Modern Engineering Thermodynamics

196 CHAPTER 6: First Law Open System Applications
EXAMPLE 6.10 (Continued )
1
Nitrogen
tank
m R
System boundary
2
m D
FIGURE 6.20
Example 6.10, tank alone.
Assumptions
1. _Q = 0 (tank is insulated).
2. _W = 0 (no work is done on or by the tank itself).
3. T T = T 2 = 70.0°F = constant (discharge is isothermal).
4. Neglect changes in KE and PE of the flow streams and the tank.
5. Treat N 2 as an ideal gas.
Then, the general energy rate balance becomes
_Q
0
−
_W
0
+ _m R ðh 1 + V1 2 /2g c + gZ 1 /g c Þ
⎵
0
− _m D ðh 2 + V 2 2 /2g c + gZ 2 /g c Þ
⎵
0
= dðmuÞ T
/dt + dðKE + PEÞ T
/dt
or
0
_m R h 1 − _m D h 2 = dðmuÞ T
/dt = _m T u T + m T _u T
Now, _u T = du T /dt = d(c v T T )/dt = 0, since T T = constant. The conservation of mass law gives
_m T = _m R − _m D :
Then, the energy rate balance becomes
or
_m R h 1 − _m D h 2 = ð _m R − _m D Þu T
_m R / _m D = ðh 2 − u T Þ/ðh 1 − u T Þ
Now,
h 2 = c p T 2 = c p T T
and
u T = c v T T
Then,
_m R / _m D = c p − c v TT / cp T 1 − c v T T = R/ cp ðT 1 /T T Þ− c v
= 1/ c p /R
ðT 1 /T T Þ− c v /R
Since
c p /R = c p / c p − c v = k/ ðk
− 1Þ
and
c v /R = c v /ðc p − c v Þ = 1/ðk − 1Þ