$\frac{\partial u}{\partial x}+\frac{\partial \upsilon }{\partial y}=0\text{,}$

(9.114)

${\rho }_{\text{n}\text{f}}\left(u\frac{\partial u}{\partial x}+\upsilon \frac{\partial u}{\partial y}-U_{\infty }\frac{d{U}_{\infty }}{dx}\right)={\mu }_{\text{n}\text{f}}\frac{{\partial }^{2}u}{\partial y^2}+\frac{{\mu }_{\text{n}\text{f}}}{K}\left(U_{\infty }-u\right)\text{,}$

(9.115)

${\left(\rho C_{\text{p}}\right)}_{\text{n}\text{f}}\left(u\frac{\partial T}{\partial x}+\upsilon \frac{\partial T}{\partial y}\right)=\frac{\partial }{\partial y}\left(k_{\text{n}\text{f}}^{*}\frac{\partial T}{\partial y}\right),k_{\text{n}\text{f}}^{*}=k_{\text{n}\text{f}}(1+\varepsilon \theta )$

(9.116)

$\begin{array}{l}y=0:u=U_{\text{w}}\left(x\right),v=v_w,T=T_{\text{w}}\\ y\to \infty :u\to U_{\infty }\left(x\right),T\to T_{\infty }\end{array}$

(9.117)

${\rho }_{\text{n}\text{f}}={\rho }_{\text{f}}(1-\phi )+{\rho }_{\text{s}}\phi$

(9.118)

${\mu }_{\text{n}\text{f}}=\frac{{\mu }_{\text{f}}}{{(1-\phi )}^{2.5}}$

(9.119)

${\left(\rho C_{\text{p}}\right)}_{\text{n}\text{f}}={\left(\rho C_{\text{p}}\right)}_{\text{f}}(1-\phi )+{\left(\rho C_{\text{p}}\right)}_{\text{s}}\phi$

(9.120)

$\frac{k_{\text{n}\text{f}}}{k_{\text{f}}}=\frac{k_{\text{s}}+(n-1)k_{\text{f}}-(n-1)\phi \left(k_{\text{f}}-k_{\text{s}}\right)}{k_{\text{s}}+(n-1)k_{\text{f}}+\phi \left(k_{\text{f}}-k_{\text{s}}\right)}$

(9.121)

Table 9.5

Models of nanofluid based on different formulas for thermal conductivity and dynamic viscosity

Model	Shape of nanoparticles	Thermal conductivity	Dynamic viscosity
I	Spherical	$\frac{k_{\text{n}\text{f}}}{k_{\text{f}}}=\frac{k_{\text{s}}+2k_{\text{f}}-2\phi \left(k_{\text{f}}-k_{\text{s}}\right)}{k_{\text{s}}+2k_{\text{f}}+\phi \left(k_{\text{f}}-k_{\text{s}}\right)}$	${\mu }_{\text{n}\text{f}}=\frac{{\mu }_{\text{f}}}{{(1-\phi )}^{2.5}}$
II	Cylindrical (nanotubes)	$\frac{k_{\text{n}\text{f}}}{k_{\text{f}}}=\frac{k_{\text{s}}+\left(1/2\right)k_{\text{f}}-\left(1/2\right)\phi \left(k_{\text{f}}-k_{\text{s}}\right)}{k_{\text{s}}+\left(1/2\right)k_{\text{f}}+\phi \left(k_{\text{f}}-k_{\text{s}}\right)}$	${\mu }_{\text{n}\text{f}}=\frac{{\mu }_{\text{f}}}{{(1-\phi )}^{2.5}}$

$u=\frac{\partial \psi }{\partial y}\text{and}\upsilon =-\frac{\partial \psi }{\partial x}$

(9.122)

$\eta ={\left(\frac{a}{\upsilon }\right)}^{1/2}y,f\left(\eta \right)=\frac{\psi }{{\left(a\upsilon \right)}^{1/2}x},\theta \left(\eta \right)=\frac{T-T_{\infty }}{T_{\text{w}}-T_{\infty }}$

