Time required to Heat Up Boilers  V.Ganapathy

Flue gas at a temperature of Tg1 enters the evaporator(superheater/economizers not considered in this model),which is initially at a temperature of t.Then the transient heating of the boiler may be represented by the equations:

M_c dt/dz =W_gC_pg (T_g1-T_g2)(1-h) = UADT                (1)  where

M_c= water equivalent of boiler(mass of steel x specific heat of steel + mass of water x specific heat of water + mass of insulation x specific heat of insulation).Include drums,tubes,casing.
dt/dz = rate of change of boiler temperature , deg F/h
W_g     = gas flow,lb/h
C_pg   = gas specific heat,Btu/lbF
T_g1,T_g2 = boiler gas inlet and exit temperatures,F
U =  overall heat transfer coefficient,Btu/ft²hF
A  = surface area, ft²
t = initial temperature of boiler system,F
z = time for heating, h
DT =log-mean temperature difference,F = (T_g1- t)-(T_g2 -t)/ ln[ (T_g1- t)/(T_g2 -t)]                                      ₍₂₎
h = heat loss,fraction

Simplifying (1) and (2) we have:

ln[ (T_g1- t)/(T_g2 -t)] = UA/W_gC_pg  (1-h)   or

T_g2 = t + (T_g1 -t)/ e ^{UA/WgCpg  (1-h)}     =  t + (T_g1-t)/K                      (3)

Substituting for Tg2 in (1),we have:

M_c dt/dz =W_gC_pg (T_g1-t)(1-h)(K-1)/K                    (4)

or

dt/(T_g1-t) = [W_gC_pg /M_c] (1-h)[(K-1)/K] dz                 (5)

Integrating between an initial temperature of t₁ to t_2,the time to heat up the boiler is given by:

ln[(T_g1-t₁)/(T_g1-t₂)] = [W_gC_pg /M_c] (1-h)[(K-1)/K] z                (6)

The above equation gives an idea of the time required to heat up the boiler from a temperature of t₁ to t₂. Note that once the boiler is brought up to boiling conditions,additional terms must be used on the left hand side of equation (1) to consider steam generation and rate of pressure build up,which are not given here.

Example:A water tube waste heat boiler of weight 50,000 lb and containing 30,000 lb of water is initially at 100 F. 130,000 lb/h of flue gases at 1400 F enter the unit. Assume:
Gas specific heat = 0.3 Btu/lbF
Steel specific heat =0.12 Btu/lbF
Surface area =21,000 ft²
Overall heat transfer coefficient = 8 Btu/ft²hF
Estimate the time to bring up the boiler to 212 F. Neglect casing heat loss .

Solution:
Estimate K = e ^{UA/WgCpg  (1-h)}   = e ^{(8x21,000/130,000/0.3)}   = 74
M_c =50,000x0.12 +30,000x1 = 36,000
From (6),
ln[(1400-100)/(1400-212)] =0.09 = 130,000x0.3x73/(74x36,000)z  or z =  0.084 h  = 5.1 min

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