5 Efficiency of Heat Engines at Max imum

Power

5.1 maximum power output

The thermal efficiency of a Carnot cycle operating between high temperature ( T H ) and low temperature (T,) reservoirs is given by

Efficiency of an internally reversible heat engine when producing

This cycle is extremely idealised. It requires an ideal, reversible heat engine (internally reversible) but, in addition, the heat transfer from the reservoirs is also reversible (externally reversible). To achieve external reversibility it is necessary that the tempera- ture difference between the reservoirs and the engine is infinitesimal, which means that the heat exchanger surface area must be very large or the time to transfer heat must be long. The former is limited by size and cost factors whilst the latter will limit the actual power output achieved for the engine. It is possible to evaluate the maximum power oufpuf achievable from an internally reversible (endoreversible) heat engine receiving heat irreversibly from two reservoirs at TH and T,. This will now be done, based on Bejan (1988).

Assume that the engine is a steady-flow one (e.g. like a steam turbine or closed cycle gas turbine): a similar analysis is possible for an intermittent device (e.g. like a Stirling engine). A schematic of such an engine is shown in Fig 5.1.

The reservoir at TH transfers heat to the engine across a resistance and it is received by the engine at temperature T,. In a similar manner, the engine rejects energy at T2 but the cold reservoir is at T,. It can be assumed that the engine itself is reversible and acts as a Carnot cycle device with

This thermal efficiency is less than the maximum achievable value given by eqn (5.1) because T,/T, > T,/T,. The value can only approach that of eqn (5.1) if the temperature drops between the reservoirs and the engine approach zero.

86 Eficiency of heat engines at maximum power

Fig. 5.1 Internally reversible heat engine operating between reservoirs at TH and T,

The heat transfer from the hot reservoir can be defined as

Q H = uHAH(TH - T I )

where U H = heat transfer coefficient of hot reservoir (e.g. kW/m2 K)

and AH = area of heat transfer surface of hot reservoir (e.g. m2) QH = rate of heat transfer (e.g. kW).

(5.3)

The heat transfer to the cold reservoir is similarly

By the first law Qc = UcAc(T2 - Tc) (5.4)

w = Q H - QC (5.5) Now, the heat engine is internally reversible and hence the entropy entering and leaving it must be equal i.e.

This means that

It is possible to manipulate these equations to give W in terms of TH, Tc, UHAH, and the ratio T2/Tl = z. From eqn (5.4)

and, from eqn (5.6),

Eficiency of an internally reversible heat engine 87

Hence

Rearranging gives

and hence

w

t l+- ( 2:;) (5.10)

(5.11)

Thus the rate of work output is a function of the ratio of temperatures of the hot and cold reservoirs, the ratio of temperatures across the engine and the thermal resistances. The optimum temperature ratio across the engine (t) to give maximum power output is obtained when

- 0 a w at --

Differentiating eqn (5.1 1) with respect to t gives

a w 1 (1 - t) (t - TJT,) - (t - TJT,) f1 - t) 7. t2

1+-

Hence aw/& = o when t = 00 or t2 = Tc/TH. Considering only the non-trivial case, for maximum work output

(5.12)

This result has the effect of maximising the energy flow through the engine while maintaining the thermal efficiency (qh = 1 - T2/T1) at a reasonable level. It compromises between the high efficiency ( qh = 1 - Tc/TH) of the Carnot cycle (which produces zero energy flow rate) and the zero efficiency engine in which T2 = TI (which produces high energy flow rates but no power).

88 EfSiciency of heat engines at maximum power

Hence the efficiency of an internally reversible, ideal heat engine operating at maximum power output is

An example will be used to show the significance of this result.

Example Consider a heat engine is connected to a hot reservoir at 1600 K and a cold one at 400 K. The heat transfer conductances (UA) are the same on both the hot and cold sides. Evaluate the high and low temperatures of the working fluid of the internally reversible heat engine for maximum power output; also calculate the maximum power.

