Thermal expansion of tank and contents.

AI Thread Summary
The discussion focuses on calculating the final temperature of a steel tank containing oil, considering the thermal expansion of both the tank and the oil. Participants express difficulty in developing a formula that accounts for the expansion of both substances, as the problem involves two unknowns: the final temperature and the change in volume. Some users have attempted to isolate the variables but struggle to reach the correct answer of 74.9°C. One user eventually shares a method for solving the problem, equating the volumetric expansions of the tank and oil to find the final temperature. The conversation highlights the complexity of the problem and the need for careful consideration of both materials' expansions.
caperjay
Messages
6
Reaction score
0

Homework Statement


At 5°c, a steel tank is 6m high, 9m in diameter, and contains oil to the 5.9m level. A heating coil is put into service, causing 3.4m^3 of oil to spill over the top of the tank by the time the final temperature is reached. Calculate the final temperature, if the coefficient of expansion of the oil is 0.00041, and coefficient of expansion for steel is 0.0000165. Note: Do not neglect the expansion of the tank.


Homework Equations


Equation for volumetric expansion for solids. ΔV= V x 3∂ x ΔT
Equation for volumetric expansion fo liquids. ΔV= V x β x ΔT
V= volume ∂=coefficient of expansion, ΔT=change in temperature, β= coefficient of expansion liquid.


3. attempt

I've been working on this one for a while now, having a hard time developing a formula that takes expansion of both into consideration. I've tried solving them seperatly, and I tried using a ratio to solove the problem, but I can't get anything close to the answer, which is 74.9°c.

If someone could just get me onto the right track, i'd like to solve it myself, thanks.
 
Last edited:
Physics news on Phys.org
How are you calculating the final volume of the container? Are you are using the equation for volumetric expansion of solids? I'm not sure that's right, because the container is hollow. So rather than figuring out the change in volume of a solid steel tank, try to find the change in volume of a container made up of sheets of steel.
 
Yes that formula works for solids, as well as hollow vessels. I've used it to solve similar problems. This one is just a bit harder since there's two unknowns, the final temperature, and the change in volume. I know that the oil will expand a lot more than the vessel will, and that the oil starts off with 6.36m^3 less that the size of the tank.
I've also figured out that if the tank didn't expand that the oil would need to increase by 9.76m^3 to spill 3.4m^3, and that a final temperature of 68.37°c, would be required to achieve this.

I am just having trouble including the expansion of the vessel, or deriving a formula for this.

Thanks in advance.

Jay
 
I would love to see the answer to this question! i am stuck on it as well, 3rd's...urgh i have found a few questions in the book that throws a curve ball. I am absolutley stuck as to how to solve this with 2 unknowns, volume and temperature of the steel tank and oil. i have got to the same numbers as above but can not get the right one. maybe they made a typo...found a couple so far.
Thanks
Jaime
I need some serious help...can't let it go!
 
JaimeR said:
I would love to see the answer to this question! i am stuck on it as well, 3rd's...urgh i have found a few questions in the book that throws a curve ball. I am absolutley stuck as to how to solve this with 2 unknowns, volume and temperature of the steel tank and oil. i have got to the same numbers as above but can not get the right one. maybe they made a typo...found a couple so far.
Thanks
Jaime
I need some serious help...can't let it go!

Show what you've tried so that we can see how to help.
 
JaimeR said:
I would love to see the answer to this question! i am stuck on it as well, 3rd's...urgh i have found a few questions in the book that throws a curve ball. I am absolutley stuck as to how to solve this with 2 unknowns, volume and temperature of the steel tank and oil. i have got to the same numbers as above but can not get the right one. maybe they made a typo...found a couple so far.
Thanks
Jaime
I need some serious help...can't let it go!

Hey Jaime,

I've since solved that problem, and from what I've been told from guys at work its more of a 2nd class problem, but anyways, here's how I solved it.

Vtank + ΔVtank = Voil + ΔVoil - 3.4

381.7 + 381.7x3x0.0000012(T-5) = 375.15 + 375.15x0.00041(T-5) - 3.4

From here you just equate the problem and solve for T.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top