Thermal Conductivity and Specific Heat Capacity: Calculating the Time Required for a Volume of Water to Cool

Question

I'm a chemist by training and have experience solving differential equations, though it has been insufficient in solving the problem I have.

Here is the problem:

I brew my own beer and I'm building a cooling apparatus to cool the wort after it has boiled. The cooling apparatus consists of $1/4"$ (d) inch copper pipe (thickness $=0.04"$) that is to be submerged in the wort $(20L)$ with cool water $(10C)$ constantly running through the copper tubing. This is analogous to how a Graham condenser works to cool hot vapours. (In case the dual measurements don't give it away, I'm Canadian. Most everything is metric except our building supplies)

My question:

How long will it take to cool the wort to $37C$? To thermal equilibrium with the cool water $(10C)$ ?

Approach/Assumptions:

1. The velocity of the running water fast enough is such that its temperature is constant $dT_a=0$.

2. The specific heat capacity of water is independent of temperature $dC_s/dT=0$

3. The thermal conductivity is independent of temperature $dk/dT=0$

4. The temperature of the water is uniform.

5. The wort has identical thermal properties to pure water (in reality, the dissolved sugars would alter these values)

Approach: First, an expression for the heat required to cool the wort to a given temperature $T_w$ is derived using specific heat capacity.

$$dq=m×C_s×dT \tag1$$

$$\ \ \ \ q=mC_s(T_w-T_0)$$

$\rm q=heat \ (J)$

$\rm m=mass \ (grams)$

$\rm C_s=specific \ heat \ capacity \ of \ water (J/g)$

$\rm T_0=initial \ wort \ temperature \ (K)$

$\rm T_w= Final \ Wort \ Temperature \ (K)$

Next, using a definition I found for thermal conductivity:

$$dq/dt=k×A/s×dT \tag2$$

where ;

$\rm k=thermal \ conductivity \ of \ copper$

$$=391 W/M^2 \ K$$

$\rm A=surface \ area\ of \ the \ copper pipe$

$$=2\pi×r(h+r)$$

$$h=12'=3.69m\ \ \ \ \ $$

$$ \ \ \ r=1/8"=0.0032m$$

$$A=0.148m^2 \ \ \ \ \ \ \ \ \ \ \ $$

$\rm s=thickness \ of \ copper\ pipe$

$$= 0.04"=0.001024\ m$$

Integrating Eq. 2 with respect to $dT$ yields Eq. 3.

$$q=k×A/s(T_w(t)-T_a)dt \tag3 $$

This is where I begin to get stuck. So I know the total amount of heat that needs to be transferred to cool the wort to temperature $T_w$ (Eq. 1). I am just a bit confused on how to go about integrating Eq. 3 such that I can determine the time required to cool the wort to the desired temperature. I suspect that I need experimental data to fit a function to $T_w(t)$ as I believe $T_w(t)=Ae^{-zt}$ or something similar. Experimental data would allow me to solve $A$ and $z$.

Am I on the right path? What am I missing that would allow me to determine the time?

Answer 1

The analysis of @Puk provides only an extreme lower bound to the amount of time required to cool the wort. This is because the dominant resistance to heat transfer will reside, not within the tube wall, but in the thermal boundary layer situated between the outside of the tube wall and the bulk of the wort fluid. The heat transfer rate (J/s) through this boundary layer will be described by $\dot{Q}=hA(T_{OTW}-T)$, where $T_{OTW}$ is the outside tube wall temperature, T is the wort temperature, h is the heat transfer coefficient, and A is the heat transfer area. In terms of the temperature difference between the water and the wort, this leads to $$\dot{Q}=UA(T_{water}-T)$$where U is the overall heat transfer coefficient (including the tube wall): $$\frac{1}{U}=\frac{s}{k}+\frac{1}{h}$$Puk's analysis can be corrected by replacing k/s in his equations with U.

The magnitude of the outside heat transfer coefficient h is going to depend on the geometry of the bath and cooling coil arrangement, and whether the wort is agitated. If agitation is provided (convective heat transfer), then h will be higher and the cooling time will be less. Otherwise, one must depend on natural convection within the bath, and the cooling time will be longer. Typical values for h in various situations are provided in Transport Phenomena by Bird, et al.

Answer 2

Your Eq. 2 isn't quite right, and there seems to be an abuse of the differential notation going on. You also have the wrong dimensions for $C_s$, the unit should be $\text{J/K/g}$.

The thermal resistance of the pipe wall (assuming $s\ll r$), in analogy with how you would calculate electrical resistance, is $$\frac{s}{kA}.$$ I think you have this right in your equation. But thermal resistance is the proportionality factor between the temperature difference $\Delta T=T_w-T_a$ across the pipe wall, and the heat flow rate $-dq/dt$, so $$\Delta T=-\frac{s}{kA} \frac{dq}{dt}$$ $$ \frac{dq}{dt} = -\frac{kA}{s}\Delta T.$$ This is the same as your Eq. 2, apart from the minus sign (note that $dq/dt$ is negative) and with $dT$ replaced by $\Delta T$ ($dT$ by itself doesn't make sense in that equation). All that remains is to combine this with Eq. 1 and solve:

$$ mC_s \frac{d\Delta T}{dt} = -\frac{Ak}{s}\Delta T.$$ $$\frac{d\Delta T}{dt}+\frac{Ak}{smC_s}\Delta T=0$$ $$\Delta T(t)=(T_0-T_a)\exp\left(\frac{-Akt}{smC_s}\right).$$

The cooling time constant is $$\tau = \frac{smC_s}{Ak}.$$ In practice the wort temperature can be considered to have settled after about $5\tau$, or you can solve the above equation for your target temperature if it is higher than $T_a$.