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Browsing through some plumbing forums about pressure tanks and someone replied with this... (I should start off by mentioning it's a 40 gallon tank)

"At 2.5 GPM, your 2-3 minutes of shower time between cycles is using between 5 and 7.5 gallons. If your tank starts at 60psi and 7.5 gallons is enough to bring it down to 40psi, that means you're starting out with 15 gallons of air in your 40 gallon tank, and ending the 3-minute cycle with 22.5 gallons of air. I calculated this using Boyle's law, or P1 * V1 = P2 * V2. And of course these numbers would be different if your shower head is higher-flow and your 3 minutes of showering used more than 7.5 gallons."

Now I've wrapped my head around this for a while, but I don't see where 22.5 and 15 gallons of air come from. I feel like a piece was left out in this explanation. This is obviously not homework (well it's housework). Does anyone see what's I'm failing to?

wesley delk
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2 Answers2

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Suppose the original amount of space the air took up (at $60$ PSI) was $V$ gallons. You then remove $7.5$ gallons of water from the tank and the air pressure goes to $40$ PSI, which means that the air in the tank must take up $V+7.5$ gallons at $40$ PSI. So (assuming that the tank is airtight, the temperature does not change, no water flows into the tank, and these are absolute pressures rather than gauge pressures) Boyle's law says that $$ 60V=40(V+7.5) $$ Solving this equation, we have $$ 20V=40(7.5)\\ V=2(7.5)=15 $$ and so the air initially took up $15$ gallons of the tank.

Micah
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  • This also ignores the inflow of water from the actual outside source, which is only logical. If it's a well, the pressure has to drop far enough to trigger the pump before any water comes in. If it's from a water main, the use of a pressure tank is proof that the outside flow is very low... – DJohnM Mar 22 '18 at 03:00
  • Jeez that went off like a lightbulb. For some reason I was using 40(V-7.5) in my calculation. I'm new to stackexchange, and want to take this question further. Do I just post a new one? – wesley delk Mar 22 '18 at 15:42
  • @wesleydelk: I would post a new question, yes. Glad I could help with this part! – Micah Mar 22 '18 at 17:23
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Assuming that the change is slow enough that the temperature in the tank can stay constant, the pressure of the air in this sort of range will be inversely proportional to the volume, as your statement of Boyle's law shows.

$$P_1V_1 = P_2V_2 = k \implies P=\frac{k}{V}$$

etc. So taking the pressure change as absolute (not gauge, which is the difference to atmospheric), a decrease to $\frac 23$ the pressure means the volume of gas as increased to $\frac 32$ its original volume - half as much again as originally.

We are given the absolute increase in volume of air, which is the same as the decease in volume of water through shower use: $7.5$ gal. So the original volume must have been $15$ gal and the increased volume after the shower $22.5$ gal.

If it is later determined that the shower only used $5$ gal, this means that the air volume must have gone up from $10$ gal to $15$ gal to fulfill that $\frac 32$ volume increase requirement implied by the $\frac 23$ pressure decrease.

Joffan
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