Solution Equation (5.11) gives the work rate (power) as

For maximum power output z = (TC/TH)'/' = (400/1600)1/2. Then

(1 - -) = - - - 200 units -- w (i-t) 1 1600

UHAH - 1600

1 2 8 - (1 + 1) 2

w and, from eqn (5.10)

= 400 units -- - QH

UnAH U H A H ( ~ -

Then

(Q, - W > = 200 units Qc 1 -- --=- Qc

UHAH UcAc &AH Thus

T , = ~ [ ~ + T c } = 2 x ( 2 0 0 + 4 0 0 ) = z UcAc 1200K

and 1200

2 T2 = - = 600 K

This results in the temperature values shown in Fig 5.2 for the engine and reservoirs.

Efficiency of an internally reversible heat engine 89

7' QH = 400 units

1' Q, = 200 units

Fig. 5.2 Example of internally reversible heat engine operating between reservoirs at TH = 1600 K and T, = 400 K with U,A,/U,A, = 1

The efficiency of the Camot cycle operating between the reservoirs would have been 7;lm = 1 - 400/1600 = 0.75 but the efficiency of this engine is 7;lm = 1 - 600/1200 = 0.50. Thus an engine which delivers maximum power is si&icantly less efficient than the Camot engine.

It is possible to derive relationships for the intermediate temperatures. Equation (5.9) gives

and eqn. (5.8) defines T2 as

T2 = - Qc +Tc UC Ac

Also Q, = Q, - W , and then

w Qc - - Q, -

Qc - (t - TCIT,)

UnAn TH UHAH TH UUAH TH Substituting from eqns (5.10) and (5.11) gives

- UH AH TH

(l+Z)

( 22) Substituting in eqn (5.8) gives

UHA, T , (Z - TCIT,) T2 = + Tc

UcAc l+-

U,A, T, (Z - t2)

= Tii2{ U, AH T;'~ + U,

+ Tc - - UcAc + &AH

I UHAH + UCAC

(5.8)

(5.13)

(5.14)

90 EfJiciency of heat engines at maximum power

Similarly

T u A T ’ / ~ + U~A,T$~

Un An -+ UcAc (5.15)

To be able to compare the effect of varying the resistances it is necessary to maintain the total resistance to heat transfer at the same value. For example, let

UHAH + UcAc = 2

Then, if UHAH/UcAc = 1 (as in the previous example), UHAH = 1.

UnAJUcAc = 112. This gives Consider the effect of having a high resistance to the high temperature reservoir, e.g.

Then

Then

Q c - -- UcAc

1600 (’”-- 1/4) (1 - 1/2) = 267 units, giving W = 177.8 units (1 + 1/2)/2

= 534 units and Q, = 355.6 units w

uti AH (1 -

(QH - W ) = 133.5 units, giving Qc = 177.8 units Qc *L- 1

- - UHAH 2 ~ U H A H

Hence

1 T, = - (133.5 + 400) = 2 x 533.5 = 1067K, and T , = 533.5 K.

z

If the resistance to the low temperature reservoir is high, i.e. U,AH/UcAc = 2, the situation changes, as shown below. First,

2 4

1+1/2 3/2 3 - -

2 UnAn = ~ - - - -

giving

= 133 units (1/2 - 1/4) (1 - 1/2) -- w - 1600

UnAn (1 + 2)/2

which results in W=177.8 kW and

-- Qn - 267 units, or Q = 355.6 kW UHAH

EfSiciency of an internally reversible heat engine 91

Then 2

-- ( QH - W ) = 267 units, giving Qc = 200 units Qc 2Qc -- UcAc UHAH UHAH

Hence

T I = ' [ A + T c ] = 2 ( 2 6 7 + 4 W ) = 1334K,andT2=667K UcAc

These results are shown graphically in Fig 5.3.

