A propane tank with 75 gallons of liquid propane is at 50 F in a sealed...

A propane tank with 75 gallons of liquid propane is at 50 F in a sealed tank. The tank develops a significant leak and propane escapes and vaporizes in a short period of time. To solve this problem you must use the propane tables not the common liquid properties.

What is the temperature near the hole in the propane tank due to the escaping propane in Fahrenheit to three significant digits? . Include a minus sign if appropriate.

How many Btus of energy are needed to vaporize the entire 75 gallons of liquid propane? . Give your answer to three significant digits in Btus.

Expert Solution

When the propane is liquified at stored inside a sealed tank at 50 F, the absolute pressure exerted on gas is very high. As soon as it encounters a leak, the liquid is subjected to atmospheric pressure. At this pressure, Propane cannot exist in liquid form and hence starts evaporating. This is similar to boiling of water. The water boils at 212 F at atmospheric pressure. However if we reduce the pressure on it, say, by half, then the water would boil at much lower temperature.

Since the gas which is escaping is subjected to atmospheric pressure, its temperature will be equal to saturation temperature at atmospheric pressure.

Now corresponding to the pressure of 14.7 psia (atmospheric), from Propane tables, the saturation temperature of propane is -44.284 F.

Therefore, the temperature near the hole in the Propane tank would be -44.2 F

The liquid Propane would absorb an amount of energy equal to latent heat of evaporation at 14.7 psia.

From the Propane tables, it can be easily read at 14.7 psia as 183.018 Btu/lb.

Total volume of Propane = 75 gallons.

Density of liquid propane at 50 F is 4.29 lb/US gallon.

Total mass:

Thus the total energy needed to vaporize the entire 75 gallons would be E:

E = 58800 Btu

Any feedback is appreciated.

samet mamat answered 2 years ago

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