11.3 Rankine cycle analysis allowing for irreversible losses

Let us consider an actual Rankine cycle (one with irreversible losses). The purpose of this analysis is to find the component elements of a steam power plant in which the major irreversible losses occur and, with real examples, to estimate the order of magnitude of these irreversible losses.

For example, let us analyze the above Rankine cycle realized for the following initial steam conditions: p₁ =16 670 kPa (170 kgf/cm²), T₁ = 550 °C and p₂ = 4.0 kPa (0.04 kgf/cm²).

The actual cycle of a steam power plant will be analyzed with the aid of the three methods described in Chapter 9: efficiency, entropy calculation of power losses, and exergy analysis. First, let us investigate the irreversible losses in its real Rankine cycle, using the method of efficiencies.

In the first place we should mention the irreversible losses suffered as the steam flows through the turbine blading and nozzles, which are due to the inevitable friction in the boundary layer between the blades and steam and to other hydrodynamic phenomena.

As has already been mentioned, the process of adiabatic friction-resisted flow proceeds with an increase in entropy. The irreversible process of frictional adiabatic expansion was already shown on i-s and T-s diagrams in Fig. 8.12. When the steam at the turbine exhaust end is wet, the temperature at the end of expansion will be the same both in the reversible (T₂) and irreversible (T_2r) processes, since the process of expansion proceeds in both cases to the same pressure p₂, and in the two-phase region (wet steam) the isobar coincides with the isotherm. This can also be seen from Fig. 11.15 representing an actual process of steam expansion in a turbine on i-s and

T-s diagrams.

If the process of steam expansion in the turbine were reversible (absence of frictional and other losses), the entire heat drop would have been converted into kinetic energy, and consequently into turbine work, and in this case: But due to the presence of irreversible losses a smaller amount of work is produced by the turbine in the real process of steam expansion:

(11.11)

In accordance with Eq. 8.56, i_2r is always greater than i₂ and, consequently,

(11.12)

The difference area 1-2-2d-II-I (Fig. 11.15). In this case the internal relative efficiency of the steam turbine will be determined as follows:

(11.13)

Fig. 11.15

The internal relative efficiency of up-to-date high-power steam turbines falls within 0.85 to 0.90.

By analogy, in accordance with Eq. (11.6) the work expended to drive the pump in the case of no irreversible losses is equal to:

and for the real process in the presence of irreversible losses

the enthalpy i_5r is always greater than i₅ and, consequently,

In fact, as was mentioned in Sec. 9.3, the work transferred to the pump from an external source will always be larger in the presence of irreversible losses than the work which would be expended to compress the water in the absence of such losses.

The real adiabatic process, proceeding in a pump, is compared with a reversible process on the i-s and T-s diagrams shown in Fig. 11.16. In accordance with (9.13), the internal relative efficiency of the pump is determined in the following way:

(11.14)

The value for usually falls within 0.85 to 0.90, i.e. it is approximately equal to the internal relative efficiency of the turbine,

Calculating the losses in the cycle of a steam power plant, due to the irreversibility of processes, the losses in the pump are usually ignored: inasmuch as in the process 3-5 the enthalpy of water increases slightly compared with the heat drop through the turbine (process 1-2), the increase in water entropy resulting from the irreversibility of the process of compression in the pump, , is negligible compared with the increase in entropy due to irreversibility in the turbine, , and in other components of the plant[1].

As was shown above, the work performed in the reversible Rankine cycle in the absence of irreversible losses is determined from Eq. (11.7):

Fig. 11.16

or, which is the same,

By analogy, the work done in the real Rankine cycle will be

(11.15)

or, which is the same,

(11.15a)

Thus, taking into account Eqs. (11.13) and (11.14), it follows that

(11.16)

Equation (11.16) yields the following expression for the internal relative efficiency of the turbine-pump set:

(11.17)

As was shown in the preceding example, for the Rankine cycle with p₁ =16 670 kPa (170 kgf/cm²), T₁ = 550 °C and p₂ = 4.0 kPa (0.04 kgf/cm²), we have i₁ = 3438 kJ/kg (821.2 kcal/kg), i₂ = 1945 kJ/kg (464.5 kcal/kg), i₃ = 120 kJ/kg (28.7 kcal/kg), i₅ = 137 kJ/kg (32.7 kcal/kg) and, consequently, = 1493 kJ/kg (356.7 kcal/kg) and = 17 kJ/kg (4.0 kcal/kg). Assuming and , we find that Eqs. (11.13) and (11.14) give:

= 1269 kJ/kg (303.2 kcal/kg) and i_2r = 2169 kJ/kg (518 kcal/kg),

= 18 kJ/kg (4.4 kcal/kg) and i_5r = 139 kJ/kg (33.1 kcal/kg).

