Boiler horsepower to steam, blowdown and makeup

Steam plant · steam rate, cycles, blowdown and makeup

A 150 BHP firetube boiler carries the heating and process load of a mid-size plant, and the water-treatment file for it needs four numbers before any chemical is quoted. The nameplate gives the first: boiler horsepower, the old and stubbornly useful unit that says what a boiler makes rather than what it burns. Monday's lab work gives the next two. A drum sample drawn off the surface blowdown line reads 2,500 ppm total dissolved solids, sitting under the 3,500 ppm the ABMA table allows at this drum pressure, and a feedwater sample pulled at the deaerator outlet reads 500 ppm. The plant's own metering gives the last: only 30 % of the steam comes back as condensate, because most of it is injected into a process and never returns. Steam carries no dissolved solids with it — every grain that enters in the feedwater stays in the drum until it is deliberately drained — and that single fact is what ties all four numbers together.

150 BHP · 5 cycles5,175 lb/h steammakeup 3,623 lb/hcondensate 1,553 lb/hDA500 ppm2,500 ppmblowdown 1,294 lb/h

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given
  • BHP = 150 BHPNameplate boiler horsepower
  • TDS_b = 2,500 ppmDrum sample, total dissolved solids
  • TDS_fw = 500 ppmFeedwater sample at the deaerator outlet
  • %CR = 30 %Condensate returned to the boiler house
Determine
  1. (a)the steam the boiler makes at its nameplate rating
  2. (b)the cycles of concentration the two lab samples imply
  3. (c)the continuous blowdown those cycles require
  4. (d)the makeup needed to replace the steam that never comes back
Step 1 of 4(a) · solve for Steam production rate

One boiler horsepower is defined as 34.5 lb/h of steam from and at 212 °F — an evaporation rate, not a shaft output, and the reason the unit survives in a trade that abandoned horsepower everywhere else. At 150 BHP that is 5,175 lb/h. This is a rating, not a measurement: a boiler feeding 220 °F feedwater and making 100 psig steam delivers somewhat less, because more of every pound goes into sensible heat. Everything below is proportional to this number, so it comes first.

Rearranged for S
S=34.5  BHPS = 34.5 \; \text{BHP}
Your values, in your units
S=34.5×(150)S = 34.5 \times \left( 150 \right)
Answer
S=2.3473 t/hS = 2.3473\ \text{t/h}

Carried onward at full precision, not this rounded figure.

Open the Boiler Horsepower to Steam Rate solver →

Step 2 of 4(b) · solve for Cycles of concentration

Cycles is the simplest ratio in the whole file and the most informative: the drum runs five times as concentrated as what feeds it, so every kilogram of dissolved solids that arrives has ridden around five times before it leaves. Both samples must be the same measurement — conductivity-derived TDS against conductivity-derived TDS — and both must be neutralized if the drum carries free caustic, or the hydroxide's conductivity inflates the drum figure and the calculated cycles come out flattering.

Rearranged for COC
COC=TDSbTDSfw\text{COC} = \frac{\text{TDS}_b}{\text{TDS}_{fw}}
Your values, in your units
COC=(2,500 ppm)(500 ppm)\text{COC} = \frac{\left( 2{,}500\ \text{ppm} \right)}{\left( 500\ \text{ppm} \right)}
Converted to base units
COC=(0.25 %)(0.05 %)\text{COC} = \frac{\left( 0.25\ \text{\%} \right)}{\left( 0.05\ \text{\%} \right)}
Answer
COC=5\text{COC} = 5

Carried onward at full precision, not this rounded figure.

Open the Boiler Cycles of Concentration solver →

Step 3 of 4(c) · solve for Blowdown rate

B = S/(COC − 1) is a solids balance wearing a disguise. Feedwater is steam plus blowdown; the solids arriving in the feedwater must equal the solids leaving in the blowdown, since none leave with the steam; and that equality rearranges to exactly this. The minus one is the whole equation: at five cycles the blowdown is a quarter of the steam, at ten cycles a ninth, and at two cycles it is the entire steam rate again. Every cycle you can win back is fought for down at that end of the curve.

