2. Formulation

Consider the steady, two-dimensional pressure-gradient-driven flow of a Boussinesq fluid in an isothermal channel of width 2h inclined at an angle $\beta $ to the horizontal, as shown in figure 1. We align the x and y coordinate axes parallel and perpendicular to the slot and non-dimensionalize all lengths on the half-width h so that the sides of the channel are given by $y ={\pm} 1$. In the absence of any fluid motion, there is a hydrostatic pressure ${p_h}(x,y)$ given by the solution of

(2.1a,b)\begin{equation}\frac{{\partial {p_h}}}{{\partial x}} ={-} \rho g\sin \beta ,\quad \frac{{\partial {p_h}}}{{\partial y}} ={-} \rho g\cos \; \beta ,\end{equation}

where g is the gravitational acceleration and $\rho $ is the density of the fluid. When the fluid moves, the basic motion is given by velocity ${\boldsymbol{v}_0}(x,y)$ and pressure $\; {p_0}(x,y)$ fields, which are

(2.2a,b)\begin{equation}{\boldsymbol{v}_0}(x,y) = (1 - {y^2},0),\quad {p_0}(x,y) ={-} 2x/Re,\end{equation}

where the velocity vector ${\boldsymbol{v}_0} = ({u_0},{v_0})$ is scaled on the maximum of the $x$-velocity ${u_{max}}$ and ${p_0}$ is the pressure in excess of ${p_h}$ based on $\rho u_{max}^2$. The form of $\; {p_0}$ depends on the Reynolds number $Re \equiv {u_{max}}h/\nu $ with $\nu $ the kinematic viscosity. Given this basic flow, it is straightforward to find the associated stream function ${\varPsi _0}$, the flow rate ${Q_0}$ through the slot and the shear force (per unit length and width) ${F_{R0}}$ (${F_{L0}}$) acting on the fluid at the right (left) wall. These quantities are given by

(2.3a–c)\begin{equation}{\varPsi _0} = y - \frac{{{y^3}}}{3} + \frac{2}{3},\quad {Q_0} = \frac{4}{3}\quad \textrm{and}\quad {F_{R0}} = {F_{L0}} ={-} 2/Re.\end{equation}

We next supply sinusoidal heating to both walls so that their scaled temperatures are given by

(2.4a)\begin{gather}y ={-} 1:\quad {\theta _R}(x) = {\textstyle{1 \over 2}}R{a_{p,R}}\cos \alpha x,\end{gather}

(2.4b)\begin{gather}y = 1:\quad {\theta _L}(x) = {\textstyle{1 \over 2}}\; R{a_{p,L}}\; \cos (\alpha x + \varOmega ),\end{gather}

where the subscripts R and L denote the right and left walls, respectively. We remark that this periodic heating is applied to the surfaces throughout the domain; moreover, this profile is of wavenumber $\alpha $ (or wavelength $\lambda = 2{\rm \pi}/\alpha $). We also remark that the two heating patterns at the two walls are offset by a phase angle $\varOmega $. Here $\theta $ denotes the relative temperature defined by $T - {T_R} = \kappa \nu \theta /(g\varGamma {h^3})$, where T and ${T_R}$ denote the absolute and reference base temperature, respectively. Furthermore, $\kappa $ is the thermal diffusivity and $\varGamma $ is the thermal expansion coefficient, and we assume that fluid density variations follow the Boussinesq approximation. The appropriate Rayleigh numbers measure the intensity of the heating at the walls; on the right-hand side, we define $R{a_{p,R}} = g\varGamma {h^3}{\theta _{p,R}}/(\kappa \nu )\; $, where ${\theta _{p,R}}$ is the magnitude of the applied heating with an analogous definition for $\; R{a_{p,L}}$ relating to the left-hand edge of the slot. We mention in passing that it is well accepted that modelling buoyancy using the Boussinesq approach serves as an excellent paradigm for the underlying physics, especially when the temperature differences involved in the problem are relatively modest. For a discussion of the validity of this type of modelling, the interested reader is directed to the recent review by Mayeli & Sheard (Reference Mayeli and Sheard2021).

Figure 1. A schematic of the flow configuration in the slot defined by $|y|\le 1$. The channel is inclined to the horizontal at an angle $\beta $ while the two walls are heated by sinusoidal thermal profiles defined by (2.4). These two profiles are offset by a phase $\varOmega $.

The continuity, Navier–Stokes and energy equations govern the system. Written in terms of the dimensionless variables, we have

(2.5a,b)\begin{gather}u\frac{{\partial u}}{{\partial x}} + v\frac{{\partial u}}{{\partial y}} ={-} \frac{{\partial p}}{{\partial x}} + {\nabla ^2}u + P{r^{ - 1}}\theta \sin \beta ,\quad \frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} = 0,\end{gather}

(2.5c,d)\begin{gather}u\frac{{\partial v}}{{\partial x}} + v\frac{{\partial v}}{{\partial y}} ={-} \frac{{\partial p}}{{\partial y}} + {\nabla ^2}v + P{r^{ - 1}}\theta \cos \beta ,\quad u\frac{{\partial \theta }}{{\partial x}} + v\frac{{\partial \theta }}{{\partial y}} = P{r^{ - 1}}{\nabla ^2}\theta ,\end{gather}

where $(u,v)$ are the velocity components in the (x, y) directions, respectively, scaled on ${U_v} = \nu /h$, p is the pressure associated with fluid movement scaled with $\rho U_\nu ^2$ and $\theta $ is the temperature; finally, $Pr = \nu /\kappa $ is the Prandtl number. (The generalization of the field equations to non-Boussinesq fluids may be found in the text by Tritton (Reference Tritton1977).) The range of heating parameters used in the analysis is similar to that used in experiments reported by Inasawa et al. (Reference Inasawa, Taneda and Floryan2019, Reference Inasawa, Hara and Floryan2021) and Floryan & Inasawa (Reference Floryan and Inasawa2021). A very good agreement between experimental data and theory was reported, demonstrating that the Boussinesq approximation captures a fluid response well. The required boundary conditions are

(2.6a–d)\begin{align}u( - 1) = u( + 1) = 0,\quad v( - 1) = v( + 1) = 0,\quad \theta ( - 1) = {\theta _R}(x)\quad \textrm{and}\quad \theta ( + 1) = {\theta _L}(x).\end{align}

We decompose the flow fields according to

(2.7a–c)\begin{gather}u(x,y) = Re\,{u_0}(y) + {u_1}(x,y),\quad v(x,y) = {v_1}(x,y),\quad \theta (x,y) = {\theta _1}(x,y),\end{gather}

(2.7d,e)\begin{gather}p(x,y) = R{e^2}{p_0}(x) + Bx + {p_1}(x,y),\quad \psi (x,y) = Re\; {\psi _0}(y) + {\psi _1}(x,y),\end{gather}

so that if no heating were applied to the walls, we would have all quantities with subscript 1 identically zero. Our concern lies in determining whether applied heating can reduce the pressure gradient required to maintain a specified flow rate. Accordingly, we impose the mass flow rate constraint of the form

(2.8)\begin{equation}Q(x){|_{mean}} = {\left. {\left( {\int_{ - 1}^1 {u(x,y)\,\textrm{d}y} } \right)} \right|_{mean}} = \frac{4}{3}Re,\end{equation}

and seek information as to the size of the mean pressure gradient

(2.9)\begin{equation}{\left. {\frac{{\partial p}}{{\partial x}}} \right|_{mean}} ={-} 2Re + B,\end{equation}

where positive values of B signify a reduction of pressure losses.