Hence,

kJ/kg (53.5 kcal/kg) and kJ/kg (0.4 kcal/kg).

The magnitude of is calculated by Eq. (11.17):

Thus, we are not surprised to find that the relative internal efficiency is virtually equal to that of the turbine, . As was already mentioned [see Eq. (9.18)], this can be traced to the small amount of work done by the pump compared with turbine work. Therefore, it will be considered below that

(11.18)

The internal absolute efficiency of the cycle is

For this cycle turbine efficiency was shown in the preceding section to be 0.46, hence , i.e. 39 % of the heat transferred to the working medium in the cycle is converted into work.

A fraction of that work is lost due to mechanical losses in various components of the turbine (friction in radial and thrust bearings). Also it is expended to drive the oil pump (delivering machine oil to turbine rubbing parts) and to actuate the turbine control system. The magnitude of these work expenditures is characterized by the mechanical efficiency of the turbine, , which is the ratio of the mechanical work transferred by the turbine to the coupled electric generator (denote this work by ) to the work done by the steam as it expands in the turbine (the already known quantity ):

(11.19)

If we now determine the absolute brake thermal efficiency of the turbine plant as

(11.20)

(the work done by the pump is ignored), it is clear form Eq. (11.20) that

(11.21)

Taking into account Eqs. (11.20) and (9.2), we find that

(11.22)

or, substituting the value of from Eq. (9.4), we get:

(11.23)

The mechanical efficiency of up-to-date high-power steam turbines falls within 0.97 to 0.995. Assuming that in our example the mechanical efficiency is 0.97, we get from Eq.(11.22):

Thus, the work transferred to the electric generator is equivalent to 38% of the heat added to the working medium in the cycle.

The work is transferred to the coupling member on the shaft of the electric generator. A fraction of this work is spent to compensate for the electrical and mechanical losses occurring in the electric generator. We shall define the efficiency of the electric generator () by the ratio

(11.24)

where is the work transferred to the external consumer (the electric power transmitted into the power system).

At the present time the efficiency of high-power generators ranges from 0.97 to 0.99. Let us introduce the concept of the absolute electric efficiency of a turbogenerator set:

(11.25)

Reducing the above expression to

and allowing for Eqs. (11.21) and (11.24), we obtain:

(11.26)

or, taking into account (11.23),

(11.27)

(naturally, this efficiency does not account for the losses of heat in the boiler, steam mains, etc).

As applied to the above example, assuming the efficiency of the generator , we obtain from Eq. (11.26):

Thus, only 37% of the heat transferred to the working medium in the cycle is converted into electric power.

Inasmuch as our analysis ignored the work expended to drive the feed pump, the absolute electric efficiency of the plant, determined by the above relationship, does not allow for the work spent to drive the pump. If the efficiency is to be determined taking into account this work, then using Eq. (11.25), we find that the work spent to drive the pump, should be subtracted from .

When we speak of the amount of heat transferred to the working medium in the cycle, we mean the difference in enthalpies (i₁ — i₅), where i₁ is the enthalpy of the steam flowing into the turbine at a pressure p₁ and temperature T₁. However, it should be borne in mind that in the boiler steam is heated to a temperature exceeding T₁; the temperature of steam reduces somewhat as it moves along the steam main from the boiler to the turbine, due to the inevitable heat losses. Let us denote the temperature and enthalpy of the steam at the boiler's steam-outlet valve by and , respectively. It is clear that the efficiency of the steam main, can be determined in the following manner:

(11.28)

In modern steam power plants this efficiency usually falls within 0.98 to 0.99. The loss of heat between the turbine and condenser and from the feed line running from the condenser to the boiler is negligible and is ignored. In this example, i₁ = 3438 kJ/kg (821.2 kcal/kg) and

i_5r = 139 kJ/kg (33.1 kcal/kg). Assuming = 0.99, we obtain from Eq. (11.28):

= 3333 kJ/kg (796.1 kcal/kg). Thus = 3472 kJ/kg (829.2 kcal/kg).

Not all the heat released upon fuel combustion in the boiler furnace is spent to heat feed water and the generated steam. A fraction of this heat is lost due to the inevitable losses in the boiler unit (heat lost with the products of combustion leaving the boiler, and due to incomplete chemical combustion, unconsumed carbon and radiation). It is clear that the degree of perfection of a boiler unit can be characterized by boiler efficiency defined as[2]

(11.29)

where is the heat transferred in the boiler to feedwater and to the generated steam, and q' is the heat released upon fuel combustion in the furnace.

