**3. Case study**

In order to analyse the above theories, the consequences have been analysed on the Shahid Rajaii power plant in Qazvin of Iran. This power plant has an installed capacity of 1000 MW electrical energy, which consists of four 250 MW steam-cycle units (Rankin cycle with reheating and recycling) and has been working since 1994. The major fuel for the plant is the natural gas and is augmented with diesel fuel.

The Shahid Rajaii power plant consists of three turbines: high pressure, medium pressure, and low pressure. The 11-stage high pressure turbine has Curtis stage. The number of the stages in the medium pressure turbine is 11 reactionary stages and in the lower pressure, 2×5 reactionary stages. All of the turbines have a common shaft with a speed of 3000 rpm. The boiler is a natural circulation type in which there is a drum with no top. Other properties of the boiler are that the super-heater is three-staged and that the reheater and the economizer are both two-staged. Figure 2 illustrates the plant diagram overlooking the boiler furnace, cooling towers, attachment (circulation and discharge pumps, blowers, etc), turbine glands condenser and regulator lands, expansion valves and governors and feed water tanks.

In Table 1, properties of water and vapour in the main parts of the cycle have been shown.

Maximum losses of cycle water in this plant are 5 kg/hr, which is negligible due to the minute amount. In analyzing the cycle and drawing diagrams, it is assumed that the temperature is 30 *C* , pressure is 90*kPa* and relative humidity is 30% as environmental conditions. Other assumptions are:


With Figure 2, the conversion equation and energy balance of boiler will be written as (Jordan, 1997):


$$
\dot{m}\_1 = \dot{m}\_2 \text{ and } \dot{m}\_3 = \dot{m}\_4 \tag{19}
$$

Table 1. Properties of water and vapour in cycle

Exergy, the Potential Work 257

*w Qin net* <sup>0</sup> *gen lost*

*w net II Qin Qin E W E E*

Exerting energy and exergy balance equations for the plant cycle, and calculating the energy

As can be seen in Figure 3, under the maximum load, the exergy efficiency is 60.78% and the energy efficiency is 41.38%, relative to different minimum loads. Therefore, boiler analysis is

Using energy balance, which is the basis of exergy balance, and implementing equations (9) and (13) and assuming the warm source temperature to be 950 K, the results of exergy lost

In part one, the power plant efficiency has been calculated, overlooking the boiler combustion process and losses under different loads and Figure 3 was therefore mapped out. In order to more accurately calculate the efficiencies, it is necessary to consider the combustion process. The energy efficiency of the plan is the amount of produced net work divided by the fuel energy. In Table 2, percentage of mass for both Natural gas and Diesel

Element Natural gas Diesel fuel *C* 75.624 85 *H* 23.225 12 *O* 0 0.4 *N* 0.206 0.2 *S* 0 2.4 *Ash* 0 0 *Moisure* 0 0 *Co2* 0.945 0

> *net <sup>I</sup> f W m LHV*

where LHV is the fuel low heating value. The exergy efficiency of the plant is the amount of

Table 2. Percentage of mass for both natural gas and diesel fuel

outgoing exergy (produced net work) divided by the fuel exergy.

and efficiency of all components of the plant cycle are shown in Table 5.

*WEE net Qin w* (22)

(24)

(25)

*E E W TS* (23)

If we use the lost work law for the closed cycle:

and efficiencies, Figure 3 consequently results.

**3.1 Analysing the different elements of the cycle** 

**3.2 Energy and exergy efficiencies of the plant** 

fuel in this study has been shown.

Exergy efficiency of plant is:

done at maximum load.

Fig. 2. The diagram of flow cycle of the plant

$$
\dot{Q}\_{\rm in} = \dot{m}\_1 \left( h\_2 - h\_1 \right) + \dot{m}\_4 \left( h\_4 - h\_3 \right) \tag{20}
$$

Energy efficiency is written as (Jordan, 1997):

$$
\eta\_l = \frac{\dot{W}\_{\text{net}}}{\dot{Q}\_{\text{in}}} \tag{21}
$$

Fig. 3. The diagram of Qazvin power plant efficiency under different loads (natural gas)

*Q mh h mh h in* 12 1 44 3 (20)

(21)

*net*

*in W Q*

*I*

Fig. 3. The diagram of Qazvin power plant efficiency under different loads (natural gas)

Fig. 2. The diagram of flow cycle of the plant

Energy efficiency is written as (Jordan, 1997):

