How Does the Decay Constant Influence Water Flow in Exponential Decay Models?

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SUMMARY

The discussion focuses on the role of the decay constant (λ) in modeling water flow through a burette using the exponential decay equation V(t) = V0 e^-λt. The decay constant, measured in 1/seconds, indicates the rate at which the initial volume (V0) decreases over time, with its inverse being proportional to the half-life of the volume. As water flows out, the hydrostatic pressure decreases, which in turn affects the flow rate. The relationship between flow rate and pressure drop across a valve is also explored, emphasizing the importance of understanding these dynamics in fluid mechanics.

PREREQUISITES
  • Understanding of exponential decay models
  • Familiarity with the decay constant (λ) and half-life concepts
  • Basic knowledge of hydrostatic pressure principles
  • Experience with fluid dynamics equations, particularly flow rate equations
NEXT STEPS
  • Study the derivation of the exponential decay equation V(t) = V0 e^-λt
  • Learn about hydrostatic pressure effects on fluid flow rates
  • Explore the relationship between decay constants and half-life in various contexts
  • Investigate flow resistance in fluid systems, focusing on valve characteristics and pressure drops
USEFUL FOR

This discussion is beneficial for students and professionals in physics, engineering, and fluid dynamics, particularly those interested in the mathematical modeling of fluid flow and the effects of pressure on flow rates.

Saado
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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?
 
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The inverse of the decay constant is proportional to the half life (The time to lose half of the volume in the experiment)
 
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:
dauto said:
The inverse of the decay constant is proportional to the half life (The time to lose half of the volume in the experiment)

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.
 
256bits said:
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.

That's why I didn't say they are equal. I said they are proportional.
 
Anything on the hydro static pressure? :P
 
Saado said:
Anything on the hydro static pressure? :P
Are you asking for a derivation of that equation, with λ related to actual physical parameters?

Chet
 
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 as well and so the rate decreases. Is this right?
 
Saado said:
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 as well and so the rate decreases. Is this right?
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:

\frac{dV}{dt}=-kρgz=-\frac{kρg}{A}V

where A is the cross sectional area of the column.

Chet
 

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