(9.123)

$f^{‴}+f{f}^{″}-{f^{\prime }}^2+{\lambda }^2+\frac{K_1}{A_1.{(1-\phi )}^{2.5}}\left(\lambda -f^{\prime }\right)=0\text{,}$

(9.124)

$\frac{1}{\mathrm{Pr}}·\frac{A_1·A_3·{\left(1-\phi \right)}^{2.5}}{A_2}\left(\left(1+\varepsilon \theta \right){\theta }^{″}+\varepsilon {\left({\theta }^{\prime }\right)}^2\right)+f{\theta }^{\prime }=0\text{.}$

(9.125)

$\begin{array}{l}f(0)=\frac{-v_{\text{w}}}{{\left(a\upsilon \right)}^{0.5}}=\gamma ,f^{\prime }(0)=1,\theta (0)=1,\\ f^{\prime }\left(\infty \right)\to \lambda ,\theta \left(\infty \right)\to 0.\end{array}$

(9.126)

Here prime denote differentiation with respect to η, λ = b/a is the Velocity ratio parameter, $\mathrm{Pr}=\left({\mu }_f{\left(\rho C_p\right)}_f\right)/\left({\rho }_f{k}_f\right)$ is the Prandtl number and K₁ = μ_f/(ρ_fKa) is the Permeability parameter and A₁, A₂, A₃ are parameters having the following form:

$A_1=\left(1-\phi \right)+\frac{{\rho }_{\text{s}}}{{\rho }_{\text{f}}}\phi$

(9.127)

$A_2=(1-\phi )+\frac{{\left(\rho C_{\text{p}}\right)}_{\text{s}}}{{\left(\rho C_{\text{p}}\right)}_{\text{f}}}\phi$

(9.128)

$A_3=\frac{k_{\text{nf}}}{k_{\text{f}}}=\frac{k_{\text{s}}+2k_{\text{f}}-2\varphi \left(k_{\text{f}}-k_{\text{s}}\right)}{k_{\text{s}}+2k_{\text{f}}+\varphi \left(k_{\text{f}}-k_{\text{s}}\right)}$

(9.129)

$C_{\text{f}}=\frac{{\mu }_{\text{n}\text{f}}}{{\rho }_{\text{f}}{U}_{\text{w}}^2}{\left(\frac{\partial u}{\partial y}\right)}_{y=0},Nu=\frac{x{k}_{\text{n}\text{f}}}{k_{\text{f}}\left(T_{\text{w}}-T_{\infty }\right)}{\left(-\frac{\partial T}{\partial y}\right)}_{y=0}\text{,}$

(9.130)

${\tilde{C}}_{\text{f}}=\frac{1}{2}{C}_{\text{f}}{\mathrm{Re}}_x^{0.5}=\frac{1}{{\left(1-\phi \right)}^{2.5}}{f}^{″}(0),\tilde{N}{u}_x=N{u}_x{\mathrm{Re}}_x^{-0.5}=-\frac{k_{\text{n}\text{f}}}{k_{\text{f}}}{\theta }^{\prime }(0)\text{,}$

(9.131)

where Re_x = ρ_fU_wx/μ_f is the local Reynolds number based on the stretching velocity. The quantities ${\tilde{C}}_{\text{f}}$ and $\tilde{N}{u}_x$ are referred as the reduced skin friction coefficient and reduced local Nusselt number, respectively.

9.3.2. Effects of active parameters

Table 9.6 shows that the effects of permeability parameter (K₁), suction/injection parameter (γ), velocity ratio parameter (λ), and nanoparticle volume fraction (φ) on skin friction coefficient when ɛ = 0.1 and Pr = 6.2. Physically, negative sign of ${\tilde{C}}_{\text{f}}$ implies that the stretching tube exerts a dragging force on the fluid and positive sign implies the opposite. For both suction(γ > 0) or injection (γ < 0) and both cases of λ (λ = 0.1 or λ = 2) it can be seen that the absolute values of skin friction increases due to increase in Permeability parameter and Suction/injection parameter, while it decreases due to decrease in nanoparticle volume fraction.