Rearranged for B
B=SCOC1B = \frac{S}{\text{COC} - 1}
2.3473 t/hcarried from step 1
5 carried from step 2
Your values, in your units
B=(0.652039 kg/s)(5)1B = \frac{\left( 0.652039\ \text{kg/s} \right)}{\left( 5 \right) - 1}
Converted to base units
B=(2,347.34 kg/h)(5)1B = \frac{\left( 2{,}347.34\ \text{kg/h} \right)}{\left( 5 \right) - 1}
Answer
B=586.84 kg/hB = 586.84\ \text{kg/h}

Carried onward at full precision, not this rounded figure.

Open the Boiler Blowdown Rate from Steam Rate solver →

Step 4 of 4(d) · solve for Makeup water rate

This is the condensate side of the makeup only: the 70 % of the steam that never comes home, which must be replaced with treated water. It is not the whole makeup bill. The blowdown from part (c) leaves the plant too and has to be replaced as well, so the softener actually sees 3,622.5 + 1,293.75 = 4,916.25 lb/h. Quoting part (d) alone as "the makeup" undersizes a softener by more than a quarter, and it is a common enough slip that it is worth writing both figures on the same line of the file.

Rearranged for M
M=S(1%CR100)M = S\left(1 - \frac{\%CR}{100}\right)
2.3473 t/hcarried from step 1
Your values, in your units
M=(0.652039 kg/s)(1(30 %)100)M = \left( 0.652039\ \text{kg/s} \right) \left(1 - \frac{\left( 30\ \text{\%} \right)}{100}\right)
Converted to base units
M=(2,347.34 kg/h)(1(30 %)100)M = \left( 2{,}347.34\ \text{kg/h} \right) \left(1 - \frac{\left( 30\ \text{\%} \right)}{100}\right)
Answer
M=1.6431 t/hM = 1.6431\ \text{t/h}

Carried onward at full precision, not this rounded figure.

Open the Boiler Makeup from Condensate Return solver →

Answer

Therefore the boiler makes 5,175 lb/h of steam, the lab samples put it at 5 cycles, those cycles demand 1,294 lb/h of continuous blowdown, and 3,623 lb/h of makeup replaces the condensate that never returns — so the softener and the chemical feed are sized for 4,916 lb/h once the blowdown is added in.

Why this order

Four numbers, one physical fact: steam leaves its solids behind. Everything in the chain is that sentence rearranged. Cycles is what the solids do when they cannot leave; blowdown is the only door they can leave by; makeup is what walks in to take their place. The order matters because parts (c) and (d) both hang off part (a) — get the steam rate wrong and both flows are wrong by the same factor — while part (b) is independent of it entirely, coming from two lab samples that would read the same whether the boiler was at full fire or a third of it. That is why cycles is the number a treatment programme is controlled ON and blowdown is merely the number it is delivered THROUGH.

The plant closes on itself, which is the real check on this file. Feedwater is steam plus blowdown: 5,175 + 1,293.75 = 6,468.75 lb/h. Blowdown as a share of feedwater is 1,293.75/6,468.75 = 20.0 %, which must equal 1/COC, and 1/5 is exactly 20 % — the balance closes without a rounding crumb. Push it one step further: the makeup total, 4,916.25 lb/h, is 76 % of feedwater, and the condensate is the other 24 %, so a makeup running about 660 ppm TDS diluted to 76 % lands the feedwater at 500 ppm, the very sample the lab pulled. The whole plant is consistent. That habit — closing the loop on the numbers you were handed — is what separates a treatment file that can be defended from one that merely has figures in it. The engineering follow-through is the blowdown itself: 1,294 lb/h leaving at drum temperature is a real heat loss, roughly 100 kW at these conditions, which is why plants fight for cycles and then put a flash tank and a heat exchanger on what is left. Raise this boiler from 5 cycles to 10 and the blowdown halves to 575 lb/h — same chemistry, same limit, half the water and half the heat out the drain.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.