Page 493 - Bird R.B. Transport phenomena

§15.5 Use of the Macroscopic Balances to Solve Unsteady-State Problems 473
Insulation Fig. 15.5-6. Free batch expansion of
a compressible fluid. The sketch
Convergent shows the locations of surfaces 1
nozzle and 2.
Tank volume = V

Ambient pressure = p a

For the same region, the energy balance of Eq. (E) of Table 15.5-1 becomes

(15.5-36)

in which V is the total volume in the system being considered, and w x is the mass rate of flow
of gas leaving the system. In writing this equation, we have neglected the kinetic energy of
the fluid.
Substituting the mass balance into both sides of the energy equation gives

(dU,
Pi dt (15.5-37)

For a stationary system under the influence of no external forces other than gravity, йФ^ /dt =
0, so that Eq. 15.5-37 becomes

(15.5-38)
p\
This equation may be combined with the thermal and caloric equations of state for the fluid in
order to obtain p (p^) and T^). We find, thus, that the condition of the fluid in the tank de-
x
pends only on the degree to which the tank has been emptied and not on the rate of dis-
charge. For the special case of an ideal gas with constant C , for which dll = C dT and p =
v
v
pRT/M, we may integrate Eq. 15.5-38 to obtain
Р\Р\ У = PoPo (15.5-39)
in which у = C /C . This result also follows from Eq. 11.4-57.
p v
(b) Discharge of the gas through the nozzle. For the sake of simplicity we assume here that
the flow between surfaces 1 and 2 is both frictionless and adiabatic. Also, since w } is not far
different from w , it is also appropriate to consider at any one instant that the flow is quasi-
2
steady-state. Then we can use the macroscopic mechanical energy balance in the form of Eq.
15.2-2 with the second, fourth, and fifth terms omitted. That is,

\v\ ±dp = O (15.5-40)

Since we are dealing with an ideal gas, we may use the result in Eq. 15.5-34 to get the instan-
taneous discharge rate. Since in this problem the ratio S /S } is very small and its square is
2
even smaller, we can replace the denominator under the square root sign in Eq. 15.5-34 by
unity. Then the p outside the square root sign is moved inside and use is made of Eq. 15.5-39.
2
This gives

- 1)][(р /ро) 2/у - (15.5-41)
2
in which S is the cross-sectional area of the nozzle opening.
2