Lesson A10 calculated what the process demands. This lesson is about whether the equipment can deliver it — the physics of getting heat across a metal wall, and why a whisper of air can quietly take five hundred rupees of heat-transfer surface and make it perform like ninety.

The rate equation

Every steam heater on earth obeys Q = U · A · ΔT: heat flow equals a transfer coefficient, times surface area, times the temperature difference driving it. A is bought in stainless steel. ΔT is bought at the boiler (higher pressure, hotter steam). U — kcal per hour, per m², per °C — is where the engineering lives, because U is not a property of the metal: it is the sum of every obstacle the heat must cross.

The obstacle course

Heat leaving condensing steam for the product crosses, in order: the condensate film on the steam side (fast — condensing films run in the thousands of kcal/h·m²·°C), the metal wall (faster still — millimetres of steel barely register), any fouling (scale, product bake-on — slow), and the product-side film (usually the honest bottleneck: a few hundred for stirred liquids, less for viscous ones). Resistances add like resistors in series, and the biggest resistance rules the total. This is why polishing an already-good steam side gains nothing while a stirrer upgrade or a descale transforms performance — improve the largest resistance or you have improved nothing.

The resistance stack, to scale (jacket of the practice-plant vessel) clean: U ≈ 500 + air film (0.2 mm): still air — 0.2 mm → U collapses to ≈ 92. Same jacket, same steam, one-fifth the heat. steam film steel wall fouling product film air film segment lengths ∝ thermal resistance · carbon-steel k ≈ 37 (stainless ≈ 14) · still air k ≈ 0.022 kcal/h·m·°C the 0.2 mm air strip ≈ 40 walls of 8 mm carbon steel — 1/U = 1/500 + 0.0002 ÷ 0.022 → U ≈ 92
Why lesson A9 called air the U-value's enemy: the film's resistance is drawn to the same scale as everything else. It is not a subtle effect.

ΔT with steam: mercifully simple

Because saturated steam condenses at one temperature (lesson A3), the driving force calculation loses half its complexity — the hot side is a flat line. Against a product warming from 30 °C to 90 °C under steam supplied at 3.5 kg/cm²g — at full load the control valve drops about half a kilo even when wide, so the jacket itself sits near 3.0 kg/cm²g (143 °C) — the difference starts at 113 °C and shrinks to 53 °C, and the honest average is the log-mean:

LMTD = (ΔT₁ − ΔT₂) ÷ ln(ΔT₁/ΔT₂) = (113 − 53) ÷ ln(113/53) ≈ 79 °C

(The arithmetic mean says 83 — close here, badly wrong when the ratio widens. Use the log-mean; it costs one extra key on the calculator.)

143 90 30 temperature, °C position along the heating surface → steam condensing at 143 °C — a flat line (A3) product, 30 → 90 °C ΔT₁ = 113 °C ΔT₂ = 53 °C the driving gap shrinks along the surface its honest average is the log-mean: LMTD ≈ 79 °C (the arithmetic mean's 83 flatters it)
Why the mean needs a logarithm: the gap the heat crosses is 113 °C at one end of the surface and 53 °C at the other, and the surface between works at every value in between. With steam, only the cold side moves — half the problem is already solved.

Closing the loop on the practice plant

The vessel's jacket offers 10 m² and a clean-service U of about 500. Capacity: Q = 500 × 10 × 79 ≈ 395,000 kcal/h — against the 387,000 kcal/h warm-up duty from lesson A10. The jacket makes its 45-minute batch with barely 2% to spare. Now re-read the figure above: with the air film, capacity is ~73,000 kcal/h and the batch takes over four hours. When operators say "the vessel has gone slow", this — not the boiler — is usually the arithmetic behind it. The ₹2,000 air vent from lesson A9 is protecting a 395,000 kcal/h asset.

At site
  • Diagnose slow heating in order: air first (thermometer-vs-gauge test, A9), then condensate flooding (is the trap passing? — D5), then fouling (when did it last hold schedule?), and only then blame U or the boiler.
  • Improve the biggest resistance. Stirring/circulation on the product side often buys more than any steam-side change.
  • Descaling is heat-transfer maintenance, not housekeeping. Log batch times; a creeping trend is fouling talking.
  • Quoting or checking an exchanger: ask what U was assumed, clean and fouled, and at what fouling factor. A vendor U without a fouling basis is a brochure number.
Pin this
  • Q = U·A·LMTD. Area is capital, ΔT is boiler pressure, U is engineering.
  • Resistances add in series; the largest one governs. Improve that one.
  • With steam the hot side is isothermal — LMTD = (ΔT₁−ΔT₂)/ln(ΔT₁/ΔT₂), one line.
  • A 0.2 mm still-air film ≈ forty steel walls. Venting is heat-transfer engineering.
  • The practice plant's jacket: 395,000 kcal/h clean, ~73,000 air-blanketed. Same steel, same steam.
Steam stories

The overall-coefficient method — treating a wall and its films as resistances in series — was standardised into engineering by Wilhelm Nusselt's 1916 analysis of film condensation, still the starting point of every condenser textbook. Nusselt worked the problem with pencil, paper and a controversial assumption (a smooth, undisturbed condensate film); a century of measurement later, his "ideal" numbers remain within shouting distance of real vertical-surface condensers — one of engineering's better returns on a simplifying assumption.

FAQ

The vessel has gone slow. Where do I start?

In this order: air first — the free thermometer-vs-gauge test from A9; then condensate flooding — is the trap actually passing (D5)?; then fouling — when did the batch last hold schedule?; and only then question U, the exchanger or the boiler. The order matters because the cheap causes are also the common ones.

Will raising the steam pressure speed up my batch?

It raises ΔT, and Q follows linearly — that is a real lever. But check the resistance stack first: if an air film or scale governs U, more pressure just pays extra fuel to push heat at the same blanket. Fix the largest resistance, then buy ΔT if the batch still needs it — and confirm the product tolerates the hotter wall.

Why use the log-mean ΔT and not the simple average?

With ΔT shrinking from 113 °C to 53 °C the two differ by only a few degrees — 79 against 83. But as the ratio between inlet and outlet difference widens, the arithmetic mean overstates the driving force badly and your surface comes up short. The log-mean costs one extra key on the calculator and is honest at every ratio.

A vendor quotes U = 800 for my duty. Can I trust it?

Not without its basis. Ask what U was assumed clean and fouled, and at what fouling factor — a U without a fouling basis is a brochure number. The practice-plant jacket runs about 500 in clean stirred-liquid service; numbers well above that deserve the question "with what product film, and how long after cleaning?"

Would a better steam supply improve my exchanger?

Usually not — the condensing film already runs in the thousands while a stirred product film runs in the hundreds, so the product side governs the series total. Polishing an already-good steam side gains nothing; a stirrer upgrade, better circulation or a descale attacks the resistance that actually rules. Improve the largest resistance or you have improved nothing.

Check yourself

Five quick questions on this lesson. No marks, no records — the score is for you.