If we use the lost work law for the closed cycle:

$$
\dot{\mathcal{W}}\_{net} = \dot{E}\_{Qin} - \dot{E}\_w \tag{22}
$$

$$\left(\dot{E}\_w\right)\_{lost} = \dot{E}\_{Qin} - \dot{\mathcal{W}}\_{net} = T\_0 \dot{S}\_{gen} \tag{23}$$

Exergy efficiency of plant is:

$$
\eta\_{\rm II} = \frac{\dot{E}\_w}{\dot{E}\_{\rm Qi}} = \frac{\dot{\mathcal{W}}\_{net}}{\dot{E}\_{\rm Qi}} \tag{24}
$$

Exerting energy and exergy balance equations for the plant cycle, and calculating the energy and efficiencies, Figure 3 consequently results.

As can be seen in Figure 3, under the maximum load, the exergy efficiency is 60.78% and the energy efficiency is 41.38%, relative to different minimum loads. Therefore, boiler analysis is done at maximum load.

### **3.1 Analysing the different elements of the cycle**

Using energy balance, which is the basis of exergy balance, and implementing equations (9) and (13) and assuming the warm source temperature to be 950 K, the results of exergy lost and efficiency of all components of the plant cycle are shown in Table 5.

#### **3.2 Energy and exergy efficiencies of the plant**

In part one, the power plant efficiency has been calculated, overlooking the boiler combustion process and losses under different loads and Figure 3 was therefore mapped out. In order to more accurately calculate the efficiencies, it is necessary to consider the combustion process. The energy efficiency of the plan is the amount of produced net work divided by the fuel energy. In Table 2, percentage of mass for both Natural gas and Diesel fuel in this study has been shown.


Table 2. Percentage of mass for both natural gas and diesel fuel

$$
\eta\_l = \frac{\dot{\mathcal{W}}\_{net}}{\dot{m}\_f \times LHV} \tag{25}
$$

where LHV is the fuel low heating value. The exergy efficiency of the plant is the amount of outgoing exergy (produced net work) divided by the fuel exergy.

$$
\eta\_{\rm II} = \frac{\dot{E}\_w}{\dot{m}\_f e\_{ch,f}} \tag{26}
$$

Exergy, the Potential Work 259

As is obviously seen, the heating efficiency of the power plant changes from 36.68% to 37.45% and the exergy efficiency from 33.5% to 36.1%, when natural gas is replaced by diesel

The boiler of this plant is designed based on the natural circulation, and high pressure cold water flow furnace and the water pipes have been appointed vertically. The design pressure of the boiler is 172 kg/cm2, the design pressure of the reheating system is 46 kg/cm2 and the capacity is 840 ton/hrs. Two centrifugal fans (forced draught fan (FDF)) provide the needed

The boiler is modeled for thermodynamic analysis. The air and gas fans, discharge pumps, and generally, the utilities which are work consuming are not considered in the model.

the de-super-heaters within the control volume have been ignored. The heat losses to the surrounding environment are introduced as *Qout B*. . The energy and exergy balance of the

, , 0 *f f m B a a g g w w w w out*

, , , , , , 0, 0 *f ch f m B a x g g x g w x w w x w gen B*

In Table 3, thermodynamic properties of water and vapour for both Natural gas and Diesel

vapour

Table 3. Thermodynamic properties of incoming water and outgoing vapour for the boiler at

*T C*( ) 247.7 538 358.1 538 *P kPa* ( ) 15640 13730 3820 3660 *h kj kg* (/ ) 1075.6 3430.2 3116.2 3534.9 (/ ) *xe kj kg* 241.9 1439.3 1109.8 1345.9 *m kg h* ( /) 840000 840000 751210 751210 *T C*( ) 247.4 538 357.1 538 *P kPa* ( ) 15640 13730 3800 3640 *h kj kg* (/ ) 1074.7 3430.2 3115.5 3535.0 (/ ) *xe kj kg* 241.8 1439.0 1109.1 1346.0 *m kg h* ( /) 840000 840000 747010 747010

Indices a, g and respectively used for air, gas (combustion products) and water (vapour).

