Optimum Work

1. Optimum Work

Consider water at high pressure P 1 and low temperature T 1 (cf. Figure 6 ) that is heated to produce steam at ( P 1 ′ , T 1 ′ ) using a large thermal reservoir (such as a boiler) that exists at a

constant temperature T R,1 . The reservoir transfers heat at a rate Q ˙ R1 , to the water. The steam first delivers useful work

W ˙ u through a deformable piston–cylinder assembly followed by shaft work W ˙ Shaft through a steam turbine. The cylinder boundary deforms, producing defor- mation or boundary work P 0 dV cyl /dt against the atmosphere. The boundary is selected in such

a manner that no property gradients exist outside it, and the boundary temperature is T 0 (as a manner that no property gradients exist outside it, and the boundary temperature is T 0 (as

W ˙ cv = W ˙ u + W ˙ sh + W ˙ 0 . where W ˙ 0 denotes the atmospheric work P o dV cyl /dt.

In this context we will now determine the optimum work for a process that occurs between the same inlet (T 1 ,P 1 ) and outlet (T 2 ,P 2 ) states and which withdraws an identical amount of heat from the reservoir, Q ˙ R1 , . In our analysis we will consider the presence of sev- eral thermal energy reservoirs at various temperatures, i.e., T R,1 , T R,2 , T R,3 ,...etc. with all ir- reversibilities being maintained within the system control volume. Applying the First law for an open system,

dE cv = Q ˙ cv − W ˙ cv + me dt ˙ iTi , +− me ˙ eTe , , (36) where the control volume work expression W ˙ cv includes W ˙ u , W ˙ shaft ,P 0 dV cyl /dt, and any other

work forms. The total heat Q ˙ cv transferred from the control volume includes the various heat interactions Q ˙

˙ R1 , , Q R2 , , Q R3 , ,... with the thermal energy reservoirs, and that with the environ- ment Q ˙ 0 . Therefore,

dE cv ˙

(37) Typically Q ˙ 0 < 0 . If turbine blades get rusted over a period of time, more energy is used to

dt = Q 0 + Q ˙ R 1 , + Q ˙ R , 2 + Q ˙ R , 3 + ..... − W ˙ cv + me ˙ iTi , − me ˙ eTe ,

overcome friction than to produce work; thus, frictional heating will occur which will cause turbine exit temperature T e to increase. In order to maintain the same T e , the heat loss to the

environment ˙ Q 0,turb must have to be increased so that under steady state operation a smaller amount of work is delivered.

We have assumed that temperature gradients only exist within the boundaries of the control volume and, consequently, entropy is generated only within it. Applying the entropy balance equation (Chapter 3),

We observe that the higher the loss Q ˙ 0 , the higher ˙ σ cv for a steady state operation for fixed s i and s e . From energy balance, the higher the heat loss, the lower is the work output. Expressing Q ˙ 0 in terms of ˙ σ cv Eq. (37) can be expressed in the form

dS

Q Rj ,

cv = T 0 ( cv

−∑ N

j = 1 + ms ˙ ee − ms ˙ ii ) +

dt

T Rj ,

∑ dE j = Q ˙

cv

1 Rj , − dt − me ˙ eTe , + me ˙ iTi , − T 0 σ ˙ cv

Rewriting Eq. (39),

W ˙ cv =− dE ( − TS

0 cv )/ dt + ∑ j = 1 Q ˙ Rj , 1 − T 0 / T Rj , − m ˙ e ψ e + m ( ˙ ) i ψ i − T 0 σ ˙ cv , (40)

cv

The absolute specific stream availability or the absolute specific flow or stream availability ψ is defined as

ψ (T,P,T 0 )=e T (T,P) – T 0 s(T,P) = (h(T,P) + ke + pe) – T 0 s(T,P). (41) where the terms ψ i and ψ e denote the absolute stream availabilities, respectively, at the inlet

and exit of the control volume. They are not properties of the fluid alone and depend upon the temperature of the environment. The optimum work is obtained for the same inlet and exit states when ˙ σ cv = 0. In this case, Eq. (40) assumes the form

W ˙ cv opt =− dE (

cv − TS 0 cv )/ dt + ∑ j = 1 Q ˙ Rj , ( 1 − T 0 / T Rj , ) − m ˙ e ψ e + m ˙ i ψ i .

(42) where the term Q ˙ Rj , ( 1 − T 0 / T Rj , ) represents the availability in terms of the quality of heat en-

ergy or the work potential associated with the heat transferred from the thermal energy reser- voir at the temperature T R,j . When the kinetic and potential energies are negligible,

ψ=h–T 0 s. (43) For ideal gases, s = s 0 – R ln (P/P ref ), where the reference state is generally assumed to be at P ref

= 1 bar. Therefore, ψ=ψ 0 +RT

0 ln (P/P ref ),

(44) and 0 ψ =h 0 –T 0 s 0 . The enthalpy h (T) = h 0 (T) for ideal gases, since it is independent of pres-

sure. If the exit temperature and pressure from the control volume is identical to the envi- ronmental conditions T 0 and P 0 , i.e., the exit is said to be at a restricted dead state, in that case W ˙ cv,opt = W ˙ cv,opt 0 and the exit absolute stream availability at dead state may be expressed as

(45) Note that e T,e,0 =h 0 since ke and pe are equal to zero at dead state.

ψ e,0 =e T,e,0 (T 0 ,P 0 )–T 0 s e,0 (T 0 ,P 0 ).