Table 9.8

Effects of velocity ratio parameter and nanoparticle volume fraction on skin friction coefficient and Nusselt number when ɛ = 0.1, K₁ = 0.1, and Pr = 6.2 for Cu–Water

		${\tilde{C}}_{\text{f}}$			$\tilde{N}{u}_{\text{x}}$
		φ
γ	λ	0	0.05	0.1	0	0.05	0.1
0.5	0.1	−1.26387	−1.25665	−1.25322	3.797954	2.817711	2.272061
	0.2	−1.17913	−1.17312	−1.17025	3.810096	2.831477	2.287149
	1.7	1.552826	1.54958	1.548038	4.089999	3.123066	2.583777
	2	2.333717	2.329335	2.327254	4.151551	3.183882	2.643174
−0.5	0.1	−0.80524	−0.79792	−0.79443	0.336449	0.374097	0.383887
	0.2	−0.76547	−0.75929	−0.75635	0.350128	0.38932	0.400184
	1.7	1.165817	1.162405	1.160785	0.67334	0.717759	0.725932
	2	1.778452	1.773849	1.771663	0.740698	0.783077	0.788656

Table 9.9 shows variation in Nusselt number with different base fluids (Water Pr = 6.2 and Ethylene glycol Pr = 203.6) when K₁ = 0.1, γ = 0.5, λ = 0.1. For fixed values of nanoparticle volume fraction, selecting Ethylene glycol instead of Water as base fluid lead to decrease thermal boundary layer thickness, so, Cu–Ethylene glycol has higher Nusselt number than Cu–Water due to lower thermal conductivity. Tables  9.10 and  9.11 display the behavior of the skin friction coefficient and Nusselt number using different nanofluids. Fig. 9.17 (A) shows that temperature distribution for different types of nanofluids when M = 1, S = 0.1, n = 1, λ = 0.1, φ = 0.1 and Pr = 6.2. These tables show that by using different types of nanofluid the value of ${\tilde{C}}_f$, $\tilde{N}{u}_x$ and temperature profile change. This means that the nanofluids will be important in the cooling and heating processes. Choosing Titanium oxide as the nanoparticle leads to the maximum amount rate of hate transfer, while selecting Silver leads to the minimum amount of it. Also, choosing Alumina as the nanoparticle leads to the maximum amount skin friction coefficient, while selecting Silver leads to the minimum amount of it. In Fig. 9.17B, the Nusselt numbers versus the volume fraction of nanoparticle is shown when different models of nanofluid based on different formulas for thermal conductivity and dynamic viscosity is used. For all amount of λ, model II for nanotubes has higher Nusselt number than model I for spherical shaped nanoparticles due to lower thermal conductivity.

$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0$

(9.132)

${\rho }_f\left(u\frac{\partial u}{\partial x}+\nu \frac{\partial u}{\partial y}+2\Omega w\right)=-\frac{\partial p^{*}}{\partial x}+\mu \left(\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}\right)\text{,}$

(9.133)

${\rho }_f\left(u\frac{\partial v}{\partial y}\right)=-\frac{\partial p^{*}}{\partial y}+\mu \left(\frac{{\partial }^{2}v}{\partial x^2}+\frac{{\partial }^{2}v}{\partial y^2}\right)\text{,}$

(9.134)

${\rho }_f\left(u\frac{\partial w}{\partial x}+\nu \frac{\partial w}{\partial y}-2\Omega w\right)=\mu \left(\frac{{\partial }^{2}w}{\partial x^2}+\frac{{\partial }^{2}w}{\partial y^2}\right)\text{,}$

(9.135)

