0
$\begingroup$

I know this is answerable but I'm not sure I know where to start:

You have a pump (constant volume flow) supplying the primary side of a heat exchanger with hot water (flow). The temperature of the flow is adjusted by mixing in cold return with a three way mixer (valve). The temperature of the medium on the secondary side out of the HX $t_{so}$ is measured, as is the flow temperature (primary side in) $t_{pi}$. The goal is a specific value $T_{so}$ for $t_{so}$ (lowercase letters: actual measurements, uppercase letters: setpoints).

I see two principal control strategies (and within each the question if one uses PID controls or simpler schemes):

  • control valve directly for $t_{s}$
  • control valve for a given $T_{f}$, if $t_{s} > T_{s}$ for a given length of time, adjust $T_{f}$ downwards and vice versa

My question is: If there's a sudden jumps in temperature or flow on the inlet of the secondary side - say 10K in 5 seconds, or flow drops to 75% within 2 s, and I have a certain parameters - say $t_s$ must never be higher then $T_s + 5K$ or $t_s = T_s +_- 0.5 K$ 60 s after the disturbance is registered, how do I know how fast acting the three way mixer must be (from 0-100% in 120s?)? The numbers are of course just indicative.

Analytically, the required $t_{pi}$, can be calculated iterativly using the following ($\dot m$ mass flows on primary and secondary sides, $c$ respective heat capacities):

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

NMech provided a closed 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:

$\endgroup$
1
$\begingroup$

you can have a look at

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.