The system (2.3)–(2.7) was solved by expressing the velocity components using a stream function $\psi $ defined in the usual manner, i.e. $\; u = \partial \psi /\partial y$, $v ={-} \partial \psi /\partial x$, then eliminating pressure and using Fourier expansions in the x direction and Chebyshev expansions in the y direction. A description of the algorithm and the benchmarking of its accuracy are described by Hossain et al. (Reference Hossain, Floryan and Floryan2012); the reader is referred to that paper for extended details. The pressure field was normalized by bringing the mean value of its periodic component to zero while the mean Nusselt number $N{u_{av}}$ was evaluated using

(2.10a)\begin{equation}N{u_{av}} ={-} {\lambda ^{ - 1}}\int_{{x_0}}^{{x_0} + \lambda } {{{\left. {\frac{{\partial \theta }}{{\partial y}}} \right|}_{y ={-} 1}}\textrm{d}\kern0.7pt x} .\end{equation}

With this definition, a positive value of $N{u_{av}}$ means that the right wall is losing energy. The cost of moving energy within a wall to create the local cold and hot spots is difficult to assess, but it contributes to the overall energy costs. The quantity of energy that has to be moved along the wall can be determined by evaluating the horizontal heat fluxes between the hot and cold wall segments. These fluxes can be quantified in terms of heat leaving each wall per half-heating wavelength and expressed by the horizontal Nusselt number defined as

(2.10b)\begin{equation}N{u_{h,R}} ={-} 2{\lambda ^{ - 1}}\int_{ - \lambda /4}^{\lambda /4} {{{\left. {\frac{{\partial \theta }}{{\partial y}}} \right|}_{y ={-} 1}}\textrm{d}\kern0.7pt x} - N{u_{av}}.\end{equation}

This is not the only possible way to estimate the energy cost. Various alternatives have been suggested in the literature; the interested reader is directed to the papers by Maxworthy (Reference Maxworthy1997), Siggers, Kerswell & Balmforth (Reference Siggers, Kerswell and Balmforth2004), Hughes & Griffiths (Reference Hughes and Griffiths2008) and Winters & Young (Reference Winters and Young2009).

We remark that the shear forces acting on the fluid at the right and left walls are given by

(2.11a,b)\begin{equation}{F_R} ={-} {\lambda ^{ - 1}}\int_0^\lambda {{{\left. {\frac{{\partial u}}{{\partial y}}} \right|}_{y ={-} 1}}\textrm{d}\kern0.7pt x} ,\quad {F_L} = {\lambda ^{ - 1}}\int_0^\lambda {{{\left. {\frac{{\partial u}}{{\partial y}}} \right|}_{y ={+} 1}}\textrm{d}\kern0.7pt x} ,\end{equation}

and heating-induced changes of these forces are given as

(2.12a,b)\begin{equation}\Delta {F_R} = {F_R} - R{e^2}{F_{R0}},\quad \Delta {F_L} = {F_L} - R{e^2}\; {F_{L0}},\end{equation}

with negative $\Delta {F_R}$ and $\Delta {F_L}$ corresponding to a reduction of these forces. The total (buoyancy) body force per unit length is given by

(2.13)\begin{equation}{F_b} = {\lambda ^{ - 1}}P{r^{ - 1}}\int_{ - 1}^1 {\int_0^\lambda {\theta \,\textrm{d}\kern0.7pt x\,\textrm{d}y} } ,\end{equation}

with its x and y components given by ${F_{bx}} = {F_b}\sin \beta $ and ${F_{by}} = {F_b}\cos \beta $, respectively.

It is noted that the above formulation is restricted to two-dimensional steady flows. There is always the possibility that the underlying structure may become time-dependent or that a secondary flow is set up. Some previous computations have addressed the question of the possibility of secondary structures in flows within horizontal channels (Hossain & Floryan Reference Hossain and Floryan2013b, Reference Hossain and Floryan2015b). Those findings suggest that no secondary flow appears unless $\; R{a_{p,R}}$ is at least 2500. The results reported below are restricted to smaller Rayleigh numbers, so the complications that secondary structures would introduce ought not to arise.

3. Flow properties when one wall is heated

To begin our account of the effect of patterning on the pressure losses experienced along the channel, we consider the situation in which only one of the walls is heated. Within our problem, several parameters could be varied, but it is helpful to note that several inherent symmetries can be exploited to reduce the task. First, we need not be concerned about whether the flow is net upwards or downwards. If we transform $\beta \to \beta + {\rm \pi}$ then the problem specification is unchanged if we switch the sign of $\theta $ and translate the coordinate $x \to x + ({\rm \pi}/\alpha )$. We conclude that we can restrict the inclination of the channel $\beta \in [0,{\rm \pi}]$. Furthermore, if one applies the transformations $(x,y,u,v,p,\theta ,\beta ) \mapsto (x, - y,u, - v,p,\theta ,{\rm \pi} - \beta )$ the governing equations are invariant while the boundary conditions swap over. This means that when it comes to considering the problem when only one wall is heated, we can examine the case when the patterning is applied to the right-hand wall so $R{a_{p,R}} \ne 0$ and $\; R{a_{p,L}} = 0$. We note in passing that these various symmetry properties have been confirmed by extensive numerical testing when the slot is horizontal (Hossain & Floryan Reference Hossain and Floryan2014, Reference Hossain and Floryan2015a).

The forms of the pressure losses as functions of the inclination angle $\beta $ and pattern wavenumber $\alpha $ are illustrated in figure 2 for three values of the Reynolds number $Re = 1$, 5 and 10; the interest in these particular values will become apparent as we proceed. It is noted that a reduction in the pressure losses can be achieved when $0 \le \beta \le {\rm \pi}/2$; further, the critical inclination at which the reduction ceases to be possible is only weakly dependent on $\alpha $ when the Reynolds number is small (see figure 2a). As $Re$ is increased so the size of the parameter domain corresponding to a pressure reduction expands and does so especially at relatively long heating wavelengths; this is evident in figure 2(b,c). The heating leads to self-pumping at small $Re$ when an opposite pressure gradient must be applied to slow the fluid if the flow rate is to be maintained.

Figure 2. The variation of the pressure-gradient correction B as a function of heating wavenumber $\alpha $ and slot inclination $\beta $ when only the right wall is heated with $R{a_{p,R}} = 500$. In the three cases, the Reynolds number $Re = $ (a) 1, (b) 5 and (c) 10. The grey shading denotes parameter combinations corresponding to a reduction in pressure losses (B > 0). The thick dotted, dashed-dotted and dashed lines identify those conditions for which there are 10 %, 50 % and 100 % reductions of pressure losses, respectively. The circles mark the parameter choices used in figure 3, while the vertical red lines indicate the values adopted in figure 4. Finally, the horizontal blue lines identify the flow conditions used in figure 7.

The forms of the flow and temperature fields as functions of the channel orientation $\beta $ and at Reynolds number up to 50 are shown in figure 3. (We point out that the flow conditions assumed for the three middle rows of this figure are marked with circles in figure 2.) The flow topology arises from a competition between the uniform isothermal Poiseuille flow in the positive x direction and the natural convection driven by the applied periodic heating. Natural convection alone generates counter-rotating rolls, as illustrated in the first row of figure 3. The structure of these rolls does not change significantly as the orientation of the conduit is varied; it is seen that the fluid flows downward from cold spots and upward toward hot spots. The superposition of a small component of Poiseuille flow generates a thin stream tube that gently meanders between the rolls. As $Re$ grows, we see an enlargement of the stream tube together with a slight reduction in the intensity of the rolls adjacent to the unheated wall. Further increase in $Re$ leads to an ongoing thickening of the stream tube and the complete elimination of the upper rolls. At the point when these upper rolls completely disappear, those next to the lower (heated) wall have only undergone a relatively mild weakening, as observed in the $Re = 10$ results shown in figure 3. Nevertheless, as the Reynolds number grows further, the lower rolls continue to shrink until they are eventually destroyed. Once $Re = 50$, all that remains is a structure reminiscent of an isothermal Poiseuille flow; this is evident in the bottom row of results shown in figure 3. Finally, it is interesting to note that the qualitative forms of the flow topology seem to be largely independent of the inclination of the channel, and variations in $\beta $ only result in relatively minor changes in the global flow properties.

Figure 3. The flow and temperature fields for one-wall heating with $R{a_{p,R}} = 500$ and a wavenumber $\alpha = 1$. The five rows of results correspond to $Re = 0$, 1, 5, 10 and 50, respectively. The downward black arrows show the direction of gravity force, while the red arrows indicate the direction of pressure-gradient force. The background colour illustrates the thermal field. The parameter choices used when $Re = 1,\; 5$ and 10 are marked in figure 2 by red circles, in figure 5 by green circles and in figure 7 by blue circles. In all the plots, the temperature has been normalized with its maximum ${\theta _{max}}$.

