Spec'ing the power of heater for an atypical heat exchanger problem

Homework Statement: Calculate the power of heating source required in a heat exchanger
Relevant Equations: Heat transfer for LMTD heat exchanger.

I have a real-world problem whereby:

Water is flowing at a constant flow rate of 10 mL/min through a PVC tube, inner radius of 1.25mm and outer radius of 2mm, that is surrounded by an aluminium shell, inner radius of 2mm and outer radius of 4mm. The length of the tube is 0.2m. Water enters at a temperature of 20 degC and must exit at a temperature of 37 degC. The aluminium shell is heated by an external heat source with an unknown power rating, and the temperature of the aluminium is regulated by a temperature feedback loop (thermocouple). The outside environmental temperature is at room temperature. Determine the power of the heat source required.

I have made the following assumptions to simplify the problem:

1. The temperature of the aluminium can be assumed to remain constant, as the temperature feedback loop is designed to keep it a fixed temperature. Additionally, the thermal heat transfer and losses to the surroundings are considered to have negligible impact on the temperature of the aluminium.
2. Heat transfer between the heat source and the aluminium can also be assumed to be 100% efficient (to simplify the problem).

And I have illustrated the problem in the diagram below:



Which I have simplified with my above assumptions to:



I approached this problem by assuming that it works similarly to a parallel flow heat exchanger, using the LMTD method as below:



I then solved to find U:



Before finding the theoretical rate of energy transfer needed to raise the temperature of the flowing water:



Knowing U, Q and A, I rearranged the LMTD heat transfer equation to solve for the temperature of the aluminium:



I would like to know:

1. If this method for solving the problem is theoretically valid, or if you would have approached it another way.
2. How you might go about determining the minimum power of the heater required to keep the aluminium at this temperature. This will need to balance the power supplied with losses to the environment and heating the fluid.

Thanks!

Here is my take on this.

$$q=q_{in}+q_{out}$$where ##q## is the heat generate per unit length of heater, ##q_{in}## is the inward heat flow per unit length of heater, and ##q_{out}## is the outward heat from per unit length of heater. $$q_{out}=2\pi r_3h_0(T_3-T_{\infty})$$$$q_{in}=2\pi k_{Al}\frac{(T_3-T_2)}{\ln{r_3/r_2}}=2\pi k_{PVC}\frac{(T_2-T_1)}{\ln{r_2/r_1}}=2\pi r_1h_{in}(T_1-T)$$where ##T_{\infty}## is the surroundings temperature and T is the fluid temperature. Solving the above set of equations for ##q_{in}## gives $$q_{in}=2\pi r_1 U_{inner}[T^*+T_{\infty}-T]$$where $$T^*=\frac{q}{2\pi r_3h_0}$$with ##T^*## a constant temperature parameter.

The differential heat balance is $$\dot{m}C\frac{dT}{dx}=2\pi r_1 U_{inner}[T^*+T_{\infty}-T]$$

OK so far?

Hi Chester, I understand your approach and follow what you have done, however I have not yet quite worked through the derivation of Q(in) in terms of T*, T(atm) and T. I will have another go later this evening.

It's not my intention to make the problem harder than it already is. But are you sure you can consider the aluminum tube to have a uniform temperature distribution?
Do you have a few thermocouples along the length? Do you have separate resistances and you control how much power goes into each of them?

My plan was to just remain in the background and try to refresh what I know about these kinds of problems from the conversation but I found that first assumption a little strange. It feels it'd be possible to obtain an answer without it although it'd be harder. Although maybe it's irrelevant. Having an estimate, a control loop, and available power to expend you'd be perfectly OK.

Juanda said:

It's not my intention to make the problem harder than it already is. But are you sure you can consider the aluminum tube to have a uniform temperature distribution?
Do you have a few thermocouples along the length? Do you have separate resistances and you control how much power goes into each of them?

My plan was to just remain in the background and try to refresh what I know about these kinds of problems from the conversation but I found that first assumption a little strange. It feels it'd be possible to obtain an answer without it although it'd be harder. Although maybe it's irrelevant. Having an estimate, a control loop, and available power to expend you'd be perfectly OK.

The assumption is that the temperature gradient in the radial direction is much larger than the temperature gradient in the axial direction. So axially heat conduction along the layers is typically neglected compared to the radial heat conduction. That is all that is being done here.

jackcarroll434 said:

Hi Chester, I understand your approach and follow what you have done, however I have not yet quite worked through the derivation of Q(in) in terms of T*, T(atm) and T. I will have another go later this evening.

