Heat Exchanger Design (Log Mean Temp Difference or Epsilon NTU Method)

Question

I am trying to design two heat exchanger interfaces (shell and tube) that transfers heat to and from an intermediate heat pipe. The picture below gives a general idea of the problem I am trying to solve, although the system would look differently. That is, the red pipe would transfer heat to the yellow pipe in a separate shell and tube heat exchanger (HX-1), and the yellow tube would then transfer heat to the blue pipe in another separate shell and tube heat exchanger (HX-2). All different colored fluids are different. Here we will assume that the end of the yellow pipe is connected to it's inlet. (Like a loop)

Specifically I know the inlet and outlet temperatures of the blue and red fluids, along with both of their mass flow rates. I also have the temperature (outlet temperature of the first heat exchanger and inlet temperature of the second heat exchanger) of the yellow fluid to be $340 ^\circ C$ as well. The unknown parameters I would like to find are listed below:

• Mass flow rate of the yellow fluid
• Inlet temperature of yellow fluid to HX-1 (or outlet temperature of yellow fluid to HX-2 since they're assumed the same)
• Required contact surface area between the red and yellow fluid
• Required contact surface area between the yellow and blue fluid.

Of main importance is the required surface areas. I am not sure if I should apply the log-mean temperature difference or the epsilon-NTU method. I am also pretty sure there will be some iteration/numerical methods involved.

Does anyone have experience in this, and would it be possible to get a rough step-by-step procedure to finding these values?

Apologies for the vagueness of the question, I've been stuck on this forever. I'd appreciate any help possible.

Thank you in advance.

Answer 1

Background

Total energy lost from fluid

The heat transfer rate $\dot{Q} $ if you know for a pipe with fluid $f$ (where f: Red, Yellow, Blue) the temperature at input and output, the mass rate, and the heat capacity of the material is given by:

$$\dot{Q}_f = \dot{m}_f\cdot C_{p,f}(T_{f,o}- T_{f,i}) $$

So for the Red-yellow pipe you have:

$$\dot{Q}_{ry} = -\dot{m}_r\cdot C_{p,r}(T_{r,o}- T_{r,i}) = \dot{m}_y\cdot C_{p,y}(T_{y,o1}- T_{y,i1}) $$ $$\dot{Q}_{ry} = -\dot{m}_r\cdot C_{p,r}(130-540) = \dot{m}_y\cdot C_{p,y}(340- T_{y,i1}) $$ $$\dot{Q}_{ry} = \dot{m}_r\cdot C_{p,r}(410) = \dot{m}_y\cdot C_{p,y}(340- T_{y,i1}) $$

So for the Blue-yellow pipe you have:

$$\dot{Q}_{by} = \dot{m}_b\cdot C_{p,b}(T_{b,o}- T_{b,i2}) = -\dot{m}_y\cdot C_{p,y}(T_{y,o2}- T_{y,i2}) $$ $$\dot{Q}_{by} = \dot{m}_b\cdot C_{p,b}(300- 80) = -\dot{m}_y\cdot C_{p,y}(T_{y,o2} - 340 ) $$ $$\dot{Q}_{by} = \dot{m}_b\cdot C_{p,b}(220) = -\dot{m}_y\cdot C_{p,y}(T_{y,o2} - 340 ) $$

Exchange due to conductive heat transfer

At that point you need the logarithmic mean temperature difference $\Delta T_{lm}$. Essentially, it estimates an equivalent temperature that you can use for calculating the heat transfer between and exchange surface A, with a coefficient of conductivity of k. Then the heat transfer rate is :

$$\dot{Q} = kA\cdot\Delta T_{lm} $$

Where:

$$\Delta T_{lm} = \frac{\Delta T_1-\Delta T_2}{\ln (\Delta T_1/\Delta T_2)}$$

The $\Delta T_{lm}$ has different definitions of $\Delta T_1$ and $\Delta T_2$ for parallel and counterflow exchangers.

• for counterflow (Red-yellow)
  • $\Delta T_1 = T_{r,i}-T_{y,o1} = 540 -340 = 200$ : temperature difference at one exit (at the center of the drawing)
  • $\Delta T_2 = T_{r,o}-T_{y,i1} = 340 -T_{y,i1} $ : temperature difference at other Exit (bottom of the drawing).

So: $$\dot{Q}_{ry} = k_{ry}A_{ry} \frac{200-(340 -T_{y,i1} )}{\ln \frac{200}{340 -T_{y,i1}}} = k_{ry}A_{ry} \frac{-140 +T_{y,i1} }{\ln \frac{200}{340 -T_{y,i1}}} $$

• for parallel flow (Blue-yellow)
  • $\Delta T_1 = T_{y,i2}-T_{b,i} = 340 - 80 =260$ : temperature difference at one exit (at the center of the drawing)
  • $\Delta T_2 = T_{y,o2}-T_{b,o} = T_{y,o2} - 300$ : temperature difference at other Exit (bottom of the drawing).

