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| Save my spa! (Posted on 2018-06-27) |
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True story: For safety, I had put chlorine powder in my spa water and my guests were coming soon. But, when I used my water testing kit, I realized I had over-chlorinated. I had thrown in about n times too much powder, and so I needed to dilute the water by adding fresh water, and fast. Would I better off first draining out some of the over-chlorinated water and then filling the spa back up with fresh water and mixing, or would it be faster emptying and filling at the same time? The garden hose fills the spa at the same rate that the drain faucet at the bottom empties the spa. Assuming that the mixing of fresh water and chlorinated water is instantaneous and that the drain rate does not change with water height, which way is faster and how does the answer depend on n?
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Submitted by Steven Lord
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Solution:
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Suppose it takes T minutes to fill the spa.
Compute t1, time for drain and refill as two steps: If n times too much chlorine was added, then one would empty out (n-1)/n of the chlorinated water and add fresh water back in. These two steps in total would take in minutes:
t1 = 2 x T (n-1) / n.
Compute t2, time for drain and fill simultaneously: We see that the change in concentration in time is proportional (via constant k) to the current concentration n, where now n is a function of t. We call the initial concentration n0.
dn/dt = k n, with k an unknown constant.
dn/n = k dt
Integration of both sides gives:
ln n = -k t + c (where c is a constant of integration).
Applying the constraints to c, we get:
n(t) = n0 exp(-k t).
The instantaneous dilution as t--> 0 must approach 1/T as so little mixing occurs.
Setting d/dt(n0-n0 exp(-kt))/n0|t=0 = 1/T
gives k=1/T
n(t)=n0 exp(-t / T)
Setting n(t2)=1,
t2 = T ln (n0).
Setting t1 = t2, to see where the methods take the same time,
2 (n-1)/n = ln(n).
This occurs when n= 4.92. When n < 4.92, the simultaneous method is faster, and, when n > 4.92, the two step method is faster. This is independent of the time it takes to fill.
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