The efficiency of modern boiler units ranges from 0.89 to 0.93, depending on the kind of fuel fired: the small values of the efficiency are characteristic, for instance, of lignites (brown coal), and highest, of natural gas. Assuming a boiler unit efficiency of 0.91 in this example, we obtain from Eq. (11.29): = 3333 kJ/kg (796.1 kcal/kg) and q' = 3663 kJ/kg (874.8 kcal/kg).

It follows from Eqs. (11.28) and (11.29) that

(11.30)

is the quantity of heat which should be obtained in the boiler by burning fuel so that the enthalpy of the working fluid at the boiler steam valve would amount to

The brake thermal efficiency of the entire heat power plant, , must be defined as the ratio of the amount of work transferred to the external consumer (electric power transmitted into the power system), , to the quantity of heat released upon fuel combustion in the boiler furnace, q':

(11.31)

This relationship can take the following form:

Inasmuch as

(here and below i_5r is assumed to be smaller than i₅), taking into account Eqs. (11.25), (11.27) and (11.30), we obtain:

(11.32)

or, which is the same,

(11.33)

This equation is a special case of Eq. (9.20) derived in Chapter 9. Assuming in this example that = 0.99 and = 0.91, we obtain with the aid of Eq. (11.32):

Thus, this heat power plant, operated on the Rankine cycle, converts 33% of the heat released upon fuel combustion in the boiler unit furnace into work transferred to an external consumer. In other words, of the amount of heat q' = 3663 kJ/kg (874.8 kcal/kg) released upon fuel combustion (per 1 kg of steam generated), 1207 kJ/kg (283.3 kcal/kg) is converted into electric power.

It will be recalled that in the above example the thermal efficiency of the reversible cycle is 0.46 (and the thermal efficiency of the reversible Carnot cycle, = 0.63). But due to the presence of irreversible losses the efficiency of the real heat power plant, operated on this cycle, reduces to = 0.33 (i.e. by more than 26% in relation to ). Thus, rather great losses are suffered in actual heat power plants due to irreversibility.

Let us return to Eq. (11.25). Allowing for Eq. (11.2), we find that Eq. (11.25) yields:

One kg of steam, whose enthalpy is i₁ at the turbine exit, does work in the plant which is transferred to an external consumer.

But if D kg of steam flow into the turbine per hour, the amount of electric power generated per hour (or, in other words, the electric power of the generator) will be

(11.34)

It follows from Eq. (11.30) that to generate 1 kg of steam at the necessary initial conditions, the following amount of heat should be released in the boiler furnace:

Correspondingly, to generate D kg of steam per hour, the following amount of heat should be released in the furnace:

(11.35)

If we denote the calorific, or heating, value of fuel (i.e. the amount of fuel released when 1 kg of fuel is burned) by then the quantity of fuel B which must be burned per hour in the boiler furnace to release heat Q' will be

(11.36)

Taking Eq. (11.35) into account, we find that the hourly consumption of fuel in the boiler of the steam power plant is

(11.37)

Replacing in Eq. (11.37) the product using Eq. (11.34), we obtain:

(11.38)

or, allowing for Eq. (11.32),

(11.39)

This relationship gives the amount of fuel B required to ensure the generation of a given electric power (N) with the aid of a steam-turbine power plant operating with an efficiency ; the required fuel consumption will naturally depend on the calorific value of fuel, i.e. on its quality.

The electric power (capacity) of a heat power plant is usually expressed in killowatts (kW). If N is expressed in kW, and in kJ/kg, then to determine the amount of fuel B in kg/h, relation (11.39) should be presented in the following form:

(11.39a)

but if N is expressed in kW and the calorific value in kcal/kg, Eq. (11.39) takes the following form:

(11.39b)

inasmuch as 1 kW ^ 860 kcal/h. A more accurate value is given in Table 2.1.

In practice, the performance of an electric station is often estimated by the specific fuel consumption which is the quantity of fuel burned to generate 1 kW-h of electric power:

(11.40)

The higher the efficiency of a thermopower plant, the smaller the specific fuel consumption.

Knowing the values of turbine efficiency and the efficiencies of all main component elements of the plant, we can determine the loss of heat in each of these elements.

The heat released upon fuel combustion, q', will be assumed as 100%. The losses of heat in the boiler will amount to

(11.41)

In this example = 0.91. Consequently, the ratio amounts to 9 %. The losses of heat in pipelines (steam mains) are

(11.42)

Allowing for Eq. (11.28), we obtain:

(11.43)

inasmuch as

(11.44)

we have

(11.45)

Here = 0.99 and = 1 %. Consequently, as the boiler and pipeline losses amount to 10% of the quantity of heat released upon fuel combustion, the amount of heat

q₁ = = constitutes 90% of q'.