$\begin{array}{l}u\frac{\partial T}{\partial x}+v\frac{\partial T}{\partial y}+w\frac{\partial T}{\partial z}=\alpha \left(\frac{{\partial }^{2}T}{\partial x^2}+\frac{{\partial }^{2}T}{\partial y^2}+\frac{{\partial }^{2}T}{\partial z^2}\right)+\\ +\frac{{\left(\rho c_{\text{P}}\right)}_{\text{p}}}{{\left(\rho c_{\text{P}}\right)}_{\text{f}}}\left[\begin{array}{l}{D}_B\left\{\frac{\partial C}{\partial x}·\frac{\partial T}{\partial x}+\frac{\partial C}{\partial y}·\frac{\partial T}{\partial y}+\frac{\partial C}{\partial z}·\frac{\partial T}{\partial z}\right\}\\ +(D_{\text{T}}/T_{\text{c}})\left\{{\left(\frac{\partial T}{\partial x}\right)}^2+{\left(\frac{\partial T}{\partial y}\right)}^2+{\left(\frac{\partial T}{\partial z}\right)}^2\right\}\end{array}\right],\end{array}$

(9.136)

$u\frac{\partial C}{\partial x}+v\frac{\partial C}{\partial y}+w\frac{\partial C}{\partial z}=D_{\text{B}}(\frac{{\partial }^{2}C}{\partial x^2}+\frac{{\partial }^{2}C}{\partial y^2}+\frac{{\partial }^{2}C}{\partial z^2})+\left(\frac{D_{\text{T}}}{T_0}\right)\left\{\frac{{\partial }^{2}T}{\partial x^2}+\frac{{\partial }^{2}T}{\partial y^2}+\frac{{\partial }^{2}T}{\partial z^2}\right\}$

(9.137)

Here u, v, and z are the velocities in the x, y, and z directions respectively. Also p^* is the modified fluid pressure. The absence of $\frac{\partial p*}{\partial z}$ in Eq. (9.135) implies that there is a net cross-flow along the z-axis. The relevant boundary conditions are:

$\begin{array}{c}\hfill u=ax,v=0,w=0,T=T_{\text{h}},C=C_{\text{h}}aty=0\hfill \\ \hfill u=0,v=v_0,w=0,T=T_0,C=C_{0}aty=+h\hfill \end{array}$

(9.138)

$\begin{array}{l}\eta =\frac{y}{h},u=axf\text{'}(\eta ),\nu =-ahf(\eta ),w=axg(\eta )\\ \theta \left(\eta \right)=\frac{T-T_{\text{h}}}{T_0-T_{\text{h}}},\phi \left(\eta \right)=\frac{C-C_{\text{h}}}{C_0-C_{\text{h}}}\end{array}$

(9.139)

$-\frac{1}{\rho h}\frac{\partial p^{*}}{\partial \eta }=a^{2}x\left[f\text{'}-ff\text{'}\text{'}-\frac{f\text{'}\text{'}\text{'}}{R}+\frac{2K_{\text{r}}}{R}g\right]\text{,}$

(9.140)

$-\frac{1}{\rho h}\frac{\partial p^{*}}{\partial \eta }=a^{2}h\left[ff\text{'}+\frac{1}{R}f\text{'}\text{'}\right]\text{,}$

(9.141)

$g\text{'}\text{'}-R\left(f\text{'}g-fg\text{'}\right)+2K_{\text{r}}f\text{'}=0$

(9.142)

$R=\frac{a{h}^2}{\upsilon },K_{\text{r}}=\frac{\Omega h^2}{\upsilon }$

(9.143)

$f\text{'}\text{'}\text{'}-R\left[f{\text{'}}^2-ff\text{'}\text{'}\right]-2{K_{\text{r}}}^{2}g=A$

(9.144)

$f^{iv}-R\left(f\text{'}f\text{'}\text{'}-ff\text{'}\text{'}\right)-2K_{\text{r}}g\text{'}=0$

(9.145)

