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In books, they just mention the temperature profile is exponential. Can someone help me come up with the exact temperature profile (the relation) for the hot and cold fluids in a double pipe parallel flow heat exchanger.

Assumptions that are to be taken:

  1. Heat exchanger is completely insulated from the surroundings and the heat transfer only occurs between the two fluids.

  2. Specific heats are assumed to be constant with temperature

  3. No axial conduction in the tube takes place, and hence the temperature in the axial direction is constant and varies only in the radial direction.

  4. Steady-state operation of the heat exchanger

  5. Flow inside the heat exchanger is thermally fully developed.

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  • $\begingroup$ Your articulation of assumption 3 is incorrect. The temperature variation in the axial direction is obviously not constant. It is just that the temperature gradient in the axial direction is negligible compared to the temperature gradient in the radial direction, so the heat conduction is considered exclusively in the radial direction. $\endgroup$ Mar 20, 2021 at 12:21
  • $\begingroup$ In a previous thread, I gave you the differential equations for the coupled axial temperature variations of the hot and cold fluids in a double pipe parallel flow heat exchanger. They consist of two coupled 1st order ordinary differential equations. Are you asking for a hint on how to solve these equations? $\endgroup$ Mar 20, 2021 at 12:26
  • $\begingroup$ I learnt about two cases in internal flow through pipes with heat transfer- 1) constant wall temperature (along the pipe) with varying heat flux 2) constant heat flux with varying wall temperature (along the pipe). I suppose both the conditions can occur in a HX (or can't they). So going by 1) I took the assumption that temperature variations in the axial direction can be neglected. $\endgroup$ Mar 20, 2021 at 12:49
  • $\begingroup$ Also please correct me but, if we don't neglect temperature variations in the axial direction then automatically there will be a heat transfer in that direction. Shouldn't the temperature variations also be neglected if we're neglecting the heat transfer in a particular direction? $\endgroup$ Mar 20, 2021 at 12:49
  • $\begingroup$ No. It is just that the temperature gradient in the axial direction corresponds to axial heat conduction that is negligible compared to the axial convection of energy. You do know the difference between negligible and zero, right? So neglecting axial conduction does not mean that we are neglecting all axial energy transport. $\endgroup$ Mar 20, 2021 at 13:23

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So you have that a heat flows between the two fluids, proportional to $T_1(x)-T_0(x)$ , over some time $\delta t$: while in that same time the fluid moves forward by an amount $v~\delta t$. With suitably integrated “linear” specific heat capacities $c_{0,1}$ and some shared energy transfer coefficient $\kappa$ this means $$ c_1 [ T_1(x+v_1\delta t) - T_1(x) ] = -\kappa [ T_1(x)-T_0(x) ]\delta t\\ c_0 [ T_0(x+v_0\delta t) - T_0(x) ] = -\kappa [ T_0(x)-T_1(x) ]\delta t $$ and expanding the term on the left to first order we get a matrix equation $$ \frac{\mathrm d\phantom t}{\mathrm d x}\begin{bmatrix}T_0\\T_1\end{bmatrix}=-\kappa \begin{bmatrix} 1/(v_0 c_0)&-1/(v_0 c_0)\\ -1/(v_1 c_1)&1/(v_1 c_1) \end{bmatrix}\begin{bmatrix} T_0\\T_1 \end{bmatrix}. $$ To analyze this system is to analyze this matrix. Let me simplify, $$ \mathbf M = -\kappa \begin{bmatrix} 1/(v_0 c_0)&-1/(v_0 c_0)\\ -1/(v_1 c_1)&1/(v_1 c_1) \end{bmatrix} = \begin{bmatrix} -\alpha&\alpha\\ \beta&-\beta \end{bmatrix} $$ There is one honking obvious eigenvector,$ \begin{bmatrix}1\\1 \end{bmatrix}$. It has eigenvalue 0, and this says that when the two temperatures are the same there is no heat transfer and you are in a steady state. This matrix is, in other words, a projection!

Because the trace $-\alpha-\beta$ is the sum of the eigenvalues and we already know one eigenvalue to be zero, that actually is the other eigenvalue. The direction is then easy to solve: the other eigenvector is $\begin{bmatrix} \alpha\\ -\beta \end{bmatrix}.$

Once you have done this you have diagonalized the equations, and the rest is linearity. So we decompose the initial condition into the diagonalizing basis for $\mathbf M$ as $$ \begin{bmatrix}T_0(0)\\ T_1(0) \end{bmatrix} =\frac{\beta T_0(0) +\alpha T_1(0)}{\alpha +\beta} \begin{bmatrix}1\\1 \end{bmatrix} + \frac{T_0(0)-T_1(0)}{\alpha +\beta} \begin{bmatrix}\alpha\\-\beta \end{bmatrix} $$ and then we use linearity to study how each of these two terms evolves over $x$, since we can reconstruct the full evolution as a sum of the independent evolutions for linear differential equations.

Well, we said that for the first term $\mathrm d\vec T/\mathrm dx=0$ so that term stays constant independent of $x$. But for the second term we have the equation$$ {\mathrm d\vec T\over\mathrm dx}=-(\alpha+\beta) \vec T,\\ \vec T(x) = \vec T(0) \exp(-(\alpha+\beta)x). $$ Combining the two, yes, we have an exponential decay of temperature along the length of the heat exchanger, when run in this parallel configuration. The first term is the steady state, and the second term is the part that is exponentially decaying.

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