Work Producing and Consumption Devices

3. Work Producing and Consumption Devices

In order to change a system to a desired end state from a specified initial state, energy must be transferred across its boundaries. In work producing or absorbing devices, this energy transfer is in the form of work. Since W (for zero irreversibility) differs from W (for realistic processes), the value of the availability efficiency η avail is instructive in assessing the overall

system design.

a. Open Systems: For a work producing device W opt =W max and

η avail = W/W max . (76) Availability or exegetic efficiency is a measure of deviation of an actual process from an ideal

reversible process for the prescribed initial and final states.The maximum value of the avail- ability efficiency is unity and the presence of irreversibilities reduces that value. The overall irreversibility

I=W max –W=W max (1– W/W max )=W max (1– η avail ). (77) If the end state of a working fluid emanating from a work–producing device, e.g., a

gas turbine, is at a higher temperature or pressure than that of its ambient, the fluid still con- tains the potential to perform work. Therefore, it is useful to define the availability efficiency considering the optimum work (which is based on the assumption that the optimal end state is

a dead state). In that case W max,0 =W opt,0 , and

η avail,0 = = W/W max,0 = (W/W max )(W max /W max,0 )= η avail (W max /W max,0 ).

(78) Note that W max ≤W max,0 and hence η avai1,0 ≤η avail. For a work–consuming device such as a com-

pressor, η avail =W min /W.

(79) If the exit state from a work-producing device is the dead state, then the availability efficiency

is. This ratio informs us of the extent of η avail,0 = (work output) ÷ (input exergy),

(80) and the conversion of the input exergy into work, but gives no indication as to whether the

exergy is lost as a result of irreversibility or with the availability leaving along with the exit flow.

For a work–consuming device such as a compressor η avail = |W min |/|W|, η re 1,0 = |W min,0 |/|W|

b. Closed Systems For processes involving work output from a closed system (which is usually expan- sion work such as that obtained during the gas expansion in an automobile engine) W u,opt = W u,max , and

η avail = |W u |/|W u,max |. (81) Likewise, for processes during which work is done on a closed system (which is usually com-

η avail = |W u |/|W u,min |, and (82) η avail,0 = η avail W/W u u,min,0 .

(83) The isentropic efficiency is not same as the availability efficiency, since isentropic

work can involve an end state that is different from a specified end state, while the determina- tion of optimum work is based on the specified end state. These differences are illustrated in the example below.

j. Example 10 An adiabatic steady–state, steady–flow turbine expands steam from an initial state characterized by 60 bar and 500ºC (State 1) to a final state at 10 kPa at which the quality x= 0.9 (state 2). Is the process possible? Determine the turbine work output. What would have been the quality x 2s at the exit and isentropic work output for the

same initial conditions for the same P 2 = 10 kPa?

Determine the work output if the final state is to be reached through a combination of

a reversible adiabatic expansion process that starts at the initial state followed by re- versible heat addition until the final state is reached. Determine the maximum possible (optimum) work. Calculate the availability efficiency based on the actual inlet and exit states and avail-

ability efficiency based on the optimum work. Solution

Applying the generalized entropy balance equation

dScv dt m s / = ˙ ii − ms ˙ ee + QT ˙/ b +σ ˙ .

(A) Under adiabatic steady state steady flow conditions, d/dt = 0, ˙ mm i = ˙ e (steady), and

Q =0. Therefore, Eq. (A) assumes the form ˙ s 2 –s 1 = σ. The specific entropies s 1 (60 bar, 773 K) = 6.88 kJ kg –1 K –1 , and s 2 (0.1 bar, x = 0.9) =

–1 K × 0.65 + 0.9 × 8.15 = 7.4 kJ kg –1 σ = 0.52 kJ kg . The process is irreversible, since σ > 0.

0.1 –1 K –1 so that

Applying the energy conservation equation for an adiabatic (q = 0) steady–state, steady–flow process

(C) The specific enthalpies h (60 bar, 773 K) = 3422.2 kJ kg –1 1 ,h 2 (0.1 bar, x = 0.9) = 0.1 × 191.83 + 0.9

–w = h 2 –h 1 .

× 2584.7 = 2345.4 kJ kg –1 so that w = –(2345.4 – 3422.2) = 1076.8 kJ kg –1 .

For an isentropic process the end state s 2s =s (= 6.88 kJ kg –1 K –1 1 ) with the final pres- sure P 2s =P 2 (although the quality of the steam differs at these two states). Therefore,

s 2s = 6.88 = (1–x 2s ) × 0.65 + x 2s × 8.15, i.e., x 2s = 0.83.

