Question 8.13: The velocity profile in the steady isothermal laminar flow o...
The velocity profile in the steady isothermal laminar flow of an incompressible Newtonian fluid contained between concentric cylinders in which the inner cylinder is rotating and the outer cylinder is stationary is given by
V =\frac{ωR_1^{2} }{R_2^{2}-R_1^{2}} (\frac{R_2^{2}}{x} – x) for R_1 ≤x≤R_2
where ω is the angular velocity of the inner cylinder, and x is measured radially outward. Determine the rate of entropy production due to laminar viscous losses for SAE-40 engine oil at 20.0°C in the gap between cylinders of radii 0.0500 and 0.0510 m when the inner cylinder is rotating at 1000. rev/min. The viscosity of the oil is 0.700 N·s/m² and the length of the cylinder is 0.100 m.
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.}