Further, since in the general form the internal absolute efficiency of the cycle is expressed by the relationship

(11.46)

the amount of heat transferred to the low-temperature source during the cycle (i.e. the heat rejected in the condenser) is

(11.47)

or, which is the same,

(11.48)

consequently,

(11.49)

Thus, taking into account formulas (11.30) and (11.2), we get:

(11.50)

In our example = 0.39 and, consequently, = 55%.

Therefore, the heat lost in the boiler and pipelines and transferred to the low-temperature source amounts to 9 + 1 + 55 = 65% of the heat released in the boiler furnace upon fuel combustion. The remaining 35% of that heat is converted into turbine work. From the equation

we obtain:

(11.51)

or, which is the same,

(11.52)

It is clear that the mechanical losses in the turbine amount to

(11.53)

thus, taking Eq. (11.52) into account, we get:

(11.54)

The mechanical efficiency was assumed = 0.97; for this value of it follows from Eq. (11.54) that = 1%.

It follows from Eq. (11.19) that the mechanical work transferred to the turbine-generator shaft is

or, allowing for Eq. (11.52),

(11.55)

(in this example = 34%).

Finally, the electric and mechanical losses in the generator are

(11.56)

Substituting into this expression the value of from Eq. (11.55), we get:

(11.57)

Since in this example the efficiency of the generator is 0.98, we have = 1 %. The work transferred to the external consumer (into the network) is equal to

(11.24a)

Thus, taking equation (11.55) into account, we get:

(11.58)

or, which is the same,

(11.59)

The heat flows of this thermopower plant are illustrated in Fig. 11.17 and plotted in accordance with the results of the above analysis. This diagram, showing the sources of the main heat losses, illustrates well the ideas outlined in this section.

Fig. 11.17

Let us analyze the same Rankine cycle with the aid of the entropy calculation method of the availability loss.

As has been shown in Sec. 9.4, the loss of availability of a system is equal to the sum of the losses in each of the n elements of the system (plant) considered:

It will be recalled that the loss of availability in an element of the system is determined by the relationship

(11.60)

where T₀ is the temperature of surroundings, and the increase in entropy in this system element due to the irreversible processes proceeding in it.

It should be emphasized that in Secs. 11.2 and 11.3 the brake thermal efficiency of the steam turbine plant was compared with the thermal efficiency of a "standard" cycle, i.e. of the reversible Carnot cycle, , realized in the same temperature interval as the Rankine cycle (T₁= 550 °C, T₂ = 28.6 °C in this example). Strictly speaking, it is not quite right to compare this cycle with the Carnot cycle; here the high-temperature source is represented by the furnace gases at a temperature T_h.t. equal to about 2000 °C, and the low-temperature source is the water used as a coolant in the condenser. The temperature of this water is equal to that of the surroundings, T_l.t, and ranges from zero to 20 °C. Therefore, in principle, the effectiveness of real cycles should be compared with the thermal efficiency of a reversible Carnot cycle realized for this temperature interval (T_h.t. to T_l.t). If the upper temperature of the Carnot cycle (550 °C in this example) is lower than that of the high-temperature source and the cycle's lower temperature is higher than that of the low-temperature source, such a Carnot cycle will be irreversible. However, since in actual steam power plants the upper temperature of the working medium is always far below the temperature in the boiler furnace[4], in practice real cycles are compared with reversible Carnot cycles, realized in the interval of temperatures characteristic of the working medium used in this real cycle. In other words, the real cycle is compared with a Carnot cycle which is reversible internally and irreversible externally (see Sec. 9.4). But from the position of the analysis of the availability of a system, as was shown in Sec. 9.4, the actual cycle should be compared with an externally reversible Carnot cycle.

Fig. 11.18

Fig. 11.18 shows the real Rankine cycle on a T-s diagram[5], in which 1-2r represents the process of adiabatic steam expansion in the turbine, taking into account the irreversible frictional losses; 2r-3 is the isobaric-isothermal process of heat rejection in the condenser; 3-5r the adiabatic process in the feed pump, taking into account the irreversible frictional losses; 5r-4-6-0 the isobaric process[6] of heat addition to the cooling water (5r-4), vapour-water mixture (4-6) and to superheated steam (6-0) in the boiler; the curve 0-1 represents in an exaggerating way the drop in steam temperature from to T₁ and the pressure drop from to p₁, taking place on the way from the boiler to the turbine because of losses in the steam main. Because of those losses, when heat is rejected, the entropy of the working medium and the theoretical and actual work of the cycle diminish, as can be seen from Fig. 11.18.

The dotted isothermal line, shown on this diagram, corresponds to the temperatures of the high-temperature source (temperature in the furnace, T_max= T_f).