$f^{iv}-R\left(f\text{'}f\text{'}\text{'}-ff\text{'}\text{'}\right)-2K_{\text{r}}g\text{'}=0$

(9.146)

$g\text{'}\text{'}-R\left(f\text{'}g-fg\text{'}\right)+2K_{\text{r}}f\text{'}=0$

(9.147)

${\theta }^{″}+\mathrm{Pr}Rf{\theta }^{\prime }+Nb{\phi }^{\prime }{\theta }^{\prime }+Nt{{\theta }^{\prime }}^2=0\text{,}$

(9.148)

${\phi }^{″}+R·Scf{\phi }^{\prime }+\frac{Nt}{Nb}{\theta }^{″}=0\text{,}$

(9.149)

$\begin{array}{l}f=0,f\text{'}=1,g=0,\theta =1,\phi =1at\eta =0\\ f=\lambda ,f\text{'}=0,g=0,\theta =0,\phi =0at\eta =1\end{array}$

(9.150)

$\begin{array}{l}\lambda =\frac{v_0}{(ah)},Pr=\frac{\mu }{{\rho }_{\text{f}}\alpha },Sc=\frac{\mu }{{\rho }_{\text{f}}D},\\ Nb=\frac{{(\rho c)}_{\text{p}}{D}_{\text{B}}(C_{\text{h}})}{\left({(\rho c)}_{\text{f}}\alpha \right)}\\ Nt=\frac{{(\rho c)}_{\text{p}}{D}_{\text{T}}(T_{\text{H}})}{\left[{(\rho c)}_{\text{f}}\alpha T_{\text{c}}\right]}·\end{array}$

(9.151)

${\tilde{C}}_{\text{f}}=\left(\frac{Rx}{h}\right)C_{\text{f}}=f^{″}\left(0\right),Nu=-\theta \text{'}(0)$

(9.152)

9.4.2. Effects of active parameters

$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0\text{,}$

(9.153)

${\rho }_{\text{nf}}\left(\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial v}+v\frac{\partial u}{\partial y}\right)=-\frac{\partial p}{\partial x}+{\mu }_{\text{nf}}\left(\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}\right)\text{,}$

(9.154)

${\rho }_{\text{nf}}\left(\frac{\partial v}{\partial t}+u\frac{\partial v}{\partial v}+v\frac{\partial v}{\partial y}\right)=-\frac{\partial p}{\partial y}+{\mu }_{\text{nf}}\left(\frac{{\partial }^{2}v}{\partial x^2}+\frac{{\partial }^{2}v}{\partial y^2}\right)\text{,}$

(9.155)

$\frac{\partial T}{\partial t}+u\frac{\partial T}{\partial x}+v\frac{\partial T}{\partial y}=\frac{k_{\text{nf}}}{{\left(\rho C_{\text{p}}\right)}_{\text{nf}}}\left(\frac{{\partial }^{2}T}{\partial x^2}+\frac{{\partial }^{2}T}{\partial y^2}\right)\text{.}$

(9.156)

$\begin{array}{l}{\rho }_{\text{nf}}=(1-\phi ){\rho }_{\text{f}}+\phi {\rho }_{\text{p}}\\ {\left(\rho C_{\text{p}}\right)}_{\text{nf}}=(1-\phi ){\left(\rho C_{\text{p}}\right)}_{\text{f}}+\phi {\left(\rho C_{\text{p}}\right)}_{\text{p}}\end{array}$

(9.157)