*in out mh W mh mh m h m h Q* **<sup>B</sup>** (32)

*in out m e W me m e m e m e TS* (33)

> Hot reheated vapour

. Also the heaters and

Cold reheated vapour

Their effect is the total work which enters the control volume , ( ) *Wm B*

boiler referring to the equations (1) and (13) are written as:

The energy and exergy efficiencies of the boiler are defined as:

fuel property Feed water Super heated

fuel. Therefore the exergy efficiency change is greater than that of energy efficiency.

**3.3 Analysing the boiler** 

air for the combustion.

fuel have been summarized.

maximum load

**Natural Gas Diesel Fuel** 

Where, *ch*. *<sup>f</sup> e* is the specific chemical exergy of the fuel.

The natural gas and diesel fuel consumption are respectively 50010 kg/hr and 59130 kg/hr under the maximum load. The low heating values of the natural gas and diesel fuel are 41597 kj/kg and 48588 kj/kg. Assuming the natural gas as a perfect gas and using the tables of standard chemical exergy, the chemical exergy of the natural gas is calculated as (Szargut et al., 1988):

$$e^{0}\_{\text{ch.NG}} = \sum Y\_i e^{0}\_{\text{ch. i}} = 50403 \frac{k\dot{j}}{kg} \tag{27}$$

And implementing the Szargut method, the chemical exergy of diesel fuel is calculated to be 45540 kj/kg (Szargut et al., 1988):

$$\begin{aligned} e\_{ch.ol} &= \left( LHV + \dot{m}\_{in\_f} h\_{fg} \right) \\ &\times \left[ 1.0401 + 0.1728 \frac{H}{C} + 0.0432 \frac{O}{C} + 0.2196 \frac{S}{C} \left( 1 - 2.0628 \frac{H}{C} \right) \right] \end{aligned} \tag{28}$$

 Where LHV is the fuel's low heating value, *fg h* and *mm*, *<sup>f</sup>* are the vaporization temperature of the hot water and the mass of the moisture content and S/C, H/C and O/C are the mass ratio of sulphur, hydrogen and oxygen, to carbon, respectively. The electric power needed for the attachments of the boiler such as fans and pumps is 3.83 MW or 4.28 MW for natural gas and diesel fuels. Feed water pumps and the condenser and other helping elements of the plant also respectively use 9.926 MW and 70.06 MW of the electrical energy. So the produced net work will be:

$$
\dot{W}\_{\text{net},\text{NG}} = 263.53 - 6.926 - 3.83 = 252.774 \,\text{MW} \tag{29a}
$$

$$
\dot{W}\_{\text{net},oil} = 261.95 - 7.06 - 4.28 = 250.61 \,\text{MW} \tag{29b}
$$

The heating and exergy efficiencies of the plant using the two fuels will be:

$$
\eta\_{I, \text{NG}} = \frac{252.774}{\left(50010 / \text{ } 3600\right) \times 11605 \left(4.1868\right)} = 37.45\% \tag{30a}
$$

$$\eta\_{\text{l\\_oil}} = \frac{250.610}{\left(59130 / \text{\textdegree 3600} \right) \times 41597} = 36.68\,\text{\textdegree} \tag{30b}$$

$$
\eta\_{\rm II,NG} = \frac{252.774}{\left(50010 \,/\, 3600\right) \times 50403} = 36.10\,\text{\AA}\tag{31a}
$$

$$
\eta\_{\text{ill\\_oil}} = \frac{250.610}{\left(59130 / \text{\AA} 600\right) \times 45540} = 33.50\% \tag{31b}
$$

*<sup>w</sup> II f ch f E m e*

The natural gas and diesel fuel consumption are respectively 50010 kg/hr and 59130 kg/hr under the maximum load. The low heating values of the natural gas and diesel fuel are 41597 kj/kg and 48588 kj/kg. Assuming the natural gas as a perfect gas and using the tables of standard chemical exergy, the chemical exergy of the natural gas is calculated as (Szargut

> . . 50403 *ch NG i ch i kj e Ye*

And implementing the Szargut method, the chemical exergy of diesel fuel is calculated to be

1.0401 0.1728 0.0432 0.2196 1 2.0628

 Where LHV is the fuel's low heating value, *fg h* and *mm*, *<sup>f</sup>* are the vaporization temperature of the hot water and the mass of the moisture content and S/C, H/C and O/C are the mass ratio of sulphur, hydrogen and oxygen, to carbon, respectively. The electric power needed for the attachments of the boiler such as fans and pumps is 3.83 MW or 4.28 MW for natural gas and diesel fuels. Feed water pumps and the condenser and other helping elements of the plant also respectively use 9.926 MW and 70.06 MW of the electrical energy. So the

