Reply to Pantokratoras, A. Comment on “Khan et al. Impact of Irregular Heat Sink/Source on the Wall Jet Flow and Heat Transfer in a Porous Medium Induced by a Nanofluid with Slip and Buoyancy Effects. Symmetry 2022, 14, 2212”

Governing Equations

With the help of all the assumptions written in [1], the leading governing equations in terms of partial differential equations (PDEs) are as follows:

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

(1)

$$u{\displaystyle \frac{\partial u}{\partial x}}+v{\displaystyle \frac{\partial u}{\partial y}}={\epsilon }_a^2{\displaystyle \frac{{\mu }_{eff}}{{\rho }_{nf}}}{\displaystyle \frac{{\partial }^{2}u}{\partial y^2}}-{\epsilon }_a^2{\displaystyle \frac{{\mu }_{nf}}{{\rho }_{nf}K\left(x\right)}}u+{\displaystyle \frac{g{\left(\rho \beta \right)}_{nf}}{{\rho }_{nf}}}\left(T-T_{\infty }\right),$$

(2)

$${\left(\rho c_p\right)}_{nf}\left(u{\displaystyle \frac{\partial T}{\partial x}}+v{\displaystyle \frac{\partial T}{\partial y}}\right)=k_{nf}{\displaystyle \frac{{\partial }^{2}T}{\partial y^2}}+Q^{AAA},$$

(3)

subject to the boundary conditions (BCs)

$$\left.\begin{array}{l}u={\displaystyle \frac{4}{\sqrt{x}}}B+{\gamma }_1\left(x\right){\mu }_{nf}{\displaystyle \frac{\partial u}{\partial y}},\hspace{0.17em}\hspace{0.17em}v=0,\hspace{0.17em}T=T_w\left(x\right)=T_{\infty }+T_0{\left({\displaystyle \frac{x}{l}}\right)}^{-2},\hspace{0.17em}\hspace{0.17em}\mathrm{at} y=0,\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}u\to 0,\hspace{0.17em}\hspace{0.17em}T\to T_{\infty }\hspace{0.17em}\hspace{0.17em}\mathrm{as}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y\to \infty .\end{array}\right\}$$

(4)

According to [1], the velocity of the surface of the wall jet is taken to be the sum of the stretching velocity, $u_w={\displaystyle \frac{4}{\sqrt{x}}}B$, plus slippery velocity, $u_{slip}={\gamma }_1\left(x\right){\mu }_{nf}{\displaystyle \frac{\partial u}{\partial y}}$, and is mathematically defined as $u\left(x,0\right)=u_w+u_{slip}$, where $B$ is the stretching rate factor. Also, ${\gamma }_1\left(x\right)$ represents the slip velocity, which is mathematically defined as $A{x}^{3/4}$, where $A$ is the positive arbitrary constant. Similarly, $K\left(x\right)=K_0{x}^{3/2}$ and ${\epsilon }_a$ signify the permeability coefficient of the porous medium and the porosity parameter, respectively.

To calculate the Units: The distinct arbitrary constants in the above equations are used to balance the respective equations. Therefore, the unit of the stretching rate factor is calculated from the given expression such as $u_w={\displaystyle \frac{4}{\sqrt{x}}}B$. Thus, ignoring the numerical constant such as

$$\begin{array}{l}{u}_w={\displaystyle \frac{1}{\sqrt{x}}}B\\ \Rightarrow \left[{\displaystyle \frac{L}{T}}\right]={\displaystyle \frac{B}{\sqrt{L}}}\\ \Rightarrow B={\displaystyle \frac{\left[L^{3/2}\right]}{\left[T\right]}}.\end{array}$$

(5)

Similarly,

$$\begin{array}{l}{u}_{slip}={\gamma }_1\left(x\right){\mu }_{nf}{\displaystyle \frac{\partial u}{\partial y}}\\ \Rightarrow {\displaystyle \frac{\left[L\right]}{\left[T\right]}}={\gamma }_1{\displaystyle \frac{\left[M\right]\left[L\right]\left[T\right]}{\left[L^2\right]\left[T^2\right]}}{\displaystyle \frac{\left[L\right]}{\left[L\right]\left[T\right]}}\\ \Rightarrow {\displaystyle \frac{\left[L\right]}{\left[T\right]}}={\gamma }_1{\displaystyle \frac{\left[M\right]}{\left[L\right]\left[T^2\right]}}\\ \Rightarrow {\gamma }_1={\displaystyle \frac{\left[L^2\right]\left[T\right]}{\left[M\right]}}.\end{array}$$

