The effectiveness NTU method is used for determining the heat transfer rate for a heat exchanger, with known inlet temperatures and surface area.

It makes use of effectiveness which is defined as

$$\epsilon = \frac{Q_{actual}}{Q_{max}}$$

What the term in the denominator truly represents?

I read that maximum rate of heat transfer is obtained in a counter flow heat exchanger with given inlet temperatures and heat capacity rates. Why is this so?

  • $\begingroup$ Have you checked any books on Heat Transfer? Simonson is a possibility. $\endgroup$
    – Solar Mike
    Commented Aug 16, 2021 at 8:03
  • $\begingroup$ I must check then. Thanks for the suggestion. $\endgroup$ Commented Aug 16, 2021 at 8:17

2 Answers 2


$q_{max }$ is the maximum heat that could be transferred between the fluids per unit time, in the ideal case that the temperature difference between the input and output of a flow is equal to the maximum possible (see below).

$q_{max }$ is calculated as the product of :

  • the maximum temperature difference i.e. the hot inlet $T_{h,i}$ minus the the cold inlet temperature $T_{c,i}$ $$\Delta T_{max} = T_{h,i}- T_{c,i}$$
  • the minimum product of mass rate and heat capacity
    $$C_{min}= \min\left(\dot{m}_h\cdot c_{p,h}, \dot{m}_c\cdot c_{p,c}\right)$$

so $q_{max }$ is given by: $$q_{max } = C_{min}\Delta T_{max}$$

The reason that is done, because in the best case scenario, the maximum heat is transferred when the hot reaches the inlet cool temperature (and vice versa). Beyond that point there is no exchange if there is no temperature difference.

The minimum $\dot{m}\cdot c_{p}$ is selected because depending on the properties, -in the generic case - only one of the flows will reach the limiting temperature (be that cold or hot). That will be depended on the mass flow, and on the material heat capacity properties (i.e. how much energy is required to increase by 1 deg the temperature of 1 kg).

  • $\begingroup$ The last two paragraphs of the answer consider a counter flow heat exchanger, I believe. Say I have a parallel flow double pipe heat exchanger and inlet temperatures of the hot and cold fluids are 100 and 20 (in Celsius) then no matter how long I make that heat exchanger the exit temperature of the cold can never reach 100 and exit temprature of the hot can never reach 20. What we will be doing in that case? $\endgroup$ Commented Aug 16, 2021 at 11:03
  • 1
    $\begingroup$ It's the same irrespective of the type of flow.. The good thing about the NTU method is that it can be applied to other flows apart from parallel and counter ( Like cross or mixture or parallel and counter). $\endgroup$
    – NMech
    Commented Aug 16, 2021 at 11:14
  • $\begingroup$ So is it like, if I were to have a (say) cross flow heat exchanger, with just a tube, inside which a fluid flows and external to which another fluid flows perpendicular to the tube, then with some given inlet temperatures, mass flow rates and specific heats, the maximum rate of heat transfer would be obtained when I make the fluids to flow in opposite directions (counter), even though my given HX is a cross flow one? $\endgroup$ Commented Aug 16, 2021 at 13:07
  • 1
    $\begingroup$ Even in a cross flow or any type of flow, the maximum possible temperature difference for at least one of the flows is going to be $T_{h,i} - T_{c,i}$. It does matter if it really occurs at any point of the heat exchanger. $\endgroup$
    – NMech
    Commented Aug 16, 2021 at 13:24
  • $\begingroup$ Oh I see. Can you suggest some sources where this concept has been dicussed? $\endgroup$ Commented Aug 16, 2021 at 13:35

We use $\Delta T_{max}$ because it is the difference between the warmest and the coldest temperatures in the system. An infinite counter flow heat exchanger would approach that, but this is an upper-bound valid in any other configuration.

To understand why we use $C_{min}$, I used the following reasoning:

The maximum rate the hot fluid can give is $C_h \Delta T_{max}$.

Likewise, the maximum heat transfer rate the cold fluid can take is $C_c \Delta T_{max}$.

Since these quantities must be equal, we take the smallest of them, $C_{min}=\min(C_h, C_c)$.


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