T, 120c

T* 600 T

1067 & K

533 k K

133

66;

T, = 400 K T Fig. 5.3 The effect of heat transfer parameters on engine temperatures for a heat engine operating

between reservoirs at TH = 1600 K and Tc = 400 K

Comparing the power outputs based on the same total resistance, i.e. UHAH + UcAc = 2, gives the following table.

UHAHluCAc Q H W Qc

1 400 200 200 112 355.6 177.7 177.7 2 355.6 177.7 177.7

It can be seen that the optimum system is one in which the high and low temperature resistances are equal. In this case the entropy generation (per unit of work) of the universe is minimised, as shown in the table below.

UHAHIUCAC QJTH QHIT, ASHIW QcIT, QcITc AScIW ZASlW

1 -0.25 +0.333 +4.17 x -0.333 + O S +4.17 x +8.34 x 112 -0.225 +0.333 +6.23 x -0.333 0.4443 +6.24 x +0.00125 2 -0.225 +0.266 +2.48 x -0.266 0.4443 +10.00 x +0.00125

92 EfSiciency of heat engines at maximum power

5.2 Efficiency of combined cycle internally reversible heat engines when producing maximum power output One way of improving the overall efficiency of power production between two reservoirs is to use two engines. For example, a gas turbine and steam turbine can be used in series to make the most effective use of the available temperature drop. Such a power plant is referred to as a combined cycle gas turbine, and this type of generating system was introduced in Chapter 3 in connection with pinch technology. These plants can be examined in the following way, based on the two heat engines in series shown in Fig 5.4.

Fig. 5.4 Two internally reversible engines in series forming a combined cycle device operating between two reservoirs at TH and T,

In this case the product UA will be replaced by a ‘conductivity’ C to simplify the notation. Then

Also, for E,

(5.16) (5.17) (5.18)

(5.19)

and for E,

Let z,

(5.20) _ - Q 2 Qc -- 7 . 3 7-4

= T2/T, and t2 = T4/T,; then rearranging eqns (5.16), (5.17) and (5.18) gives

T2 = Q2/c2 i- T3

= TH - QH/CH (5.21) (5.22)

Eficiency of combined cycle internally reversible heat engine 93

and

T4 = Q C K C + Tc

Also

and

Similarly

From eqns (5.16) and (5.26)

1 Q 1 Q c Q H = C n T H - - L + - -+Tc [ T I ( C2 t 2 { T c })] Rearranging eqn (5.27) gives

which can be written as

The power output of engine E , can be obtained from eqn (5.29) as

In a similar manner Wc/CHTH can also be evaluated as

(5.23)

(5.24)

(5.25)

(5.26)

(5.27)

(5.28)

(5.29)

(5.30)

(5.31)

(5.32)

94 EfSiciency of heat engines at maximum power

It can be seen from eqns (5.30) and (5.32) that it is possible to split the power output of the two engines in an arbitrary manner, dependent on the temperature drops across each engine. The ratio of work output of the two engines is

w, 1 (1 -rl)

w c r1 (1 -r2) - (5.33)

This shows that if the temperature ratios across the high temperature and the low temperature engines are equal (i.e. t, = r2 ) , the work output of the low temperature engine will be

ci/, = T1WH (5.34)

Hence the work output of the low temperature engine will be lower than that of the high temperature engine for the same temperature ratio. The reason comes directly from eqns (5.30) and (5.31), which show that the work output of an engine is directly proportional to the temperature of the ‘heat’ at entry.

