Modern Engineering Thermodynamics

Problems 197
we have
_m R / _m D = ðk − 1Þ/ ½kT ð 1 /T T Þ− 1Š
In part b, we are given that T T = 70.0°F = 530. R and T 1 = 200.°F = 660. R.
Table C.13a gives us the specific heat ratio for nitrogen as k = 1.40. Therefore, our resultant equation gives
_m R / _m D = ð1:40 − 1Þ/1:40 ½ ð660:/530:
Þ− 1Š = 0:538
Thus, 53.8% of the discharge mass must be recycled to keep the tank temperature constant.
To find the rate of recycle heat transfer required in part b, we must analyze the heater and compressor as a separate system
(Figure 6.21).
m R
m R
2
1
Heater
Compressor
W C = −3.00 hp
System
boundary
Q H = ?
FIGURE 6.21
Example 6.10, heater and compressor as a separate system.
In general, _m R is not constant in time in this system. We assume that a feedback control system exists that automatically
scales down _W C and _Q H as _m R goes to zero and the tank empties. We also assume that the internal, kinetic, and potential
energies of the entire system are constant in time, so that we have a steady state, steady flow, single-inlet, single-outlet system.
Under these conditions, the modified energy rate balance becomes
_Q H − _W C + _m R ðh 1 − h 2 Þ = 0
where we again neglect any changes in flow stream specific kinetic and potential energies. Using the ideal gas relationship
for specific enthalpy gives
Table 13a gives c p = 0.248 Btu/(lbm · R) for nitrogen. Then,
_Q H = _m R c p ðT 2 − T 1 Þ+ _W C
_Q H = ð0:500 lbm/sÞ½0:248 Btu/ðlbm.RÞŠð660: − 530: RÞ
+ ð−3:00 hpÞ½550 ft.lbf/ ðs.hpÞŠ½1 Btu/ ð778:16 ft.lbfÞŠ
= 14:0 Btu/s
SUMMARY
In this chapter, we investigate a series of open system examples and carry out a first law analysis of them using
the energy balance or the energy rate balance. The primary purpose of this is to illustrate the material presented
in Chapter 4. Several new equations are introduced in this chapter. Some of the more important equations are
these.
1. The one-dimensional mass flow equation:
dm
= ð _mÞ
dt
flow stream
= ðρAVÞ flow stream
(6.1)
flow stream