It is already well known that reducing pressure losses is possible in a horizontal channel. This phenomenon arises owing to a combination of various effects. First, reducing the direct contact between the walls and the stream lessens the frictional resistance. There is also a propulsive effect of rotation in the separation bubbles driven by horizontal density gradients. Finally, bubbles reduce the effective cross-sectional area available to the stream (Hossain et al. Reference Hossain, Floryan and Floryan2012; Floryan & Floryan Reference Floryan and Floryan2015; Hossain & Floryan Reference Hossain and Floryan2016). If the slot is inclined away from horizontal, all these effects remain operative, but now a buoyancy force acting along the channel also comes into play. The streamwise average of this additional force is zero when the channel is horizontal, but once $\beta \ne 0$, a flow-induced break in periodic symmetry generates a streamwise component (Floryan et al. Reference Floryan, Benayoun, Panday and Bassom2022a,Reference Floryan, Wang, Panday and Bassomc, Reference Floryan, Wang and Bassom2023d), which affects the pressure loss. A significant reduction in losses is feasible when $0 < \beta < {\rm \pi}/2$ for which the buoyancy force assists in the flow direction while a large increase occurs for ${\rm \pi}/2 < \beta < {\rm \pi}$; now the buoyancy force acts against the flow direction. These cases are illustrated in figure 3. We also note in connection with figure 3 that an increase in $Re$ eliminates the separation bubbles by the stage $Re = 50$, and then the drag reduction mechanism is turned off.

The interplay between the various forces that affect the fluid motion is illustrated in figure 4. These results portray how the relative importance of the pressure gradient, the wall shear and the buoyancy forces evolve as the Reynolds number and inclination angle vary. Figure 4(a) shows the cumulative effect of the three forces. This demonstrates a reduction in the pressure gradient when $\beta $ is relatively small; as $Re$ grows, the range of $\beta $ over which $B > 0$ diminishes. It is remarked that the pressure-gradient correction is not zero when the channel is horizontal (Floryan & Floryan Reference Floryan and Floryan2015) or vertical (Floryan et al. Reference Floryan, Wang and Bassom2023d); it is just much smaller than when the channel is inclined. The results shown in figure 4(c,d) help explain the mechanisms that underpin the heating-induced pressure-gradient reduction. Changes in the shear force on the heated wall propel fluid forward when $\beta < \sim {\rm \pi}/2$ but oppose fluid movement otherwise. Interestingly, the shear force at the isothermal surface exhibits a contrasting behaviour in as much that it resists fluid movement for $\beta < \sim {\rm \pi}/2$ but supports it otherwise. Typically, the magnitudes of the forces generated at the heated wall are larger. Further insight is gained from figure 4(b), which shows the variations in the x component of the buoyancy force. It is seen that the size of this force is significantly larger than that of the shear forces, so it dominates. It must be remembered that this force cannot affect the pressure losses when the slot is horizontal since its streamwise-averaged component is precisely zero. In practice, then, for an inclined channel, it is the buoyancy mechanism that is the principal player in determining the pressure reduction, but as $\beta \to 0$ the shear forces increasingly assume this role. We also note that all the forces associated with heating seem to decrease with $Re$, as heating-induced flow modifications gradually weaken and completely disappear once the stream becomes sufficiently strong to wash out the separation bubbles.

Figure 4. Flow properties as functions of the inclination angle $\beta $ when $R{a_{p,L}} = 0$, $R{a_{p,R}} = 500$ and $\alpha = 1$ with various Reynolds numbers in the interval $Re \in [0,50]$. (a) The pressure correction $B$. (b) The x component of the buoyancy force ${F_{bx}}$. Also illustrated are the heating-induced changes in the shear forces acting on the fluid at the (c) right (heated) wall and (d) left (isothermal) wall. The flow conditions when $Re = 1,\; 5,\; 10$ are marked by the vertical red lines in figure 2.

The dependence of the pressure-gradient correction parameter B on the Reynolds number $Re$ and the heating wavenumber $\alpha $ is considered in figure 5. These results demonstrate how a drag reduction is replaced by a drag increase as the inclination angle $\beta $ increases. As $\beta $ grows, the range of drag-reducing wavenumbers shrinks and moves toward long-wavelength modes, an effect we commented upon in our discussion of figure 2. We notice that when $\beta = 0$ or ${\rm \pi}/4$, the most effective heating wavenumber, at least as far as drag reduction is concerned, appears to be $\alpha \approx 1$. The most significant pressure reduction occurs at longer wavelengths $\alpha \approx 0.3$ for a vertical slot, but the decrease in B is also smaller. Our patterned heating no longer possesses drag-reducing properties once $\beta = 3{\rm \pi}/4$. Our results thus far demonstrate that pressure-gradient losses can be significantly mitigated and, perhaps, even eliminated if a judicious combination of parameters is made.

Figure 5. The pressure-gradient correction parameter B as a function of $\alpha $ and $Re$ when $R{a_{p,L}} = 0$ and $R{a_{p,R}} = 500.$ The four plots correspond to slot inclination angles $\beta = $ (a) 0, (b) ${\rm \pi}/4$, (c) ${\rm \pi}/2$ and (d) $3{\rm \pi}/4$. The grey shading indicates those parameter combinations that lead to a reduction in the pressure loss. The thick dotted, dashed-dotted and dashed lines identify conditions leading to 10 %, 50 % and 100 % reductions in pressure losses, respectively. The green vertical lines identify the flow conditions used in figure 6, while the green circles show those in figure 3. The blue lines denote the conditions adopted in figure 7, while the brown lines and circles identify the parameter combinations used in figure 8.

Detailed information about the effects of the Reynolds number can be gleaned from the data displayed in figure 6. It is clear that the pressure-gradient correction approaches a constant, inclination-dependent limit as $Re \to 0$, corresponding to the pure natural convection that Floryan et al. (Reference Floryan, Benayoun, Panday and Bassom2022a) considered. Our analysis here is based on the assumption of the prescribed flow rate constraint (2.8), which, in the limit $Re \to 0$ corresponds to a zero flow rate. As heating generates a net longitudinal flow, its elimination necessitates the imposition of a pressure gradient, as suggested by figure 6. The pressure-gradient correction approaches a limit that approaches zero as $Re \to \infty $ irrespective of the inclination angle. This limit is reached when $Re$ exceeds about 100, and the separation bubbles have been completely destroyed. Furthermore, the flow loses any dependence on the heating and acquires characteristics akin to an isothermal Poiseuille flow. We emphasize that obtaining a significant reduction in pressure losses for smaller Reynolds numbers is possible.

Figure 6. The pressure-gradient correction B as a function of the Reynolds number Re when $R{a_{p,L}} = 0, R{a_{p,R}} = 500$ and $\alpha = 1$. Plotted are the values of B for the eight inclinations $\beta = j{\rm \pi}/8$ with $j = 0,\textrm{ }1, \ldots ,\textrm{ }7$. The flow conditions for $Re = 1,\; \textrm{ }5,\textrm{ }\; 10$ are marked using blue lines in figure 2, while the circles illustrate the flow conditions used in figure 3. The various levels of grey shading indicate parameter combinations that reduce pressure losses; the borders between the shadings depict the combinations for which 10 %, 50 % or 100 % pressure-gradient reductions are possible.

Detailed information concerning the effects of varying the heating wavenumber can be inferred from figure 7. Heating seems to have the most marked influence on the pressure losses when $\alpha = O(1)$; naturally, the details depend on the inclination angle. The magnitude of these losses decreases in both the long- and short-heating-wavelength limits. In particular, it may be shown that $B \propto {\alpha ^2}$ as $\alpha \to 0$ while $B \propto {\alpha ^{ - 3}}$ when $\alpha \to \infty $, at least when the slot is inclined, and the details of these limits are discussed in §§ 5 and 6. Horizontal and vertical channels are somewhat special cases, as the results in figure 7 suggest. It seems that heating always reduces the pressure losses in horizontal channels irrespective of the heating wavenumber but only reduces losses in vertical channels if the wavelength is sufficiently long. There are cases of heating wavenumbers that are so effective that an opposite-signed pressure gradient must be imposed to prevent flow acceleration.

Figure 7. The pressure-gradient correction B when $R{a_{p,L}} = 0$ and $R{a_{p,R}} = 500$ for the four channel inclinations $\beta = j{\rm \pi}/4$ with $j = 0,1,2,3$. Shown are the results for two Reynolds numbers: (a) $Re = 1$ and (b) $Re = 10$. The flow conditions used in this figure are shown using blue lines in figure 5. The various degrees of grey shading indicate those parameter combinations that reduce the pressure losses; the borders between the different shadings correspond to 10 %, 50 % and 100 % pressure-gradient reduction. The circles identify the conditions adopted in figure 8.

