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Suppose we have a liquid nitrogen tank hooked up to a freezer. The freezer is replenished with 37.2 L of LN2 every 5 days from the tank. Additionally, the tank loses LN2 at a rate of 1.5% per day. The tank begins with 180.0 L of LN2.

How would I go about solving for the volume in the tank as a function of time?

To me, this looks like a continuous compound interest problem, but I can't seem to figure out how to factor in the volume change from the regular intervals.

I attempted to adapt the answer given in Continually Compounded Interest + Addition to Principal, changing rate and deposit to be negative. It didn't seem to line up correctly though.

I know this can be done simply recursively by subtracting from the previous value, but I would really like to figure out how to make this a continuous function of time.

J. W. Tanner
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user208795
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1 Answers1

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Let $V_0 = 180 L$ and $y = (1 - 0.15) = 0.985$ and F = 37.2 L

$$V_1 = V_0y$$ $$V_4 = V_0y^4$$ $$V_5 = V_0y^5 - F$$ $$V_9 = (V_0y^5 - F)y^4 = V_0y^9 - Fy^4$$ $$V_{10} = V_0y^{10} - F(y^5 + 1)$$

Follow this pattern for n is a multiple of 5 we have

$$Vn = V_0y^n - F(y^{n-5} + y^{n-10} +...+ y^0)$$ $$= V_0y^n - \frac{F(y^n - 1)}{y^5 - 1}$$

For n is not a multiple of 5 we have $$V_n = V_0y^n - F(y^{n-5} + y^{n-10} + ... + y^{n-5[\frac{n}{5}]})$$

Based on the above formulae we have $V_{19} = 37.48L$ and $V_{20} = -0.28L$

Hence at the beginning of the 20th day there is not enough liquid nitrogen to feedback the freezer.

KY Tang
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