The temperature of the surroundings is assumed to be equal to the temperature of the cooling water, T₀. For this thermopower plant it is assumed that T_f = 2000 °C and T₀ = 10 °C.

Let us calculate the loss of availability in each component of the power plant, per kilogram of the working medium.

(I) Boiler unit. As before, the quantity of heat released upon fuel combustion in the boiler furnace is denoted by q'. The loss of availability in the boiler has two causes: first, a fraction of the heat q' is lost, and second, the heat realized in the boiler furnace upon fuel combustion is added to the working medium at a considerable difference between the temperature of the gases T_f and the working medium (it changes in the process of heat addition from T_5r to ).

The loss of availability due to the loss of heat is calculated in the following way. The magnitude of the heat loss is calculated from Eq. (11.41):

On the basis of the considerations outlined in Sec. 3.6 it is clear that the increase in system entropy, due to the transfer of heat from the furnace at a temperature T_f to the surroundings at a temperature T₀, will amount to

(11.61)

It follows, in accordance with Eqs. (11.41) and (11.60), that the loss of system availability due to the irreversibility of the process will amount to

(11.62)

In this example,

Let us now determine the loss of system availability resulting from the irreversibility of the process in which the heat released in the furnace is added to the working medium.

The quantity of heat added to the working medium as it is being heated in the boiler is equal, according to equation (11.29), to

As this amount of heat is transferred to the working medium, the entropy of the high-temperature source (burning fuel) diminishes by

(11.63)

(as a first approximation the temperature in the boiler furnace, T_f, is assumed constant), and the entropy of the working medium will increase bv

(11.64)

(see Fig. 11.18). Inasmuch as in the process of heating the temperature of the working medium increases from T_5r to , the change in the entropy of the working medium cannot be calculated with the aid of relationships of the type

which is valid provided in the process of heat addition the temperature of the working medium remains constant. The quantity , however, can be determined with the aid of the Steam Tables, T-s or i-s diagrams plotted on the basis of known values of , and T_5r.

The entire change in system entropy due to the irreversibility of the process of heat addition to the working medium will be

(11.65)

and, accordingly, the loss of system availability during this process is

(11.66)

For the example considered, let us find and , with the aid of Steam Tables. We find the magnitude of knowing the enthalpy of steam in this state = 3472 kJ/kg (829.2 kcal/kg) and steam pressure p₁ = 16 670 kPa (170 kgf/cm²); from the Steam Tables we find that

= 6.5029 kJ/(kg-K) = 1.5532 kcal/(kg-K) (correspondingly,= 562 °C). Knowing

i_5r = 139 kJ/kg (33.1 kcal/kg), we find for the same pressure

s_5r = 0.4241 kJ/(kg-K) = 0.1013 kcal/(kg-K) [T_5r = 29.5 °C].

In accordance with Eq. (11.60), we obtain:

The entire loss of availability, due to the irreversibility of the processes developing in the boiler, is

(11.67)

For this cycle

(II) Steam main. In accordance with Eq. (11.45), the losses of heat in the steam main amount to

Because of these heat losses, the temperature of the steam in the steam main drops from at the entrance to T₁ at the exit.

Inasmuch as the difference between and T₁ is not very great, it may be assumed that the temperature of the steam flowing through the steam main is

The increase in system entropy due to the transfer of heat in the steam main from the steam at a temperature to the surroundings at a temperature T₀ amounts to

(11.68)

It follows that the loss of system availability due to this process amounts to:

(11.69)

In this example = 562 °C and T₁ = 550 °C. Consequently, = 556 °C, and from Eq. (11.69) it follows that

(III) Turbine-generator unit. In the course of frictional adiabatic expansion of steam in a turbine entropy increases. Let us calculate the increase in steam entropy due to the irreversibility of the process of steam expansion in the turbine, In accordance with

Eq. (8.69),

If the exhaust steam (at the turbine exit) is wet (Fig. 11.18), then T₂ = T_2r and Eq. (8.69) yields:

(11.70)

It follows from Eq. (11.70) that

(11.71)

Further, it is easy to obtain from Eq. (11.13) that

(11.72)

From the above relation, we get from Eq. (11.71):

(11.73)

or, which is the same,

(11.73a)

From this it follows that

(11.74)

As can be seen from Eqs. (11.72) and (11.74), the loss of work due to friction appearing as the steam flows through the turbine is greater than the loss of availability This is explained by the fact that the work lost on account of friction, equal to i_2r — i₂, turns into heat and is removed from the turbine at a temperature of T₂ > T₀. It is clear that, in principle, a fraction of that heat can again be converted into work in a cycle realized in the temperature interval from T₂ to T₀. But if T₂ = T₀, the loss of work is equal to the loss of availability.