The evolution of flow structures when $\alpha = O(1)$ has already been discussed in connection with figure 3. We observed that the structures changed slightly as the inclination angle altered. To complement these findings, in figure 8, we present evidence as to the form of the flow structures in the small- and large-wavenumber limits. There is a dramatic change in the flow topology at long wavelengths as the slot is inclined at an angle increasing from horizontal. The long-wavelength structures can be analysed, and the pertinent results are summarized in § 5. There, we point out that the limit $\beta \to 0$ is a special case, and this conclusion is supported by the flow structures displayed in the first column of rows 1 and 3 of figure 8. On the other hand, at short wavelengths, boundary layers form near the heated walls. The asymptotic solutions described in § 6 show that the details of the boundary layers differ if the channel is horizontal or vertical, but the modifications are somewhat technical so that the overall flow topology does not change significantly.

Figure 8. The flow and temperature fields for flows with heating intensities $R{a_{p,L}} = 0$ and $R{a_{p,R}} = 500$. Shown are the structures that correspond to relatively long-wavelength $(\alpha = 0.1)$ and short-wavelength $(\alpha = 10)$ modes and at the two Reynolds numbers Re = 1 and Re = 10. The background colour illustrates the temperature field. The black arrows show the direction of gravity force, while the red arrows denote the direction of pressure-gradient force. The flow conditions used in these plots are marked in figures 5 and 7. In all the plots the temperature has been normalized with its maximum ${\theta _{max}}$.

One last aspect of the problem that warrants attention is the effect of the heating intensity. The change in the pressure gradient is proportional to $Ra_{p,R}^2$ for relatively weak heating; this result is evident in the results contained in figure 9. In the opposite case of intense heating, we have evidence of saturation, and the magnitude of B does not grow indefinitely as $R{a_{p,R}}$ increases, at least in the case of small Reynolds numbers (see figure 9a). We see that the case of a vertical channel is somewhat a special case in as much that there is a transition from a reduction to a growth in the pressure losses, which occurs at $R{a_{p,R}} \approx 400$ (Floryan et al. Reference Floryan, Wang and Bassom2023d). At larger Reynolds numbers, the behaviour is rather more intricate. Now, a switch between reduced and enhanced pressure-gradient losses is possible in slots that no longer need to be vertical.

Figure 9. The pressure-gradient correction B for a heating wavenumber $\alpha = 1$ and eight inclination angles $\beta = j{\rm \pi}/8$ for $j = 0,\textrm{ }1, \ldots ,\textrm{ }7$. Two Reynolds numbers are considered: (a) $Re = 1$ and (b) $Re = 10$. The different levels of shading indicate parameter combinations that reduce pressure losses, and the borders between the colours correspond to cases of 10 %, 50 % and 100 % pressure-gradient reduction.

One may inquire about the stability properties of these flows and whether a transition to secondary states might occur. Previous work on analysing flows in horizontal channels has shown that the periodic Rayleigh number must reach values in excess of 2500 for any instability to appear, irrespective of the heating wavenumber (Hossain & Floryan Reference Hossain and Floryan2013b, Reference Hossain and Floryan2015b). The results summarized in figure 9 demonstrate a saturation effect associated with an increase in heating intensity above $R{a_{p,R}} = 2000$. The upshot is that such large intensities are of little interest in developing practical drag-reducing techniques. Strictly, one should check whether the intensity of the critical heating decreases as the inclination angle increases. However, since the channel-transverse component of the buoyancy force drives convective instabilities, and this component drops as the inclination angle grows, so the critical heating intensity is expected to increase with the inclination angle.

The above analysis raises a natural question about the energy costs of reducing pressure losses. The reduction of energy delivered through the pressure gradient (per unit length) is $\Delta {E_{pres}} = {\textstyle{4 \over 3}}ReB$ with variations of B given in figure 2. The heat energy delivered to the system has two parts. The first one involves heat flow between the walls quantified by the average Nusselt number $N{u_{av}}$, and it represents the true energy cost. The second one represents the energy flow between the hot and cold sections of the wall. This heat flow averages out to zero over a wavelength, but there must be a cost associated with maintaining temperature variations along the wall. Some information about it is provided by evaluating heat flow leaving/entering the wall per half-wavelength, which is quantified by the horizontal Nusselt number $N{u_{h,R}}$. Variations of $N{u_{av}}$ and $N{u_{h,R}}$ for the same conditions as in figure 2 are displayed in figure 10. While $N{u_{h,R}}$ exceeds by an order of magnitude $N{u_{av}}$, as noted in Hossain & Floryan (Reference Hossain and Floryan2013a), it is not a good measure of energy cost as this heat is recovered over the next half of the heating wavelength, i.e. it averages out to zero. The actual energy cost of maintaining temperature variations along the wall is a function of the details of the construction of the heating system and cannot be determined unless the wall composition is specified and conduction within the wall is accounted for. This cost is expected to be much smaller than the heat flow over half a wavelength. The magnitude of energy fluxes can be estimated based on the data from figures 2 and 10. Consider $Re = 5$ (the middle columns in these figures) and take the middle of the drag-reducing zone, which gives $B \approx 5$. The energy saving due to the reduction of pressure losses is $\mathrm{\sim }33$, the energy cost due to the heat flow between the walls is $\mathrm{\sim }20$ and the amount of energy leaving the wall and re-entering it is $\mathrm{\sim }300$. This underlines that wall construction dominates the energy cost, but an analysis of possible wall constructions and their energy efficiencies is outside the scope of our analysis. If this energy cost can be eliminated, or at least significantly reduced, it might be possible to develop a system where energy cost of loss reduction is smaller than the energy saved by loss reduction. It should be remarked that the cost could be acceptable even if energy cost of reduction significantly exceeds the reduction itself as one may be able to use waste energy as a heating source, making the energy cost largely immaterial.

Figure 10. The variations of the average Nusselt number $N{u_{av}}$ and the horizontal Nusselt number $N{u_{h,R}}$ (see (2.10)) as functions of the heating wavenumber $\alpha $ and the slot inclination angle $\beta $ when only the right wall is heated with $R{a_{p,R}} = 500$.

4. Heating applied to both sides of the channel

We now briefly discuss the situations that can arise when both sides of the slot are heated. Our observations so far suggest that changes in the friction at the heated wall are larger than at the isothermal wall, thereby reducing pressure losses. When both walls are heated, the reduction of losses is expected to be enhanced as the separation bubbles on both walls are directly driven by the heating applied at each. This additional heating introduces two further parameters into the problem: the intensity of the heating of the second surface and the phase difference $\varOmega $ between the two patterns. An in-depth study of all possible parameter combinations would lead to an unwieldy plethora of results. To focus matters, we concentrate on understanding the role played by the phase difference when equal heating is applied to the two walls so $R{a_{p,L}} = R{a_{p,R}}$. Calculations show that the details of the pressure losses depend strongly on the phase $\varOmega $ and the channel orientation $\beta $, and some sample results are illustrated in figure 11. The value of the correction B seems to depend only weakly on the phase $\varOmega $ when the channel is horizontal. This behaviour is perhaps illusory, as there is some variation with $\varOmega $, which is largely hidden on the scale used to plot figure 11. Hossain & Floryan (Reference Hossain and Floryan2016) discuss this problem in some detail. It is recalled that the streamwise component of the mean buoyancy force is zero when the slot is horizontal, but this no longer holds once the channel is inclined. This induces much larger changes in the value of B and, as the offset $\varOmega $ is adjusted, significant swings in the magnitude of B arise and, moreover, it can change sign.

Figure 11. The pressure-gradient correction B as a function of $\varOmega $ and $\beta $ when $R{a_{p,L}} = R{a_{p,R}} = 500$ and $\alpha = 1$. The three plots correspond to a Reynolds number $Re = $ (a) 1, (b) 5 and (c) 10. The shading indicates those parameter combinations that lead to a reduction in pressure losses. The thick dotted, dashed-dotted and dashed lines identify the conditions leading to 10 %, 50 % and 100 % reduction of pressure losses.

