T change transient convection system

[florentina1980]

Hi, here is my question

A big cubic plastic bag of 2500L containing water is stored in a room initially at 4°C. The cubic plastic bag is standing up thanks to a metal box of 1 cm thickness.
What will be the raise of temperatue of the water within the bag following 4h at 21°C (sudden raise of room T from 4 to 21°C) and assuming that water was at 4°C before the raise of temperature.

Is the problem easy to solve? What are the equations?
Any assumptions to simplify the problem is welcomed, since I do not want to solve my problem using finite elements.

Thanks a lot for your help.

 

[zekeman]

Is this homework?

 

[florentina1980]

Of course not!

I am trying to evaluate if we have to scrap our buffer solution, since we are producing vaccine.

I need to evaluate the impact of that temperature change on our buffer solution.

I did not find any simple equation in the literature.

 

[corus]

Wouldn't it be easier just to put the bag into a room for 4 hours and measure the temperature?

Hoping to say Tata

 

[florentina1980]

no, it's not possible under a GMP pharmaceutical environement...testing is not that easy to perform.
Is this problem relatively complex to solve (would it require finite elements or CFD calculation?)

 

[ione]

The problem is not easy.

This is a simplified solution of the problem (lump model for transient analysis that assumes there is a negligible internal resistance to heat flow, or that is to assume heat conduction through the water mass is so fast that at each time the whole mass is at the same temperature, though temperature changes at each time).
The temperature profile follows the equation below, with boundary conditions that at time t = 0 water temperature T(0) is equal to Ti = 4°C and at a time t --> ? water temperature T(?)--> Tf = 21 °C.

T(t) = Tf + (Tf - Ti)*exp[-A*h*t/(m*cp)]

Where:

T(t) = water temperature at time t
Ti = starting temperature = 4°C
Tf= final temperature = 21 °C
A = surface area of plastic bag
h = convection coefficient (approx 10 W/[m^2*°C])
m = water mass (2500 kg)
cp = water specific heat (4.185 kJ/[kg*°C])
t = time

Be aware this approach will give you a rough estimate as conduction through the water mass is not negligible.

 

[cloa]

Couldn`t you do an equivalent (obviously not the same) test in the work office?

 

[IRstuff]

Is the metal box around the bag? Your statement is unclear: "bag is standing up thanks to a metal box"

In any case, what is the intent of the calculation? How accurately do you need the answer? There are sufficient differences in what the natural convection coefficient is to make a fairly substantial variance in the potential answer.

Are you trying to warm it up, or trying to keep it from warming up?

Additionally, what is the thermal radiation environment and what the emissivity of the box/bag?

These questions are the stuff of endless controversy and discussion among analysts, and a simple measurement would cut to the chase much more quickly and effectively.

TTFN

FAQ731-376

 

[IRstuff]

I think that you had previously posted this question and it got deleted because it was posed using homework-like language.

You can calculate the effective time constant, based on the thermal mass and the apparent convection and radiative coefficients. You'll find that the time constant can be pretty much anything you want, depending on what you use as inputs.

TTFN

FAQ731-376

 

[florentina1980]

Thanks for your answers!

Again, this is not homework, it is a real situation I am facing up in a production environnement, and if I was using homework-like words, this comes from assumptions I made to explain the situation as simple as possible.

As a process engineer, I have to evaluate the degradation of products within 1500L plastic bag surrounded by metal box following an HVAC shutdown. By first, I was thinking to evaluate the T reached within the bag after 4h of HVAC shutdown. and then calculate product degradation (this is another issue)
I assumed that
- T of the surrounding environement reached suddenly 21°C
- I have a cubic bag of 1500L containing water-like liquid

The calculation is an approximation: I would like to know if during the 4h breakdown, T reached within the liquid is more closer to 4 or 21°C??
With the equation you gave me, it seems that the effect of breakdown is negligeable (T around 7°C). Does is seem ok?

 

[ione]

florentina1980,

There was an error in the equation I have previously posted, that you have for sure discovered

T(t)=Tf - (Tf - Ti)*exp[-A*h*t/(m*cp)]

If we are talking only about natural convection [h = 10 W/(m^2*°C)] the temperature will be closer to the starting temperature (4 °C) than to the new ambient temperature (21 °C) after 4 hours (Note you will have a temperature gradient inside your liquid mass, which will become colder going from the outer surface towards the centre).

 

[florentina1980]

Thanks a lot Ione, your help was useful, I will present this approximation of the situation to say that ther is no dramatic impact.

 

[zekeman]

"If we are talking only about natural convection [h = 10 W/(m^2*°C)] the temperature will be closer to the starting temperature (4 °C) than to the new ambient temperature (21 °C) after 4 hours (Note you will have a temperature gradient inside your liquid mass, which will become colder going from the outer surface towards the centre)"

Ione,

Not to nitpick a good solution but

1) your lumped system solution is more accurate than you think since internal convection will dominate the internal and external heat transfer process, making the temperature inside more evenly distributed than you suggest.

2)I get a time constant(I assumed a cubical envelope and 5 sides convecting heat) of 6 hours; from which, after 4 hours the math shows a temperature of 12 degrees.Please check this against your calculations.

 

[ione]

Zekeman,

Your comments are always appreciated and I take them in regard.

Further to the temperature reached after 4 hours (14400 s) I’ve found a different result

Cubic container (L = 1.145 m; V = 1,5 m^3=1500 l; A=7.87 m^2)
Mass = 1500 kg
Specific heat = 4186 J/(kg*°C)
Heat transfer coefficient h = 5 W/(m^2*°C)

T(14400)=21-(21-4)*EXP(-7,87*14400*5/(4186*1500))=5.47 °C


With h = 10 W/(m^2*°C)

T(14400)=6.81 °C

 

[zekeman]

Ione,

You're right. I used 2500cc instead of 2500liters for the bag so my time constant was out by a factor of 10.
Sorry to inconvenience you.