As part of an heat exchanger problem, we are given two unknowns $T_{cs}$ and $T_{fs}$ solving the following equations: $\forall (Q,U,F, m_c,m_f, C_c,C_f, \Sigma , T_{ce},T_{fe}) \in \mathbb{R_+^*}^{10}, Q = U \Sigma F \frac{T_{ce}-T_{fs} -T_{cs} +T_{fe}}{ln(\frac{T_{ce}-T_{fe}}{T_{cs}-T_{fs}})} = m_c \cdot C_c (T_{ce} - T_{cs}) = m_f C_f (T_{fs} - T_{fe})$

Side note: f means cold (froid in french), c means hot (chaud in french), s means outlet (sortie in french), e means inlet (entrée in french).

Then, I thought, oh that's easy, let's isolate the two variables and solve the system.

We introduce the change of variables:

  • $x = T_{cs}$
  • $y = T_{fs}$
  • $A = \frac{m_c \cdot C_c}{U \Sigma F} \in \mathbb{R^*_+}$
  • $B = \frac{m_f \cdot C_f}{U \Sigma F} \in \mathbb{R^*_+}$

We get the following system (two equations of two unknowns):

$\frac{T_{ce} - y - x + T_{fe}}{ln(\frac{T_{ce}-y}{x - T_{fe}})} = A (T_{ce} - x) = B(y-T_{fe})$

I introduce another change of variable so that it's more comfortable for math people (these are constants):

  • $\Gamma = T_{ce}$
  • $\Lambda = T_{fe}$

We get:

$$\frac{\Gamma + \Lambda - y - x}{ln(\frac{\Gamma-y}{x - \Lambda})} = A (\Gamma - x) = B(y-\Lambda)$$

\begin{cases} \frac{\Gamma + \Lambda - y - x}{ln(\frac{\Gamma-y}{x - \Lambda})} = A (\Gamma - x) \\ \frac{\Gamma + \Lambda - y - x}{ln(\frac{\Gamma-y}{x - \Lambda})} = B (y - \Lambda) \end{cases}

Therefore my question is: $\forall (A,B,\Lambda, \Gamma) \in \mathbb{R_+^*}^4,$ solve: \begin{cases} \frac{\Gamma + \Lambda - y - x}{ln(\frac{\Gamma-y}{x - \Lambda})} = A (\Gamma - x) \\ \frac{\Gamma + \Lambda - y - x}{ln(\frac{\Gamma-y}{x - \Lambda})} = B (y - \Lambda) \end{cases}

My attempt was to find an exponential change of variable that would work but it didn't worked out well (in heat exchangers, the profile is an exponential). I also tried all the methods I knew for non-linear systems of equations without much results.

Remark: You might think that the coefficients are wrong because we have terms with the form $\Gamma - y$ inside the logarithm and $y - \Lambda$ in the RHS, and that I just messed up the index of my temperatures. Well the LHS corresponds to a logarithmic mean between the temperature at the origin of space $x=0$ and the end of the exchanger $x=L$. Here, we have a counter-flow exchanger, so the inlet of the hot fluid is the outlet of the cold fluid and vice-versa. While on the RHS, it corresponds to a change of enthalpy between $x=0$ and $x=L$.

Thanks in advance


1 Answer 1


$$ \frac{\Gamma + \Lambda - y - x}{\log(\frac{\Gamma-y}{x - \Lambda})} = A (\Gamma - x)\tag 1$$ $$\frac{\Gamma + \Lambda - y - x}{\log(\frac{\Gamma-y}{x - \Lambda})} = B (y - \Lambda) \tag 2$$

Since the lhs are the same $$A (\Gamma - x)= B (y - \Lambda)\implies y=\frac{A (\Gamma -x)}{B}+\Lambda$$ Plug this values in $(1)$ to get $$(x-\Gamma ) \left(\frac{\frac{A}{B}-1}{\log \left(\frac{-A \Gamma +A x+B \Gamma -B \Lambda }{B (x- \Lambda) }\right)}+A\right)=0$$ Excluding $x=\Gamma$, then $$\color{blue}{x=\Lambda+\frac{e^{\frac{1}{B}} (A-B) (\Gamma -\Lambda )}{A e^{\frac{1}{B}}-B e^{\frac{1}{A}}}}$$ $$\color{blue}{y=\Gamma-\frac{e^{\frac{1}{A}} (A-B) (\Gamma -\Lambda )}{A e^{\frac{1}{B}}-B e^{\frac{1}{A}}}} $$

  • $\begingroup$ Thank you very much, I didn't thought of combining the two RHS. $\endgroup$ Commented Feb 26, 2019 at 6:42
  • 1
    $\begingroup$ @PackSciences. You are welcome ! Notice that using $X=x-\Lambda$, $Y=\Gamma-y$ makes the problem simpler (in fact, it is what I did). $\endgroup$ Commented Feb 26, 2019 at 6:50

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