A better understanding of possible flow responses can be gained from the data displayed in figure 12. Here, we consider the quantity ${B_{comp}} = ({B_2} - {B_1})/{B_1}$ which gives a very rough estimate of the comparison of the system performance, whether only one wall or both walls are heated. In this definition, we denote by ${B_j}$ the pressure-gradient correction observed when j walls are heated (j = 1, 2). This then naturally leads to a division of the parameter space into four regions (i)–(iv) defined by (i) ${B_1} > 0,\textrm{ }{B_2} > 0$; (ii) ${B_1} < 0,\textrm{ }{B_2} > 0$; (iii) ${B_1} < 0,\textrm{ }{B_2} < 0$; and (iv) ${B_1} > 0,\textrm{ }{B_2} < 0$. We have conditions for improved pressure corrections with two-wall heating. It seems that within (ii), there is also an advantage in using two-wall heating as it generates a positive pressure-gradient correction, while the same conditions give a negative pressure-gradient correction for one-wall heating. Overall, the advantage of the two-wall heating seems to increase with $Re$.

Figure 12. A comparison of the effectiveness of the one-wall and two-wall heating as functions of $\varOmega $ and $\beta $ when $R{a_{p,L}} = R{a_{p,R}} = 500$ and $\alpha = 1$ for Reynolds numbers $Re = $ (a) 1, (b) 5 and (c) 10. Shown plotted is the quantity${B_{comp}} = ({B_2} - {B_1})/{B_1}\;\textrm{in}\;\textrm{which}\;{B_j}$ denotes the pressure-gradient correction observed when j walls are heated (j = 1, 2). The parameter space is divided into four regions (i)–(iv) as defined in the text.

In the next two sections, we analytically examine the cases of long-wavelength and short-wavelength heating. This serves twin objectives: on the one hand, they provide insight into the mechanisms that underpin the flow structure; on the other hand, comparison with the numerical simulations in appropriate limits gives us additional confidence in the accuracy of the calculations.

5. Long-wavelength heating

We begin a discussion of the analytic solution for the flow by considering long-wavelength heating $(\alpha \ll 1)$. To that end, we introduce the stretched coordinate $X = \alpha x$, which brings the governing equations and thermal boundary conditions to the following form:

(5.1a)\begin{gather}\alpha [Re(1 - {y^2}) + {u_1}]{u_{1X}} + {v_1}( - 2Re\,y + {u_{1y}}) ={-} \alpha {p_{1X}} + {\alpha ^2}{u_{1XX}} + {u_{1yy}} + \frac{{{\theta _1}}}{{Pr}}\sin \beta ,\end{gather}

(5.1b)\begin{gather}\alpha [Re(1 - {y^2}) + {u_1}]{v_{1X}} + {v_1}{v_{1y}} ={-} {p_{1y}} + {\alpha ^2}{v_{1XX}} + {v_{1yy}} + \frac{{{\theta _1}}}{{Pr}}\cos \beta ,\end{gather}

(5.1c)\begin{gather}\alpha [Re(1 - {y^2}) + {u_1}]{\theta _{1X}} + {v_1}{\theta _{1y}} = \frac{1}{{Pr}}({\alpha ^2}{\theta _{1XX}} + {\theta _{1yy}}),\end{gather}

(5.1d)\begin{gather}\alpha {u_{1X}} + {v_{1y}} = 0,\end{gather}

with the thermal boundary conditions becoming

(5.2a,b)\begin{equation}{\theta _1}(X, - 1) = {\textstyle{1 \over 2}}R{a_{p,R}}\cos X,\quad {\theta _1}(X,1) = {\textstyle{1 \over 2}}R{a_{p,L}}\cos (X + \varOmega ).\end{equation}

Notice that we have incorporated the phase shift in the heating profiles and allowed different magnitudes of heating at the two walls. It is helpful for the subsequent calculations to define the three related quantities

(5.3a–c)\begin{equation}{R_1} = R{a_{p,R}} + R{a_{p,L}}\cos \varOmega ,\quad {R_2} = R{a_{p,R}} - R{a_{p,L}}\cos \varOmega \quad \textrm{and}\quad {R_3} = R{a_{p,L}}\sin \varOmega .\end{equation}

We seek solutions that assume the structure

(5.4)\begin{align}({u_1},{v_1},{p_1},{\theta _1}) = {\alpha ^{ - 1}}(0,0,{\hat{P}_{ - 1}},0) + ({\hat{U}_0},0,{\hat{P}_0},{\hat{\theta }_0}) + \alpha ({\hat{U}_1},{\hat{V}_0},{\hat{P}_1},{\hat{\theta }_1}) + {\alpha ^2}({\hat{U}_2},{\hat{V}_1},{\hat{P}_2},{\hat{\theta }_2}) + \cdots ,\end{align}

where all the unknowns are functions of X and y. Given the form of the boundary conditions, we are led to the leading-order solutions

(5.5a,b)\begin{equation}{\hat{U}_0} ={-} \frac{1}{{24Pr}}y(1 - {y^2})({R_2}\cos X + \; {R_3}\sin X)\sin \beta ,\quad {\hat{V}_0} ={-} \frac{1}{{96Pr}}{(1 - {y^2})^2}({R_3}\cos X - \; {R_2}\sin X)\sin \beta ,\end{equation}

(5.5c,d)\begin{equation} {\hat{\theta }_0} = \; \; \frac{1}{4}\; [({R_1} - {R_2}y)\cos X - {R_3}(1 + y)\sin X]\; \quad \textrm{and}\quad {\hat{P}_{ - 1}} = \frac{1}{{4Pr}}({R_1}\sin X + \; {R_3}\cos X)\sin \beta.\nonumber\\ \end{equation}

At $O(\alpha )$ the thermal field consists of mean and $X$-dependent parts. It may be shown that

(5.6)\begin{align}{\hat{\theta }_1}(X,y) ={-} {\textstyle{1 \over 4}}PrRe[{F_1}(y)\sin X + \; {F_2}(y)\cos X] + {\textstyle{1 \over {384}}}[{F_3}(y)\sin 2X + {F_4}(y)\cos 2X + {F_5}(y)]\sin \beta ,\end{align}

where the polynomials ${F_1}(y)$–${F_5}(y)$ are defined to be

(5.7a)\begin{gather}{F_1}(y) = {\textstyle{1 \over {60}}}(1 - {y^2})[5({y^2} - 5){R_1} - y(3{y^2} - 7){R_2}],\end{gather}

(5.7b)\begin{gather}{F_2}(y) = {\textstyle{1 \over {60}}}\; {R_3}(1 - {y^2})(3{y^3} + 5{y^2} - 7y - 25),\end{gather}

(5.7c)\begin{gather}{F_3}(y) = {\textstyle{1 \over {60}}}(1 - {y^2})(3{y^4} - 2{y^2} - 17)(R_3^2 - R_2^2) - {\textstyle{1 \over {30}}}\; y(1 - {y^2})(7 - 3{y^2})(R_3^2 + {R_1}{R_2}),\end{gather}

(5.7d)\begin{gather}{F_4}(y) = {\textstyle{1 \over {30}}}\; (1 - {y^2})(3{y^4} + 3{y^3} - 2{y^2} - 7y - 17){R_2}{R_3} - {\textstyle{1 \over {30}}}\; y(1 - {y^2})(3{y^2} - 7){R_1}{R_3},\end{gather}

(5.7e)\begin{gather}{F_5}(y) ={-} {\textstyle{1 \over {30}}}y(1 - {y^2})(7 - 3{y^2})({R_1} + {R_2}){R_3}.\end{gather}

The remaining quantities satisfy the equations

(5.8a)\begin{gather}{\hat{U}_{1yy}} - {\hat{P}_{0X}} + \frac{1}{{Pr}}{\hat{\theta }_1}\sin \beta = Re(1 - {y^2}){\hat{U}_{0X}} - 2Rey{\hat{V}_0} + {\hat{U}_0}{\hat{U}_{0X}} + {\hat{V}_0}{\hat{U}_{0y}},\end{gather}

(5.8b,c)\begin{gather}{\hat{P}_{0y}} = \frac{1}{{Pr}}{\hat{\theta }_0}\cos \beta \quad \textrm{and}\quad {\hat{U}_{1X}} + {\hat{V}_{1y}} = 0.\end{gather}

Strictly, we only need to ascertain the parts of the solutions proportional to $\sin X$ and $\cos X$; the nonlinear terms in (5.8a) will generate terms proportional to $\sin 2X$ and $\cos 2X$ but these do not contribute to calculating the orders that must be considered. It then follows that the relevant parts of the solution of (5.8) are

