# Steam Speed from a Heat Ruptured 120 Gallon Water Tank?

PROBLEM: A 120 gallon tank (26" diameter, 60" height) containing 40 Gallons of water is heated up until the tank ruptures from overpressure at 125 PSI (maximum tank rating). How to calculate the initial escaped steam velocity at the ruptured tank, assuming the gash is 2 feet long in the vertical direction? Also how to figure out the "wind speed" at 15 feet away?

MY THOUGHTS: We know at the time of rupture the tank is full of steam pressurized at 125 PSI, but there is a limited air and water molecules inside the tank, so this burst of steam can only travel a finite distance. What law of physics may govern this?

Twigg
Gold Member
If the tank is rated to 125 PSI, I doubt it will rupture at that pressure. I would guess that 125 PSI is well below the pressure at which the tank wall starts to yield. That pressure rating probably has a hefty safety factor to account for real world effects like metal fatigue, corrosion, and manufacturing variability. Not to mention most tanks for pressurized anything have relief valves for the worst case scenario. Supposing the valve failed, if you want a better guess of the pressure at rupturing, find the thickness of the thinnest part of the tank wall (probably the relief valve itself or some other non-uniform feature) and use the hoop stress equation to find the pressure at which that feature would reach its tensile strength.

I think this paper discusses the relevant fluid physics. The methods discussed there will only be approximate because a tank rupturing isn't a very good point source.

seeker11
That pressure rating probably has a hefty safety factor to account for real world effects like metal fatigue, corrosion, and manufacturing variability. Not to mention most tanks for pressurized anything have relief valves for the worst case scenario. Supposing the valve failed, if you want a better guess of the pressure at rupturing, find the thickness of the thinnest part of the tank wall (probably the relief valve itself or some other non-uniform feature) and use the hoop stress equation to find the pressure at which that feature would reach its tensile strength.
All very astute and pragmatic observations! This tank is actually designed to operate only at 38 PSI, it supplies low level pressure in water pipes to prevent backflow. It is built with 16 gauge cold rolled steel wall, so 125 PSI is already 3X of its normal operating range. As you can see, the tank wall just split in two.

What is the best way to estimate the distance the steam can travel before it dissipates? Like if a giant balloon pops, how far away can you still "feel" the burst of escaping air?

Baluncore
How to calculate the initial escaped steam velocity at the ruptured tank, assuming the gash is 2 feet long in the vertical direction?
The speed of the expanding wavefront will be determined by the speed of sound in the steam.

The volume of steam produced when the pressure is released will depend on the volume of hot water in the tank.

The energy released by the expansion will continue outwards through the air, well beyond the volume of steam generated.

seeker11
anorlunda
Staff Emeritus
A steam explosion is unlike an air balloon popping because of the dominant thermal effects. That's even more so if a tank of hot water flashes to steam.

The hot plume will rise until reaching equilibrium with the ambient air, capping the plume in a mushroom cloud. (No atomic explosion needed.)

Lnewqban and seeker11
Twigg
Gold Member
Holy smokes! I hope no one got hurt.

Hard to say how far away you could "feel" the shock. It's easier to say at what point you would experience X pressure above atmospheric. My guess is that you could "feel" an excess pressure of 0.01psi, because I estimate the human body has a cross sectional area around 1000 sq. in. and that yields 10lbs of force, which sounds feel-able.

Per the paper I linked, the pressure should drop as roughly ##R^{-3}## up to a pressure of 1.5atm (37psia). As the crudest approximation, I use Boyle's law for a spherical volume around the 120gal tank. 120gal is 231 cu. in., and 125psig is 140psia. That gives $$\frac{37psia}{140psia} = \frac{231in^3}{\frac{4}{3}\pi r^3}$$ which gives ##r \approx 29in##. The paper also says the pressure drops roughly like ##R^{-2}## at 0.5atm (22psia), so we can do another approximate calculation: $$\frac{22psia}{37psia} \approx (\frac{29in}{r'})^2$$ which gives ##r' \approx 45in##. Then we can do one more approximate step from 0.5atm to 0.01psi using the large distance approximation of ##\frac{1}{R}##: $$\frac{14.71psia}{22psia} = \frac{45in}{r''}$$ which gives ##r''\approx 67in##.

That's a *real* rough guess for the record. It doesn't include the thermal energy of the steam, condensation, or the effect of reflections off the walls. What I did would be more appropriate for a compressed air tank, but maybe someone here can correct for these inaccuracies.

Lnewqban and seeker11
That's a *real* rough guess for the record. It doesn't include the thermal energy of the steam, condensation, or the effect of reflections off the walls. What I did would be more appropriate for a compressed air tank, but maybe someone here can correct for these inaccuracies.
Twigg, fortunately nobody was hurt...Apparently we are in the presence of big brains that I only read stories about.

Yes this rough estimation is what we are looking for. So according to the back-of-the-envelope calculation, there is likely to be less than 0.01 PSI of pressure at a distance of 6 feet from the tank immediately after the rupture, that feels about right for us math-challenged.

Would it be possible then to estimate the amount of pressure the tank needs to be under to generate 5 pound per square feet of force at 15 feet away from the tank at the time of rupture? Would this be physically possible?

