MATERIAL BALANCE OF MULTIPLE BOILERS IN PARALLEL

INTRODUCTION

In most boiler houses there is more than one boiler, connected in parallel.

As definition, using the term "Boilers in parallel" we mean the connection of more than one boilers, having common feedwater (FW), producing steam of the same pressure in common steam consumption circuit. Each one of them is working independently, according to the steam demand and has its own blow-down control

There is a common hot well, condensate return circuit and a common make-up treatment.

Such a system, including three boilers in presented at the draw below.

Three boilers in parallel.

If the system is operating under negligible impurities, means using demineralized make-up water and without any condensate contamination, the material balance has not special meaning.

But if impurities are present, like in hard or soft make-up, or condensate contamination conditions, the distribution of the impurities along the boilers is not the same.

Lets take as an example a boiler house, operating three boilers in parallel, and consider that boiler #3 has been over concentrated. All other are working in perfect operating conditions

If we try to increase the blow-down in boiler #3 (BD3) as to deconcentrate it, the increase of the blow-down will result to the increase of its feed-water FW3, As long as the feed waters of the other boilers (FW1 και FW2 ) will remain constant, will be increased the overall feed-water FWολ.

The total condensate return C will remain constant, so it will be increased the make-up quantity as to cover the increased overall feed-water.

The increase of the MU will increase the X factor, meaning the concentration of the impurities into the FW. So the maximum permissible NF for all boilers will be decreased.

Finally, trying to deconcentrate Boiler #3 will overconcentrate Boiler #1 and Boiler #2.

THE MATHEMATICAL MODEL

Τhe n boilers system can be considered as one boiler, having total feed-water F which equals to:

F = F1 + F2 + .... + Fn

Produces steam:

S = S1 + S2 + .... + Sn

And rejects blow-down:

B = B1 + B2 + .... + Bn

Calling as Ki the ratio Ki = Si/S1, (Κ1 = 1)

then

Fi = Bi+Si και NFi = Fi/Bi

and

Si = Fi-Βi = BiΝFi - Bi = Bi(NFi - 1)

We conclude that:

Fi - Fi/NFi = Ki Si

So::

Fi = S1 *Ki/[1-(1/NFi)]

And the sum:

n

F =S1 * Σ [Ki/(1-(1/NFi))]

i=1

Because

Bi = Fi/NFi = S1*Ki/(NFi-1)

Has to be:

n n

B = Σ Bi = S1 * Σ Ki/NFi - 1

i=1 i=1

resulting to:

n Ki

Σ ------- F i=1 NFi - 1

NFολ = ----------- = ------------------- Β n Ki

Σ --------- i=1 1-(1/NFi)

Because: n n K1

F = Σ Fi = S1* Σ --------- i=1 i=1 1-(1/NF2)

Will result to:

F n ki (NF1 - 1) 1

----- = Σ------------- = ----- B1 i=1 1 - 1/NFi λ

Defining NFi = Bi/Fi, if we will increase BDi by:

∆BDi = α * BDi

Under constant steam production:

F'i = Fi + ∆BDi

then:

Fi + α * BDi Fi/BD1 + α NFi + α

NF'i = F'i/B'i = --------------- =--------------------- = --------

BDi + α * BDi 1-α 1 + α

And

NFi = [NFi + α]/[1 + α]

At the other boilers the new NF is equal to the previous, NF'i = NFi

If bi, fi the initial concentrations of an impurity, after increase of BDi to BD'i will be changed the bi to b'i and fi to fi

NF'i= b'i/f'i = [NFi + α]/[1 + α]

F'i = b'i * [1 + α] / [NFi + α]

The new material balance of the impurity into the hotwell will be:

F' * F = m'* M' + c'* c and

F'= F + ∆Βi

M'= M + ∆Bi

m'= m

c'= c'

C = C

then: F'*(F +∆Bi) = m*(M + ∆Bi) + c'C

and: F' (F + α BDi) = m (M + αBDi) + c'c

Dividing both sides by F

F' [1-α*BDi/F] = m*M/F + c'*C/F + α*m*BDi/F

f

If i = 1:

F'(1+α*λ) = m*α*λ + f

ή f' = [f + m*α*λ]/[1 + α*λ]

And f'= b'i * [1 + α]/[NFi + α] then:

f' = [f + m * α * λ]/[1 + α λ] =[b'i + b'iα]/[NF1 + α]

The f, m are known, and λ comes from the water analysis and b'1 the permissible concentration we want for boiler #1.

So we can calculate the unknown α by the procedure:

N * NF1 + α m λ NF1 + α m λ NF1 + αf * α2 mλ - b'1 - α b'1 - α λ b'1 - α2 λ b1 = 0

α2(m λ - b'1 λ) + α (m * NF1 * λ + f - b'1 - λ b'1) + (b1 - b'1) = 0

Dividing by m (m has not to be =0)

α2λ [1 - (b'1)/m) + α [ NF1 * λ + f/m - b'1/m (1+λ)] +(b1/m - b'1/m) = 0

α2λ (N C'1 -1) + α [ NC'1 (1+λ) - (NF1 * λ + f/m)] - (NC1 - NC'1) = 0