(5.9a) \begin{align} {{\hat{U}}_1}(X,y) & = \dfrac{1}{{10\;080}}Re{R_3}\left( {{F_6} - \dfrac{{{F_7}}}{{Pr}}} \right)\cos X\sin \beta \; + \dfrac{1}{{10\;080}}Re\; \left( {{F_8} + {R_2}\dfrac{{{F_7}}}{{Pr}}} \right)\sin X\sin \beta\nonumber \\ & \quad + \dfrac{1}{{480Pr}}(1 - {y^2})[{R_3}(5{y^2} - 20y + 1)\cos X + ({R_2}(1 - 5{y^2})\nonumber \\ & \quad + 20{R_1}y)\sin X]\cos \beta + {U_{M1}}(y) + \cdots ,\end{align}

(5.9b) \begin{align} {{\hat{V}}_1}(X,y) & = \dfrac{1}{{10\;080}}Re{R_3}\left( {{F_9} \,{-}\, \dfrac{{{F_{10}}}}{{Pr}}} \right)\sin X\sin \beta \,{+}\, \dfrac{1}{{10\;080}}Re\left( {{F_{11}} - {R_2}\dfrac{{{F_{10}}}}{{Pr}}} \right)\cos X\sin \beta\nonumber \\ & \quad - \dfrac{1}{{480Pr}}{(1 - {y^2})^2}[{R_3}(y + 5)\sin X - (5{R_1} - {R_2}y)\textrm{cos }X]\cos \beta + \cdots , \end{align}

(5.9c) \begin{align} {{\hat{P}}_0} & = \dfrac{{17}}{{210}}Re({R_3}\sin X - \; {R_1}\cos X)\sin \beta\nonumber \\ & \quad + \dfrac{1}{{40Pr}}[({R_2}(1 \,{-}\, 5{y^2}) \,{+}\, 10{R_1}y)\cos X \,{-}\, {R_3}(5{y^2} \,{+}\, 10y \,{-}\, 1)\sin X]\cos \beta \,{+}\, \cdots\!, \end{align}

in which the polynomials ${F_6}(y)$–${F_{11}}(y)$ are given by

(5.10a)\begin{gather}{F_6}(y) = (1 - {y^2})(3{y^5} + 7{y^4} - 18{y^3} - 98{y^2} + 31y + 19),\end{gather}

(5.10b)\begin{gather}{F_7}(y) = y({y^2} - 1)(5{y^4} - 16{y^2} + 19),\end{gather}

(5.10c)\begin{gather}{F_8}(y) = ({y^2} - 1)[(3{y^5} - 18{y^3} + 31y){R_2} - (7{y^4} - 98{y^2} + 19){R_1}],\end{gather}

(5.10d)\begin{gather}{F_9}(y) ={-} {\textstyle{1 \over 8}}{({y^2} - 1)^2}(3{y^4} + 8{y^3} - 22{y^2} - 152y + 51),\end{gather}

(5.10e)\begin{gather}\; {F_{10}}(y) = {\textstyle{1 \over 8}}{({y^2} - 1)^2}(5{y^4} - 18{y^2} + 29),\end{gather}

(5.10f)\begin{gather}{F_{11}}(y) ={-} {\textstyle{1 \over 8}}{({y^2} - 1)^2}[(3{y^4} - 22{y^2} + 51){R_2} - 8y({y^2} - 19){R_1}].\end{gather}

Furthermore, the mean part of the streamwise velocity ${\hat{U}_{M1}}(y)$ is another polynomial, but this consists only of terms of odd degree, so the mass flux across the slot $- 1 \le y \le 1$ is zero.

We now move on to consideration of the $O({\alpha ^2})$ terms. If we suppose that the pressure gradient

(5.11)\begin{equation}{p_X} = A{\alpha ^2} + \cdots ,\end{equation}

then the $O({\alpha ^2})$ terms in the streamwise momentum and energy equations are

(5.12a)\begin{align}Re(1 - {y^2}){\hat{U}_{1X}} - 2Rey{\hat{V}_1} + {\hat{U}_0}{\hat{U}_{1X}} + {\hat{U}_1}{\hat{U}_{0X}} + {\hat{V}_0}{\hat{U}_{1y}} + {\hat{V}_1}{\hat{U}_{0y}} ={-} (A + \cdots ) + {\hat{U}_{0XX}} + {\hat{U}_{2yy}} + \frac{1}{{Pr}}{\hat{\theta }_2}\sin \beta ,\end{align}

(5.12b)\begin{equation}Re(1 - {y^2}){\hat{\theta }_{1X}} + {\hat{U}_0}{\hat{\theta }_{1X}} + {\hat{U}_1}{\hat{\theta }_{0X}} + {\hat{V}_0}{\hat{\theta }_{1y}} + {\hat{V}_1}{\hat{\theta }_{0y}} = \frac{1}{{Pr}}({\hat{\theta }_{0XX}} + {\hat{\theta }_{2yy}}).\end{equation}

If we denote the mean part of ${\hat{\theta }_2}(x,y)$ by ${\hat{\theta }_{2M}}(y)$ then (5.12b) furnishes an expression for ${({\hat{\theta }_{2M}})_{yy}}$ that can be integrated twice, subject to the requirement that ${\hat{\theta }_{2M}}$ vanishes at $y ={\pm} 1$. This result is substituted into (5.12a) for the mean function ${\hat{U}_{2M}}(y)$ which can be found subject to ${\hat{U}_{2M}}({\pm} 1) = 0$; the pressure gradient is then adjusted to ensure the mean mass flux is zero. After some algebraic manipulation and simplification, we find that

(5.13) \begin{align}\begin{aligned} A & = \dfrac{1}{{12\,600\; Pr}}({Ra_{p,R}^2 - Ra_{p,L}^2} )\sin 2\beta + \dfrac{{{{\sin }^2}\beta }}{{227\,026\,800P{r^2}}}Re\\ & \quad \times \{ (Ra_{p,L}^2 + Ra_{p,R}^2 - 2R{a_{p,L}}\; R{a_{p,R}}\cos \varOmega )(51 - 790\; Pr)\\ & \quad + [5876(Ra_{p,L}^2 + Ra_{p,R}^2) + 9620\; R{a_{p,L}}R{a_{p,R}}\cos \varOmega ]P{r^2}\} , \end{aligned}\end{align}

and the associated pressure gradient is $B = A{\alpha ^2}$.

The reader may note that when heating is applied at the right wall only, the above expression simplifies to the form

(5.14)\begin{align}{\alpha ^2}\left\{ {\frac{1}{{227\,026\,800P{r^2}}}Re\,Ra_{p,R}^2(51 - 790Pr + 5876P{r^2}){{\sin }^2}\beta + \frac{1}{{12\,600\; Pr}}Ra_{p,R}^2\sin 2\beta } \right\}.\end{align}

The first term is always positive for any $Pr$ while the second term becomes negative for $\beta \in ({\rm \pi}/2,{\rm \pi})$ and for $\beta \in (3{\rm \pi}/2,2{\rm \pi})$ with precise boundaries of these intervals being discussed later in the presentation. It is noted that a vertical $(\beta = {\rm \pi}/2)$ has the flow moving upwards. This wall begins to serve as the upper wall with a further increase of $\beta $ with the flow directed to the left. It becomes horizontal when $\beta = {\rm \pi}$, with the flow then moving to the left and the upper wall being heated. A further increase of $\beta $ results in the flow travelling downwards with the upper wall now heated. It becomes vertical at $\beta = 3{\rm \pi}/2$ with the flow now downward and the left wall heated. The channel returns to a horizontal position at $\beta = 2{\rm \pi}$, with the lower wall being heated and the flow moving to the right. Reduction of pressure losses is possible only for the channel orientation between vertical with the right wall heated and being horizontal with the upper wall heated and between being vertical with the left wall heated and horizontal with the lower wall heated. The above equation also shows that the solution for the inclined channel in the limit of $\beta \to 0$ (horizontal channel) has a different structure with $A = 0({\alpha ^4})$ (Floryan & Floryan Reference Floryan and Floryan2015) rather than $A = 0({\alpha ^2})$ for the inclined channel. Details of this transition are omitted from this discussion.