(6)

In addition,

$$\begin{array}{l}{\gamma }_1=A{x}^{3/4}\\ \Rightarrow {\displaystyle \frac{\left[L^2\right]\left[T\right]}{\left[M\right]}}=A\left[L^{3/4}\right]\\ \Rightarrow A={\displaystyle \frac{\left[L^{5/4}\right]\left[T\right]}{\left[M\right]}}.\end{array}$$

(7)

The SI unit of the permeability coefficient of the porous medium $K\left(x\right)$ is $\left[L^2\right]$. To calculate the unit for the arbitrary constant $K_0$, the below expression is used as follows:

$$\begin{array}{l}K\left(x\right)=K_0{x}^{3/2}\\ \Rightarrow \left[L^2\right]=K_0\left[L^{3/2}\right]\\ \Rightarrow K_0=\left[L^{1/2}\right].\end{array}$$

(8)

Principle of Homogeneity (POH):

It is stated that in any physically meaningful equation, all terms must have the same dimensions (or units). This means that you can only add, subtract, or equate quantities that are dimensionally consistent. Thus, apply the same role to the continuity equation and the momentum equation, which are as follows:

Continuity Equation

$$\begin{array}{l}{\displaystyle \frac{\partial u}{\partial x}}+{\displaystyle \frac{\partial v}{\partial y}}=0.\\ {\displaystyle \frac{\partial u}{\partial x}}\to {\displaystyle \frac{\left[L/T\right]}{\left[L\right]}}=\left[T\right].\\ {\displaystyle \frac{\partial v}{\partial y}}\to {\displaystyle \frac{\left[L/T\right]}{\left[L\right]}}=\left[T\right].\end{array}$$

(9)

Here, u and v are the velocity components (meter/second) and (x & y) are the Cartesian coordinates (meter). Since all terms have the same dimension [T] (time), the continuity equation is dimensionally homogeneous.

Momentum Equation

$$\begin{array}{l}u{\displaystyle \frac{\partial u}{\partial x}}+v{\displaystyle \frac{\partial u}{\partial y}}={\epsilon }_a^2{\displaystyle \frac{{\mu }_{eff}}{{\rho }_{nf}}}{\displaystyle \frac{{\partial }^{2}u}{\partial y^2}}-{\epsilon }_a^2{\displaystyle \frac{{\mu }_{nf}}{{\rho }_{nf}K\left(x\right)}}u+{\displaystyle \frac{g{\left(\rho \beta \right)}_{nf}}{{\rho }_{nf}}}\left(T-T_{\infty }\right)\\ u{\displaystyle \frac{\partial u}{\partial x}}\to \left[L/T\right]{\displaystyle \frac{\left[L/T\right]}{\left[L\right]}}={\displaystyle \frac{\left[L\right]}{\left[T^2\right]}},\\ v{\displaystyle \frac{\partial u}{\partial y}}\to \left[L/T\right]{\displaystyle \frac{\left[L/T\right]}{\left[L\right]}}={\displaystyle \frac{\left[L\right]}{\left[T^2\right]}},\\ {\epsilon }_a^2{\displaystyle \frac{{\mu }_{eff}}{{\rho }_{nf}}}{\displaystyle \frac{{\partial }^{2}u}{\partial y^2}}={\epsilon }_a^2{\displaystyle \frac{{\upsilon }_{eff}}{{\rho }_{nf}/{\rho }_f}}{\displaystyle \frac{{\partial }^{2}u}{\partial y^2}}\to \left[L^2/T\right]{\displaystyle \frac{\left[L/T\right]}{\left[L^2\right]}}={\displaystyle \frac{\left[L\right]}{\left[T^2\right]}},\\ {\epsilon }_a^2{\displaystyle \frac{{\mu }_{nf}}{{\rho }_{nf}K\left(x\right)}}u={\epsilon }_a^2{\displaystyle \frac{{\mu }_{nf}/{\mu }_f}{{\rho }_{nf}/{\rho }_{f}K\left(x\right)}}{\upsilon }_{f}u\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\epsilon }_a^2{\displaystyle \frac{{\mu }_{nf}/{\mu }_f}{{\rho }_{nf}/{\rho }_f}}{\displaystyle \frac{{\upsilon }_f}{K\left(x\right)}}u\to \left[L^2/T\right]{\displaystyle \frac{\left[L/T\right]}{\left[L^2\right]}}={\displaystyle \frac{\left[L\right]}{\left[T^2\right]}},\\ {\displaystyle \frac{g{\left(\rho \beta \right)}_{nf}}{{\rho }_{nf}}}\left(T-T_{\infty }\right)=g{\displaystyle \frac{{\beta }_{nf}}{{\beta }_f}}{\beta }_f\left(T-T_{\infty }\right)\to {\displaystyle \frac{\left[L/T^2\right]\left[K\right]}{\left[K\right]}}={\displaystyle \frac{\left[L\right]}{\left[T^2\right]}}.\end{array}$$