The power output of a combined cycle power plant is the sum of the power of the individual engines, hence

This may be reduced to

wcc CH TH

- t,r21

The efficiency of the combined cycle is defined by

- 1 - vth=-- wcc

Q H

Equation (5.36) shows that the expression for

(5.35)

(5.36)

the efficiency of the combined cycle engine at maximum power output is similar to that for the efficiency of a single heat engine, except that in this case the temperature ratio of the single engine is replaced by the product of the two temperature ratios. If the combined cycle device is considered to be two endoreversible heat engines connected by a perfect conductor (i.e. the resistance between the engines is zero; C, = m) then T3 = T,, and eqn (5.36) becomes

(5.37)

The efficiency given in eqn (5.37) is the same efficiency as would be achieved by a single endoreversible heat engine operating between the same two temperature limits, and what would be expected if there was no resistance between the two engines in the combined cycle plant.

Eflciency of combined cycle internally reversible heat engine 95

To determine the efficiency of the combined cycle plant, composed of endoreversible heat engines, producing maximum power output requires the evaluation of the maxima of the surface WE plotted against the independent variables z1 and t2. It is difficult to obtain a mathematical expression for this and so the maximum work will be evaluated for some arbitrary conditions to demonstrate the necessary conditions. This will be done based on the following assumptions:

temperatures: TH = 1600 K, conductivities: CH = C2 = Cc = 1

T, = 400 K

1' Qn AI Qn

'' Q i $c

T3 -

T4 - r

l' Qc

I' Qc

96 Eficiency of heat engines at maximum power

While these assumptions are arbitrary, it can be shown that the results obtained are logical and general. The variation of maximum power output with temperature ratios across the high and low temperature engines is shown in Fig 5.5. It can be seen that the maximum power occurs along a ridge which goes across the base plane. Examination shows that, in this case, this obeys the equation zlzz = 0.5 = .ITc'/7;1. Hence, the efficiency of a combined cycle heat engine operating at maximum power output is the same as the efficiency obtainable from a single heat engine operating between the same two reservoirs. This solution is quite logical, otherwise it would be possible to arrange heat engines as in Fig 5.6, and produce net work output while transferring energy with a single reservoir.

Thus the variation of z1 with z2 to produce maximum power output is shown in Fig 5.7: all of these combinations result in the product being 0.5.

2 06

6 055 ' 0 5 ' 0 5 055 06 065 0 7 075 0 8 085 09 095 1

Temperature ratio of high temperature engine

Fig. 5.7 Variations in temperature ratio of high and low temperature heat engines, which produce maximum power output

5.3 Concluding remarks

A new method for assessing the potential thermal efficiency of heat engines has been introduced. This is based on the engine operating at its maximum power output, and it is shown that the thermal efficiency is significantly lower than that of an engine operating on a Carnot cycle between the same temperature limits. The loss of efficiency is due to external irreversibilities which are present in all devices producing power output.

It has been shown that a combined cycle power plant cannot operate at a higher thermal efficiency than a single cycle plant between the same temperature limits. However, the use of two cycles enables the temperature limits to be widened, and thus better efficiencies are achieved.

PROBLEMS

1 Explain why the Carnot cycle efficiency is unrealistically high for a real engine. Introducing the concept of external irreversibility, evaluate the efficiency of an endoreversible engine at maximum power output.

Problems 97

Consider a heat engine connected to a high temperature reservoir at TH = 1200 K and a low temperature one at T, = 300 K. If the heat transfer conductances from the reservoirs to the engine are in the ratio (UA),/(UA), = CH/CC = 2, evaluate the following:

0 the maximum Carnot cycle efficiency; 0 the work output of the Carnot cycle; 0 the engine efficiency at maximum power output; 0 the maximum power output, W / C ~ ; 0 the maximum and minimum temperatures of the working fluid at maximum power

Derive all the necessary equations, but assume the Carnot efficiency, 17 = 1 - T,/TH.

output.

[0.75; 0; 0.5; 100; 1000 K; 500 K]

A heat engine operates between two finite reservoirs, initially at 800 K and 200 K respectively. The temperature of the hot reservoir falls by 1 K for each kJ extracted from it, while the temperature of the cold reservoir rises by 1 K for each kJ added. What is the maximum work output from the engine as the reservoir temperatures equalise? Is the equalisation temperature for maximum work a higher or lower limit of the equalisation temperature?

