Decay constant for water flow

  1. When modelling exponential decay in class we did a water flow through a burette experiment. We were given the equation V(t)= V0 e^-λt and ln(V0/V)=λt Where lambda is the decay constant, V0 is the initial volume and V is the volume at any time t. What does the decay constant actually tell you in this situation? I know it's measured in 1/seconds but what does it show you?
     
  2. jcsd
  3. The inverse of the decay constant is proportional to the half life (The time to lose half of the volume in the experiment)
     
  4. Thank you. A follow up question. As the volume decreases as the water flows out, the rate at which the water flows out also decreases. Is this because there is less hydro-static pressure on the water?
     
    Last edited: Apr 9, 2014
  5. Not quite the formal definition of the decay constant, but of the half life time constant, which uses the exponential of 2, and not e.

    From the equation V(t)= Vo e^-λt , or V(t)/Vo = e^-λt,
    one can see that if the exponent λt = 1,
    then,
    V(t)/Vo = 1/e = 0.3678 ..

    In other words the initial value Vo has decayed to 1/e of its value after one decay constant.
     
  6. That's why I didn't say they are equal. I said they are proportional.
     
  7. Anything on the hydro static pressure? :P
     
  8. Chestermiller

    Staff: Mentor

    Are you asking for a derivation of that equation, with λ related to actual physical parameters?

    Chet
     
  9. No. I'm just wondering if hydro static pressure has an effect on the rate of flow of water. I'd assume that at peak volume, the hydro static pressure is higher and so the rate is fast and as the volume decreases, the pressure does so aswell and so the rate decreases. Is this right?
     
  10. Chestermiller

    Staff: Mentor

    Suppose you have a valve at the bottom of the column, and the characteristic of this valve is that Q = k(P-P0), where Q is the volume rate of flow out the valve, and P-P0 is the pressure drop across the valve. Assume that this is the dominant flow resistance in the system. Also, the pressure at the bottom of the column is P0+ρgz. Then a mass balance on the column gives:

    [tex]\frac{dV}{dt}=-kρgz=-\frac{kρg}{A}V[/tex]

    where A is the cross sectional area of the column.

    Chet
     
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