We can also comment on the Nusselt number form. If we take the expression for ${\hat{\theta }_{2M}}$ and evaluate the derivative at $y ={-} 1$ we find that

(5.15) \begin{align}{\left. {\frac{{\textrm{d}{{\hat{\theta }}_{2M}}}}{{\textrm{d}y}}} \right|_{y ={-} 1}} &= \frac{1}{{45\,360}}Re(Ra_{p,R}^2 - Ra_{p,L}^2)(1 - 10Pr)\sin \beta - \frac{1}{{12\,600}}(9Ra_{p,R}^2 + 9Ra_{p,L}^2\nonumber\\ &\quad + 17R{a_{p,L}}R{a_{p,R}}\cos \varOmega )\cos \beta .\end{align}

Figure 13 compares the asymptotic predictions against some numerical simulations of the complete governing system. We plot the form of $B = A{\alpha ^2}$ as given by (5.13) and the Nusselt number $N{u_{av}}$ predicted by (5.15). The comparison shows excellent agreement between the analytical and numerical findings, at least for wavenumbers $\alpha < 0.1$.

Figure 13. Variations of the numerically determined (${B_n}$, solid black line) and the analytically determined (${B_a}$, dotted black line) pressure-gradient corrections, and of the numerically determined ($N{u_{av,n}}$ solid red line) and the analytically determined ($N{u_{av,a}}$, dotted red line) average Nusselt numbers as functions of the heating wavenumber $\alpha $ as $\alpha \to 0$ for $Re = 1$, $R{a_{p,R}} = 400$, $R{a_{p,L}} = 0$. The dashed-dotted lines identify the differences $\mathrm{\Delta }B = |{B_a} - {B_n}|$ (black line) and $\mathrm{\Delta }Nu = |N{u_{av,n}} - N{u_{av,a}}|$ (red line). These results suggest that the leading-order solutions are indeed in error by $O({\alpha ^4})$.

6. Short-wavelength heating

Here, we consider the opposite limit of short-wavelength heating. When $\alpha $ is large, most of the interesting motion occurs near the edges of the channel, so let us consider appropriate scales so that $X = \alpha x$ and $Y = \alpha (y + 1)$ near the right-hand wall. The governing equations written in these scales become

(6.1a)\begin{gather}{\alpha ^{ - 1}}[Re(2\alpha Y - {Y^2}) + {\alpha ^2}{u_1}]{u_{1X}} + {\alpha ^{ - 1}}{v_1}[2Re(\alpha - Y) + {\alpha ^2}\; {u_{1Y}}]\nonumber\\ ={-} \alpha {p_{1X}} + {\alpha ^2}({u_{1XX}} + {u_{1YY}}) + P{r^{ - 1}}{\theta _1}\sin \beta ,\end{gather}

(6.1b)\begin{gather}{\alpha ^{ - 1}}[Re(2\alpha Y \,{-}\, {Y^2}) \,{+}\ {\alpha ^2}{u_1}]{v_{1X}} \,{+}\ \alpha \; {v_1}{v_{1Y}} ={-} \alpha \; {p_{1Y}} \,{+}\ {\alpha ^2}({v_{1XX}} \,{+}\ {v_{1YY}}) + P{r^{ - 1}}{\theta _1}\cos \beta ,\end{gather}

(6.1c)\begin{gather}{u_{1X}} + {v_{1Y}} = 0,\end{gather}

(6.1d)\begin{gather}{\alpha ^{ - 1}}[Re(2\alpha Y - {Y^2}) + {\alpha ^2}{u_1}]{\theta _{1X}} + \alpha \; {v_1}{\theta _{1Y}} = {\alpha ^2}P{r^{ - 1}}({\theta _{1XX}} + {\theta _{1YY}}),\end{gather}

where the subscript 1 refers to flow modulations caused by the heating. We solve these equations subject to periodic heating on the edges of the slot so that

(6.2)\begin{equation}{\theta _1}(X, - 1) = {\textstyle{1 \over 2}}R{a_{p,R}}\cos X,\quad \; {\theta _1}(X,1) = {\textstyle{1 \over 2}}R{a_{p,L}}\cos (X + \varOmega ).\end{equation}

When $\alpha \; {{''}}1$ we seek solutions that assume the structure

(6.3) \begin{align}{u_1} = {\alpha ^{ - 2}}\left( {\sum\limits_0^\infty {{\alpha^{ - j}}{{\hat{U}}_j}} } \right),\quad {v_1} = {\alpha ^{ - 2}}\left( {\sum\limits_0^\infty {{\alpha^{ - j}}{{\hat{V}}_j}} } \right),\quad {p_1} = {\alpha ^{ - 1}}\left( {\sum\limits_0^\infty {{\alpha^{ - j}}{{\hat{P}}_j}} } \right),\quad {\theta _1} = \left( {\sum\limits_0^\infty {{\alpha^{ - j}}{{\hat{\varTheta }}_j}} } \right),\end{align}

where all the unknowns are functions of X and Y. Leading-order terms in the thermal equation give

(6.4)\begin{equation}{\hat{\varTheta }_{0XX}} + {\hat{\varTheta }_{0YY}} = 0\quad \Rightarrow \quad {\hat{\varTheta }_0}(X,Y) = {\textstyle{1 \over 2}}R{a_{p,R}}\textrm{ exp}( - Y)\cos X,\end{equation}

to satisfy the boundary condition on $Y = 0$ and to decay as $Y \to \infty $. Next-order equations show that ${\hat{\varTheta }_1}(X,Y) \equiv 0$; we also have ${\hat{U}_1} = {\hat{V}_1} = {\hat{P}_1} \equiv 0$. At $O({\alpha ^{ - 2}})$ we find that

(6.5)\begin{align}2Y\,Re{\hat{\varTheta }_{0X}} = P{r^{ - 1}}({\hat{\varTheta }_{2XX}} + {\hat{\varTheta }_{2YY}})\quad \Rightarrow \quad {\hat{\varTheta }_2}(X,Y) = {\textstyle{1 \over 4}}PrReR{a_{p,R}}Y(Y + 1)exp ( - Y)\sin X,\end{align}

while the O(1) terms in the two momentum and continuity equations give

(6.6a–c)\begin{equation}0 ={-} {\hat{P}_{0X}} + {\hat{U}_{0XX}} + {\hat{U}_{0YY}} + P{r^{ - 1}}{\hat{\varTheta }_0}\sin \beta ,\quad 0 ={-} {\hat{P}_{0Y}} + {\hat{V}_{0XX}} + {\hat{V}_{0YY}} + {\Pr ^{ - 1}}{\hat{\varTheta }_0}\cos \beta ,\quad {\hat{U}_{0X}} + {\hat{V}_{0Y}} = 0,\end{equation}

whose solution is

(6.7a,b)\begin{gather}{\hat{U}_0} ={-} \frac{{R{a_{p,R}}\; }}{{16Pr}}Y(2 - Y)exp ( - Y)\sin (X - \beta ),\quad {\hat{V}_0} = \frac{{R{a_{p,R}}\; }}{{16Pr}}{Y^2}exp ( - Y)\cos (X - \beta ),\end{gather}

(6.7c)\begin{gather}{\hat{P}_0} = \frac{{R{a_{p,R}}\; }}{{8Pr}}[(2Y + 3)\cos X\cos \beta + (2Y + 1)\sin X\sin \beta ].\end{gather}

We next move back to the energy equation at $O({\alpha ^{ - 3}})$. We have

(6.8)\begin{equation}{\hat{U}_0}\; {\hat{\varTheta }_{0X}} + {\hat{V}_0}{\hat{\varTheta }_{0Y}} - Re\; {Y^2}{\hat{\varTheta }_{0X}} = \frac{1}{{Pr}}({\hat{\varTheta }_{3XX}} + {\hat{\varTheta }_{3YY}}),\end{equation}

which, on substituting the leading-order results, becomes

(6.9) \begin{align}{\hat{\varTheta }_{3XX}} + {\hat{\varTheta }_{3YY}} &= {\textstyle{1 \over {32}}}Ra_{p,R}^2Y[(1 - Y)\cos \beta - \textrm{cos}(2X - \beta )]\textrm{exp}({ - 2Y} )\nonumber\\&\quad + {\textstyle{1 \over 2}}PrReR{a_{p,R}}{Y^2}\textrm{exp}( - Y)\sin X.\end{align}