[200 kJ; lower]

Closed cycle gas turbines operate on the infernally reversible Joule cycle with an efficiency of

where rp = pressure ratio of the turbine, and K = ratio of specific heats, c,/c, = 1.4. This equation significantly overestimates the efficiency of the cycle when external

irreversibilities are taken into account. Shown in Fig P5.3 is a T-s diagram for a closed

I, ,,’p2 = 12 bar

_,*’ p, = 1 bar

Entropy, S

Fig. P5.3 Temperature-entropy diagram for Joule cycle

98 Eficiency of heat engines at maximum power

cycle gas turbine receiving energy from a high temperature reservoir at TH = 1200 K and rejecting energy to a low temperature reservoir at T, = 400 K.

(i) Evaluate the Carnot efficiency and compare this with the Joule efficiency: explain why the Joule efficiency is the lower.

(ii) Calculate the ratio of the gas turbine cycle work to the energy delivered from the high temperature reservoir (QH for the Carnot cycle). This ratio is less than the Joule efficiency - explain why in terms of unavailable energy.

(iii) Calculate the external irreversibilities and describe how these may be reduced to enable the Joule efficiency to be achieved.

(iv) What are the mean temperatures of energy addition and rejection in the Joule cycle? What would be the thermal efficiency of a Carnot cycle based on these mean temperatures?

[0.667; 0.508; 0.4212; ZH/cP=79.9; 994 K; 489 K]

4 Explain why the Carnot cycle overestimates the thermal efficiency achievable from an engine producing power output. Discuss why external irreversibility reduces the effective temperature ratio of an endoreversible engine.

Show that the thermal efficiency at maximum power output of an endoreversible engine executing an Otto cycle is

where TH = maximum temperature of the cycle and T, = minimum temperature of the cycle.

5 The operating processes of a spark-ignition engine can be represented by the Otto cycle, which is internally reversible and gives a thermal efficiency of

where r = volumetric compression ratio; K = ratio of specific heats, cp/c,. An Otto cycle is depicted in Fig P5.5, and the temperatures of the two reservoirs

associated with the cycle are shown as TH and T,. The thermal efficiency of a Carnot cycle operatirig between these twc. reservoirs is q = 1 - T,/TH. This value is signifi- cantly higher than that of the Otto cycle operating between the same reservoirs. Show the ratio of net work output for the Otto cycle to the energy transferred from the high temperature reservoir for the Carnot cycle, QH, is

where T2 and T3 are defined in Fig P5.5.

Problems 99

Entropy, S

Fig. P5.5 Temperature-entropy diagram for Otto cycle

Explain why this value of differs from that of the Otto cycle, and discuss the significance of the term

T3 - T2 T3 In ( TdT2 )

Evaluate the entropy change required at the high temperature reservoir to supply the Otto cycle in terms of the entropy span, s3 - s2, of the Otto cycle.

[As, = C , T ~ ( ~ * ~ ~ ~ ~ - l)/&i]

6 It is required to specify an ideal closed cycle gas turbine to produce electricity for a process plant. The first specification requires that the turbine produces the maximum work output possible between the peak temperature of 1200 K and the inlet tempera- ture 300 K. The customer then feels that the efficiency of the turbine could be improved by

(i) incorporating a heat exchanger; (ii) introducing reheat to the turbine by splitting the turbine pressure ratio such

that the pressure ratio of each stage is the square root of the compressor pressure ratio.

Evaluate the basic cycle efficiency, and then evaluate separately the effects of the heat exchange and reheat. If these approaches have not increased the efficiency, propose another method by which the gas turbine performance might be improved (between the same temperature limits), without reducing the work output; evaluate the thermal efficiency of the proposed plant. The ratio of specific heats may be taken as K = 1.4.

[OS; 0.4233; 0.5731 ]