In this cycle , and

T₂ = 28.6 °C. Substituting these values into Eq. (11.74), we obtain:

Account should also be taken of the losses of availability due to mechanical losses in the turbine and electrical losses in the electric generator (let us denote these losses by and respectively).

It follows from Eq. (11.19) that the mechanical losses in the turbine,

(11.75)

are determined in the following manner:

(11.76)

Thus, taking Eqs. (11.11) into account, we get:

(11.77)

or

(11.78)

By analogy, the mechanical and electrical losses in the generator,

(11.79)

are determined in accordance with Eq. (11.56):

.

Thus, taking into account Eqs. (11.11) and (11.19), we get:

(11.80)

or

(11.81)

The losses and are transferred in the form of heat to component elements of the turbine and generator. This heat is transferred at a constant temperature, since the plant is in steady-state operation. Assuming, as a first approximation, that this temperature is close to that of the surroundings, T₀, we find that the increase in system entropy, due to losses in the turbine and generator, can be determined from the formulas

(11.82)

and

(11.83)

Thus, for the quantities and we obtain:

(11.84)

and

(11.85)

As was already mentioned, for T₂ = T₀ the loss of system availability is equal to the loss of work.

In this example = 0.97, = 0.98 and, consequently, = 38 kJ/kg (9.1 kcal/kg) and = 25 kJ/kg (5.9 kcal/kg).

The total loss of availability traced to the irreversibility of the processes proceeding in the turbine-generator unit can be presented in the following manner:

(11.86)

in this example

kJ/kg (65.2 kcal/kg).

(IV) Condenser. In accordance with Eq. (11.48), the heat removed from the steam undergoing an isobaric-isothermal process in the condenser amounts to

Assuming the rate of flow of cooling water through the condenser to be so high that its temperature T₀ remains practically constant in the condenser, we obtain:

(11.87)

and, accordingly,

(11.88)

In our example = 2169 kJ/kg (518.0 kcal/kg), = 120 kJ/kg (28.7 kcal/kg) and

kJ/kg (30.2 kcal/kg).

(V) Feed pump. The increase in entropy due to the irreversibility of the adiabatic process in the feed pump is calculated in the following way. From Eq. (11.14),

we find that the supplementary increase in entropy because of the heat released due to friction amounts to

(11.89)

Since the temperatures T_5r and T₅ (Fig. 11.16) differ little-from each other (in this example

T₅ = 29.0 °C and T_5r = 29.5 °C[7]), we can write

(11.90)

where

It follows from Eq. (11.90) that the increase in the entropy of water in the pump,
∆, amounts to

(11.91)

Thus, taking Eq. (11.89) into account, we get:

(11.92)

or, which is the same,

(11.92a)

where the theoretical work of the pump.

It follows that

(11.93)

In this example = 29.2 °C, and so

kJ/kg (0.4 kcal/kg).

The values of the losses of availability (power losses) in individual elements of this cycle are given in Table 11.1.

Table 11.1 Losses of availability in a steam power plant

Plant elements	Reasons for loss	Availability loss		Availability loss factor
		kJ/kg	kcal/kg
Boiler unit	Boiler heat losses	288	68.9	8.9
	Irreversible heat exchange	1305	311.7	40.5
Steam main	Heat losses	22	5.3	0.7
Turbine-generator	Irreversible process of steam expansion in turbine	210	50.2	6.5
	Mechanical losses in turbine	38	9.1	1.2
	Mechanical and electrical losses in generator	25	5.9	0.8
Condenser	Irreversible heat exchange	126	30.2	3.9
Pump	Irreversibility of process	2	0.4	0.1
Total		2016	481.7	62.6

The total loss of availability for the entire cycle realized in the plant,

(11.94)

amounts to 2016 kJ/kg (481.7 kcal/kg).

The maximum work which could be obtained from the heat q' involved in the system "high-temperature source—working medium—low-temperature source" represents, as has been already noted in Sec. 9.4, the work of the reversible Carnot cycle realized in the temperature interval between T_f and T₀.

Since heat q' is added to the plant, in accordance with Eq. (9.26), we have

where

In our example

and since q' = 3663 kJ/kg (874.8 kcal/kg), we have:

kJ/kg (770.0 kcal/kg).

From Eq. (9.33),

it follows that in this example the work transferred from the steam power plant (denoted by l_p) is

kJ/kg (288.3 kcal/kg).

This result coincides, of course, with the quantity l_p determined previously with the aid of the method of thermal efficiencies (page 401).

In this example, the availability loss factor, determined from Eq. (9.35), is

and amounts to

and the relative availability factor (or the degree of thermodynamic perfection) for the plant,

is equal to 1—0.63 = 0.37.