(10)

Hence, all terms have the same dimension ${\displaystyle \frac{\left[L\right]}{\left[T^2\right]}}$ (meter/time*time), and the momentum equation is dimensionally homogeneous. Similarly, follow the same concept for the energy equation in which each term has the same dimension ${\displaystyle \frac{\left[K\right]}{\left[T\right]}}$ (Kelvin/time), and the energy equation is also dimensionally homogeneous.

Similarity Transformations

$$\xi =y\sqrt{{\displaystyle \frac{B}{{\alpha }_f}}}{x}^{-3/4},\psi =4\sqrt{B{\alpha }_f}{x}^{1/4}F,G={\displaystyle \frac{T-T_{\infty }}{T_w-T_{\infty }}}.$$

(11)

The above Equation (11) is the same as Equation (10) of reference [1]. In the published paper, there is a typo, or missing arbitrary constant or stretching rate factor. The rest of the errors or mistakes in the expression depend on this typo, so please see the below corrections.

Errors and Responses

Many errors exist in the above paper [1].

Reply: Dear reviewer, the errors were raised due to a typo in the similarity transformations in Equation (10) of the published paper (see, [1]), and the rest of the equations depend on it. Please see the corrections below and respond to your queries.

1st error

Equation (10) in [1] presents the dimensionless similarity variable $\xi$ as follows:

$$\xi ={\left({\alpha }_f^2{x}^3\right)}^{-1/4}y$$

(12)

where ${\alpha }_f\hspace{0.17em}\left(m^2{\sec }^{-1}\right)$ is the fluid thermal diffusivity and $x,y(m)$ are the Cartesian coordinates. From Equation (12), it is found that the units of $\xi$ are $m^{-3/4}{\sec }^{1/2}$ and the parameter $\xi$ is dimensional and wrong.

Reply: Dear reviewer, there is a typo in Equation (12). The pseudo-similarity variable $\xi$ is equal to $y\sqrt{{\displaystyle \frac{B}{{\alpha }_f}}}{x}^{-3/4}$. Using individual units of each term and Equation (5), the above expression becomes:

$$\begin{array}{l}\xi =y\sqrt{{\displaystyle \frac{B}{{\alpha }_f}}}{x}^{-3/4}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\Rightarrow \left[L\right]\left[{\displaystyle \frac{L^{3/4}/\sqrt{T}}{L/\sqrt{T}}}\right]\left[L^{-3/4}\right]=1.\end{array}$$

Thus, it is dimensionless.

2nd error

Equation (10) in [1] appears as the following equation:

$$\psi ={\left({\alpha }_f^{2}x\right)}^{1/4}F(\xi )$$

(13)

where $F(\xi )$ is the dimensionless stream function and $\psi (m^2{\sec }^{-1})$ is the dimensional stream function. Equation (13) is wrong because the units of the LHS are $m^2{\sec }^{-1}$, whereas the units of the RHS are $m^{5/4}{\sec }^{-1/2}$.

Reply: Dear reviewer, using the above corrected similarity transformations of Equation (11), the stream function is defined as $\psi =4\sqrt{B{\alpha }_f}{x}^{1/4}F\left(\xi \right)$. Thus, plugging the unit of each term and the stretching rate factor given in Equation (5) into the right-hand side of the above expression, we obtain the following:

$$\begin{array}{l}\psi =4\sqrt{B{\alpha }_f}{x}^{1/4}F\left(\xi \right)\\ \psi =\left[\sqrt{{\displaystyle \frac{L^{3/2}}{T}}\times {\displaystyle \frac{L^2}{T}}}\right]\left[L^{1/4}\right]\\ \psi =\left[{\displaystyle \frac{L^{7/4}}{T}}\right]\left[L^{1/4}\right]=\left[L^2/T\right].\end{array}$$

Thus, the left-hand side unit is equal to the right-hand side. It is dimensionless.

