10.1 Q = UAΔT, LMTD, and Heat Exchangers

10.1 Q = UAΔT, LMTD, and Heat Exchangers

Heat does not jump from a blast freezer to the evaporative condenser. It crosses metal, films, and whatever is stuck on those films. CIRO Heat Flow items treat that crossing as a rate equation, not a slogan. If you can name each symbol, you can predict what a dirty coil, a starved pump, or a tight approach will do to suction and discharge pressure.

The rate equation

Q = U A ΔT

Rearranged, the same sentence is the operator's diagnostic:

ΔT = Q / (U A)

For a fixed load and a fixed area, anything that lowers U must raise ΔT. On an evaporator that means refrigerant saturation temperature is pulled down relative to the room or process. On a condenser it means saturation temperature is pushed up relative to leaving water or outdoor wet bulb. You feel those shifts as suction and discharge psig.

U is not a single film coefficient. Resistances add in series:

1/U = 1/h_warm + x/k_wall + R_fouling + 1/h_cold

You will not be asked to invent a metal conductivity. You will be asked what happens when R_fouling grows. Oil film on a flooded chiller, ice on a unit cooler, scale on condenser tubes, and a dry evaporative-condenser coil (no water) are all extra resistance. Metal thickness barely moves from year to year. Fouling does.

A is not the footprint of the penthouse. It is the heat-transfer surface that still participates. A plugged circuit, a frozen slab covering half the fins, or a condenser bay isolated for repair is lost area. Lost A is the same math as a lower U: required ΔT rises.

ΔT is the argument people mix up. Three different differences live in the engine room:

1. Local difference at one point along a tube (T_hot − T_cold at that station).
2. Logarithmic mean temperature difference (LMTD) — the correct average of the two end differences when the two fluids (or one fluid and a two-phase refrigerant) change temperature along the exchanger.
3. Operator TD — a plant shortcut such as room temperature minus evaporator saturation, or condensing saturation minus leaving water (approach).

CIRO will name which one it wants. Do not feed operator TD into Q = UAΔT when the stem just computed LMTD, and do not call LMTD the approach.

Worked example — fouling raises required ΔT

A 20-ton coil must move Q = 20 × 12,000 = 240,000 Btu/h. Fin-and-tube area A = 2,000 ft². In a clean, wet, fully fed condition the coil's LMTD is 8°F.

U_clean = Q / (A × LMTD) = 240,000 / (2,000 × 8) = 15 Btu/h·ft²·°F

After a week of product ice and a light oil film, U falls to 12 Btu/h·ft²·°F (a 20% hit — directional teaching numbers, not a manufacturer rating).

To hold the same 20 tons:

LMTD_dirty = 240,000 / (12 × 2,000) = 10°F

If the room is still 0°F, a clean LMTD of 8°F might have meant a saturation temperature near −8°F. A dirty 10°F LMTD pulls saturation toward −10°F. The compressor now sees a lower suction pressure, higher specific volume, and less mass per cubic foot — the capacity problem of Section 10.3. Operators who only chase the TXV miss that U fell. The coil did not forget how to boil ammonia; the driving difference had to grow because the surface got worse.

If load falls (empty dock) while U stays dirty, ΔT can look 'normal' again. Judge U against Q, not against a single suction gauge in isolation.

LMTD, not the arithmetic mean

When a single-phase fluid changes temperature while the other stream is at a known temperature (two-phase ammonia is nearly isothermal at Tsat), the two end differences are:

• ΔT1 = difference at one end of the exchanger
• ΔT2 = difference at the other end

LMTD = (ΔT1 − ΔT2) / ln(ΔT1 / ΔT2)

The arithmetic mean (ΔT1 + ΔT2)/2 is always larger than LMTD when ΔT1 ≠ ΔT2. Using it overstates Q. The error is small when the two ends are close and ugly when they are not. If either end difference is zero, LMTD is zero: you have no driving force at that end, and the formula's logarithm blows up. That is the physics of trying to condense at the leaving-water temperature or at the wet bulb.

