Figure 6 two components of stresses on a fluid particle [5]
Forces on the left and right faces are denoted as F1
\(F_{1}=\left[\left(P-\frac{\partial P}{\partial x}\frac{\text{δx}}{2}\right)-\left(\sigma_{\text{xx}}-\frac{\partial\sigma_{\text{xx}}}{\partial x}\frac{\text{δx}}{2}\right)\right]\delta y\delta z+\left[-\left(P+\frac{\partial P}{\partial x}\frac{\text{δx}}{2}\right)+\left(\sigma_{\text{xx}}+\frac{\partial\sigma_{\text{xx}}}{\partial x}\frac{\text{δx}}{2}\right)\right]\text{δyδz}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\)
\(F_{1}=\left[\left(-\frac{\partial P}{\partial x}\frac{\text{δx}}{2}\right)+\left(\frac{\partial\sigma_{\text{xx}}}{\partial x}\frac{\text{δx}}{2}\right)\right]\delta y\delta z+\left[-\left(\frac{\partial P}{\partial x}\frac{\text{δx}}{2}\right)+\left(\frac{\partial\sigma_{\text{xx}}}{\partial x}\frac{\text{δx}}{2}\right)\right]\text{δyδz}\)
\(F_{1}=\left[\left(-\frac{\partial P}{\partial x}\text{δx}\right)+\left(\frac{\partial\sigma_{\text{xx}}}{\partial x}\text{δx}\right)\right]\text{δyδz}\)
\(F_{1}=\left[-\frac{\partial P}{\partial x}+\frac{\partial\sigma_{\text{xx}}}{\partial x}\right]\text{δxδyδz}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left(3.37\right)\)
Forces on the front and back faces are denoted as \(F_{2}\)
\(F_{2}=\left(\tau_{\text{yx}}+\frac{\partial\tau_{\text{yx}}}{\partial y}\frac{\text{δy}}{2}\right)\delta x\delta y-\left(\tau_{\text{yx}}-\frac{\partial\tau_{\text{yx}}}{\partial y}\frac{\text{δy}}{2}\right)\text{δxδy}\)
\(F_{2}=\frac{\partial\tau_{\text{yx}}}{\partial y}\text{δxδyδy}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left(3.38\right)\)
Forces on the top and bottom faces are denoted as \(F_{3}\)
\(F_{3}=\left(\tau_{\text{yx}}+\frac{\partial\tau_{\text{yx}}}{\partial z}\frac{\text{δy}}{2}\right)\delta x\delta y-\left(\tau_{\text{yx}}-\frac{\partial\tau_{\text{yx}}}{\partial z}\frac{\text{δy}}{2}\right)\text{δxδy}\)
\(F_{3}=\frac{\partial\tau_{\text{zx}}}{\partial z}\text{δxδyδy\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left(3.39\right)\)
Then, the net forces in the x – direction is equal to the sum of all of the three forces;
\(F_{x}=F_{1}+F_{2}+F_{3}=\left\{\left[-\frac{\partial P}{\partial x}+\frac{\partial\sigma_{\text{xx}}}{\partial x}\right]+\frac{\partial\tau_{\text{yx}}}{\partial y}+\frac{\partial\tau_{\text{zx}}}{\partial z}\right\}\text{δxδyδy}\)
\(F_{x}=\left[\frac{\partial\left(-P+\sigma_{\text{xx}}\right)}{\partial x}+\frac{\partial\tau_{\text{yx}}}{\partial y}+\frac{\partial\tau_{\text{zx}}}{\partial z}\right]\text{δxδyδy}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left(3.40\right)\)
Now, the work done in the x – direction which is denoted as \(W_{x}\)will be the product of the force by the velocity;
\(W_{x}=\left[\frac{\partial\left[u\left(-P+\sigma_{\text{xx}}\right)\right]}{\partial x}+\frac{\partial{[u\tau}_{\text{yx}}]}{\partial y}+\frac{\partial{[u\tau}_{\text{zx}}]}{\partial z}\right]\text{δxδyδy}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left(3.41\right)\)
Similarly, the surface stresses components in the y and z direction is given by the mathematical formula indicated below;
\(W_{y}=\left[\frac{\partial{[v\tau}_{\text{xy}}]}{\partial x}+\frac{\partial\left[v\left(-P+\sigma_{\text{yy}}\right)\right]}{\partial y}+\frac{\partial{[v\tau}_{\text{zy}}]}{\partial z}\right]\text{δxδyδy}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left(3.42\right)\)
\(W_{z}=\left[\frac{\partial{[w\tau}_{\text{xz}}]}{\partial x}+\frac{\partial{[w\tau}_{\text{yz}}]}{\partial y}+\frac{\partial\left[w\left(-P+\sigma_{\text{zz}}\right)\right]}{\partial z}\right]\text{δxδyδy}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left(3.43\right)\)
Now, it is the time to collect the terms that contains the pressure together as indicated below;
\(-\frac{\partial\left[\text{uP}\right]}{\partial x}-\frac{\partial\left[\text{vP}\right]}{\partial x}-\frac{\partial\left[\text{wP}\right]}{\partial x}=-div\left(P\mathbf{u}\right)\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\)
The total work done on the fluid particle will be;
\(W=-div\left(P\mathbf{u}\right)+\left[\frac{\partial\left(u\sigma_{\text{xx}}\right)}{\partial x}+\frac{\left(\text{uτ}_{\text{yx}}\right)}{\partial y}+\frac{\partial\left(\text{uτ}_{\text{zx}}\right)}{\partial z}+\frac{\partial\left(v\sigma_{\text{yy}}\right)}{\partial x}+\frac{\left(\text{vτ}_{\text{xy}}\right)}{\partial y}+\frac{\partial\left(\text{vτ}_{\text{zy}}\right)}{\partial z}+\frac{\partial\left(w\sigma_{\text{zz}}\right)}{\partial x}+\frac{\left(\text{wτ}_{\text{yz}}\right)}{\partial y}+\frac{\partial\left(\text{wτ}_{\text{xz}}\right)}{\partial z}\right]\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left(3.44\right)\)
Now let us find the Energy flux due to conduction heat transfer