Modern Engineering Thermodynamics

8.2 Systems Undergoing Reversible Processes 251
For the case of constant heat flux and isothermal system boundaries (i.e., _q and T b both constant), Eqs. (8.5)
and (8.6) reduce to
Closed system reversible processes :
Entropy balance ðSBÞ: S 2 − S 1 = ms ð 2 − s 1 Þ = 1 Q 2
T b
rev
(8.7)
and
Q
Entropy rate balance ðSRBÞ: _S = m_s =
_
where 1 Q 2 and _Q are the total heat transfer and total heat transfer rate, respectively, and T b is again the absolute
temperature of the system boundary (assumed isothermal here).
If the system boundary temperature is not constant, then the exact analytical dependence of _q on system boundary
temperature T b , system boundary area A, and process time t must be known before the integrals in Eqs. (8.5)
and (8.6) can be evaluated. This information must be provided in the problem statement (or determined by
measurement or hypothesis in the case of real engineering situations).
T b
rev
(8.8)
EXAMPLE 8.1
Water in the amount of 2.00 kg undergoes a reversible isothermal expansion from a saturated liquid at 50.0°C toasuperheated
vapor at 50.0°C and 5.00 kPa. Determine the heat and work transports of energy for this process.
Solution
First, draw a sketch of the system (Figure 8.1).
1 Q 2 = ?
1 W 2 = ?
Water
m = 2.00 kg
x 1 = 0
T 1 = 50.0°C
State 1
Isothermal process
State 2
Superheated vapor
m = 2.00 kg
T 2 = 50.0°C
p 2 = 5.00 kPa
FIGURE 8.1
Example 8.1.
The unknowns here are 1 Q 2 and 1 W 2 . The material is the water inside the cylinder (a closed system). The thermodynamic
states are:
The basic equations here are the energy balance, EB,
and the entropy balance, SB,
Reversible and
ƒƒƒƒ!
Isothermal
State 1
State 2
T 1 = 50:0 °C T 2 = T 1 = 50:0 °C
x 1 = 0
p 2 = 5:00 kPa
1Q 2 − 1 W 2 = mðu 2 − u 1 Þ + KE 2 − KE 1 + PE 2 − PE 1
Z Z
ms ð 2 − s 1 Þ =
τ Σ
0 ðassume the system is stationaryÞ
_q
T b
dA dt
rev
Since this process is isothermal, the system boundary absolute temperature T b is a constant and
Z Z
_q
dA dt = 1 Q 2
τ Σ T b rev
T b rev
(Continued )