ENTROPY METHOD AS CRITERIA FOR ANALYSIS A STEAM POWER PLANT

In this paper a theoretical analysis of South Baghdad and Dura power plant is carried out according to second law of thermodynamic depending on entropy (irreversibility coefficient or lost work) method instead of exergy (availability) method. In the used entropy method. The power plant is divided into main blocks ( boiler, turbine, condenser


INTRODUCTION
With increasing fuel prices and the possibility of diminishing supplies in the years a head, the importance of developing systems which make efficient use of energy is apparent.The second law o f thermodynamic method of analysis is particularly suited for furthering the goal of more efficient energy use, for it identifies the locations, types, and the true magnitudes of energy resources waste and loss, such method can also be used to guide steps taken to reduce inefficiencies.
According to this second law different criteria are defined for analysis the performance of power plants based on the concept of exergy (availability).If all of these criteria are used, they must all give the same results.Although availability pinpointed the real losses of a steam power plant, it is difficult, complex and can not gives direct relationship between component losses and overall efficiency of plant.Thus, the criteria for selecting the best procedure to evaluate thermodynamic analysis should be, best ease of use, best degree of correspondence with the viewpoint and background of intended users and greatest breadth of application.On these grounds, the entropy method (lost work) approach was believed to be superior to other approaches in common use (Seader 1986).
The purpose of this work is to analyze performance of South Baghdad and Dura power plants according to 2 nd law of thermodynamic depended on the concept of entropy method instead of exergy method.Then compare the results with that which obtained previously by other researches (Hashe m and Murad 1998) and (Mathure et.al. 2000) depending on exergy method.

EXERGY METHOD ANALYSIS OF THE THERMAL POWER PLANT
The processes in steam turbine plant are steady flow processes .Where the general form of the exergy value was calculated from the following formula (Has hem and Murad 1998), (Moran 1982) and (Yunus 1994) : For exergy analysis the plant was divided in to the following main blocks: steam boiler, steam turbine, steam condenser, feed water heaters and feed water pump.The exergy losses in each b lock can be defined as follows (Hashe m and Murad 1998) and (Mathure et.al. 2000) :

Steam Boiler
The total exergy losses of boiler are given, This total exergy losses of boiler are divided into three main losses,

a-Combustion Losses
Where, h ex = C pg .T ex , s ex = C pg ln (T ex / T o )

c-Heat Transfer Losses
Exergy losses due to heat transfer (E Lht ), Thus, the total exergy losses of boiler can be calculate from, E Lb = Eq .(2)+ Eq .(3) + Eq .(4) And the second law efficiency of boiler can be calculated as, And 2 nd law efficiency of turbine, Where, T maxsteam temperature at condenser inlet T minsurrounding temperature Whereas, exergy losses through the condenser is equal to the exergy losses by steam.

feed wate r pump
Exergy losses = W inputexergy increasing of working fluid (12 ) Effectiveness of pump = (exergy increase of working fluid)/(W input ) According to 2 nd law steam cycle efficiency is given as, So from Eq. ( 5) η IIb = m s .∆ew /E f , then the 2 nd law efficiency of the power plant is given as, The schematic diagram of one unit of South Baghdad power plant is shown in Fig. 1 .The description data of its steam cycle is given in Table.1

ENTROPY METHOD
According to (Gashteen and Varkevker 1986) and (Yunus 1994) irreversibility losses (lost work) are given as, But for reversible engine, Sub. Eq. ( 16) in Eq. ( 18), then For example for the arbitrary system shown below the irreversibility losses can be calculated as follows, If there is no heat transfer across the boundary ( From Eq.( 21) irreversibility losses can be calculated depending only on the change of entropy.This is why this method was given name of entropy by (Gashteen 1963) .Irreversibility coefficient (Ω) for each component of the power plant is equal to, Ω i = Φ i /(exergy input to the plant) ( And the overall irreversibility coefficient for any power plant, By using entropy method to analyze thermal power plant, the plant was divided into main blocks and irreversibility losses for each block was calculated according to Eqs.( 20) or (21) and then irreversibility coefficient for each block was calc ulated from Eq. ( 22) .The overall irreversibility coefficient of the plant was found from Eq. ( 23) and then from Eq. ( 24) the thermal efficiency of plant can be calculated (Gashteen and Varkevker 1986).
The main blocks of thermal power plants and their irreversibility coe fficient are as the following,