$\begin{array}{l}{k}_{\text{nf}}=1+\frac{3\left(\frac{k_{\text{p}}}{k_{\text{f}}}-1\right)\phi }{\left(\frac{k_{\text{p}}}{k_{\text{f}}}+2\right)-\left(\frac{k_{\text{p}}}{k_{\text{f}}}-1\right)\phi }+5\times {10}^4{g}^{\prime }(\phi ,T,d_{\text{p}})\phi {\rho }_{\text{f}}{c}_{\text{p,f}}\sqrt{\frac{{\kappa }_{\text{b}}T}{{\rho }_{\text{p}}{d}_{\text{p}}}}\\ g^{\prime }\left(\phi ,T,d_{\text{p}}\right)=\left(a_6+a_7\ln \left(d_{\text{p}}\right)+a_8\ln (\phi )+a_9\ln (\phi )\ln \left(d_{\text{p}}\right)+a_{10}\ln {\left(d_{\text{p}}\right)}^2\right)\\ +\ln \left(T\right)\left(a_1+a_2\ln \left(d_{\text{p}}\right)+a_3\ln (\phi )+a_4\ln (\phi )\ln \left(d_{\text{p}}\right)+a_5\ln {\left(d_{\text{p}}\right)}^2\right)\\ R_{\text{f}}+d_{\text{p}}/k_{\text{p}}=d_{\text{p}}/k_{\text{p,eff}},R_{\text{f}}=4\times {10}^{-8}\text{K}{\text{m}}^2\text{/W}\end{array}$

(9.158)

${\mu }_{\text{nf}}=\frac{{\mu }_{\text{f}}}{{\left(1-\phi \right)}^{2.5}}+\frac{k_{\text{Brownian}}}{k_{\text{f}}}\times \frac{{\mu }_{\text{f}}}{Pr}$

(9.159)

$\begin{array}{l}u=0,v=-v_{\text{w}}=-A{a}^{•},T\left(a\right)=T_{0}aty=a(t),\\ u=0,v=0;aty=0.\end{array}$

(9.160)

where $A=v{}_{\text{w}}/a^{•}$ is the measure of wall permeability, and T₀ is the temperature of the porous plate (y = a) which has the same temperature as that of the incoming coolant.

$\psi =\frac{vx}{a}F\left(\eta ,t\right)\text{,}$

(9.161)

$\psi =\frac{{\upsilon }_{\text{nf}}x}{a}F\left(\eta ,t\right)\text{,}$

(9.162)

$F_{\eta \eta \eta \eta }+\frac{A_1}{A_2}\alpha \left(\eta F_{\eta \eta \eta }+3F_{\eta \eta }\right)+F{F}_{\eta \eta \eta }-F_{\eta }{F}_{\eta \eta }-\frac{A_1}{A_2}{a}^2{{\upsilon }_f}^{-1}{F}_{\eta \eta t}=0\text{,}$

(9.163)

$\alpha =\frac{a{a}^{•}}{{\upsilon }_{\text{f}}}$

(9.164)

$A_1=\frac{{\rho }_{\text{nf}}}{{\rho }_{\text{f}}},A_2=\frac{{\mu }_{\text{nf}}}{{\mu }_{\text{f}}}\text{,}$

(9.165)

$F_{\eta }\left(0\right)=0,F\left(0\right)=0,F_{\eta }\left(1\right)=0,F\left(1\right)=\frac{A_1}{A_2}R\text{,}$

(9.166)

$f=\frac{F}{R},l=\frac{x}{a}$

(9.167)

$\alpha =\frac{a{a}^{•}}{{\upsilon }_{\text{f}}}=\frac{a_0{a_0}^{•}}{{\upsilon }_{\text{f}}}=cte\text{,}$

(9.168)

where a₀ and ${a_0}^{•}$ denote the initial channel height and expansion ratio, respectively. Integrating Eq. (9.168) with respect to time, the similar solution can be achieved. The result is

$\frac{a}{a_0}=\sqrt{1+2{\upsilon }_{\text{f}}\alpha t{a}_0^{-2}}$

(9.169)

For a physical setting in which the injection coefficient A is constant [26]. Since $v_{\text{w}}=A{a}^{•}$, an expression for the injection velocity variation can be determined. From Eqs. (9.168) and (9.169), it is clear that