This will all make sense once you see the context.

Baluncore
How far will the steam reach if it flash boils and then cools to STP ?
A 120 gal(us) tank of water will weigh 120 * 3.785 = 454.2 kg.
The molecular weight of water is 18 g/mole.
454.2 kg / 0.018 kg = 25233. mole.
The volume of steam at STP will be 25233 mole * 22.4 litre / mole = 565219.2 litre = 565 m3.
Once it cools to STP the radius of the sphere of expanded steam will be; 5.129 m.

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Lnewqban and Twigg
Twigg
Gold Member
The 6ft result was bothering me... seems way too short. I just caught the mistake: ##231in^3## is one gallon not 120gal! Whoops! The steps are all there though if you want to correct that.

Baluncore gives the correct way to calculate how far the steam would spread overall.

As far as getting 5psig at 15ft: $$r'' = \frac{22psia}{20psia} \times r' = 1.1 r'$$ $$r' = \sqrt{\frac{37psia}{22psia}} \times r = 1.3r$$ $$r'' = 1.4r$$ or $$r = 0.7r''$$ Given 15ft, $$r = 0.7 \times 15ft = 10.5ft = 126in$$ $$\frac{37psia}{p} = \frac{27720in^3}{\frac{4}{3}\pi (126in)^3}$$ yields a whopping $$p = 11200psia$$

Either my math is horribly wrong or that's definitely not physically possible

seeker11
jrmichler
Mentor
This tank is actually designed to operate only at 38 PSI, it supplies low level pressure in water pipes to prevent backflow. It is built with 16 gauge cold rolled steel wall
Was that tank in a fire? If it was, then the sheet metal tank wall above water level would heat up fast, while the water would only heat a little. The air temperature inside would increase to a temperature higher than the water temperature and lower than the steel temperature. The air pressure would increase with air temperature. Steel gets weaker at high temperatures. Here is a plot from MIL-HDBK-5J showing how the strength of a particular low alloy steel varies with temperature. It does not apply to the alloy used in your water tank, but does show the trend:

The air pressure rises while the tank wall weakens until it fails. It is likely that you had an air, not steam, explosion. When a pressure vessel fails by excess pressure, the failure mode typically starts as a small crack that spreads at the speed of sound in the steel. The air inside rapidly departs the tank, and quickly expands until reaching ambient pressure.

seeker11 and Twigg
The 6ft result was bothering me... seems way too short. I just caught the mistake: ##231in^3## is one gallon not 120gal! Whoops! The steps are all there though if you want to correct that.
The amount of water inside the 120 gallon tank could NEVER EXCEED 40 gallons, since the tank is used to apply slight static pressure to prevent water backflow. Given that, I need to replace ##231in^3## with ##(40 * 231)in^3## and re-calculate?

Twigg, to make sure I understand ##11200psia## .....does this means the tank needs to be able to withstand 5.6 ton of pressure per square inch in order to achieve 5 psi at a distance of 15 feet? Assuming the calculation is correct.

Was that tank in a fire? If it was, then the sheet metal tank wall above water level would heat up fast, while the water would only heat a little.....The air pressure rises while the tank wall weakens until it fails. It is likely that you had an air, not steam, explosion. When a pressure vessel fails by excess pressure, the failure mode typically starts as a small crack that spreads at the speed of sound in the steel. The air inside rapidly departs the tank, and quickly expands until reaching ambient pressure.
Yes, the tank was heated by a source outside the tank, but it's not clear what type. The fact that the tank showed a big vertical crack down the middle, does this tell us anything about the type of heat that caused the tank overpressure? Meaning was it a "fast heat" (going to high temperature in a split second), or a "slow heat" (going to high temperature in minutes). I suppose if it's "fast heat", then the container wall will likely fail before the water could turn into steam (high thermal mass of water), so resulting in an air explosion as jrmichler suggested. If it's "slow heat", then maybe water has a chance to boil and over pressure the tank, in which case maybe the valves should break off first as Twigg initially suggested?

Baluncore
Meaning was it a "fast heat" (going to high temperature in a split second), or a "slow heat" (going to high temperature in minutes).
There was only a small amount of water in the tank when it was heated by the external fire.
The water in the tank boiled and built up steam pressure, until the tank ruptured, and the remaining liquid water flashed to steam. The fire took some time to heat the tank. It was not an instantaneous process, until the tank ruptured.

There are signs of burning localised to the area surrounding the split in the tank. The coincident position of the burn and the split, suggest that the temperature of that wall rose significantly above the boiling point of water, which weakened the steel.

Internal pressure in a thin walled cylinder causes a hoop stress that is twice the axial stress. That explains why a longitudinal split is seen in the walls of cylinders that rupture due to internal pressure.

Lnewqban and seeker11
There was only a small amount of water in the tank when it was heated by the external fire.
The water in the tank boiled and built up steam pressure, until the tank ruptured, and the remaining liquid water flashed to steam. The fire took some time to heat the tank. It was not an instantaneous process, until the tank ruptured.
Baluncore, one thing that troubled us is if it's a slow burning fire then we would expect to see more melting of the surrounding items, such as the red rubber hoses, or the yellow insulation sheets. Another strange thing is the skinny cylinder immediately to the left showed some charring from the top? The only thing that is clear is the WR-360 got roasted by something really hot.