3rd error

Equation (11) in [1] appears as follows:

$$u={\displaystyle \frac{4}{\sqrt{x}}}F\prime (\xi )$$

(14)

where $u(m{\sec }^{-1})$ is the horizontal fluid velocity and $F\prime (\xi )$ is the dimensionless horizontal velocity. Equation (14) is wrong because the units of the LHS are $m{\sec }^{-1}$, whereas the units of the RHS are $m^{-1/2}$.

Reply: Dear reviewer, keep the above Equation (11) in mind and use the x-component of the stream function, which is defined as $u={\displaystyle \frac{\partial \psi }{\partial y}}$. Taking the partial change in the stream function with respect to y, we obtain the velocity component in the x-direction as $u={\displaystyle \frac{4B}{\sqrt{x}}}F\prime \left(\xi \right)$. Plugging the unit of each term and Equation (5) into the left-hand and right-hand sides, we obtain

$$\begin{array}{l}u={\displaystyle \frac{4B}{\sqrt{x}}}F\prime \left(\xi \right)\\ \Rightarrow \left[L/T\right]=\left[{\displaystyle \frac{L^{3/2}/T}{L^{1/2}}}\right]\\ \Rightarrow \left[L/T\right]=\left[L/T\right].\end{array}$$

It is dimensionless.

4th error

Equation (11) in [1] appears as follows:

$$\upsilon =-\sqrt{{\alpha }_f}{x}^{-3/4}\left(F(\xi )-3\xi F\prime (\xi )\right)$$

(15)

where $\upsilon (m{\sec }^{-1})$ is the vertical velocity. Equation (15) is wrong because the units of the LHS are $m{\sec }^{-1}$, whereas the units of the RHS are $m^{1/4}{\sec }^{-1/2}$.

Reply: Dear reviewer, keep Equation (11) in mind and use the y-component of the stream function, which is defined as $v=-{\displaystyle \frac{\partial \psi }{\partial x}}$. Taking the partial change in the stream function with respect to x, we obtain the velocity component in the y-direction as $v=-\sqrt{B{\alpha }_f}{x}^{-3/4}\left(F\left(\xi \right)-3\xi F\prime \left(\xi \right)\right)$. Thus,

$$\begin{array}{l}v=-\sqrt{B{\alpha }_f}{x}^{-3/4}\left(F\left(\xi \right)-3\xi F\prime \left(\xi \right)\right)\\ v=\left[\sqrt{{\displaystyle \frac{L^{3/2}}{T}}\times {\displaystyle \frac{L^2}{T}}}\right]\left[L^{-3/4}\right]\\ v=\left[L/T\right]\end{array}$$

It is dimensionless.

5th error

The dimension mixed convection parameter is as follows:

$$\lambda ={\displaystyle \frac{g{\beta }_f{T}_0}{4}}$$

(16)

where $g\hspace{0.17em}\hspace{0.17em}(m{\sec }^{-2})$ is the gravitational acceleration, ${\beta }_f\hspace{0.17em}\hspace{0.17em}(K^{-1})$ is the thermal expansion coefficient, and $T_0\hspace{0.17em}\hspace{0.17em}(K)$. From Equation (16), it is found that the unit of $\lambda$ is $m{\sec }^{-2}$.

Note: The person has written the unit of $T_0\hspace{0.17em}(m^{2}K)$ incorrectly. The correct unit is $T_0\hspace{0.17em}(K)$.

Reply: Dear reviewer, considering Equation (11) and the temperature at the wall surface in the momentum equation, we obtain the following dimensionless mixed convection parameter:

$$\lambda ={\displaystyle \frac{g{\beta }_f{T}_0{l}^2}{4B^2}}.$$

(17)

Therefore, plugging the unit of each term into the above expression, we obtain

$$\begin{array}{l}\lambda ={\displaystyle \frac{\left[L/T^2\right]\left[K^{-1}\right]\left[K\right]\left[L^2\right]}{4\left[L^{3/2}{T}^{-1}\right]\left[L^{3/2}{T}^{-1}\right]}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\displaystyle \frac{\left[L/T^2\right]\left[L^2\right]}{4\left[L^3{T}^{-2}\right]}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}={\displaystyle \frac{1}{4}}\left[1\right]\end{array}$$

Thus, it is dimensionless.