, 263.53 6.926 3.83 252.774 *<sup>W</sup>*

, 261.95 7.06 4.28 250.61 *<sup>W</sup>*

,

50010 / 3600 11605 4.1868

59130 / 3600 41597

50010 /3600 50403

59130 / 3600 45540

The heating and exergy efficiencies of the plant using the two fuels will be:

,

,

,

 

*HOS H CCC C*

*net NG MW* (29a)

*net oil MW* (29b)

252.774 37.45%

250.610 36.68%

252.774 36.10%

250.610 33.50%

*I NG* (30a)

*I oil* (30b)

*II NG* (31a)

*II oil* (31b)

0 0

Where, *ch*. *<sup>f</sup> e* is the specific chemical exergy of the fuel.

. ,

*ch oil in f fg e LHV m h*

et al., 1988):

45540 kj/kg (Szargut et al., 1988):

produced net work will be:

.

(26)

*kg* (27)

(28)

As is obviously seen, the heating efficiency of the power plant changes from 36.68% to 37.45% and the exergy efficiency from 33.5% to 36.1%, when natural gas is replaced by diesel fuel. Therefore the exergy efficiency change is greater than that of energy efficiency.

### **3.3 Analysing the boiler**

The boiler of this plant is designed based on the natural circulation, and high pressure cold water flow furnace and the water pipes have been appointed vertically. The design pressure of the boiler is 172 kg/cm2, the design pressure of the reheating system is 46 kg/cm2 and the capacity is 840 ton/hrs. Two centrifugal fans (forced draught fan (FDF)) provide the needed air for the combustion.

The boiler is modeled for thermodynamic analysis. The air and gas fans, discharge pumps, and generally, the utilities which are work consuming are not considered in the model. Their effect is the total work which enters the control volume , ( ) *Wm B* . Also the heaters and the de-super-heaters within the control volume have been ignored. The heat losses to the surrounding environment are introduced as *Qout B*. . The energy and exergy balance of the boiler referring to the equations (1) and (13) are written as:

$$
\dot{m}\_f \mathbf{h}\_f + \dot{W}\_{\dot{m},B} + \dot{m}\_d \mathbf{h}\_a - \dot{m}\_g \mathbf{h}\_g + \sum\_{in} \dot{m}\_w \mathbf{h}\_w - \sum\_{out} \dot{m}\_w \mathbf{h}\_w - \dot{Q}\_{out,\mathbf{B}} = \mathbf{0} \tag{32}
$$

$$
\dot{m}\_f e\_{ch,f} + \dot{W}\_{\dot{m},B} + \dot{m}\_a e\_{x,g} - \dot{m}\_g e\_{x,g} + \sum\_{\text{int}} \dot{m}\_w e\_{x,w} - \sum\_{\text{out}} \dot{m}\_w e\_{x,w} - T\_0 \dot{S}\_{\text{gen},B} = 0 \tag{33}
$$

Indices a, g and respectively used for air, gas (combustion products) and water (vapour). The energy and exergy efficiencies of the boiler are defined as:

In Table 3, thermodynamic properties of water and vapour for both Natural gas and Diesel fuel have been summarized.


Table 3. Thermodynamic properties of incoming water and outgoing vapour for the boiler at maximum load

Exergy, the Potential Work 261

*out ww ww in*

*f dry air a in B mh mh*

, ,

Natural gas ( 29.77) *Mdry gas* Diesel fuel

Enthalpy Exergy Enthalpy Exergy

,, , *out wxw wxw in*

*f ch f in B dry air x a me me me w m e*

*m LHV m h w*

In Table 4, the calculations of the enthalpy and exergy of the dry combustion gases, vapour and wet combustion gases have been summarized. Combustion gases do not have CO. Since the chemical exergy of CO is high, the exergy of the combustion products is negligible.