$\frac{a}{a_0}=\frac{v_{\text{w}}\left(0\right)}{v_{\text{w}}\left(t\right)}=\sqrt{1+2{\upsilon }_{\text{f}}\alpha t{a}_0^{-2}}$

(9.170)

$f^{iv}+\frac{A_1}{A_2}\left(\alpha \left(\eta f^{‴}+3f^{″}\right)+R\left(f{f}^{‴}-f^{\prime }{f}^{″}\right)\right)=0\text{,}$

(9.171)

$f^{\prime }\left(0\right)=0,f\left(0\right)=0,f^{\prime }\left(1\right)=0,f\left(1\right)=1\text{,}$

(9.172)

$T=T_0+{\sum C_m\left(\frac{x}{a}\right)}^m{q}_m\left(0\right)$

(9.173)

$\frac{A_3}{A_4}\left(-\alpha \mathrm{Pr}\left(m{q}_m+\eta {q_m}^{\prime }\right)+\text{Pr}R\left(m{f}^{\prime }{q}_m-f{q_m}^{\prime }\right)\right)={q_m}^{″}$

(9.174)

$A_3=\frac{{\left(\rho C_{\text{p}}\right)}_{\text{nf}}}{{\left(\rho C_{\text{p}}\right)}_{\text{f}}},A_4=\frac{k_{\text{nf}}}{k_{\text{f}}}$

(9.175)

$q_m\left(0\right)=1,q_m\left(1\right)=0$

(9.176)

$T_w=T_0+C_m{\left(\frac{x}{a}\right)}^m{q}_m\left(0\right)$

(9.177)

$Nu=-\frac{k_{\text{nf}}}{k_{\text{f}}}\frac{\partial T}{\partial \eta }/\left(T_{\text{w}}-T_0\right)=-{q_m}^{\prime }\left(0\right)$

(9.178)

9.5.2. Numerical method

Let $x_1=\eta ,x_2=f,x_3=f^{\prime },x_4=f^{″},x_5=f^{‴},x_6=q_m,x_7={q_m}^{\prime }$. We obtain the following system:

$\left(\begin{array}{l}{x_1}^{\prime }\\ {x_2}^{\prime }\\ {x_3}^{\prime }\\ {x_4}^{\prime }\\ {x_5}^{\prime }\\ {x_6}^{\prime }\\ {x_7}^{\prime }\end{array}\right)=\left(\begin{array}{c}1\\ x_3\\ x_4\\ x_5\\ -\frac{A_1}{A_2}\left(\alpha \left(x_1{x}_5+3x_4\right)+R\left(x_2{x}_5-x_3{x}_4\right)\right)\\ x_7\\ \frac{A_3}{A_4}\left(-\alpha Pr\left(m{x}_6+x_1{x}_7\right)+PrR\left(m{x}_3{x}_6-x_2{x}_7\right)\right)\end{array}\right)$

(9.179)

$\left(\begin{array}{l}{x}_1\\ x_2\\ x_3\\ x_4\\ x_5\\ x_6\\ x_7\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\\ u_1\\ u_2\\ 1\\ u_3\end{array}\right)$

(9.180)

The previous nonlinear coupled ODEs along with initial conditions are solved using fourth Order Runge-Kutta integration technique. Suitable values of the unknown initial conditions u₁, u₂, u₃, and u₄ are approximated through Newton’s method until the boundary conditions at $f^{\prime }\left(1\right)=0,f\left(1\right)=1,q_m\left(1\right)=0$ are satisfied. The computations have been performed by using MAPLE. The maximum value of x = 1, to each group of parameters is determined when the values of unknown boundary conditions at x = 0 do not change to a successful loop with error less than 10^–6. Recently, researchers utilized new numerical methods for simulating nanofluid flow and heat transfer [29–41].

9.5.3. Effects of active parameters

$E=\frac{Nu\left(\phi =0.04\right)-Nu\left(\text{basefluid}\right)}{Nu\left(\text{basefluid}\right)}\times 100$

(9.181)