UA T ( 1 − T 2 )( 1 − C )

UA T ( 1 − T 2 )( 1 − C )

WC h ph ( T 1 − T 2 ) = Q =

(A.16)

ln 

− CT ( 1 − T 2 ) } /T ( 2 − t 1 ) 

 { T 1 −− t 1 CT ( 1 − T 2 ) } 

UA ( 1 − C )

 { T 1 −− t 1 CT ( 1 − T 2 ) } 

UA ( 1 C )

 = K (A.18)  

 = exp 

 WC h ph  T 1 ( 1 − C ) −+ t 1 CT 2 = K ( T 2 − t 1 ) (A.19)

2 − t 1 ) + t 1 − CT / 2 ] ( 1 − C ) (A.20)

From (A.14a), it may be shown that

= ( K − 1 ) WC h ph ( T 2 − t 1 Q ) (A.21)

Now K, T 2 ,t 1 are known or can be reasonably estimated as the tube geometry, and surface areas are known. Hence, T 1 may be estimated from this and then t 2 and then Q.

Example A.3

Let us apply this procedure to the boiler discussed in Example 3.7 and check the overall boiler performance and efficiency. Start from the economizer end as the field measure- ments are generally more accurate at the cold end.

Solution

Economizer T 2 = economizer exit gas temperature = 154°C and t 1 = feed water in = 116°C

WC h ph =

C = 123 000 0 2815 , × .

WC

c pc

372 Appendix A: Boiler Design and Performance Calculations

U = 40.6 kcal/m 2 h °C and A = 2395 m 2 (from Table 3.13)

=  . 40 6 2395 1 0 3393 × ( − . K  exp )  = . 6 391

From (A.20), T 1 = [6.391 × (154 − 116) + 116 − .34 × 154]/(1 − 0.34) = 464°C, which agrees with the calculations in Chapter 3. One may also compute Q and then t 2 from T 1 ,T 2 and then obtain the LMTD and check the Q.

Q duty of economizer = = 123 000 0 2815 , × . × ( 464 154 − ) = . 10 73 10 × 6 kca ll/h

The specific heat of water = 1.044 kcal/kg °C from the earlier text. Hence, t 2 = 116 + 10,730,000/97,740/1.044 = 221°C

Evaporator Performance Let us use Equation A.13b.

 ln ( T 1 − t )    = UA

  ( T 2 − t )  ( g pg  WC )

In our case, U = 105 kcal/m 2 h °C, A = 322 m 2 ,C pg = 0.3024, t s = 256°C. Hence,

ln  ( T 1 − 256 )    =

 ( T 11 − 256

exp ( . 0 909 ) = . 2 482 or T 1 = 777 ° C

Duty of evaporator = 123,000 × (777 – 466) × 0.3024 = 11.57 × 10 6 kcal/h Primary Superheater W h C ph /W c C pc = C = 123,000 × 0.3163/(96,740 × 0.8) = 0.502 (C pg and C pc values taken from

Example 3.10). About 3260 kg/h spray was used.

U = 105 kcal/m h C A 2 ° = 77 m 2 , K = exp  ( 105 77 × × ( − 1 0 502 ) / 123 000 0 31 , × . 663 )  =   . 1 109

T 1 =  . 1 109 × ( 777 256 − ) + 256 0 502 777 − . ×  / ( − 1 0 50 . 22 ) = 891 ° C Q = duty = 123 000 0 3163 , × . × ( 891 777 − ) = . 4 435 10 × 6 kcal/h

Steam temperature at exit = 256 + 4.435 × 10 6 /(96,740 × 0.8) = 313°C

Final Superheater Gas exit temperature = 891°C and the steam inlet temperature after spray = 286°C.

W h C ph /W c C pc = C = 123,000 × 0.3244/100,000 × 0.629 = 0.634 U = 109.3 kcal/m 2 h °C, A = 103 m 2 ; K = exp[(109.3 × 103 × (1 − 0.634)/(123,000 × 0.3244)] =

exp(0.103) = 1.108

Appendix A: Boiler Design and Performance Calculations 373

T 1 = [1.108 × (891 − 286) + 286 −.634 × 891]/(1 − 0.634) = 1070°C

Q = 123,000 × 0.3244 × (1070 − 891) = 7.14 × 10 6 kcal/h

Exit steam temperature = 286 + 7.14 × 10 6 /100,000/0.629 = 400°C

Screen Section The screen section is nothing but an evaporator. Due to its location and function,

namely, to shield or protect the superheater, it is called the screen section. The calcula- tion procedure is the same as that for the evaporator.

ln[(t g1 −t s )/(t g2 −t s )] = UA/(W g C pg ) U = 117.8 kcal/m 2 h °C; A = 105 m 2 ,C pg = 0.3328 kcal/kg °C ln[(t g1 − 256)/(1,071 − 256)] = 117.8 × 105/(123,000 × 0.3328) = 0.302 or t g1 = 1359°C Screen duty Q = 123,000 × 0.3328 × (1,359 − 1,071) = 11.79 × 10 6 kcal/h

The total energy absorbed by steam = 64.68 MM kcal/h (Table 3.13). Hence, furnace duty = (64.68 − 11.79 − 7.14 − 4.435 − 11.57 − 10.73)10 6 = 19.01 × 10 6 kcal/h From Chapter 2 on furnace calculations, the furnace duty may be estimated as follows:

Q f =W f × LHV × (1 − casing loss − unaccounted loss) − W g ×h e

W g = total flue gas flow = 123,000 kg/h

Casing loss and unaccounted loss = 0.01

LHV = 11,653 kcal/kg

Hence, Furnace duty = W f × 11,653 × 0.99 − 123,000 × (412 − 1.4) + 117,000 × 1.3 = 19.01 × 10 6

(412 and 1.4 kcal/kg refer to enthalpies of the flue gas at the furnace exit gas temperature and at the reference or ambient temperature. Enthalpy values of flue gas are taken from Appendix F.)

Hence, W f = 6,012 kg/h (air flow = 117,000 kg/h and flue gas flow = 123,000 kg/h). This fuel flow agrees with the data provided by the boiler supplier of 6024 kg/h. Hence, efficiency on LHV basis = 64.68 × 10 6 /(6,012 × 11,653) = 92.32% agrees closely with the data provided earlier. Hence, the boiler performance is reasonable. The plant engineer may also apply the methods discussed in Chapter 2 on the estima- tion of furnace exit gas temperature and see based on the heat input and effective furnace area whether the furnace exit gas temperature is close to that estimated as earlier.

Thus, we have arrived at the furnace exit gas temperature from the boiler rear end mea- surements. This method may also be used to model the furnace exit gas temperature as a function of boiler load and net effective heating surface. As furnace modeling is a complex

374 Appendix A: Boiler Design and Performance Calculations

process, substantiating the model with field data and such back calculations will ensure that the model for that type of furnace, fuel, excess air, burner location is reasonably accu- rate. One may also evaluate the efficiency from the oxygen measurements and exit gas temperature and see how it compares with design or proposal values. Thus, there are several tools for a process engineer to evaluate a boiler performance independently and challenge the supplier if significant differences arise. For example, if the boiler can oper- ate only at, say, 75% initially for various plant restrictions, then how can one prove that the boiler performance is satisfactory? The aforementioned procedure can be applied. The fuel gas flow may be measured and compared with the calculated. One may work from the burner end as discussed in Example 3.7, or from the economizer exit as described earlier. If there are large variations, then the design may be challenged.