Dry combustion gases 384.8 10.15 409.0 17.42 Vapour 740.4 314.30 801.6 417.19 Wet combustion gases 474.3 48.15 472.3 50.34 Table 4. Thermodynamic properties of dry and wet combustion gases and vapour outgoing

Two important processes happen in the boiler: Combustion and heat transfer. Therefore the internal exergy losses of the boiler 0 , ( ) *T Sgen B* are the losses of both exergy and heat transfer. Of course, there is a small exergy losses caused by friction which is calculated in the exergy caused by heat transfer. In the boiler, the exergy losses caused by friction have two different reasons; one is the pressure losses of the combusting gases (at most 4.4 kPa) which is to be neglected and the other is the pressure losses of the actuating fluid. These kinds of exergy losses, as was previously mentioned in the first part, are negligible compared to the losses due to heat transfer. Thus, there will be no specific analysis of friction; these two kinds of

The result of calculating the last four equations is briefly shown in Figure 4.

exergy losses together called the exergy losses due to heat transfer.

 

,

(34)

(35)

( 30.36) *Mdry gas*

,

,

*II B*

Fuel

from pre-heater

**3.4 The boiler processes** 

Fig. 5. The model of the boiler furnace

*I B*

Fig. 4. The diagram of boiler exergy losses and efficiency

Fig. 4. The diagram of boiler exergy losses and efficiency

$$\eta\_{l\_{\perp}B} = \frac{\sum\_{\text{out}} \dot{m}\_w h\_w - \sum\_{\text{in}} \dot{m}\_w h\_w}{\dot{m}\_f LHV + \dot{m}\_{dry\,air} h\_a + \dot{w}\_{\dot{m}\_a B}} \tag{34}$$

$$
\eta\_{II,B} = \frac{\sum\_{out} \dot{m}\_w e\_{x,w} - \sum\_{in} \dot{m}\_w e\_{x,w}}{\dot{m}\_f e\_{ch,f} + \dot{w}\_{in,B} + \dot{m}\_{dry\ air} e\_{x,a}} \tag{35}
$$

In Table 4, the calculations of the enthalpy and exergy of the dry combustion gases, vapour and wet combustion gases have been summarized. Combustion gases do not have CO. Since the chemical exergy of CO is high, the exergy of the combustion products is negligible.


Table 4. Thermodynamic properties of dry and wet combustion gases and vapour outgoing from pre-heater

The result of calculating the last four equations is briefly shown in Figure 4.

### **3.4 The boiler processes**

Two important processes happen in the boiler: Combustion and heat transfer. Therefore the internal exergy losses of the boiler 0 , ( ) *T Sgen B* are the losses of both exergy and heat transfer. Of course, there is a small exergy losses caused by friction which is calculated in the exergy caused by heat transfer. In the boiler, the exergy losses caused by friction have two different reasons; one is the pressure losses of the combusting gases (at most 4.4 kPa) which is to be neglected and the other is the pressure losses of the actuating fluid. These kinds of exergy losses, as was previously mentioned in the first part, are negligible compared to the losses due to heat transfer. Thus, there will be no specific analysis of friction; these two kinds of exergy losses together called the exergy losses due to heat transfer.

Fig. 5. The model of the boiler furnace

Exergy, the Potential Work 263

All the measured data such as temperature, pressure, flow, etc and also the calculated magnitude of the exergy efficiency and losses were assumed under some specific environmental conditions. But the plant is not always under these specific conditions, and in fact there are some conditions under which the efficiency changes. These are divided into two parts: internal conditions such as excess air and moisture content and the external

An increase in the amount of air in the combustion process can easily decrease the adiabatic temperature of the flame so that the adiabatic exergy of the products is reduced and the exergy losses of the combustion increase. On the other hand, the combustion products have a lower temperature and greater mass flow as they flow inside the boiler, which leads to

The correction factor for the boiler exergy efficiency caused by the internal and external conditions is done using the equation (35). For instance, this factor due to the excess air is:

Fig. 6. The diagram of the exergy loss and efficiency

**3.6 The correction factor of the boiler efficiency** 

lower exergy losses in the heart transfer process.

(environmental) conditions such as temperature and humidity.

### **3.4.1 Combustion**

To study the furnace, we assume that the combustion is in an isolated control volume. According to Figure 5, using the energy balance, we will get the enthalpy of the combustion gases:

$$
\dot{Q}\_{\text{out,comb}} = \dot{m}\_f h\_f + \dot{m}\_a h\_a + \dot{m}\_r h\_r - \dot{m}\_p h\_p,\\
\dot{m} = 0 \tag{36}
$$

By determining the temperature of the combustion products, using the iteration method, we can write the exergy balance equation of the furnace in order to determine the chemical exergy losses of the furnace.

