Modern Engineering Thermodynamics

10.10 Modified Availability Rate Balance (MARB) Equation 335
■ Case 4 Isothermal Boundaries
If the system has a constant (isothermal) system boundary temperature, T b , then all the heat transports can
be combined into a single net heat transport term, ∑ _Q i = _Q net = _Q, where we drop the subscript net for
i
convenience. Then the (net) heat transport of availability for isothermal boundaries can be written as
∑ n
i=1
1 − T
0
T bi
_Q i = 1 − T
0
T b
_Q net = 1 − T
0 _Q (10.25)
T b
■ Case 5 Constant Volume
If the volume of space enclosed by the system boundary is constant in time, then dV/dt = 0, and the loss or gain
of available energy associated with the moving environment vanishes. Then, p 0 ðV 2 − V 1Þ = 0 in Eq. (10.18), and
p 0 ðdV/dtÞ = 0 in Eq. (10.19). Note that Eq. (10.22) tells us that, if the system is at a steady state, then it also
must have a constant volume.
■
■
10.10 MODIFIED AVAILABILITY RATE BALANCE (MARB) EQUATION
Now, we can refer to the open system availability rate balance for a steady state, steady flow, single-inlet,
single-outlet, constant volume, isothermal boundary system (or SS, SF, SI, SO, CV, IB system) as:
The modified availability rate balance for a SS, SF, SI, SO, CV, IB open system
1 − T 0
T b
_Q − _W + _m½ða f Þ in − ða f Þ out Š − _I = 0 (10.26)
where Eq. (10.15) requires that _I ≥ 0.
EXAMPLE 10.7
A horizontal pipe carrying superheated steam at 5000. psia and
1000.°F suddenly develops a small crack. Steam enters the
crack at 50.0 ft/s and passes through it in a steady state, adiabatic
_Q = 0 , aergonic _W = 0 process to exit at 14.696 psia
with a velocity of 300. ft/s. Determine the specific flow availabilities
at the inlet and outlet of the crack, and calculate the
irreversibility per unit mass of steam exiting the crack. Take the
local environment (ground state) to be saturated liquid water
at 70.0°F.
Solution
First, draw a sketch of the system (Figure 10.12).
300. ft/s
5000. psia
50.0 ft/s
1000.°F
Ground state:
x 0 = 0.00
T 0 = 70.0°F
The unknowns are the specific flow availabilities at the inlet and
outlet of the crack and the irreversibility per unit mass of steam
exiting the crack. The material is steam, and this is an open
system. The data at the inlet and outlet flow stations are
FIGURE 10.12
Example 10.7.
Station 1 Station 2
p 1 = 5000: psia p 2 = 14:696 psia
T 1 = 1000:°F ?
h 1 = 1363:4 Btu/lbm
s 1 = 1:3990 Btu/lbm.R
(Continued )