8.3 Heat Exchangers and LMTD

8.3 Heat Exchangers and LMTD

Quick Answer: Duty Q̇ = ṁ C_p ΔT (sensible). Mean driving force for many exchangers is the log mean temperature difference (LMTD). Counter-current usually beats co-current. Multipass units use Q̇ = U A F (LMTD) with correction F ≤ 1.

Sections 8.1–8.2 built mechanisms and overall U. This section closes Domain C heat transfer for typical UPDA items: energy balance on both streams, LMTD, flow arrangement, and basic shell-and-tube / double-pipe awareness used across Qatar process plants.

Energy Balance on a Heat Exchanger

At steady state, neglecting heat loss to ambient and shaft work:

Heat lost by hot stream = heat gained by cold stream = Q̇

Sensible Heat Only (No Phase Change)

Q̇ = ṁ_h C_p,h (T_h,in − T_h,out) = ṁ_c C_p,c (T_c,out − T_c,in)

Exam checks:

1. Close the energy balance—do not invent a duty inconsistent with both streams.
2. Use consistent units (°C differences are fine for ΔT; C_p in kJ/(kg·K) with ṁ in kg/s → Q̇ in kW).
3. If a stream condenses or boils at constant pressure, use ṁ × λ (latent heat) instead of ṁ C_p ΔT for that portion.

Worked Duty Example

Cold oil is heated from 35 °C to 75 °C at ṁ_c = 2.4 kg/s with C_p,c = 2.1 kJ/(kg·K).

Q̇ = 2.4 × 2.1 × (75 − 35) = 2.4 × 2.1 × 40 = 201.6 kW

Hot fluid: ṁ_h = 1.8 kg/s, C_p,h = 2.8 kJ/(kg·K), T_h,in = 120 °C. Find T_h,out (no losses).

201.6 = 1.8 × 2.8 × (120 − T_h,out) 201.6 = 5.04 × (120 − T_h,out) 120 − T_h,out = 40.0 T_h,out = 80.0 °C

LMTD Definition and Formula

For a exchanger with constant U and approximately constant C_p streams, integration along the unit yields:

Q̇ = U A ΔT_lm

Log mean temperature difference:

ΔT_lm = LMTD = (ΔT_a − ΔT_b) / ln(ΔT_a / ΔT_b)

where ΔT_a and ΔT_b are the local hot-minus-cold temperature differences at the two ends of the exchanger.

Never use (ΔT_a + ΔT_b)/2 as a substitute when the exam is testing LMTD—unless the two end differences are essentially equal.

Counter-Current vs Co-Current End Differences

Counter-current (counterflow): hot and cold flow in opposite directions.

• End A: hot inlet meets cold outlet → ΔT_A = T_h,in − T_c,out
• End B: hot outlet meets cold inlet → ΔT_B = T_h,out − T_c,in

Co-current (parallel flow): both enter at the same end.

• Inlet end: ΔT_in = T_h,in − T_c,in
• Outlet end: ΔT_out = T_h,out − T_c,out

Classic feasibility: In pure co-current flow, T_c,out cannot exceed T_h,out. Counter-current can allow T_c,out > T_h,out while still keeping local ΔT positive everywhere if designed properly.

Using the numerical temperatures above (T_h 120→80 °C, T_c 35→75 °C):

Counter-current ends:

ΔT_A = 120 − 75 = 45 °C ΔT_B = 80 − 35 = 45 °C

LMTD_cf = 45 °C (equal ends)

Co-current ends:

ΔT_in = 120 − 35 = 85 °C ΔT_out = 80 − 75 = 5 °C

LMTD_co = (85 − 5) / ln(85/5) = 80 / ln(17) = 80 / 2.833 ≈ 28.2 °C

Same terminal temperatures, much smaller co-current LMTD → needs more area (or higher U) for the same duty.

Worked LMTD Numerical Example (Original Numbers)

Problem. A double-pipe exchanger cools a hot process liquid from 155 °C to 95 °C. Cooling water heats from 28 °C to 46 °C. Assume pure counter-current flow, steady state, no heat loss. U = 420 W/(m²·K). Hot stream: ṁ_h = 1.25 kg/s, C_p,h = 2.40 kJ/(kg·K).