So: $$\dot{Q}_{by} = k_{by}A_{by} \frac{260-(T_{y,o2} - 300 )}{\ln \frac{260}{T_{y,o2} - 300}} = k_{by}A_{by} \frac{560 -T_{y,o2} }{\ln \frac{260}{T_{y,o2} - 300}} $$

Equality between change in heat capacity and conductivity transfer rate

At this point it is useful to remind that the heat transfer rate due to change in heat capacity in the red fluid is equal to the heat transmitted conductively between the red-yellow interface. Therefore you can write:

• for the red-yellow interface:

$$\dot{Q}_{ry} = \dot{m}_r\cdot C_{p,r}(410) = k_{ry}A_{ry} \frac{-140 +T_{y,i1} }{\ln \frac{200}{340 -T_{y,i1}}} $$

or simply (in $\color{red}{\text{red}}$ color I am highlighting unknown quantities:

$$\dot{m}_r\cdot C_{p,r}(410) = k_{ry}\color{red}{A_{ry}} \frac{-140 +\color{red}{T_{y,i1}} }{\ln \frac{200}{340 -\color{red}{T_{y,i1}}}} $$

• Similarly for the blue-yellow interface, ultimately you get:

$$\dot{m}_b\cdot C_{p,b}(220) = k_{by}\color{red}{A_{by}}\frac{560 -\color{red}{T_{y,o2}} }{\ln \frac{260}{\color{red}{T_{y,o2}} - 300}} $$

As you can see you have seemingly two equations with four unknowns. However there still a dependence from the equations right at the top:

$$\dot{m}_r\cdot C_{p,r}(410) = \color{green}{\dot{m}_y}\cdot C_{p,y}(340- \color{red}{T_{y,i}}) $$ $$\dot{m}_b\cdot C_{p,b}(220) = -\color{green}{\dot{m}_y}\cdot C_{p,y}(\color{red}{T_{y,o2}} - 340 ) $$

So essentially you have 1 degree of freedom (3 equation and 4 unkwowns).

How to solve

As I said just above you have 3 equation and 4 unkwowns. To sum them up:

$$\begin{cases} \dot{m}_r\cdot C_{p,r}(410) = k_{ry}\color{red}{A_{ry}} \frac{-140 +\color{red}{T_{y,i1}} }{\ln \frac{200}{340 -\color{red}{T_{y,i1}}}}\\ \dot{m}_b\cdot C_{p,b}(220) = k_{by}\color{red}{A_{by}}\frac{560 -\color{red}{T_{y,o2}} }{\ln \frac{260}{\color{red}{T_{y,o2}} - 300}}\\ \dot{m}_r\cdot C_{p,r}(410) = \color{green}{\dot{m}_y}\cdot C_{p,y}(340- \color{red}{T_{y,i1}}) \\ \dot{m}_b\cdot C_{p,b}(220) = -\color{green}{\dot{m}_y}\cdot C_{p,y}(\color{red}{T_{y,o2}} - 340 ) \end{cases}$$

assuming you know:

• $\dot{m}_r$ the mass rate of the red fluid
• $C_{p,r},C_{p,y},C_{p,b}$: the heat capacities
• $k_{ry},k_{by}$: the coefficient of heat conductivity

Then one possible way is the following:

1. You set a temperature for the input for the yellow material $T_{y,i1}$. Then you can find:
2. You can now find $A_{ry}$ from the 1st equation and assuming you know the contact geometry the required length $L_{ry}$ for the red yellow interface.
3. you can find $m_{y}$ from the 3rd equation
4. you can find $T_{y,o2}$ you can use the 4th equation, since you just calculated $m_{y}$.
5. you can find $A_{ry}$ from the 1st equation and assuming you know the contact geometry the required length $L_{by}$ for the blue yellow interface.
   $$ kA\cdot\frac{\Delta T_1-\Delta T_2}{\ln (\Delta T_1/\Delta T_2)}= \dot{Q} = m_s\cdot C_{p,s}(T_{s,o}-T_{s,i}) $$

Similar problem (only a single counterflow). A similar problem can be found at this link

Answer 2

Edit to add: The OP states the applicaiton is a heat pipe, so this answer assumed the yellow fluid is a liquid and moves by free convection. It turns out the yellow fluid will undergo a phase change, which changes a lot.

This is an outline for a solution that should get you going:

Step one - find an equation describing the convection of the yellow medium.

$\dot{m}_{yellow}$ is determined by $T_{in}$, $T_{out}$ (340°C) and the shape and physical properties of the heat exchanger, how they are situated to each other, etc. To define this function $\dot{m}_{yellow}(T_{in},T_{out})$ is the first, and likely most difficult, step.

Step two - find your $\dot{m}_{yellow}$

Find a value that satisfies $$\dot{m}_{yellow}(T_{in},T_{out}) * c_{p, yellow} * (T_{out} - T_{in}) = Q_{red}$$

If your function for $\dot{m}_{yellow}$ is a line (IMO unlikely) there will be many possible values, else it should be only one.

Size heat exchangers

for the $\dot{m}_{yellow}$ value and the other values found above.

Simulate the resulting system

with the fixed inlet temperatures for the blue and red fluids and some arbitrary starting temperatures, to see if the system indeed approaches the values you assumed. Here the NTU method is probably the best approach.