Consider a single CNT with its axis oriented along a unit vector n, defined in spherical coordinates by angles $\theta$ (polar angle from the z) and $\varphi$ (azimuthal angle in the xy-plane):

(2) $$n=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$$

The CNT’s thermal conductivity along its axis is k_CNT, while perpendicular to the axis, it is typically negligible (assumed zero for simplicity, as CNT radial conductivity is orders of magnitude lower). The heat flow in the nanofluid is driven by a temperature gradient $\nabla$T, assumed along the x-axis for convenience:

(3)$$\nabla T=-\frac{dT}{dx}{\mathrm{e}}_x;{\mathrm{e}}_x=(1,0,0)$$

The heat flux q_CNT through the CNT depends on the component of the temperature gradient along the CNT’s axis. The projection of $\nabla$T onto n is:

(4)$$(\nabla T\cdot \mathrm{n})\mathrm{n}=-\frac{dT}{dx}\left({\mathrm{e}}_x\cdot \mathrm{n}\right)\mathrm{n}$$

Where:

(5)$${\mathrm{e}}_x\cdot \mathrm{n}=\sin \theta \cos \varphi$$

The heat flux along the CNT’s axis is proportional to the axial component of the temperature gradient:

(6)$$q_{\mathrm{CNT}}=-k_{\mathrm{CNT}}(\nabla T\cdot \mathrm{n})\mathrm{n}=k_{\mathrm{CNT}}\frac{dT}{dx}(\sin \theta \cos \varphi )\mathrm{n}$$

We are interested in the x-component of the heat flux, as it contributes to the effective thermal conductivity in the direction of $\nabla$T:

(7)$$q_{\mathrm{CNT},x}=q_{\mathrm{CNT}}\cdot e_x=k_{\mathrm{CNT}}\frac{dT}{dx}(\sin \theta \cos \varphi )(n\cdot e_x)=k_{\mathrm{CNT}}\frac{dT}{dx}{(\sin \theta \cos \varphi )}^2$$

For a 3D random orientation, the CNTs are equally likely to point in any direction. The probability density of orientations is uniform over the unit sphere, with the differential solid angle element:

(8)$$d\mathrm{\Omega }=\sin \theta d\theta d\varphi$$

The total solid angle of the sphere is:

(9)$${\int }_0^{2\pi }𝑑\varphi {\int }_0^{\pi }\sin \theta d\theta =2\pi \cdot {[-\cos \theta ]}_0^{\pi }=2\pi \cdot (1-(-1))=4\pi$$

The probability density function is thus $\frac{1}{4\pi }$, and the average of any function f($\theta$,$\varphi$) over all orientations is:

(10)$$⟨f⟩=\frac{1}{4\pi }{\int }_0^{2\pi }𝑑\varphi {\int }_0^{\pi }f(\theta ,\varphi )\sin \theta d\theta$$

For the x-component of the heat flux, we need the average of (sin$\theta$cos$\varphi$)²:

(11)$$⟨{(\sin \theta \cos \varphi )}^2⟩=\frac{1}{4\pi }{\int }_0^{2\pi }𝑑\varphi {\int }_0^{\pi }{(\sin \theta \cos \varphi )}^2\sin \theta d\theta =\frac{1}{4\pi }{\int }_0^{2\pi }{\cos }^2\varphi d\varphi \cdot {\int }_0^{\pi }{\sin }^3\theta d\theta$$

Evaluate the $\varphi$-integral:

(12)$${\cos }^2\varphi =\frac{1+\cos 2\varphi }{2}$$

(13)$${\int }_0^{2\pi }{\cos }^2\varphi d\varphi ={\int }_0^{2\pi }\frac{1+\cos 2\varphi }{2}𝑑\varphi =\frac{1}{2}{\left[\varphi +\frac{\sin 2\varphi }{2}\right]}_0^{2\pi }=\frac{1}{2}(2\pi +0)=\pi$$

Evaluate the $\theta$-integral:

(14)$${\int }_0^{\pi }{\sin }^3\theta d\theta ={\int }_0^{\pi }\sin \theta (1-{\cos }^2\theta )𝑑\theta$$

Use substitution: u = cos$\theta$, du = –sin$\theta$ d$\theta$;

Limits: $\theta =0$➔ u = 1, $\theta =\pi$ ➔ u = –1.

$${\int }_0^{\pi }{\sin }^3\theta d\theta ={\int }_1^{-1}(1-u^2)(-du)={\int }_{-1}^1(1-u^2)𝑑u={\left[u-\frac{u^3}{3}\right]}_{-1}^1=\left(1-\frac{1}{3}\right)-\left(-1+\frac{1}{3}\right)=\left(1-\frac{1}{3}\right)-\left(-1+\frac{1}{3}\right)=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}$$

Combine:

(15)$$⟨{(\sin \theta \cos \varphi )}^2⟩=\frac{1}{4\pi }\cdot \pi \cdot \frac{4}{3}=\frac{1}{4}\cdot \frac{4}{3}=\frac{1}{3}$$

This result is the average projection factor for the x-direction. Due to isotropy, the same factor applies for the y- or z-directions:

(16)$$⟨{(\sin \theta \sin \varphi )}^2⟩=⟨{\cos }^2\theta ⟩=\frac{1}{3}$$

The average x-component of the heat flux is:

(17)$$⟨q_{\mathrm{CNT},x}⟩=k_{\mathrm{CNT}}\frac{dT}{dx}⟨{(\sin \theta \cos \varphi )}^2⟩=k_{\mathrm{CNT}}\frac{dT}{dx}\cdot \frac{1}{3}$$

The effective thermal conductivity k_eff-CNT relates the average heat flux to the temperature gradient:

(18)$$⟨q_{\mathrm{CNT},x}⟩=k_{\mathrm{eff}-\mathrm{CNT}}\frac{dT}{dx}$$

Equating:

(19)$$k_{\mathrm{eff}-\mathrm{CNT}}{\displaystyle \frac{dT}{dx}}=k_{\mathrm{CNT}}{\displaystyle \frac{dT}{dx}}\cdot {\displaystyle \frac{1}{3}}$$ $$k_{\mathrm{eff}-\mathrm{CNT}}={\displaystyle \frac{1}{3}}{k}_{\mathrm{CNT}}$$

Physical Interpretation: The $\frac{1}{3}$ factor arises because, in 3D, the CNT’s axis is equally likely to align with any direction, and only one-third of its conductivity contributes to any given axis (e.g., x-direction) on average. This is a consequence of the isotropic averaging over the unit sphere, where the squared cosine of the angle between the CNT axis and the heat flow direction averages to $\frac{1}{3}$. In contrast, the $\frac{1}{2}$ factor in 2D assumes a higher contribution because the CNTs are constrained to a plane, increasing their alignment with the heat flow direction.