6th error

The term $e^{\xi }$ appears in Equations (5), (13) and (14) in [1]. In mathematics, the $e^x$ with x dimensional is meaningless. Considering that $\xi \hspace{0.17em}\hspace{0.17em}\left(m^{-3/4}{s}^{1/2}\right)$ is dimensional, Equations (5), (13) and (14) in [1] are wrong.

Reply: Dear reviewer, because of the 1st error response, the similarity variable $\xi$ is dimensionless. Thus, $e^{\xi }$ in Equations (5), (13) and (14) in [1] appears to be dimensionless and is found correctly.

7th error

The velocity slip parameter ${\Sigma }_a={\displaystyle \frac{A{\mu }_f}{\sqrt{{\alpha }_f}}}$ appears in the dimensionless Equation (15) in [1]. The parameter A is included in ${\gamma }_1\left(x\right)=A{x}^{3/4}$ and the parameter ${\gamma }_1\left(x\right)$ is included in Equation (4) in [1]. From these equation, it is found that the units of ${\Sigma }_a$ are $m^{-3/4}{s}^{1/2}$. Therefore Equation (15) in [1] is wrong because in a dimensionless equation, all terms must be dimensionless.

Reply: Dear editor, using the above arbitrary constant in Equation (4) of [1] and the velocity slip conditions, it is written more simply as:

$$u={\displaystyle \frac{4B}{\sqrt{x}}}+{\displaystyle \frac{{\mu }_{nf}}{{\mu }_f}}A{x}^{3/4}{\mu }_f{\displaystyle \frac{\partial u}{\partial y}}.$$

(18)

Let Equation (18) be rewritten as follows:

$$u=u_w+\hspace{0.17em}{u}_{slip}.$$

(19)

Now, ignoring the other numerical constants and calculating the units of A and B, we obtain the following:

$$\begin{array}{l}u={\displaystyle \frac{4B}{\sqrt{x}}}\\ \left[L/T\right]={\displaystyle \frac{B}{{\left[L\right]}^{1/2}}}\\ B\Rightarrow \left[L^{3/2}{T}^{-1}\right].\end{array}$$

(20)

Similarly, Equations (7) and (19) are used in Equation (18) above. Thus, ignoring the other constants and plugging the units of each term into Equation (18) RHS, we obtain

$$\begin{array}{l}u\Rightarrow \left[L^{3/2}{T}^{-1}{L}^{-1/2}\right]+\left[L^{2}T{M}^{-1}\right]\left[M{T}^{-1}{L}^{-1}\right]\left[T^{-1}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\Rightarrow \left[L{T}^{-1}\right]+\left[L{T}^{-1}\right]\end{array}$$

Thus, each term added on the right-hand side of Equation (18) has the same dimension as the LHS. Thus, the equation is dimensionally correct. In addition, the velocity slip parameter is defined as follows:

$${\Sigma }_a=A{\mu }_f\sqrt{{\displaystyle \frac{B}{{\alpha }_f}}}.$$

(21)

Plugging the units of A and B in Equation (21) gives

$$\begin{array}{l}{\Sigma }_a=\left[T{L}^{5/4}{M}^{-1}\right]\left[M{T}^{-1}{L}^{-1}\right]\left[\sqrt{{\displaystyle \frac{L^{3/2}{T}^{-1}}{L^2{T}^{-1}}}}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\Rightarrow \left[L^{5/4}\right]\left[L^{-1}\right]\left[\sqrt{L^{-1/2}}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\Rightarrow \left[L^{1/4}\right]\left[L^{-1/4}\right]=\left[1\right]\end{array}$$

(22)

Thus, it is dimensionless.

8th error

In Equation (12) in [1], the parameter $K_a={\displaystyle \frac{{\upsilon }_f{\epsilon }_a^2}{K_0}}$ appears. However, the parameter $K_0$ is unknown.

Reply: As per the 1st comment [2], this parameter will be $K_a={\displaystyle \frac{{\upsilon }_f{\epsilon }_a^2}{B{K}_0}}$. Plugging the values of each term, we achieve

$$\begin{array}{l}{K}_a={\displaystyle \frac{\left[\frac{L^2}{T}\right]}{\left[\frac{L^{3/2}}{T}\right]\left[L^{1/2}\right]}}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\Rightarrow {\displaystyle \frac{\left[L^2\right]}{\left[L^2\right]}}=\left[1\right]\end{array}$$

Thus, it is dimensionless.