This equation admits the solution

(6.10) \begin{align}\begin{aligned} {{\hat{\varTheta }}_3}(X,Y) & ={-} {\textstyle{1 \over {24}}}PrReR{a_{p,R}}Y(2{Y^2} + 3Y + 3)\; exp ( - Y)\sin X\\ & \quad + {\textstyle{1 \over {512}}}Ra_{p,R}^2Y(1 + 2Y)exp ( - 2Y)\cos (2X - \beta )\\ & \quad + {\textstyle{1 \over {256}}}Ra_{p,R}^2[1 - (1 + 2Y + 2{Y^2})exp ( - 2Y)]\cos \beta . \end{aligned}\end{align}

The streamwise momentum equation at $O({\alpha ^{ - 5}})$ gives

(6.11)\begin{equation}- {Y^2}Re{\hat{U}_{0X}} + {\hat{U}_0}{\hat{U}_{0X}} - 2YRe{\hat{V}_0} + {\hat{V}_0}{\hat{U}_{0Y}} ={-} {\hat{P}_{3X}} + {\hat{U}_{3XX}} + {\hat{U}_{3YY}} + \frac{{{{\hat{\varTheta }}_5}}}{{Pr}}\sin \beta .\end{equation}

Again, we are interested in the mean parts of this equation. If we suppose that ${\hat{P}_{3M}} = AX$ then the mean part of ${\hat{U}_3}$, call it ${\hat{U}_{3M}}(y)$, satisfies

(6.12)\begin{equation}\cdots ={-} \boldsymbol{A} + \hat{\boldsymbol{U}}_{3\boldsymbol{M}}^{\boldsymbol{^{\prime\prime}}} + \frac{1}{{\boldsymbol{Pr}}}{\hat{\boldsymbol{\varTheta }}_{3\boldsymbol{M}}}\sin \boldsymbol{\beta }.\end{equation}

We do not need to solve this explicitly, for it is sufficient to note that for large Y we have

(6.13)\begin{equation}\hat{U}_{3M}^{^{\prime\prime}} \sim \left[ {A - \frac{{Ra_{p,R}^2\; \; \; }}{{512Pr}}\sin 2\beta } \right] \Rightarrow {\hat{U}_{3M}} \sim \left[ {A - \frac{{Ra_{p,R}^2\; \; \; }}{{512Pr}}\sin 2\beta } \right]\frac{1}{2}{Y^2}.\end{equation}

We point out that all the terms on the left-hand side of (6.12) decay exponentially for large Y, so they cannot contribute to the leading-order form of the mean flow solution at large values.

This step completes the analysis of the wall layer. We see that the $O({\alpha ^{ - 3}})$ component of ${\theta _1}$ tends to a constant as $Y \to \infty $ (see (6.10)) while the $O({\alpha ^{ - 5}})$ component ${\hat{U}_{3M}}(y)$ grows quadratically. Then, across the bulk of the slot where $y = O(1)$, we have that

(6.14a–c)\begin{equation}{u_1} = {\alpha ^{ - 3}}\tilde{U}(y) + \cdots ,\quad {\theta _1} = {\alpha ^{ - 3}}\tilde{\varTheta }(y) + \cdots ,\quad {p_1} = {\alpha ^{ - 3}}Ax + \cdots ,\end{equation}

and these satisfy

(6.15)\begin{equation}0 ={-} A + \frac{{{\textrm{d}^2}\tilde{U}}}{{\textrm{d}{y^2}}} + \frac{1}{{Pr}}\tilde{\varTheta }\sin \beta ,\quad \frac{{{\textrm{d}^2}\tilde{\varTheta }}}{{\textrm{d}{y^2}}} = 0.\end{equation}

The form of (6.10) shows that for matching, we require

(6.16)\begin{equation}\tilde{\varTheta } \to \frac{{Ra_{p,R}^2\; }}{{256}}\cos \beta \quad \textrm{as}\;y \to - 1.\end{equation}

An exactly parallel calculation at the left wall shows that we must also demand

(6.17)\begin{equation}\tilde{\varTheta } \to - \frac{{Ra_{p,L}^2\; }}{{256}}\cos \beta \quad \textrm{as}\;y \to 1.\end{equation}

Hence,

(6.18)\begin{equation}\tilde{\varTheta }(y) = {\textstyle{1 \over {512}}}[Ra_{p,R}^2 - Ra_{p,L}^2 - (Ra_{p,L}^2 + Ra_{p,R}^2)y]\cos \beta ,\end{equation}

so that

(6.19)\begin{equation}\frac{{{\textrm{d}^2}\tilde{U}}}{{\textrm{d}{y^2}}} = A - \frac{1}{{1024\; Pr}}[Ra_{p,R}^2 - Ra_{p,L}^2 - (Ra_{p,L}^2 + Ra_{p,R}^2)y]\sin 2\beta .\end{equation}

There is no need to solve this completely – if we integrate twice and impose that $\tilde{U}({\pm} 1)\,{=}\,0$, this requires that the constant term in the above equation disappears. Hence,

(6.20)\begin{equation}A = \frac{1}{{1024\; Pr}}(Ra_{p,R}^2 - Ra_{p,L}^2)\sin 2\beta .\end{equation}

In conclusion, we see that to maintain a constant mass flux through the channel, we need to impose a pressure-gradient correction $B = {\alpha ^{ - 3}}A$, where A is given by (6.20). This correction does not depend on $Re$ at the leading order of approximation. If only the right wall is heated, a reduction of pressure losses can be achieved only for $\beta \in (0,{\rm \pi}/2)$ or for $\beta \in ({\rm \pi},3{\rm \pi}/2)$. The reader may note that the solution changes in the limits of $\beta \to 0$ (a horizontal channel) and $\beta \to {\rm \pi}/2$ (a vertical channel). Repeating the analysis for a horizontal channel shows that in this case $A = O({\alpha ^{ - 7}})$ (Floryan & Floryan Reference Floryan and Floryan2015) while for a vertical channel $A = O({\alpha ^{ - 5}})\; $ (Floryan et al. Reference Floryan, Wang and Bassom2023d).

To deduce the Nusselt number, we start by looking at the solution (6.12). It would be expected that the leading-order Nusselt number would be

(6.21)\begin{equation}{\alpha ^{ - 2}}{\left. {\frac{{\textrm{d}{{\hat{\varTheta }}_{3M}}}}{{\textrm{d}Y}}} \right|_{Y = 0}},\end{equation}

but it can easily be verified that this quantity actually vanishes. The next-order problem is given by

(6.22)\begin{equation}\frac{1}{{Pr}}({\hat{\varTheta }_{4XX}} + {\hat{\varTheta }_{4YY}}) = 2Y\,Re{\hat{\varTheta }_{2X}},\end{equation}

and we can determine the mean part of the solution. If this mean component is denoted as ${\hat{\boldsymbol{\varTheta }}_{4\boldsymbol{M}}}(\boldsymbol{Y})$ then

(6.23)\begin{equation}{\hat{\varTheta }_{4M}} ={-} {\textstyle{1 \over {512}}}Y(Ra_{p,L}^2 + Ra_{p,R}^2)\cos \beta \end{equation}

by matching directly with the core-layer solution. Hence, the Nusselt number becomes

(6.24)\begin{equation}{\alpha ^{ - 3}}{\left. {\frac{{\textrm{d}{{\hat{\varTheta }}_{4M}}}}{{\textrm{d}Y}}} \right|_{Y = 0}} ={-} {\alpha ^{ - 3}}\left\{ {\frac{1}{{512}}(Ra_{p,L}^2 + Ra_{p,R}^2)\cos \beta \; } \right\}.\end{equation}

The above discussion shows that to maintain a constant mass flux through the channel exposed to the heating, we need to impose a pressure-gradient correction $B = {\alpha ^{ - 3}}A$, where A is given by (6.20). The resulting heat flow between the walls is given by (6.24). In figure 14, we compare our asymptotic predictions with some numerical simulations; excellent agreement is observed between the two cases for $\alpha > 10$.

Figure 14. The form of the pressure-gradient correction B (black lines) and the average Nusselt number $N{u_{av}}$ (red lines). Solid lines denote numerically determined values, while the dotted lines indicate the analytical values as given by (6.20) and (6.24). Calculations performed with parameter values $Re = 1$, $R{a_{p,R}} = 400$ and $R{a_{p,L}} = 0$. The dashed-dotted lines denote the differences $\mathrm{\Delta }B$ and $\mathrm{\Delta }Nu$ between the numerical and analytical predictions.