Applying the First law to the process 1–2s,

–w 12s =h 2s –h 1 , i.e.,

(D)

h = 0.17

× 191.83 + 0.83 × 2584.7 = 2177.9 kJ kg , and

2s 2s

sequently, the steam leaves the turbine with a relatively higher enthalpy at the conclu-

sion of process 1–2. Therefore, w 12 <w 12s .

The adiabatic or isentropic efficiency is η=w 12 /w 12s = 0.865. The infinitesimal enthalpy change dh = δq – δw. One could react state 2 by using an

isentropic process first to P2= 10 kpa and x2s =0.83 and then adding heat at con- stant T 2 to state 2 to obtain the quality x 2 = 0.9. Since the paths 1–2s and 2s–2 are re- versible, δq = T ds. Hence,

T ds – w = dh. (E) Integrating the equation appropriately, we have

s 1 Tds + ∫ s 2 s Tds w − 12 −− s 2 = h 2 − h 1 .

The path 1–2 s involves no entropy change so that

T (s 2 –s 2s )–w 12 =h 2 –h 1 .

Hence, – w 1–2s–2 = 2345.4 – 3422.2 – 318.8 × (7.4 – 6.88) – 1242.6 kJ kg –1 . Since work is path–dependent and the paths 1–2 and 1–2s–2 are different, it is incorrect to

write w 1–2s–2 as w 12 . The work w 1–2s–2 is larger than the answer obtained in part 0 of

s –2 the re- versible heat added q 2s–2,rev = T (s 2 –s 2s ) = 165.8 kJ kg . A portion of this heat is con- verted into additional work. We have not, however, given any information on what source is used to add the heat. The heat addition process involves an interaction with a source other than the ambient. We will now use an availability analysis to determine the maximum work output that is possible in the absence of entropy generation while maintaining the same initial and final states. Simplifying the availability balance equation for this situation, the opti- mum work

the solution, since the process 1–2 s –2 is reversible. During the process 2

(F) where –1 ψ

w opt = ψ 1 – ψ 2 ,

1 =h 1 –T 0 s 1 = 3422.2 – 298 × 6.88 = 1371.9 kJ kg . Likewise, ψ 2 = 2345.4

– 298 –1 × 7.4 = 140.2 kJ kg , and w

opt = 1231.7 kJ kg .

The availability efficiencies η avail =w 12 /w opt = 0.874, and

η avail,0 =w 12 /w opt,0 , where

w opt,0 = ψ 1 – ψ 0 and ψ 0 =h 0 –T 0 s 0 . The dead state for the working fluid is that of liq- uid water at 298 K, 1 bar, so that h 0 = 104.89 kJ kg –1 and s 0 = 0.3674 kJ kg –1 K –1 , and ψ 0 = 104.89 – 298

× 0.3674 = –4.6 kJ kg –1 .

Consequently, w opt,0 = 1376.5 kJ kg –1 and η avail,0 = 0.78. Remarks

We see that w 12 <w opt <w 1–2s–2 <w opt,0 . For the optimum work the only outside inter- action that occurs is with the ambient that exists at the temperature T 0 while W 1-2s-2 is We see that w 12 <w opt <w 1–2s–2 <w opt,0 . For the optimum work the only outside inter- action that occurs is with the ambient that exists at the temperature T 0 while W 1-2s-2 is

output of w –1 1–2s = 1244.3 kJ kg . Then we can use a portion of this work to operate a Carnot heat pump that absorbs heat from the ambient (at T 0 ) and adds 165.8 kJ kg –1 of heat to the process 2s–2. The Carnot heat pump must operate between 319 K and 298 K. Therefore, the Carnot COP = 318.8

÷ (318.8 – 298) = 15.3. Since 165.8 kJ kg –1 of heat is required, a work input of 165.8

÷ 15.3 = 10.8 kJ kg –1 is necessary. This 10.8 kJ kg –1 of work is subtracted from w 1–2s and, consequently, w opt = 1244.3 – 10.8 = 1233.5

kJ kg –1 , which is essentially the same answer as that based on the above availability analysis. For the optimum process 1-2s-2, entropy generation is zero.

The availability calculation does not explain why a final state x 2 = 0.9 at P 2 = 0.1 bar is reached instead of the state x 2s = 0.83 at P 2 = 1 bar. It only provides information as to what the optimum work could have been had the inlet and exit states been fixed. In

a cyclic process, all the states are normally fixed. In a power plant all the states are normally fixed to maintain a steady state. A power plant operator must monitor the exit conditions, optimum work, and the entropy generation as the plant equipment de- grades over time.