The simplest way to do the algebra is to express each of the temperature differences in terms of ##q_{in}## or ##a_{out}## and then add all the temperature differences for ##q_[in]## together to eliminate the intermediate boundary temperatures and get ##T-T_{infty}## in terms of ##q_{in}##

As per your suggestion, I found
$$ T_1 = \frac{q_{in}}{2 \pi r_1 h_i} + T$$ $$ T_2 = \frac{q_{in} \ln \frac{r_2}{r_1}}{2 \pi k_{pvc}} + \frac{q_{in}}{2 \pi r_1 h_i} + T $$ $$ T_3 = \frac{q_{in} \ln \frac{r_3}{r_2}}{2 \pi k_{al}} + \frac{q_{in} \ln \frac{r_2}{r_1}}{2 \pi k_{pvc}} + \frac{q_{in}}{2 \pi r_1 h_i} + T$$

Therefore, ## T_3 = q_{in} R_T + T ##, and as ## UA = \frac{1}{R_t} ##, we can say that $$ T_3 = \frac{q_{in}}{UA} + T $$

Substituting ## T_3 ## into ## q_{out} ## you get:
$$ q_{out} = 2 \pi r_3 h_o (\frac{q_{in}}{U A} + T - T_{\infty})$$
Rearranging this I end up with $$ q_{in} = 2 \pi r_1 U_{inner}(\frac{q_{out}}{2 \pi r_3 h_o} + T_{\infty} - T) $$
Noting that I get ##q_{out}## here rather than ##q## like you did. Do you see an obvious error in my approach?

jackcarroll434 said:

As per your suggestion, I found
$$ T_1 = \frac{q_{in}}{2 \pi r_1 h_i} + T$$ $$ T_2 = \frac{q_{in} \ln \frac{r_2}{r_1}}{2 \pi k_{pvc}} + \frac{q_{in}}{2 \pi r_1 h_i} + T $$ $$ T_3 = \frac{q_{in} \ln \frac{r_3}{r_2}}{2 \pi k_{al}} + \frac{q_{in} \ln \frac{r_2}{r_1}}{2 \pi k_{pvc}} + \frac{q_{in}}{2 \pi r_1 h_i} + T$$

Therefore, ## T_3 = q_{in} R_T + T ##, and as ## UA = \frac{1}{R_t} ##, we can say that $$ T_3 = \frac{q_{in}}{UA} + T $$

Substituting ## T_3 ## into ## q_{out} ## you get:
$$ q_{out} = 2 \pi r_3 h_o (\frac{q_{in}}{U A} + T - T_{\infty})$$
Rearranging this I end up with $$ q_{in} = 2 \pi r_1 U_{inner}(\frac{q_{out}}{2 \pi r_3 h_o} + T_{\infty} - T) $$
Noting that I get ##q_{out}## here rather than ##q## like you did. Do you see an obvious error in my approach?

No. This is what I got. Now substitute ##q_{out}=q-q_{in}##.

Chestermiller said:

No. This is what I got. Now substitute ##q_{out}=q-q_{in}##.

If I substite ##q_{out}=q-q_{in}## then: $$ q_{in}=2 \pi r_1U_{inner}( \frac{q-q_{in}}{2 \pi r_3 h_o} + T_\infty - T)$$
$$q_{in}= \frac{r_1U_{inner}q}{r_3h_0} - \frac{r_1U_{inner}q_{in}}{r_3h_0} + 2 \pi r_1U_{inner}T_\infty - 2 \pi r_1U_{inner}T$$
Unless the term ## \frac{r_1U_{inner}}{r_3h_0}## is insignificantly small and can be considered 0, in which case the ## q## term would also be considered 0, then I do not see how to get a single ##q_{in}## term on the LHS of the equation.

From your next-to-last equation,
$$\frac{q_{in}}{2\pi r_1U_{inner}}+\frac{q_{in}}{2\pi r_3 h_0}=\frac{q_{in}}{2\pi r_1 U}=\left[\frac{q}{2\pi r_3h_0}+T_{\infty}-T\right]$$ $$q_{in}=2\pi r_1 U\left[\frac{q}{2\pi r_3h_0}+T_{\infty}-T\right]$$$$q_{in}=2\pi r_1 U\left[T^*+T_{\infty}-T\right]$$where $$\frac{1}{U}=\frac{1}{U_{inner}}+\frac{r_1}{r_3 h_0}$$and$$T^*=\frac{q}{2\pi r_3h_0}$$

Ah I see. It makes sense now that you have introduced the new term $$U = \frac{1}{\frac{1}{U_{inner}}+\frac{r_1}{r_3h_0}}$$

So this would make the differential equation: $$\dot{m}C\frac{dT}{dx} = 2\pi r_1U[T^*+T_\infty - T]$$

I now follow

jackcarroll434 said:

Ah I see. It makes sense now that you have introduced the new term $$U = \frac{1}{\frac{1}{U_{inner}}+\frac{r_1}{r_3h_0}}$$

So this would make the differential equation: $$\dot{m}C\frac{dT}{dx} = 2\pi r_1U[T^*+T_\infty - T]$$

I now follow

What do you get for T* and for qL? What fraction of the heat from the heater went into heating the liquid?

