05.04.2016 Views

Modern Engineering Thermodynamics

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

16.8 Linear Momentum Rate Balance 673<br />

c. For entropy, X = S and x = S/m = s. Entropy is not conserved, so σ s ≠ 0: The non-mass flow entropy flux is identified in<br />

Chapter 7 as J s = _q /T b : Then, for an isothermal (i.e., one-dimensional) boundary, Eq. (16.31) reduces to<br />

where<br />

is the net non-mass flow entropy transport rate, and<br />

Q<br />

ðdS/dtÞ sys = _ +∑ _ms −∑ _ms + _S P<br />

T b in out<br />

_Q /T b = ∑<br />

in<br />

_q/T b −∑ _q/T b<br />

out<br />

Z<br />

_S p = σ s dV<br />

V<br />

is the entropy production rate. These results are identical to those developed in Chapter 9 in for the one-dimensional<br />

entropy rate balance originally presented in Eq. (9.6).<br />

Exercises<br />

22. Use Eq. (16.31) to produce a general closed system rate balance equation for an arbitrary property X. Answer:<br />

∑<br />

in<br />

Z<br />

J x A −∑ J x A +<br />

out<br />

V<br />

<br />

σ x dV =<br />

23. Apply the general closed system rate balance equation developed in Exercise 22 to produce the steady state closed<br />

system entropy rate balance equation. Answer:<br />

or<br />

∑<br />

in<br />

_q<br />

T b<br />

−∑<br />

out<br />

Z<br />

_q<br />

<br />

+ σ s dV =<br />

T b<br />

V<br />

_Q<br />

T b<br />

+ _S P = 0<br />

dS<br />

dt<br />

dX<br />

dt<br />

<br />

<br />

system<br />

system<br />

= 0<br />

where<br />

_Q<br />

= ∑<br />

T b in<br />

_q<br />

T b<br />

−∑<br />

out<br />

Z<br />

_q<br />

and _S P = σ s dV<br />

T b<br />

V<br />

24. Use Eq. (16.31) to produce an availability rate balance for a steady state closed system. Answer:<br />

<br />

_Q − _W +∑ _ma f −∑ _ma f + _I =<br />

dA <br />

= 0<br />

in out<br />

dt system<br />

16.8 LINEAR MOMENTUM RATE BALANCE<br />

The linear momentum rate balance (LMRB) for a one-dimensional open system can be easily developed from<br />

Eq. (16.31) by letting X = mV and x = mV/m = V. Since momentum is conserved, σ x = σ V = 0: External forces are<br />

the source of the non-mass flow momentum transport rate across the system boundary, so the one-dimensional<br />

momentum flux, J x = J V , is the external force per unit area, or<br />

Then, Eq. (16.31) gives the LMRB as 3<br />

d<br />

dt ðmVÞ open<br />

system<br />

J V = ðF ext /AÞg c<br />

= ∑ðF ext Þg c +∑ _mV −∑ _mV (16.32)<br />

net<br />

in out<br />

3 The vector nature of momentum necessarily causes Eq. (16.32) to be three dimensional. The one-dimensional restriction on this<br />

equation implies only that V and F ext are area averaged quantities over the surface of the system. However, components of V and F ext<br />

may be in each of the three coordinate directions.

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!