Taking these values into account, let us calculate the brake thermal efficiency of the given steam power plant, . In accordance with Eq. (9.50),

we obtain:

which, as could be expected, coincides with the result obtained by the method of efficiencies from Eq. (11.33).

Analyzing the loss of availability in individual elements of the steam power plant, we find that the greatest availability losses (1593 kJ/kg = 380.6 kcal/kg) occur in the boiler unit, where irreversibility is the highest due to the large difference between the temperature of the furnace gases and that of the working fluid. [The losses of availability traced only to this difference, i.e. due to the insufficient utilization of the temperature potential of the heat released upon fuel combustion, amount to = 1305 kJ/kg (3.117 kcal/kg).]

To decrease the availability loss , it is first necessary to decrease the degree of irreversibility involved in the process of heat exchange in the boiler. It is clear that important results can be obtained by diminishing the difference between the temperatures of the products of combustion and of the working medium. In turn, this temperature difference can be diminished either by reducing the temperature of the products of combustion in the furnace or by increasing the mean temperature of the working medium in the process of heat addition. It can be easily shown that the first method fails to give the desired result: with a decrease in combustion temperature in the boiler furnace the loss of availability actually decreases. However, a decrease in combustion temperature involves a drop in system availability by exactly the same value [see Eq. (9.27)].

Therefore it is clear that the availability losses of a system can be diminished only by increasing the temperature of the working medium. However, as was already mentioned, although this is thermodynamically expedient, it involves an increased cost in plant construction, which cannot be justified economically.

Considerable losses of availability take place in the turbine-generator set [they can be reduced by improving the design of the blading (nozzles and blades) and of the mechanical elements of the turbine, and by improving the generator] and in the condenser. The loss of availability in the condenser can be reduced by increasing the difference between the temperature of the condensing steam and that of the cooling water by a further lowering of condenser pressure p₂. However, it ought to be borne in mind that as was already mentioned this will call for larger condenser heat-transfer surfaces and, consequently, will involve greater expenditures in constructing the steam power plant, which are usually unjustified and cannot always be ensured with existing water-supply facilities.

The losses in the steam main are comparatively small; a further reduction involves an improvement of heat insulation and of the hydrodynamic characteristics of the pipeline.

The losses in the pump are negligible.

Let us now analyze the losses of availability in the components of the steam power plant, applying the exergy method.

It will be recalled that, as has been shown in Chapter 9, the exergy e of a flow of the working medium is determined by Eq. (9.56b),

and the exergy e_q of a heat flow q, by (9.60),

with the availability loss of the working medium flowing through the heat apparatus, to which heat q is added simultaneously, being equal [according to Eq. (9.61)] to

where is the work performed by this apparatus and transferred to the external consumer. Let us apply this relationship to each of the components of the steam power plant.

(I) Boiler unit. The water entering the boiler has a temperature T_5r at pressure p₁; the exergy of the water is

(11.95)

The boiler unit also receives heat q' from a high-temperature source (burning fuel) which is at a temperature T_f; the exergy of this heat flow is

(11.96)

The generated steam leaves the boiler at a temperature and a pressure p₁, its exergy is

(11.97)

Inasmuch as no useful work is done in the boiler, in accordance with Eq. (9.62) we have:

(11.98)

While examining the Rankine cycle, it was assumed that p₀ = 98 kPa (1 kgf/cm²) and

T₀ = 10 °C = 283.15 K. At these steam conditions the enthalpy and entropy of water are, respectively, i₀ = 42 kJ/kg (10.1 kcal/kg) and s₀ = 0.1511 kJ/(kg-K) [0.0361 kcal/(kg-K)].

We find with the aid of Steam Tables the value of s_5r for the boiler unit of this power plant:

s_5r = 0.4241 kJ/kg [0.1013 kcal/(kg-K)], we find from Eqs. (11.95) to (11.97) that

kJ/kg (4.5. kcal/kg),

kJ/kg (765.8 kcal/kg),

kJ/kg (389.5 kcal/kg).

Allowing for these values, we obtain from Eq. (11.98) the magnitude of the loss of availability in the boiler unit:

kJ/kg (380.8 kcal/kg).

(II) Steam main. Steam enters the steam main with conditions and and leaves it with conditions p₁ T₁. It is clear that the exergy of steam at the steam-main entry is equal to the exergy

of steam at the boiler exit, or steam value:

and the exergy of steam at the steam-main exit is

(11.99)

The loss of steam availability in the steam main (the steam does no useful work in the pipeline) amounts to

(11.100)

In this example = 1631 kJ/kg (389.5 kcal/kg) and

kJ/kg (384.3 kcal/kg); the entropy of steam, s = 6.4619 kcal/kg-K = 1.5434 kcal/(kg-K), is found from Steam Tables. The loss of availability clue to heat losses in the steam main is equal to

kJ/kg (5.3 kcal/kg).

