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Decay constant for water flow 
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#1
Apr914, 02:30 PM

P: 33

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
Apr914, 02:39 PM

Thanks
P: 1,948

The inverse of the decay constant is proportional to the half life (The time to lose half of the volume in the experiment)



#3
Apr914, 02:50 PM

P: 33

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 hydrostatic pressure on the water?



#4
Apr914, 03:20 PM

P: 1,469

Decay constant for water flow
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. 


#5
Apr914, 03:29 PM

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P: 1,948




#6
Apr914, 03:56 PM

P: 33

Anything on the hydro static pressure? :P



#7
Apr914, 05:01 PM

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PF Gold
P: 5,174

Chet 


#8
Apr914, 05:04 PM

P: 33

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?



#9
Apr914, 05:14 PM

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PF Gold
P: 5,174

[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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