Counterflow versus parallel flow matters when both streams change temperature (glycol-to-glycol, oil cooler with sensible oil and a liquid thermosiphon that still has a temperature glide). Counterflow keeps a more uniform local ΔT and can allow the cold stream to leave hotter than the hot stream leaves. Parallel flow cannot. Two-phase evaporators and condensers with a flat Tsat have the same two end differences in either arrangement, so LMTD uses the same two numbers — but you still must not replace LMTD with (Tsat − T_fluid,avg).

Worked LMTD — water-cooled condenser

Ammonia condenses at a constant 95°F in a shell-and-tube unit. Cooling water enters at 75°F and leaves at 85°F.

• Range (water temperature rise) = 85 − 75 = 10°F
• Approach (condensing sat minus leaving water) = 95 − 85 = 10°F
• End difference at the water inlet = 95 − 75 = 20°F
• End difference at the water outlet = 95 − 85 = 10°F

LMTD = (20 − 10) / ln(20/10) = 10 / 0.693 = 14.4°F

Arithmetic mean = (20 + 10)/2 = 15.0°F — about 4% high. If U = 180 Btu/h·ft²·°F and A = 400 ft²:

Q_LMTD = 180 × 400 × 14.4 = 1,036,800 Btu/h ≈ 86.4 tons of rejection

Using 15°F would invent 1,080,000 Btu/h. The extra 43,200 Btu/h is not in the water; it is an arithmetic lie.

Wider spread, worse lie: ends of 30°F and 10°F give LMTD = 20 / ln(3) = 18.2°F versus arithmetic 20°F (~10% high).

Evaporator TD versus LMTD

Evaporator TD = room (or entering process) temperature − refrigerant saturation temperature.

A +34°F dock coil at 20°F saturated ammonia has TD = 14°F. A 0°F freezer with −10°F Tsat has TD = 10°F. A glycol chiller with 20°F entering glycol and 10°F Tsat has a process TD of 10°F. That is how operators and many coil selections talk.

It is not LMTD if the air or glycol leaves colder than it entered. The coil sees a smaller difference at the leaving end.

Worked air unit. Room 34°F, air off the coil 28°F, ammonia Tsat = 22°F.

• Operator TD = 34 − 22 = 12°F
• On-coil difference = 34 − 22 = 12°F
• Off-coil difference = 28 − 22 = 6°F
• LMTD = (12 − 6) / ln(12/6) = 6 / 0.693 = 8.7°F

If someone multiplies UA by 12°F instead of 8.7°F, they overstate coil capacity by about 38%. On a CIRO table item, that is the difference between a coil that holds the room and a coil you are about to starve or over-defrost.

Process evaporators use the same split: process TD from entering fluid (or setpoint) to Tsat; LMTD from both end differences if the stem gives in and out temperatures. Flooded vessels with tiny fluid ΔT have LMTD almost equal to TD. DX air coils with a large air drop do not.

Where the equation lives in the plant

• Evaporators — air units, plate-and-frame, shell-and-tube chillers, jacketed tanks. Q is the refrigerating load. ΔT is LMTD or the stated TD to Tsat.
• Condensers — evaporative, water-cooled, air-cooled. Q is heat of rejection (evaporator heat plus compressor work from Chapter 9), not evaporator tons alone.
• Oil coolers — thermosiphon, water, or liquid injection as a heat exchanger between hot oil and a sink. Same UAΔT; a fouled thermosiphon is a U problem, not a personality trait of the screw.
• Intercoolers and cascade evaporators — later chapters; the rate equation does not change.

Manufacturers often hide U and A inside a rated THR at a stated TD. You still think in Q = UAΔT: if the nameplate assumes 15°F wet-bulb approach and you only have 8°F to the day's wet bulb, the condenser cannot reject nameplate heat unless Tsat rises or load falls.

Exam traps

• Treating operator TD as LMTD on a coil with a large air or fluid temperature change.
• Using arithmetic mean of the end differences and calling it LMTD.
• Raising ΔT in your head when U fell, then blaming the expansion valve for the suction drop.
• Using condenser approach (one end) as if it were the LMTD for Q = UAΔT.
• Forgetting that condenser Q is rejection, not evaporator tons.
• Mixing °F TD with psig; convert through the saturation table before you talk pressure.