What's the setpoint of the temperature on the primary side of a heat exchanger?

Question

I have a heat exchanger with known flows on priamry and secondary sides - $\dot m_p$, $\dot m_s$, known temperature in $t_{si}$ and a desired temperature out on the secondary side $t_{so}$. How to find the required $t_{pi}$ analytically? So far I've always done these calculations numerically with the excel solver.

This is where I get stuck in solving analytically - from the energy balance we find:

$$ t_{pi}=t_{po} - \frac{\dot m_s c_s}{\dot m_p c_p} * (t_{so} - t_{si})$$

but we need $t_{po}$, when I try this (from the heat transfer equation) I think I'm stuck:

$$ t_{po} = \frac{kA \Delta T_m}{\dot m_p c_p} + t_{pi} $$

because to solve I'd need to extract my temperatures from the $ln$ in the denominator of $\Delta T_m$ and that's where I'd rather look in a textbook than do the math myself. Except my textbook (Perry) doesnt have that detail. $A$ is known and $k$ can be estiamted, so is known for the purpose of this questions.

So, for given mass flows, and desired outlet temperature on the secondary side, what is the required inlet temperature on the primary side?

The ultimate problem I want to solve is that I want to ask a question about HX control and I feel it would help to have this equation.

Answer 1

Background

Normally the process would go line this. Let's say the primary is hot, and the secondary is the cooler fluid.

The heat transfer rate $\dot{Q} = -\dot{Q}_p = \dot{Q}_s $. I.e.:

• The cooler fluid (s) gains $$\dot{Q} = m_s\cdot C_{p,s}(T_{s,o}- T_{s,i}) $$

• The hot fluid (p) loses: $$\dot{Q} = - m_p\cdot C_{p,p}(T_{p,o}- T_{p,i}) $$

Therefore the change in temperature is :

$$ T_{p,o} = T_{p,i}- \frac{m_s\cdot C_{p,s}}{m_p\cdot C_{p,p}}(T_{s,o}- T_{s,i}) $$

Normally, if you get to that point then you need to calculate the length of the exchangerm and therefore the $A$ but since you know it I'll press on.

Here, I'm going to assume counterflow. At that point you need the logarithmic mean temperature difference $\Delta T_{lm}$. (please note that for different type of flows parallel, cross etc you need to change this).

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

where:

• $\Delta T_1 = T_{p,i}-T_{s,o}$ : temperature difference at one exit
• $\Delta T_2 = T_{p,o}-T_{s,i}$ : temperature difference at other Exit

Then you can apply:

$$ kA\cdot\frac{\Delta T_1-\Delta T_2}{\ln (\Delta T_1/\Delta T_2)}= \dot{Q} = m_s\cdot C_{p,s}(T_{s,o}-T_{s,i}) $$

$$ kA\cdot\frac{T_{p,i}-T_{s,o}-T_{p,o}+T_{s,i}}{\ln \left(\frac{T_{p,i}-T_{s,o}}{T_{p,o}-T_{s,i}}\right)}= \dot{Q} $$

Iterative solution

Here you can, solve with respect to $T_{p,i}$ $$ T_{p,i}-T_{s,o}-T_{p,o}+T_{s,i}= \frac{\dot{Q}}{kA\cdot} \ln \left(\frac{T_{p,i}-T_{s,o}}{T_{p,o}-T_{s,i}}\right) $$

$$ T_{p,i}= \frac{\dot{Q}}{kA} \ln \left(\frac{T_{p,i}-T_{s,o}}{T_{p,o}-T_{s,i}}\right) +T_{s,o}+T_{p,o}-T_{s,i}$$

From that point the easiest think of Excel is to iterate in order to find the solution (set a guess for $T_{p,i}$ apply it on the right hand, get a new $T_{p,i}'$, which you plug into the equation until $T_{p,i}-T_{p,i}' \rightarrow 0 $ ).

close form solution

The other option if you need a close form solution (which is probably what you are asking), you can look at the Lambert W function. In that case, you start from:

$$ kA\cdot\frac{\Delta T_1-\Delta T_2}{\ln (\Delta T_1/\Delta T_2)}= \dot{Q}$$

After replacing $\Delta T_1, \Delta T_2 $,the solution for $T_{p,i}$ takes the following form:

$$ T_{p,i} = T_{s,o} - \frac{ Q }{A k} ProductLog[-\frac{A k(T_{p,o}-T_{s,i}) e^{-\frac{A k }{Q}(T_{p,o}-T_{s,i}) } }{Q}]$$

Where:

• $ProductLog[x]$ is the Lambert W function.

However Excel does not have this function builtin (at least to my knowledge). So you need to either do it another language/system (e.g. octave, or python) or find a macro for lambert w.