Could a "fast heat" scenario explain the state of the tank?

Baluncore
Another strange thing is the skinny cylinder immediately to the left showed some charring from the top?
I wondered about that thinner cylinder. Did it contain fuel gas? Has it burnt off hoses or blown out a fusible plug designed to relieve pressure during a fire?
Was the split tank mounted above that thin container?
Was the split tank mounted vertically or horizontally? How far has the split tank moved?
What is a WR-360?

seeker11
I wondered about that thinner cylinder. Did it contain fuel gas? Has it burnt off hoses or blown out a fusible plug designed to relieve pressure during a fire? Was the split tank mounted above that thin container?
Was the split tank mounted vertically or horizontally? How far has the split tank moved? What is a WR-360?
The skinny cylinder on the immediate left of the split tank is an EMPTY heavy duty gas cylinder, it was used to store propane, but it contained just air at the time of fire, and it does not appear to be damaged. Both the split tank (WR-360) and the skinny cylinder stood vertically along the wall, next to each other. After the incident, the split tank did not really "move", its wall just split and it slumped over and leaned to the right (as in the picture).

The WR-360 is the "split tank", it is described HERE ...it is a lightly pressurized tank (38 PSI) with a rubber bladder inside that may contain up to 40 gallons of water. It is connected to a well water pipe system, when the piping water pressure is low, it pushes water into the system to prevent backflow. The tank is built to withstand a maximum pressure of 125 PSI.

As far as getting 5psig at 15ft: $$r'' = \frac{22psia}{20psia} \times r' = 1.1 r'$$
Twigg, I apologize, I am trying to follow the math....shouldn't the above r ratio have a square or cubic relationship? I'm trying to match it to below, both have some type of exponent...

...I use Boyle's law for a spherical volume around the 120gal tank. 120gal is 231 cu. in., and 125psig is 140psia. That gives $$\frac{37psia}{140psia} = \frac{231in^3}{\frac{4}{3}\pi r^3}$$ which gives ##r \approx 29in##. The paper also says the pressure drops roughly like ##R^{-2}## at 0.5atm (22psia), so we can do another approximate calculation: $$\frac{22psia}{37psia} \approx (\frac{29in}{r'})^2$$

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Twigg
Gold Member
In my first calculation, I knew the tank pressure, and was calculating the distance at which you'd feel a given "wind pressure". In this second calculation, I knew the distance and "wind pressure", and I was calculating the corresponding tank pressure, so I did the same calculation in reverse. Started with a 1/r trend, then 1/r^2 as I got closer in, and then 1/r^3.

seeker11
Baluncore
The only thing that is clear is the WR-360 got roasted by something really hot.
That fuel gas cylinder must be the prime suspect. It is burnt heavily at the top, but not on the sides, and it was placed with the top close to where the heated and split patch on the WR-360 would have been. Can you examine the fusible plug at the top of that gas cylinder?

That fuel gas cylinder must be the prime suspect. It is burnt heavily at the top, but not on the sides, and it was placed with the top close to where the heated and split patch on the WR-360 would have been. Can you examine the fusible plug at the top of that gas cylinder?
Unfortunately, all we have are the pictures.

If the WR-360 was in a propane explosion, will it rupture as it appear in the picture? I guess the air will instantaneously reach the combustion temperature, which weakens the 1/16 inch steel wall, and the inside pressure differential will create the rupture?

Baluncore
If the WR-360 was in a propane explosion, will it rupture as it appear in the picture?
The WR-360 exploded due to internal pressure. I see no way it could possibly have filled with propane-air mix, and then been ignited.

A fuel gas cylinder should be stored in a ventilated area, not enclosed inside a building structure. That is also true for “empty” cylinders. A child could have opened, and then not completely closed the propane valve. There may have been a propane fire at some stage, but there was no propane explosion. If a general fire in the area caused the fusible plug in the propane cylinder to melt, the propane would escape and burn like a blowtorch. If the propane cylinder was almost empty, then heating the cylinder could drive out the remaining gas.

There was a general fire in that space. If the space was poorly ventilated, it may well have burned quietly, without sufficient oxygen, reaching high temperatures over a long period of time. The plastic on the floor was relatively cool while the thick smoke at the top was very hot. That “stratified” profile would explain the variation in damage. The steam explosion of the WR-360 would have ventilated the area by bursting the enclosure, which may have removed the source of heat and so extinguished the fire.

There is insufficient evidence or information to come to any conclusion. All we can do is keep expanding the possibilities.

Lnewqban
Lnewqban
Gold Member
Note the deformation of the weld line between the cap and the cylindrical wall.
It seems to me that a pre-explosion fire weakened that area of the tank, as a shape of triangular metal got peeled back from there by escaping steam.

That is not the typical shape of metal failure induced purely by internal pressure.

https://en.m.wikipedia.org/wiki/Steam_explosion

Baluncore