# Closed System

#### Jaybee1

If I have a semi-rigid hot water heater tank with a volume of 55 gals, which was initially dead state conditions (14.7 psia, 75 °F). The contents are then heated until the tank ruptures at 300 psia. How do determine the temperature at the time of rupture and the fractional increase in volume (%)

I'm just confused with the word semi-rigid. I would appreciate some help on how to go about this. Thank you

#### THERMO Spoken Here

Jaybee1, I agree - not clear. Perhaps "semi-rigid" means rigid until not rigid, that is ruptured.
But the author implies tank volume change with the event. I can't help. What text did this premise?

However, just by way of interest, check this out. Pressure Cooker

topsquark

#### studiot

Well I would use International Steam Tables for this.
Ordinary Physics gas law calculations are not accurate enough for water which is an exceptional substance.

So to begin, the initial pressure and temperature are irrelevant.
The water is heated in a semi rigid container so any steam generated will be saturated vapour confined by the liquid (which will be at the same pressure).
Effectively semi rigid means that the vessel undergoes negligable expansion, but does not pretend to be ideal.
My tables give a temperature of 417 F for a saturated water substance at 300 psi.

To calculate the expansion % after bursting, the table gives the specific volume of the water at this temperature and pressure as 0.018896 cuft /lb and for vapour
1.5442 cuft / lb.

You would have to decide if you are going to say all the water turns to steam on release at bursting.
Then the ration of these two figures will give the expansion %

#### Jaybee1

Well I would use International Steam Tables for this.
Ordinary Physics gas law calculations are not accurate enough for water which is an exceptional substance.

So to begin, the initial pressure and temperature are irrelevant.
The water is heated in a semi rigid container so any steam generated will be saturated vapour confined by the liquid (which will be at the same pressure).
Effectively semi rigid means that the vessel undergoes negligable expansion, but does not pretend to be ideal.
My tables give a temperature of 417 F for a saturated water substance at 300 psi.

To calculate the expansion % after bursting, the table gives the specific volume of the water at this temperature and pressure as 0.018896 cuft /lb and for vapour
1.5442 cuft / lb.

You would have to decide if you are going to say all the water turns to steam on release at bursting.
Then the ration of these two figures will give the expansion %

This makes a lot sense. Thanks a lot

#### Woody

I think that by semi-rigid, they are implying that the tank will expand to match the volume of the water.
in other words the tank is rigid enough to hold the water (until it ruptures)
but not rigid enough to compress the water, and thus alter your calculations of increase of the volume of the water due to the temperature

#### Jaybee1

Well I would use International Steam Tables for this.
Ordinary Physics gas law calculations are not accurate enough for water which is an exceptional substance.

So to begin, the initial pressure and temperature are irrelevant.
The water is heated in a semi rigid container so any steam generated will be saturated vapour confined by the liquid (which will be at the same pressure).
Effectively semi rigid means that the vessel undergoes negligable expansion, but does not pretend to be ideal.
My tables give a temperature of 417 F for a saturated water substance at 300 psi.

To calculate the expansion % after bursting, the table gives the specific volume of the water at this temperature and pressure as 0.018896 cuft /lb and for vapour
1.5442 cuft / lb.

You would have to decide if you are going to say all the water turns to steam on release at bursting.
Then the ration of these two figures will give the expansion %
@studiot.
But the question is asking for a fractional increase (%) in volume just before rupture. Won't that be a Sat. mixture. And if I need to find the exergy just before rupture, won't it be Saturated mixture internal energy values also?

#### studiot

How do determine the temperature at the time of rupture and the fractional increase in volume (%)
Not just before rupture.

I don't know if you copied the exact words of the question (or all of it), but It was poorly phrased an open to interpretation.

#### Jaybee1

Not just before rupture.

I don't know if you copied the exact words of the question (or all of it), but It was poorly phrased an open to interpretation.
@studiot
See below the whole thing. Thanks for your input. You have been so helpful
Assume a 55 gal. hot water heater is initially filled with liquid water at dead state conditions (14.7 psia, 75 °F). The contents are then heated until the tank ruptures at 300 psia. Determine: a) the temperature of the water when the system ruptures (F°), b) the fractional increase in tank volume (%) just before rupture, c) the exergy of the water just prior to rupture (BTU), d) if just 1% of the exergy is converted to kinetic energy, what initial speed and maximum altitude would the tank obtain after rupture (ft/s, ft)

#### studiot

Well I am still not impressed with the question but let us try again.

Attacheded is a short form steam table.

At 1atm it says the specific volume is 0.01672 cuft/lb

At 300psi it gives .018890 cuft/lb and a temperature of 417F

So the % increase in volume of water is (.01890 - .01672) / .01672 *100 % which is about 13%.

Can you read the energy table of the water under these conditions and complete the last part ?
You will need to convert 55 gals to lbs (US gals I presume) to get total energy as asked for

This is why UK gallons were better than US ones - a little ditty I learned in primary school

A pint of pure water
weighs a pound and a quarter

So a UK gallon weighs 10 lbs.