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Modern Engineering Thermodynamics

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8.3 Systems Undergoing Irreversible Processes 269<br />

then<br />

ðσ W<br />

Þ vis<br />

= μ T<br />

<br />

dV 2 ω 2 R 4<br />

=<br />

1 μ<br />

dx<br />

ðR 2 2 − R2 1Þ 2 T<br />

Equation (7.69) gives the entropy production rate for viscous effects as<br />

_S P<br />

<br />

W<br />

vis<br />

Z<br />

=<br />

ðσ W<br />

V<br />

Þ vis<br />

dV<br />

<br />

R 2 2<br />

2<br />

x + 1 2<br />

For the differential volume element dV, we use the volume of an annulus of thickness dx, ordV = 2πLx dx (Figure 8.15).<br />

Annular differential<br />

volume<br />

d = 2πLxdx<br />

A<br />

x<br />

L<br />

x + dx<br />

FIGURE 8.15<br />

Example 8.13, annular differential volume.<br />

Putting these expressions for ðσ W Þ vis and dV into Eq. (7.70) and carrying out the integration gives<br />

ðS P Þ Wvis<br />

= 2πLω2 R 4 1 μ <br />

ðR 2 2 − R2 1Þ 2 T 2R2 2 ln R 2<br />

+ R4 2<br />

R 1<br />

2R 2 1<br />

− R2 1<br />

2<br />

<br />

where μ = 0.700 N · s/m 2 , L = 0.100 m, and<br />

<br />

ω = 1000: rev<br />

min<br />

<br />

2π rad<br />

rev<br />

R 1 = 0.0500 m, R 2 = 0.0510 m, and T = 20.0°C = 293.15 K.<br />

Substituting these values into the preceding formula gives<br />

<br />

1 min<br />

60 s<br />

<br />

_S P = 2:08 W/K<br />

W<br />

vis<br />

<br />

= 104:7 rad/s<br />

In this example, we have a reasonably high entropy production rate. This is due to the very small gap between the cylinders<br />

and the high viscosity of the engine oil. Check for yourself to see that _S P<br />

W-vis ! 0asR 2 becomes much larger than R 1 ,or<br />

as μ → 0. Conversely, check to see that _S P<br />

W-vis ! ∞ as R 1 ! R 2 or as μ → ∞. Some of these elements are explored in the<br />

following exercises.<br />

Exercises<br />

23. Determine the entropy production rate _S P in Example 8.13 if the oil is changed from SAE-40 motor oil with a viscosity of<br />

μ =0.700N·s/m 2 to SAE-10 motor oil with μ =0.150N·s/m 2 at 20.0°C. Keep the values of all the other variables the<br />

same as they are in Example 8.13. Answer: _S P = 0:455 W=K:<br />

W-vis<br />

24. Recalculate the entropy production rate _S P in Example 8.13 when the bearing gap is reduced from 1.00 mm to 0.500 mm<br />

by making R 1 = 0.0500 m and R 2 = 0.0505 m. Keep the values of all the other variables the same as they are in<br />

Example 8.13. Answer: _S P = 4:13 W/K:<br />

25. Determine the entropy production rate _S P in Example 8.13 when the shaft rotation is increased to 8000. rpm.<br />

Keep the values of all the other variables the same as they are in Example 8.13. Plot the entropy<br />

(Continued )

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