$$T\_0 \dot{S}\_{gen,comb} = \dot{m}\_f e\_{dh,f} + \dot{m}\_a e\_{x,a} + \dot{m}\_r e\_{x,r} - \dot{m}\_p e\_{x,p,ad} \tag{37}$$

Therefore the combustion exergy efficiency will be:

$$
\eta\_{II,comb} = \frac{\dot{m}\_p e\_{x,p,ad}}{\dot{m}\_f e\_{ch,f} + \dot{m}\_r e\_{x,r} + \dot{m}\_a e\_{x,a}} \tag{38}
$$

#### **3.4.2 Heat transfer**

In the boilers, heat transfer to the actuating fluid is classified into four categories:


Exergy losses caused by heat transfer occur in five main parts of the boiler; the evaporator, the economizer, the super-heater, the reheaer and the air preheater. Exergy losses of the main five elements together with the furnace losses are the total losses of the boiler which were mentioned in Section 3.3.

$$T\_0 \dot{S}\_{\text{gen},Q} = T\_0 \dot{S}\_{\text{gen},B} - T\_0 \dot{S}\_{\text{gen},comb} \tag{39}$$

#### **3.5 The boiler elements**

Each element of the boiler is, in fact, a heat exchanger. Therefore, when there is more one inlet or outlet to the heat exchanger, exergy balance is written as follows:

$$T\_0 \dot{S}\_{gen, element} = \sum\_{in} \dot{m}\_{\mathcal{g}} e\_{x,\mathcal{g}} - \sum\_{out} \dot{m}\_{\mathcal{g}} e\_{x,\mathcal{g}} + \dot{m}\_w \left( e\_{x,w,in} - e\_{x,w,out} \right) \tag{40}$$

Therefore the heat transfer exergy efficiency will be:

$$\eta\_{II, element} = \frac{\dot{m}\_w \left( e\_{\mathbf{x}, w, out} - e\_{\mathbf{x}, w, in} \right)}{\sum\_{in} \dot{m}\_{\mathcal{S}} e\_{\mathbf{x}, \mathcal{S}} - \sum\_{out} \dot{m}\_{\mathcal{g}} e\_{\mathbf{x}, \mathcal{g}}} \tag{41}$$

The results of exergy losses and efficiency of each element of the boiler are shown in Figure 6.

To study the furnace, we assume that the combustion is in an isolated control volume. According to Figure 5, using the energy balance, we will get the enthalpy of the combustion

By determining the temperature of the combustion products, using the iteration method, we can write the exergy balance equation of the furnace in order to determine the chemical

, , <sup>0</sup> *Q m h m h mh mh out comb f f a a r r p p ad* (36)

*TS m e me me me* 0 , *gen comb <sup>f</sup> ch*, , , ,, *<sup>f</sup> a xa r xr <sup>p</sup> <sup>x</sup> <sup>p</sup> ad* (37)

(38)

*TS TS TS* 0, 0, 0, *gen Q gen B gen comb* (39)

(41)

, ,

*m e m e me me*

Exergy losses caused by heat transfer occur in five main parts of the boiler; the evaporator, the economizer, the super-heater, the reheaer and the air preheater. Exergy losses of the main five elements together with the furnace losses are the total losses of the boiler which

Each element of the boiler is, in fact, a heat exchanger. Therefore, when there is more one

0 , *gen element <sup>g</sup> <sup>x</sup>*, *g g <sup>x</sup>*, *<sup>g</sup> w x w in x w out* ,, ,,

*T S me me m e e* (40)

,, ,,

*g x g g x g in out*

*me me*

*w x w out x w in*

*me e*

The results of exergy losses and efficiency of each element of the boiler are shown in

, ,

In the boilers, heat transfer to the actuating fluid is classified into four categories:

*f ch f r xr a xa*

, ,, *p x p ad*

**3.4.1 Combustion** 

exergy losses of the furnace.

**3.4.2 Heat transfer** 

were mentioned in Section 3.3.

**3.5 The boiler elements** 

Figure 6.

Therefore the combustion exergy efficiency will be:

heat transfer in the pipes of the furnace walls

heat transfer in the first and secondary reheaters

Therefore the heat transfer exergy efficiency will be:

,

*II element*

,

*II comb*

heat transfer in the pipes of the first and secondary economizers

inlet or outlet to the heat exchanger, exergy balance is written as follows:

*in out*

heat transfer in the first, secondary and final super-heaters

gases:

Fig. 6. The diagram of the exergy loss and efficiency
