Heat Engines

1. Heat Engines

a. Efficiency Different heat engines employ various cyclical processes (e.g., the Rankine, Brayton, and Otto cycles) that first absorb heat and then reject it to the environment in order to produce work. The efficiency η = Sought/Bought = work output ÷ heat input = W/Q in = (Q in –Q out )/Q in

( Figure 3a ) presents an energy band diagram for a heat engine operating between two fixed–condition thermal energy reservoirs. The Carnot efficiency of an ideal heat engine η CE =

1 – T L /T H , where T L and T H , respectively, are the low and high temperatures associated with the two reservoirs. We note that even for idealized cycles involving isothermal energy reser- voirs and internally reversible processes, η CE < 1 due to Second law implications, and avail-

ability analysis tells us that work potential of heat is equal to Q in (1- T 0 /T H ). A part of Q in is converted to W opt,cyc and Q 0,opt,cyc is rejected to the ambient under ideal conditions. y.

b. Availability or Exergetic (Work Potential) Efficiency In power plants based on the Rankine cycle, heat is transferred from hot boiler gases to cooler water in order to form steam. A temperature difference exists between the gases and the water, thereby creating an external irreversibility even though the plant may be internally reversible. Therefore its work output W cyc is lower than the maximum possible work output W opt,cyc for the same heat input, and hot gas, ambient, and cold water temperatures. The Avail- ability Efficiency is defined as

η Avail = W/W opt,cyc = W/W max,cyc , where (68) W opt,cyc =W max,cyc = (W cyc +I cyc ), and

I cyc = T 0 σ cyc . Note that σ cyc refers to entropy generation in isolated system during a cyclic process. Exergetic efficiency for a cycle is a measure of deviation of an actual cycle from an

ideal reversible cycle. Equation (68) can be used to compare different cycles that operate be- tween similar thermal energy reservoirs. For a cycle operating between fixed–temperature thermal energy reservoirs

(70) the optimum cyclic process rejects a smaller amount of heat Q 0,opt,cy as compared to a realistic

W opt,cyc =W max,cyc =Q R,1 (1 – T 0 /T R,1 ), and

process. The difference Q 0,opt,cyc –Q 0,cyc = I = T 0 σ cyc is the irreversibility. Figure 12a and b illustrate the energy and availability band diagrams for a heat engine. The term Q R (1-T 0 /T R ) is

the availability associated with heat Q R ,W cyc is the availability transfer through work, and I cyc is the availability loss in the cycle.

h. Example 8

A nuclear reactor transfers heat to water in a boiler that is at a 900 K temperature, thereby producing steam at 60 bar, and 500ºC. The steam exits an adiabatic turbine in the form of saturated vapor at 0.1 bar. The vapor enters a condenser where it is con- densed into saturated liquid at 0.1 bar and then pumped to the boiler using an isen- tropic pump. Determine: The optimum work. The availability efficiency. The overall irreversibility of the cycle. The irreversibility in the boiler, turbine, and condenser.

Solution Analyzing the Rankine cycle: The turbine work is

(A) w

q 12 –w 12 =h 2 –h 1 , i.e.,

12 = 2585 – 3422 = 837 kJ kg . The heat rejected in the condenser

q 23 –w 23 =h 3 –h 2 , i.e.,

q 23 =q out = 192 – 2585 = –2393 kJ kg –1 . Likewise, in the pump

(B) Since properties for the liquid state at 4 may be unavailable, they can be otherwise

q 34 –w 34 =h 4 –h 3 ≈v 3 (P 4 –P 3 ), or

determined. The work w = –0.001

34 × (60 – 0.1) × 100 = – 6 kJ kg . From Eq. (B) h 3 = 192 kJ kg –1 (sat liquid at 0.1), and

h 4 = 192 + 6 = 198 kJ kg –1 . In the boiler

q –1

in =q 41 –w 41 =h 1 –h 4 = 3422 – 198 = 3224 kJ kg .

Therefore, the cyclical work w –1

cyc =w t –w p = 837 – 6 = q in –q out = 3224 – 2393= 831 kJ kg . The efficiency η=w cyc /q in = 831/3224 = 0.26.

Integrating the general availability balance equation over the cycle w cyc,opt =q in (1 – T 0 /T b ) = 3224 (1 – 298/900) = 2156 kJ kg –1 . The actual Rankine cycle work = 831 kJ kg –1 and the actual cycle efficiency η = 0.26.

The Carnot work is 2156 kJ kg –1 and the Carnot efficiency η Carnot = 0.67. The relative efficiency is

η Avail =w cyc /w cyc,opt = η/η Carnot = 831/2156 = 0.39. The overall irreversibility of the cyclical process is I=w cyc,opt – w = 2156 – 831 = 1325 kJ kg –1 .

An availability analysis can be performed on the various system components as fol- lows: For the turbine,

s –1

1 = 6.88 kJ kg K , and s 2 = 8.15 kJ kg K so that s 2 >s 1 . Furthermore, ψ –1

1 =h 1 –T 0 s 1 = 3422 – 298 × 6.88 = 1372 kJ kg , and, likewise, = 2585 – 298

2 × 8.15 = 156.3 kJ kg . Therefore, w t,opt = ψ 1 – ψ 2 = 1372 – 156.3 = 1216 kJ kg –1 , and

I t = 1216 – 837 = 379 kJ kg –1 . For the condenser,

3 = 191.8 – 298 × 0.649 = –1.6 kJ kg , and w

= 156.3 – (–1.6) = 157.9 kJ kg cond,opt –1 ψ 2 ψ 3 . Therefore,

I cond –1 = 157.9 kJ kg . For the pump,

4 =s 3 = 0.649 kJ kg s –1 K –1 , and ψ –1

4 =h 4 –T 0 s 4 = 198 – 298 × 0.649 = 4.6 kJ kg . Consequently, w p , opt = ψ 3 – ψ 4 = –1.6 – 4.6 = – 6.2 kJ kg –1 . Since w p = – 6.2 kJ kg –1 ,I p = 0 kJ kg –1 . For the boiler,

W b , opt ==q b (1– T 0 /T b )+ ψ 4 – ψ 1 = 2156 + 4.6 – 1372 = 789 kJ kg –1 , and

I b ==W b , opt –W b = 789 – 0 = 789 kJ kg –1 . The total irreversibility = 379+158+0+789 = 1326 kJ kg –1 is the same as that calcu- lated above.

The availability input at the boiler inlet ψ 4 = 4.6 kJ kg –1 , and

f (25 C) = 104.89-298 ×0.3674 = -4.6 kJ kg . The availability input through heat transfer in the boiler ψ 41 =q b (1 – T 0 /T b ) = 3224 × (1 – 298/2000) = 2156 kJ kg –1 . Figure 13 illustrates the exergy band diagram for the cyclic process.

ψ 0 =h 0 –T 0 s

sat

0 ≈h f (25 C) –298

×s –1

sat

W opt, cyc

W cyc

Q in

HE I cyc

Ambience

Q 0, cyc

T R,1

Q 0, opt, cyc

irreversibility, I

Optimum

(a) Actual