Steam Boiler
The total irreversibility losses of steam boiler are divided into three parts, (a) Exhaust losses (b) combustion losses (c) heat transfer losses.The area abb `a`a represents the heat released during combustion at T c .The equal area cdd `a`c represents the same heat released when combustion occurs at T c `. From Eqs.(28) and (29) it is clear that these losses are increasing when combustion temperature is increasing.Thus, Eq.(26) + Eq.(28) and Eq.(27) + Eq.(29) give the sum of irreversibility losses due to combustion and heat transfer at different combustion temperatures.From which a conclusion can be achieved that irreversibility losses due to combustion and heat transfer are given as, ) .( .Where s 1 &s 2 are the entropy of working fluid at outlet and inlet of boiler respectively Eq.( 30) is valid whatever the temperature of combustion is.Thus, the total irreversibility losses of steam boiler can be calculated as follows, Total irreversibility losses of boiler = Eq.(25) + E q. (30)

Steam Turbine
Irreversibility losses of turbine can be calculated from Eq. ( 21).

Steam Condenser
This irreversibility losses can be calculated from Eq.( 20) in which Q=m s h s

Feed Water Heaters And Pumps
This can be calculated from Eq. ( 21)

Mechanical And Generator Losses
These can be calculated as follows, Mechanical irreversibility losses =w t -η m .wt (31) Generator irreversibility losses =η m .w t -η m .ηg .w t (32) For all above components the irreversibility coefficient (Ω) for each component can be calculated by dividing Φ component by fuel exergy, which is in our calculated equal to m f .C.V.Then, The results obtained from entropy method are summarized in Table .5 and 6 for South Baghdad and Dura power plant respectively.

CONCUSION
 Both exergy and entropy methods give approximately the same results Figs. 3 and 4 ,since the second law of thermodynamic is unambiguous.
 Entropy method (irreversibility) independent on dead state except for the value of T o .Whereas the exergy is determined in relative to restricted dead state, which can be some what misleading.
 Entropy method which is simple and easy to apply is superior to exergy method.Since the former method requires only one property (entropy) to obtain the results, while the later method requires two properties (entropy and enthalpy).
 The entropy method is pinpointing the real losses in each component and giving direct relationship between them and the overall efficiency of the plant.By this the effect of inefficient components can be directly reduced to improve the performance of plant.
 The entropy method show that combustion temperature not posses influence on the irreversibility losses of the boiler.
effectiveness = (exergy gain by surrounding) /(exergy losses by steam through the condenser)

(
Mathur et.al.2000).Whereas, the schematic diagram of one unit of Dura power plant and the description data of its steam cycle are shown in Fig.2 and Table .2,respectively (Has hem and Murad 1998) .Using Tables. 1 and 2 and above equations, exergy analysis of each component was calculated for South Baghdad (Mathur et.al.2000) and Dura (Hashem and Murad 1998)and the results are summarized in Tables.3 and 4 , respectively.
(a) Exhaust losses These losses are calculated as follows, Irreversibility exhaust losses = Q -Q.η b Irreversibility exhaust losses = Q (1-η b ) (25) Where Q= (m s .Δh w )/η b (b) Combustion losses Combustion leads to appearance irreversibility losses which are depended on temperature of combustion.Thus when this temperature increases the irreversibility losses decreases as shown on T-S diagram below : ` < T c , then irreversibility losses at T c are less than that at T c`.These losses are shown on T-s diagram which are equals to area efb `a`e and egd `a`e at T c and T c` respectively.

Fig. 1 :Fig3:
Fig. 1: The Heat Cycle Of The Unit Of South Baghdad Powe r Plant  Hashem H.H, Murad N.K., (1998) Pinpointing the real losses of a steam power plant by using exergy method.(Accepted for publication) in Technical research journal.