I understand that for fully developed flow inside a tube (with constant properties), Nusselt number doesn’t change in axial direction but it is surprising that it is also independent of Reynolds number which means that the rate of heat transfer (for constant pipe diameter and fluid properties) doesn’t change with mass flow rate (and velocity). So could someone please give an explanation for this.


4 Answers 4


In general, heat transfer with internal flows - like flows in a pipe - does depend on Reynolds Number. For small enough Reynolds number, the flow is laminar and the nature of the laminar flow is such that fluid inertia has negligible influence, resulting in a constant value for the laminar Nusselt number, but only when the internal flow is "fully developed." For laminar flow in the entrance region of the pipe, the Nusselt number is not constant, and its value is different for different heat transfer configurations. A more complete explanation is below.

The Reynolds number is a ratio of inertial forces to viscous forces, VD/Lambda, where V is fluid velocity, D pipe diameter, and Lambda fluid kinematic viscosity. When inertial forces are small enough, viscosity dominates, and the transfer of momentum across minute streamlines is solely because of friction (viscosity), and it occurs on a molecular level. At that level, the momentum is "diffused" across the streamlines, in a similar way to how mass and heat are diffused, with the fundamental mechanism due to the action of molecule bumping into molecule. With larger Reynolds numbers, inertia becomes important and momentum transfer occurs at a larger length scale, varying from microscopic to that visible by the unaided eye.

With laminar flow, heat is also diffused by molecule bumping into molecule, and there's a fixed relation between momentum transfer and heat transfer, since the mechanisms for both are physically similar. And even within the phenomenon of heat transfer, the ratio of convective heat transfer to conductive heat transfer, a constant in fully-developed flow, is given by the Nusselt number, hD/k, where h is the film coefficient, D pipe diameter, and k fluid thermal conductivity. But here, the "convection" is due to the axial flow of the fluid in the pipe, not that due to the radial heat transfer. The Nusselt number relates the radial heat conduction to the enthalpy change due to the axial flow.

The film coefficient is a convenient definition, made in the simplest way because the phenomenon has a linear dependence on temperature difference, or Q = h*(Ts - Tref), where Q is the bulk heat transferred, Ts the temperature of the heated/cooled surface, and Tref is a reference temperature in the fluid, chosen for convenience. In pipe flow, this temperature difference is usually taken to be that between different pipe sections. Q is the heat transferred between the surface across the fluid to where the reference temperature is defined.

The questioner is correct to say that the heat transfer in this laminar case doesn't depend on the flow through the pipe. It's as though the fluid is solidified, with the radial heat transfer solely due to conduction. There is no radial convection of energy, i.e., energy transfer due to a transfer of mass. Only thermal energy and kinetic energy is radially transferred, and we use the term "diffusion" to describe both. The fluid is thus behaving like a solid material that moves axially along the pipe. This axial movement is what transfers the radial conduction of heat from the pipe walls to the energy (enthalpy) difference between the pipe inlet and outlet. (The heat energy transferred via the walls equals the heat energy transferred axially between entrance and exit.)

Thus, with laminar flow, there is no radial convection, but in a loose sense, we can say that the fluid acts to convect (carry) heat from the pipe walls to the pipe outlet. The fluid motion consists of a boundary layer and temperature profile, as well as a velocity profile. The Nusselt number is constant only when this temperature profile becomes fixed (fully developed). For the entrance region of the pipe, the profile is still developing, and the laminar Nusselt number is not constant, changing with axial distance.

The laminar Nusselt number also depends on the heat transfer character of the flow. For instance, it equals 4.36 with a state of constant heat flux along the length of passage and 3.66 for constant wall temperature, and that's because the shape of the temperature profile is different for these different conditions. Thus, strictly speaking, even with laminar flow, the Nusselt number is not really constant, depending on different flow and heat transfer character.

With higher Reynolds numbers, inertial effects begin to dominate over viscosity, leading to turbulence, and turbulence increases as fluid velocities (Reynold's numbers) increase. Turbulent velocity and temperature profiles can also become fully developed, but they cannot be computed from first principles. The turbulent Nusselt number is a function of Reynolds number, and different correlations serve to provide values for different flow situations. Some correlations work better than others, and it's somewhat of an art to pick the most useful ones.


The Nusselt number is about the ratio of conductive and convective heat transfer across a boundary.