Find: (1) duty Q̇, (2) required cold-water mass flow if C_p,c = 4.18 kJ/(kg·K), (3) LMTD, (4) required area A.

Step 1 — Duty from hot stream

Q̇ = 1.25 × 2.40 × (155 − 95) = 1.25 × 2.40 × 60 = 180 kW = 180,000 W

Step 2 — Cold-water flow

180 = ṁ_c × 4.18 × (46 − 28) 180 = ṁ_c × 4.18 × 18 180 = ṁ_c × 75.24 ṁ_c = 2.39 kg/s (approximately)

Step 3 — Counter-current end differences

ΔT₁ = T_h,in − T_c,out = 155 − 46 = 109 °C ΔT₂ = T_h,out − T_c,in = 95 − 28 = 67 °C

LMTD = (109 − 67) / ln(109/67) = 42 / ln(1.627) = 42 / 0.4867 ≈ 86.3 °C

Step 4 — Area

Q̇ = U A (LMTD) 180,000 = 420 × A × 86.3 A = 180,000 / (420 × 86.3) ≈ 180,000 / 36,246 ≈ 4.97 m²

Arithmetic-mean trap: (109 + 67)/2 = 88 °C, close here because the ratio is moderate—but still not the definition. When one end ΔT is small (pinch), arithmetic mean badly overestimates driving force and undersizes the exchanger.

Shell-and-Tube and Double-Pipe Awareness

When the F Correction Appears

True pure counterflow LMTD is exact only for certain arrangements (e.g., ideal double-pipe counterflow, or 1-1 pure counterflow).

Multipass shell-and-tube exchangers (e.g., 1 shell pass / 2 tube passes) mix crossflow and multipass patterns. Industry practice:

Q̇ = U A F (LMTD_counterflow)

• LMTD_counterflow is computed as if the exchanger were pure counterflow with the same four terminal temperatures
• F is a correction factor from charts (function of two dimensionless temperature ratios), 0 < F ≤ 1
• If F is low (often warned below ~0.75–0.8), temperature crosses or multipass limits—consider more shell passes or different arrangement

UPDA level: Know why F exists (flow is not pure counterflow) and that F multiplies LMTD, reducing effective driving force. You are not expected to redraw TEMA F-charts from memory.

Qualitative Design Notes

• Put fouling or corrosive fluid on the tube side when that eases cleaning/metallurgy (common heuristic—not absolute)
• Cooling water often on tube side in many plant standards
• Baffles raise shell-side h (velocity) but increase ΔP
• Approach temperature: difference between hot outlet and cold inlet (or other specified pair)—small approaches need large A

Linking U, A, LMTD, and Fouling

From Section 8.2, fouling lowers U. For fixed A and terminal temperatures (fixed LMTD), duty falls. Operators may open utility valves (change flows → new terminals and new LMTD) until limits. Designers pick A using dirty U and appropriate F × LMTD so the unit still delivers required Q̇ at end of run.

Common Traps

• Using co-current end pairs while claiming counterflow LMTD
• Arithmetic mean instead of log mean when ΔT₁ ≠ ΔT₂
• Forgetting F conceptually on multipass shells (overestimating driving force)
• Mismatched energy balance (hot Q̇ ≠ cold Q̇) before computing LMTD
• Celsius vs kelvin confusion—ΔT in °C equals ΔT in K; absolute T only needed for radiation

Exam Workflow

1. Draw streams; label all four temperatures.
2. Compute Q̇ from the fully known stream; find the missing flow or temperature on the other side.
3. Choose counter vs co-current end ΔT pairs as the stem requires.
4. Compute LMTD = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂).
5. Apply Q̇ = U A (F)(LMTD); solve for the unknown (often A or U).
6. Comment on fouling or F if the stem mentions dirty service or multipass shells.

With mechanisms (8.1), overall U and fouling (8.2), and LMTD exchangers (8.3), you have the heat half of Domain C transport. Mass transfer and separations follow in Chapter 9.