# Temperature drop in tank

Tags:
1. Apr 1, 2015

### Alex.malh

Hello all,

I'm stuck with following problem:

At our plant we have a tank filled with a liquid maintained on 80°C with a steam coil.
During the night the steam is switched off and the liquid temperature start to drop.
Tank mass not taken into account atm.

mass liquid: 20000 kg
cp liquid: 1.76kJ/(kg °C)
tank surface sides As: 82m²
tank surface top At: 7m²
Heat transfer coeff sides Us: 0.6W/(m² °C)
Heat transfer coeff top Ut: 0.8W/(m² °C)
Tstart: 80°C
Tair: 10°C
Cooling time t: 8h

I used following formulas:
Q=U x A x dT [kJ/s]
E=m x cp [kJ/°C]
Temperature drop = Q/E x t x 3600

Now the heat transfer rate Q is dependable of the dT but that changes in function of time.
The used method is simplified in that aspect and only uses the max heat transfer rate possible in given process.
in other words: this method is linear while reality is not. probably logarithmic i guess.

How can i calculate the temperature drop more accurately?
I'm thinking in direction of thermal equilibrium, get a expression of temperature in function of time of the process.

Thx.

Alex

2. Apr 1, 2015

### Delta²

Have to model the system using a differential equation in order to get more accurate predictions. Suppose T(t) is the temperature as a function of time t. An infinitesimal temperature drop dT(t) is equal to
$dT(t)=\frac{Q}{E}dt$
hence
$\frac{dT(t)}{dt}=\frac{Q}{E}$.

But Q is also a function of temperature difference T(t)-Tair(t) hence a function of time t, so we get
$Q(t)=UA(T(t)-T_{air}(t))$.

Hence the differential equation with unknown the function T(t) is:

$\frac{dT(t)}{dt}=\frac{UA}{E}(T(t)-T_{air}(t))$

if Tair(t) is constant 10C then this simplifies the equation to
$\frac{dT(t)}{dt}=\frac{UA}{E}(T(t)-10)$

3. Apr 1, 2015

### Staff: Mentor

If you don't want to go through the pain of the differential equations, you can do an iterative method with a spreadsheet.

Last edited: Apr 1, 2015
4. Apr 1, 2015

### Tom_K

I think we would need some empirical data to formulate a proper differential equation.
For example, You gave the Tstart and T air but did not give a T finish after 8 hours
Do you have any data for that?
I worked out a (very) rough calculation of how long it would take the tank to lose all of it’s heat energy above 10ºC to the ambient air, assuming the heat loss is linear.
To lose all the heat energy, E = Q
That is, cmΔT = UAt
t (in seconds) = [(1.76 kJ/kgºC) (20000 kg) (80ºC-10 ºC)] / (54.8 m^2 ºC)
= 45,000 seconds 12.5 hours
If the loss is linear in 8 hours the tankT decreases by 45ºC from 80ºC to 35ºC
Do those numbers roughly agree with any actual data you may have?

5. Apr 1, 2015

### Staff: Mentor

This approach is right on target. The only change I would make is that there should be a minus sign on the right hand side of the equation, since the contents are cooling.

Chet

6. Apr 1, 2015

### Delta²

Thanks chet for the correction. I didnt have time to complete my post. The solution to the differential equation assuming Tair is constant is
$T(t)=T_{air}+c_1e^{-\frac{UA}{E}t}$

T(0)=80C so can determine constant c1=70C easily from this initial condition.

But i dont know whether we can safely assume that Tair remains constant, depends on the cooling of the enviroment around the tank. Theoretically at least the air around the tank cant have perfectly stable temperature cause of the continuous heat transfer from the tank.

7. Apr 1, 2015

### Staff: Mentor

There might also be some significant resistance to heat transfer on the water side of the wall, in addition to the outside resistance, especially if the tank is not stirred. What do you think?

I also like Tom_K's idea of calibrating the model parameters to experimental data.

Chet

8. Apr 1, 2015

### Delta²

yes this seems to be true, depending on the type of liquid an whether the tank is stirred or not the temperature will not be the same throughout the volume of the tank.

Last edited: Apr 1, 2015