The Reynolds number is about how a fluid is moving - usually laminar (below 1700), turbulent (above 2000) or in-between aka critical. However these numbers are not absolute as laminar has been seen above 2000 in very carefully controlled conditions i.e. it is not stable.

So, while heat transfer can be affected by the type of fluid flow, those changes may not change in proportion to any change in the Reynolds number.

  • $\begingroup$ But what if we just consider laminar flow? $\endgroup$
    – moeinSj
    Apr 20, 2020 at 17:15
  • $\begingroup$ I mean the Nusselt number for laminar flow inside a pipe is constant (4.364) which means however high is the velocity (of course to the point that flow stays laminar), the heat transfer rate is not affected furthermore the case for turbulent flow is different and Nusselt number is a function of reynolds and prandtl number. $\endgroup$
    – moeinSj
    Apr 20, 2020 at 17:30

@moeinSj Regarding your interesting questions of heat transfer rate for fully-developed laminar pipe flow versus flow rate, I think one quick approach is to set up a CFD simulation and inspect the results, maybe we can gain good insights. As far as convection is concerned, the total heat transfer rates should be higher at higher flow rates, this is obvious from conservation of energy of an open system for the pipe flow.


TLDR: at any given cross-sectional slice along the length of the pipe, the Reynolds number only affects the overall amplitude of the lateral temperature variation, not its detailed functional form. Since the Nusselt number involves the quotient of the local radial temperature gradient at the wall by the average radial temperature gradient between wall and central axis, the Nusselt number depends only on the details of the functional form of the lateral temperature variation, not on its overall amplitude. Hence, Nusselt number doesn't depend on Reynolds number.

Long and algebra-heavy version: The fully-developed laminar flow in the pipe has the Poiseuille velocity profile

$$u = 2\langle u\rangle\left(1-\left(\frac{2r}{D}\right)^2\right)$$

where $\langle u\rangle$ is the mean velocity, $r$ the radial co-ordinate of a cylindrical polar co-ordinate system aligned with the pipe, and $D$ the pipe diameter.

In terms of Reynolds number,

$$u = 2\frac{\nu}{D}\,\textit{Re}\,\left(1-\left(\frac{2r}{D}\right)^2\right)$$

where $\nu$ is the kinematic viscosity of the fluid.

The temperature field is governed by the steady, axisymmetric advection-conduction equation

$$u\frac{\partial\!\! T}{\partial\! z} = \kappa\frac{1}{r}\frac{\partial\!\!}{\partial\!r}\left(r\frac{\partial\!\! T}{\partial\! r}\right)$$

where $\kappa$ is the thermal diffusivity of the fluid, and the $\partial^2\! T/\partial\! z^2$ term has been neglected on the basis of a geometrical scaling argument that temperature varies more rapidly along the radial direction than along the axial direction. Looking for solutions separable in these cylindrical polar co-ordinates, one finds

$$T = T_0+T_1\exp\left(-\frac{48}{\textit{Pr}\,\textit{Re}\,D}z-\frac{24}{D^2}r^2\right)$$

where $T_0$ and $T_1$ are arbitrary constants, $\textit{Pr} := \nu/\kappa$ is the Prandtl number of the fluid, and $z$ is the axial (i.e. along-pipe) co-ordinate.

The heat flux per length into the pipe wall is

$$q = -\mathrm{\pi} kD\left(\frac{\partial\!\! T}{\partial\! r}\right)_{r = D/2}$$ $$ = 12\mathrm{\pi} kT_1\exp\left(-\frac{48}{\textit{Pr}\,\textit{Re}\,D}z-6\right)$$

where $k$ is the thermal conductivity of the fluid. The temperature difference between wall and axis is

$$\Delta\! T := \left(T\right)_{r = 0}-\left(T\right)_{r = R/2}$$ $$= T_1\left(\exp\left(-\frac{48}{\textit{Pr}\,\textit{Re}\,D}z\right)-\exp\left(-\frac{48}{\textit{Pr}\,\textit{Re}\,D}z-6\right)\right)$$ $$= T_1\exp\left(-\frac{48}{\textit{Pr}\,\textit{Re}\,D}z\right)\left(1-\exp\left(-6\right)\right)$$

The Nusselt number is defined as

$$\textit{Nu} := \frac{q}{\mathrm{\pi}k\Delta\! T}$$ $$= \frac{12\exp\left(-6\right)}{1-\exp\left(-6\right)}$$

The dependence on Reynolds number (and everything else) has cancelled.


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