2dr/dt = 4πr^2dr/dt = -k(4πr^2), and simplifing this:

    dr/dt = - k

    Am I correct upto this?
     
  2. jcsd
  3. May 5, 2007 #2
    Looks a good start.

    But the differential equation you wrote describes the evaporation, not the motion -strictly speaking-. You should check what is exactly expected. (context of this exercice)

    My usual undestanding is that an equation of motion is based on Newton's law of motion.
    Sometimes differential equations are also called equations of motion, kind of abstraction.

    Assuming your analysis match the question, then you can easily conclude about the time needed for complete evaporation.

    What is the context of this question? Math course, chemical engineering, ... ?
     
  4. May 5, 2007 #3
    That's right I am to find the velocity....stuff that can only be found as we solve the equation of motion invoking Newton's 2nd law.It is found in a mathematical physics question paper.

    How to find that F=ma?We are given no force to equate with...Even we do not know in what irection is the drop moving...vertically under gravity or something else?
     
  5. May 6, 2007 #4

    siddharth

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    Homework Helper
    Gold Member

    IMO, a good assumption to make would be that the drop is falling under the force of gravity and buoyancy.

    If you want a better assumption, you could also include the drag force.
     
  6. May 6, 2007 #5
    I think that would be better to do.
     
  7. May 6, 2007 #6
    Neelakash,

    You should find useful information on wiki starting from this page:


    If you can assume low (relative) velocities, then the Stokes's drag would be all you need.
    Since your droplet would be decreasing in size, the drag force and the mass of the droplet would change in time.

    Finally note that this problem has some industrial applications.
    In gas conditioning towers, water sprays are often used to cool down a gas stream before further treatment.
    The small difference with your problem is that the gas is streaming.
    As you may imagine, the efficiency of the cooler tower is essentially determined by the size of the droplets.
    If the droplets are not small enough, they will not have enough time to evaporate completely in the tower. When the gases are dusty, this may produce sludges at the bottom of the tower.

    Note also that the evaporation rate will depend on the temperature and humidity of the gas. Obviously saturated gas will block any evaporation, and high temperatures will increase evaporation. If you want to know more about this, have a look for "psychrometry" on wiki or elsewhere. The psychrometric ratio will also show you a link between the heat transfer coefficient and the mass transfer coefficient (for a water vapor-air mixture). With all these information, all the details of droplet evaporation in gas stream can be simulated.
     
    Last edited: May 6, 2007
  8. May 6, 2007 #7

    Astronuc

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    Staff: Mentor

    Is one to compute the velocity of the drops boundary or the speed of the drop as it moves?

    Also keep in mind that there are two areas - surface area of the drop, and projected area, which is used in resistance to motion.
     
  9. May 6, 2007 #8
    I am getting a different idea.Should not it be like that:

    The drop evaporates and consequently its self energy decreases and that self energy is being converted into kinetic energy?Note that we are only given that dV/dt=-dA/dt and no referene to frictional drags are specified.
     
  10. May 7, 2007 #9
    You must clarify the question.
     
  11. May 7, 2007 #10
    I cut the problem and pested it over here and showed my work.That is all.I cannot edit the question.Right?
     
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