Question 8.13: The velocity profile in the steady isothermal laminar flow o...

First, draw a sketch of the system (Figure 8.14).

The unknown is the rate of entropy production due to laminar viscous losses. The material is SAE-40 engine oil at 20.0°C. Here, we use Eq. (7.72) for the viscous dissipation of mechanical work. By differentiating the velocity formula just given, we get

$\frac{dV}{dx} =- \frac{ωR_1^{2} }{R_2^{2}-R_1^{2}} (\frac{R_2^{2}}{{x^{2} } } + 1)$

then

$(σ_W)_\text{vis}=\frac{μ}{T} (\frac{dV}{dx} )^{2} = \frac{ω^{2}R^{4}_1μ}{(R^{2}_2 – R^{2}_1)^{2}T} (\frac{R^{2}_2}{x^{2} }+1 )^{2}$

Equation (7.69) gives the entropy production rate for viscous effects as

$(\dot{S} _P)_{\substack{\text{W}\\\text{vis}\\}}=\int_{\sout{V}}(σ_W)_\text{vis}d\sout{V}$

For the differential volume element $d\sout{V}$, we use the volume of an annulus of thickness $dx$, or $d\sout{V} = 2πLx dx$ (Figure 8.15).

Putting these expressions for $(σ_W)_\text{vis}$ and $d\sout{V}$ into Eq. (7.70) and carrying out the integration gives

$(S_P)_{\substack{\text{W}\\\text{vis}\\}} = \frac{2πLω^{2} R^{4}_1μ}{(R^{2}_2 – R^{2}_1)^{2}T} (2R^{2}_2\text{ln}\frac{R_2}{R_1}+\frac{R^{4}_2}{2R^{2}_1} – \frac{R^{2}_1}{2} )$

where $μ = 0.700  \text{N.s/m}^{2} , \text{L}= 0.100  \text{m}$ , and

$ω = (1000.\frac{\text{rev}}{\text{min}})(\frac{2π  \text{rad}}{\text{rev}})(\frac{1  \text{min}}{60  \text{s}}) =  104.7  \text{rad/s}$

$R_1 = 0.0500  \text{m}, R_2 = 0.0510  \text{m}$ , and $T = 20.0°\text{C }= 293.15  \text{K}$.

Substituting these values into the preceding formula gives

$(\dot{S} _P)_{\substack{\text{W}\\\text{vis}\\}}=  2.08  \text{W/K}$

In this example, we have a reasonably high entropy production rate. This is due to the very small gap between the cylinders and the high viscosity of the engine oil. Check for yourself to see that $(\dot{S} _P)_\text{W-vis}→ 0$ as $R_2$ becomes much larger than $R_1$, or as $μ → 0$. Conversely, check to see that $(\dot{S} _P)_\text{W-vis} → ∞$ as $R_1 → R_2$ or as $μ → ∞$ . Some of these elements are explored in the following exercises.

production rate $(\dot{S} _P)$ vs. shaft angular velocity $ω$ as $ω$ varies from 0 to 10,000. rpm (Figure 8.16).
Answer: $(\dot{S} _P) = 133  \text{W/K.}$