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I am having this problem that some LPG tanks don't provide fuel and other does..

I was told that the problem is whenever the ambient temperature is low (so is the gas), the pressure drops as well and hencely 2 bar (say it is the 10%) won't feed up..

But is this really the case? I think when the pressure is 10% in hot surrounding, it should have higher mass than when its surrounding is colder with the same pressure (10%-say 2 bar). So if it is really 10% it should always feed up, the difference is it has lower mass when it is hot. But if it really 10% and feed up for some tanks but not the other, so the issue is not about the temperature and pressure, maybe the installation.

What do you think?

And how do we get fed by gas after the pressure drops to 10%? do we use pump or heater for example? I was thinking to fill the tank up to 30 bars but it is not allowed to exceed 25 bars..

Another issue; say we fill the tank in the Winter up to 20 bars, in Summer we also fill it up to 20 bars but surely the mass of the filled gas in Winter is greater than in Summer. So would it last more in Winter -assuming same use?

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    $\begingroup$ The mass won't change with temperature, but the relationship between volume, pressure and temperature does, so when the volume is constant as with a tank then the pressure changes with temperature. $\endgroup$ – Solar Mike Dec 6 '20 at 13:11
  • $\begingroup$ I will ask that this question be improved. What is 10% mean? Relative to what? 10% of the initial pressure measured at the point that it is completely filled? 10% of its maximum fill pressure? 10% of the pressure needed at the exit point of the delivery tube? 10% of WHAT? Otherwise, the reasoning that follows as a proposed solution remains confusing not just because it is convoluted but also because the initial problem is ambiguously defined. $\endgroup$ – Jeffrey J Weimer Dec 8 '20 at 15:13
  • $\begingroup$ yea 10% of initial pressure.. LPG tanks have maximum pressure of 25 bar, after the the valve leaks the gas. So suppose it is filled untill the pressure reaches 20 bar, so the 10% is when it reaches 2 bar. $\endgroup$ – alFeraas Dec 8 '20 at 15:51
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LPG has to evaporate before it can flow out of the tank (the L stands for liquified). This is endothermic, and the heat has to be replaced from the surroundings through the tank wall. When drawing from the tank, the LPG is colder than the surroundings. When it's cold and the tank gets low, there isn't enough surface area to transfer heat, and the temperature drop between the surroundings and the LPG and gets bigger. Eventually, the LPG gets so cold it won't evaporate at the desired demand rate. Use bigger tanks, or connect several in parallel.

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  • $\begingroup$ What about the consumption then? Say I I consume the first day till the pressure drops to 97% in Summer, then in Winter with the same exact use, will the pressure drop 3% in the first day? (since the number of molecules should be higher in Winter) $\endgroup$ – alFeraas Dec 7 '20 at 10:16
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Pressure Drop

The gas goes through pipes and valves to get to the outlet or point of use. The gas must be pushed through the pipes and valves, meaning that it must be at a higher pressure than its outlet pressure. The difference is the pressure drop through the pipes and valves. The pressure drop is a function of the shape of the pipes and valves. Narrower pipes have lower pressure drops per length for example. The pressure drop is also a function of the vertical distance between the source and the outlet, primarily as $\Delta p = \rho g \Delta h$.

Source Pressure

Assume that you have $n$ moles of gas in a container with constant volume $V$. Assume that it is an ideal gas so that $p \bar{V} = R T$ where $p$ is its pressure, $\bar{V} = V/n$, $R$ is the ideal gas law constant, and $T$ is the temperature. As you see, when $T$ decreases, $p$ will decrease.

Summary

As the gas in the tank gets colder, its pressure $p$ drops below the value that is needed to overcome the pressure drop $\Delta p$ to reach the outlet. Alternatively, at any given temperature $T$, as the amount of gas (moles or mass) in the tank decreases, its molar volume $\bar{V}$ increases. The inverse of molar volume is molar density $\rho_n$ (mols/m$^3$). As the gas becomes less dense in the tank, $\bar{V}$ increases and the pressure $p$ decreases.

Caveats

When the gas in the tank is not ideal, the basic modification is that you must use a real gas equation of state. The above findings do not change. What changes is the temperature or density of the gas where $p$ is too low to support the required $\Delta p$.

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  • $\begingroup$ The problem is when the tank reaches 10% in hot cities it still can feed, but not in cold cities. But I think the tanks in cold cities should contain more gas, right? And if we want to get fed after the 10%, what is recommended to do? use pump, heater? $\endgroup$ – alFeraas Dec 6 '20 at 16:48
  • $\begingroup$ I cannot understand what you mean then by the 10% rule. 10% of pressure? 10% of gas amount? I can modify accordingly once you clarify accordingly, preferably in your original posting. $\endgroup$ – Jeffrey J Weimer Dec 6 '20 at 16:58
  • $\begingroup$ yes 10% of the starting pressure, say we fill the tank up to 20 bar, so how to get fed when it reaches 2 bar? $\endgroup$ – alFeraas Dec 6 '20 at 17:01
  • $\begingroup$ Please modify your original question to make this absolutely clear. 10% of STARTING pressure is not the same as 10% of amount is not the same as 10% of external pressure. Once this is clear, I can modify to explain. $\endgroup$ – Jeffrey J Weimer Dec 6 '20 at 17:10

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