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Modern Engineering Thermodynamics

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16.7 Reynolds Transport Theorem 669<br />

Then,<br />

<br />

dm T /dt = −0:532 A pffiffiffiffiffiffi<br />

m T RT os / V T T os<br />

<br />

= −0:532 A p<br />

R ffiffiffiffiffiffi <br />

T os/VT<br />

mT<br />

p ffiffiffi <br />

= − 0:532 lbm. R / ðlbf⋅sÞ<br />

4:80 × 10 −5 ft 2 ½53:34 ft.lbf/ ðlbm.RÞŠ<br />

h<br />

i<br />

× ð70:0 + 459:67 RÞ 1/2 /ð1:00 ft 3 Þ<br />

m T<br />

= − ð0:0313 s −1 Þm T<br />

or<br />

dm T<br />

= −ð0:0313Þdt<br />

m T<br />

Integrating this result from the initial mass in the tire m Ti to the mass in the tire m Tτ when the valve stem unchokes at time τ<br />

gives<br />

or<br />

Now, m Ti /m Tτ = ½p os V T /ðRT os ÞŠ i<br />

/ ½p os V T /ðRT os ÞŠ τ<br />

= p osi /p osτ ,so<br />

ln ðm Tτ /m Ti Þ = −ð0:0313Þτ<br />

τ = 31:95½ln ðm Ti /m Tτ ÞŠseconds<br />

τ = 31:95 ln ðp osi /p osτ Þ = 31:95 ln ð50:0/27:8Þ = 18:7s<br />

Exercises<br />

19. Determine the time required for the valve stem in Example 16.8 to unchoke when the tire pressure is increased from<br />

50.0 to 70.0 psia. Assume all the other variables remain unchanged. Answer: τ = 29.5 s.<br />

20. If the internal temperature of the tire in Example 16.8 is maintained at 0.00°F rather than 70.0°F during the deflation<br />

process, determine the time required for the valve stem to unchoke. Assume all the other variables remain unchanged.<br />

Answer: τ = 20.1 s.<br />

21. Suppose the valve core is not removed in Example 16.8 but instead the tire has a slow leak through a hole 1.00 × 10 –4 inches<br />

in diameter. Determine how long it would take for the leak hole to unchoke when all the other variables remain unchanged.<br />

Answer: τ = 1.64 × 10 7 s = 190. days.<br />

16.7 REYNOLDS TRANSPORT THEOREM<br />

In Chapter 2, we define a closed system as any system in which mass does not cross the system boundary, but<br />

energy (heat and work) may cross the boundary. An open system then is defined to be any system in which<br />

both mass and energy may cross the system boundary. In classical mechanics, closed and open systems are<br />

called Lagrangian and Eulerian systems, respectively; and their use in problem solving is referred to as Lagrangian<br />

analysis and Eulerian analysis.<br />

The Lagrangian analysis technique is named after the French mathematician Joseph Louis Lagrange (1736–<br />

1813). Basically, it involves solving the equations of energy and motion for a fixed mass (or closed) system. The<br />

Eulerian analysis technique is named after the Swiss mathematician Leonhard Euler (1707–1783). It involves<br />

solving the equations of energy and motion for a nonconstant spatial volume (or open system). For a given<br />

situation, usually one or the other of these techniques is easier to use, but regardless of their ease of application,<br />

both must give the same results. Therefore, we must be able to mathematically transform our governing equations<br />

back and forth between these two analysis frames. The Reynolds transport theorem is a method of carrying<br />

out this transformation.<br />

In Chapter 2, we define a simple balance equation for any extensive property X as<br />

and on a rate basis as<br />

X G = X T + X P (2.11)<br />

_X G = _X T + _X P (2.12)

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