Putting ## dx ## on the RHS and the ##T## terms on the LHS, integrating between the start and end boundary conditions of ## T=20 ^{\circ}C ## at ## x=0##, and ## T = 37 ^{\circ}C## at ## x=0.2m##, I get a value of 72.36 W for ## q ##. How does this compare to your solution?

jackcarroll434 said:

Putting ## dx ## on the RHS and the ##T## terms on the LHS, integrating between the start and end boundary conditions of ## T=20 ^{\circ}C ## at ## x=0##, and ## T = 37 ^{\circ}C## at ## x=0.2m##, I get a value of 72.36 W for ## q ##. How does this compare to your solution?

Before we do that, we have a notational issue. In your original development, you defined Uinner, which involves the summation of 4 terms. But that Uinner doesn't seem consistent with what you call Uinner in post #9, which seems to involve the summation of only the first 3 terms. I continued this version is subsequent posts, but this incorrectly seems to double count the outside resistance. Can you please resolve this?

Interesting - since your first post (post #2) I've been working on the basis that $$ U_{inner} = \frac{1}{ \frac{1}{h_i} + \frac{r_1 ln( \frac{r_2}{r_1})}{k_{pvc}L} + \frac{r_1 ln( \frac{r_3}{r_2})}{k_{alu}L} }$$
Therefore I do not think the outside resistance has been coutned twice, as it's only counted in the ##q_{out}## term.

jackcarroll434 said:

Interesting - since your first post (post #2) I've been working on the basis that $$ U_{inner} = \frac{1}{ \frac{1}{h_i} + \frac{r_1 ln( \frac{r_2}{r_1})}{k_{pvc}L} + \frac{r_1 ln( \frac{r_3}{r_2})}{k_{alu}L} }$$
Therefore I do not think the outside resistance has been coutned twice, as it's only counted in the ##q_{out}## term.

But the Uinner in post #1 includes the 4th term.

What we are now calling U in posts 10-12 involves all 4 terms, right? The 26.9 value in post #1 is for this U, right?

I get 14.5 W for the heater power. This agrees with your value of 72.36 W/m for q. MyT* is 289 K

I would ignore the value and derivation I used in post #1. Since post #2 we have modelled the total heat supplied as the sum of heat in and heat out. The heat out term (##q_{out}##) deals with the convective losses to the environment (1 term), whereas the heat in (##q_{in}##) term deals with conduction through 2 materials and convection to the fluid (3 terms).

In post #7, ##q_{out}## is derived with all all 4 resistive terms included. Note that it would probably be clearer if I had labelled ##U## as ##U_{inner}## in that equation and the one above (##T_3##). From that point onwards the equations have only been rearranged to isolate ##q_{in}##. I am not seeing where any of the resistive terms are duplicated.

For my calculation in post #13 I used: $$ U = \frac{1}{ \frac{1}{U_{inner}}+ \frac{r_1}{r_3h_0}}$$ $$ U = \frac{1}{ (\frac{1}{h_i} + \frac{r_1 ln \frac{r_2}{r_1}}{k_{pvc}} + \frac{r_1 ln \frac{r_3}{r_2}}{k_{alu}} ) + \frac{r_1}{r_3h_0}}$$
Plugging the constants in from post#1 I get ##U = 26.91##

My T* came to 287.91 K - slightly different

Chestermiller said:

MyT* is 289 K

jackcarroll434 said:

My T* came to 287.91 K - slightly different

As we used to say in the Missile & Armament Engineering Lab, "Oh well, close enough for Government work."

jackcarroll434 said:

My T* came to 287.91 K - slightly different

Watch your sig figs there. The inputs you specified in your original post are pretty loose:
h_inner =500 W/m ² K
h_outer =10 W/m ² K
k_PVC =0.15 W/mK
k_Alu =240 W/mK

By the way, how did you establish the heat transfer coefficients?

The analytical results can vary quite a bit if the assumed values are off. Experimentation is often required to get the desired results (in this case, the water outlet temp). I would at least include a means of varying the heater power such that it can be adjusted to yield the target water outlet temperature.

Chestermiller said:

I get 14.5 W for the heater power. This agrees with your value of 72.36 W/m for q. MyT* is 289 K

Thanks for the help Chester!

The heat transfer coefficients and thermal conductivity values were all taken from the internet. The heat trasnfer coefficeint for air assumes only free convection is taking place, while the heat transfer coefficient for water assumes slow laminar flow of water in a tube.

I intend to use these calculations and apply a safety factor (of 2) to account for inaccuracies in the assumed constants as well imperfect contact between the heater and the aluminium.