Applications of the Availability Balance Equation

4. Applications of the Availability Balance Equation

We now discuss various applications of the availability balance equation. An unsteady situation exists at startup when a turbine or a boiler is being warmed, and the availability starts to accumulate. Here,

d (E cv –T 0 S cv )/dt ≠ 0. If a system has a nondeformable boundary, then

W cv =W shaft , P 0 dV cyl /dt = 0, W ˙ u =0

When a system interacts only with its ambient (that exists at a uniform temperature T 0 ), and there are no other thermal energy reservoirs within the system, the optimum work is provided by the relation

(52) For a system containing a single thermal energy reservoir (as in the case of a power

cv opt , = W cv += I m ˙ i ψ i − m ˙ e ψ e − dE ( cv − TS 0 cv )/ dt .

plant containing a boiler, turbine, condenser and pump, ( Figure 8 ) or the evaporation of water from the oceans as a result of heat from the sun acting as TER), omitting the subscript 1 for the reservoir,

0 cv )/ dt = m ˙ i ψ i + Q R ( 1 − T 0 / T R ) − m ˙ e ψ e − W cv − I .

(53) For a steady state steady flow process (e.g., such as in power plants generating power

dE ( − TS

cv

under steady state conditions), mass conservation implies that ˙ m i = m ˙ e = . Fur- m ˙ thermore, if the system contains a single inlet and exit, the availability balance as-

sumes the form m N ˙( ψψ

i − e ) + ∑ j = 1 Q ˙ Rj , ( 1 − T 0 / T ˙ ˙ Rj , ) − W cv −= I 0 .

(54) On unit mass basis

ψ i − ψ e +∑ q Rj , ( 1 − T 0 / T Rj , ) − w cv −= i 0 ,

(55) where q Rj , = Q ˙ Rj , /˙, mw cv = W ˙ cv /˙, mi = ˙/˙ Im . When a system interacts only with its

ambient at T 0 and there are no other thermal energy reservoirs within the system, the optimum work is given by the relation

W ˙ cv opt , = m ˙ i ψ i − m ˙ e ψ e − dE ( cv − TS 0 cv )/ dt .

(56) In case the exit state is a restricted dead state, (e.g., for H 2 O, dead state is liquid water

at 25 °C 1 bar) W ˙

= m ˙ ψ ′+ ∑ j = 1 Q ˙ Rj , ( 1 − T 0 / T Rj , )

cv opt ,

(57) where ψ′ = ψ – ψ 0 is the specific stream exergy or specific-relative stream availability

(i.e., relative to the dead state). Since ψ 0 =h o -T o s o in the absence of kinetic and po- (i.e., relative to the dead state). Since ψ 0 =h o -T o s o in the absence of kinetic and po-

For a system containing multiple inlets and exits the availability equation is dE ( − TS

0 cv )/ dt =∑ m ˙ i ψ i + ∑ j = 1 Q ˙ Rj , ( 1 − T 0 / T Rj , ) −∑ m ˙ e ψ e − W ˙ cv − ˙ I . inlets (58) exits

cv

For a single inlet and exit system containing multiple components the expression can

be generalized as dE ( cv − TS

0 cv )/ dt =∑ m ˙ ki , ψ ki , + ∑ j = 1 Q ˙ Rj , ( 1 − T 0 / species T Rj , )

−∑ m ˙ species ke , ψ ke , − W ˙ cv − ˙ I

where ψ k =h k (T,P,X k )–T 0 s k (T,P,X k ) denotes the absolute availability of each com- ponent, and X k the mole fraction of species k. For ideal gas mixtures, ψ 0

k =h k –T 0 (s k – R ln (p k /P ref )),

since the partial pressure of the k–th species in the ideal gas mixture P k =X k P. Consider an automobile engine in which piston is moving and at the same time mass is entering or leaving the system (e.g., during the intake and exhaust strokes). In addi-

tion to the delivery of work through the piston rod W ˙ u , atmospheric work is per- formed during deformation, i.e., W ˙ 0 =P 0 dV/dt. Therefore,, W ˙ cv = W ˙ u + P dV / dt 0 cyl

and the governing availability balance equation is

ki , ψ ki , + ∑ j = 1 Q ˙ Rj , ( 1 − T 0 / species T Rj , )

dE ( − TS 0 )/ dt =∑ m ˙

cv cv

−∑ m ˙ ke , ψ ke , − W ˙ u − P dV dt o / − ˙ I species

Simplifying. dE ( cv − TS 0 cv + P dV dt o /)/ dt =∑ m ˙

ki , ψ + j 1 Q Rj , 1 species T ki , ∑ = ˙ ( − 0 / T Rj , )

−∑ m ˙ ψ

ke , ke , − W u − I species ˙

For steady cyclical processes the accumulation term is zero within the control vol-

ume, and ψ i = ψ e . Therefore,

cv cycle , += ˙ I ∑ j = 1 Q Rj , ( 1 − T 0 / T Rj , ) .

c. Example 3 Steam enters a turbine with a velocity of 200 m s –1 at 60 bar and 740ºC and leaves as saturated vapor at 0.2 bar and 80 m s –1 .The actual work delivered during the process is 1300 kJ kg –1 . Determine inlet stream availability, the exit stream availability, and

the irreversibility. Solution

× 7.519 = 1769 kJ kg –1 . Likewise, = 2609.7 + (80 2 ψ –1 e ÷2000) –298 × 7.9085 = 256 kJ kg . Therefore,

i = (h 1 +v /2g) – T 0 s 1 = 3989.2 + 20 – 298

w = 1769 – 256 = 1513 kJ kg –1 opt , and

I = 1513 – 1300 = 213 kJ kg –1 . The entropy generation

σ = 213÷298 = 0.715 kJ kg –1 K –1 . Remarks

The input absolute availability is 1769 kJ kg –1 . The absolute availability outflow is 256 kJ kg –1 . The absolute availability transfer through work is 1300 kJ kg –1 . The availability loss is 213 kJ kg –1 . The net outflow is 1769 kJ kg –1 .

d. Example 4 This example illustrates the interaction between a thermal energy reservoir, its ambi- ent, a steady state steady flow process, and a cyclical process. Consider the inflow of water in the form of a saturated liquid at 60 bar into a nuclear reactor (state 1). The reactor temperature is 2000 K and it produces steam which subsequently expands in a turbine to saturated vapor at a 0.1 bar pressure (state 2). The ambient temperature is 25ºC. The reactor heat transfer is 4526 kJ per kg of water. Assume that the pipes and turbines are rigid. What is the maximum possible work between the two states 1 and 2? If the steam that is discharged from turbine is passed through a condenser (cf. Figure

8 ) and then pumped back to the nuclear reactor at 60 bar, what is the maximum possi- ble work under steady state cyclical conditions? Assume that the inlet condition of the water into the pump is saturated liquid.

Solution If the boundary is selected through the reactor, for optimum work I = σ = 0. Under

steady state conditions time derivatives are zero, and, since the body does not deform

W u = 0, so that W cv =W shaft and ˙ m i = m ˙ e = . Therefore, m ˙

m ˙( ψψ i − e ) + Q ˙ R 1 ,

( ˙ 1 − T 0 / T R 1 , ) − W cv opt , = 0 .

(A) Dividing Eq. (A) throughout by the mass flow rate,

Reservoir R