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_{g}C_{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/ft^{2}hF

A
= surface area, ft^{2}

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_{g}C_{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_{g}C_{pg} (T_{g1}-t)(1-h)(K-1)/K
(4)

or

dt/(T_{g1}-t)
= [W_{g}C_{pg} /M_{c}] (1-h)[(K-1)/K] dz
(5)

Integrating
between an initial temperature of t_{1} to t_{2,}the time
to heat up the boiler is given by:

**ln[(T _{g1}-t_{1})/(T_{g1}-t_{2})]
= [W_{g}C_{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 t_{1} to t_{2}. 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 ft^{2}

Overall
heat transfer coefficient = 8 Btu/ft^{2}hF

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