(III) Turbine-generator unit. The conditions of the steam supplied to the turbine are p₁ and

T₁, and the parameters of the exhaust steam (at the turbine exit) are p₂ and T2r. Accordingly,

and

(11.101)

Inasmuch as the turbine-generator unit produces useful work

in accordance with Eq. (9.57), the loss of availability in the turbine-generator unit amounts to

(11.102)

This quantity accounts for the availability losses due to both the irreversible nature of steam flow through the turbine blading and to frictional losses in the turbine mechanism, and also due to mechanical and electrical losses in the generator.

For this steam power plant

kJ/kg (384.3 kcal/kg)

and

kJ/kg (30.8 kcal/kg);

the entropy of exhaust steam at the turbine exit, s_2r = 7.2063 kJ/(kg-K) = 1.7212 kcal/(kg-K), is found from Steam Tables. Taking into account that = 1207 kJ/kg (283.8 kcal/kg), we find from Eq. (11.102) that

kJ/kg (65.2 kcal/kg).

(IV) Condenser. The exergy of steam flowing from the turbine into the condenser (of the exhaust steam) is

and the exergy of the condensate leaving the condenser is

(11.103)

Inasmuch as no useful work is done in the condenser, the availability loss of the flow in the condenser is

(11.104)

Here = 129 kJ/kg (30.8 kcal/kg) and

kJ/kg (0.6 kcal/kg)

(the entropy of water in the state of saturation, s₃ = s' = 0.4178 kJ/kg = 0.0998 kcal/kg is taken from Steam Tables).

The loss of steam availability in the condenser is

kJ/kg (30.2 kcal/kg).

(V) Feed pump. The exergy of water entering the pump is

and the exergy of water at the pump discharge connection is

To drive the pump work is transferred from an external source, equal to

inasmuch as this work is expended ultimately to increase the enthalpy of feedwater, the amount of work transferred is equivalent to addition of heat. The exergy of this heat, introduced into the pump,

(11.105)

In accordance with Eq. (9.61), the loss of water availability in the feed pump amounts to

(11.106)

where is the work transferred to the pump from the outside (negative work).

In this example = 2.5 kJ/kg (0.6 kcal/kg), = 18.8 kJ/kg

(4.5 kcal/kg), = 18.4 kJ/kg (4.4 kcal/kg), and

kJ/kg (0.3 kcal/kg).

Consequently,

kJ/kg (0.5 kcal/kg).

As can be seen from the results of this analysis, the values of the losses of availability in each element of the plant found by the exergy method coincide, as was to be expected, within the accuracy of the calculation (equal to 0.1 kcal/kg), with the values determined by the entropy method.

The diagram of the flows of exergy for this steam power plant is shown in Fig. 11.19. To a certain extent this diagram resembles, although externally, the diagram of the flows of heat (Fig. 11.17). The exergy of the heat released upon combustion of fuel in the boiler furnace is assumed to be 100%; the diagram shows what fraction of the heat flow is spent as the loss of availability in each particular element of the power plant. It should be emphasized that, as can be seen from the diagram, a fraction (although negligible) of the flow of the exergy returns into the cycle, i.e. the exergy of feed water entering the feed pump. The diagram of the flows of exergy is an illustration of the above calculations.

The methods of analysis of real steam power cycles described above are also applicable to more intricate cycles considered below. We shall restrict ourselves to an analysis of internally reversible cycles realized in these plants, taking irreversibility into account.

Fig. 11.19

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[1] The estimates for and are given below

[2] Feed water enters the boiler with an enthalpy not equal to i_5r but somewhat inferior to it, due to the loss of heat in the feed line, running from the condenser to the boiler. This loss is considered negligible.

[3] The quantity is called the lowest working calorific value of fuel, reduced to the working mass of fuel and taking no account of the heat released upon condensation of water vapour (H₂0) in the combustion products.

[4] In this connection, it will be recalled that in internal combustion engines the temperature of the working medium can be considered virtually equal to that of the high-temperature source (for the combustion products are the working medium).

[5] It will be emphasized again that this representation is hypothetical, since irreversible processes cannot be represented on a diagram of state.

[6] Due to the inevitable hydraulic losses, the pressure of the working medium reduces somewhat as it flows through the flues; however, in an estimate this pressure drop can be ignored, assuming

[7] The magnitude of T_5r is found from the Steam Tables, on the basic of knowing p₁ = 170 kgf/cm²

and i_5r = 33.1 kcal/kg.