Heat exchangers

Why it matters

Heat exchangers move heat from one fluid to another without mixing them: radiators and oil coolers on machine tools, coolant chillers for grinding and EDM, condensers and boilers in power plants, quench-oil coolers in heat-treatment shops and waste-heat recuperators on furnaces. Sizing one (finding the area) or rating one (finding the outlet temperatures of an existing unit) is a routine plant-engineering job and a regular exam numerical.

Key ideas

Types by construction.

• Double-pipe (concentric tube): one fluid in the inner pipe, the other in the annulus. Simple; used for small duties.
• Shell-and-tube: a tube bundle inside a cylindrical shell, with baffles that force the shell-side fluid across the tubes (raising h) and support the tubes. The high-pressure or more fouling fluid usually goes in the tubes, which are cheaper to make pressure-tight and easier to clean.
• Plate: corrugated plates clamped together; high h, compact, easy to open and clean, limited in pressure and temperature by the gaskets.
• Compact / finned (cross-flow): fins on the gas side, as in car radiators and air coolers, because the gas-side h is low.
• Regenerators: a matrix stores heat from the hot gas and gives it to the cold gas (rotary air preheaters).

Flow arrangements.

• Parallel flow: both fluids enter at the same end. The temperature difference is large at inlet and small at outlet; the cold outlet can never exceed the hot outlet.
• Counterflow: the fluids enter at opposite ends. The temperature difference is more uniform, the LMTD is the largest for given terminal temperatures, and the cold outlet can exceed the hot outlet. For the same duty it needs the least area.
• Cross-flow and multipass: treated with a correction factor F (≤ 1) on the counterflow LMTD, read from charts, or with ε–NTU relations.

Overall heat transfer coefficient U. Heat passes through a series of resistances: hot-side convection, hot-side fouling, the wall, cold-side fouling and cold-side convection. U is the reciprocal of the total resistance per unit area. It is limited by the smallest h, so improving the poorer side (fins, higher velocity) pays most. Typical magnitudes: water–water 1000–2500 W/m²·K, water–oil 100–350, gas–gas 10–50 — take design values from a data book.

Fouling. Scale, rust, biological growth and deposits add a fouling resistance R_f (m²·K/W) on each side. It lowers U with time, so designers add area margin and plan cleaning. Fouling factors are empirical; take them from a data book or TEMA tables.

Heat capacity rate. C = ṁ·c_p (W/K). The fluid with the smaller C (C_min) undergoes the larger temperature change. In a condenser or evaporator one fluid changes phase at constant temperature, so its C is effectively infinite and C_r = C_min/C_max = 0.

Two methods.

• LMTD method: use when all four terminal temperatures are known or can be found from an energy balance — the sizing problem (find A).
• ε–NTU method: use when only inlet temperatures are known — the rating problem (find outlets for a given A). It avoids trial and error.

Effectiveness and NTU. ε = actual Q / maximum possible Q, where Q_max = C_min·(T_h,in − T_c,in) is what a counterflow exchanger of infinite area would transfer. NTU = U·A/C_min is a measure of the "thermal size" of the exchanger. ε rises with NTU but with diminishing returns.

Assumptions. Steady flow, no heat loss to surroundings, constant U and c_p, negligible kinetic and potential energy changes, no axial conduction.

Formulas

Q = ṁ_h·c_ph·(T_h,in − T_h,out) = ṁ_c·c_pc·(T_c,out − T_c,in)

• Energy balance. ṁ in kg/s, c_p in J/kg·K, T in °C or K (differences), Q in W.

1/(U·A) = 1/(h_i·A_i) + R_f,i/A_i + ln(r_o/r_i)/(2π·k·L) + R_f,o/A_o + 1/(h_o·A_o)

• Tube wall. h in W/m²·K, R_f in m²·K/W, k in W/m·K. For a thin plane wall: 1/U = 1/h_i + t/k + 1/h_o (+ fouling).

Q = U·A·ΔT_lm

• A in m², ΔT_lm in K.

ΔT_lm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂)

• Parallel flow: ΔT₁ = T_h,in − T_c,in, ΔT₂ = T_h,out − T_c,out.
• Counterflow: ΔT₁ = T_h,in − T_c,out, ΔT₂ = T_h,out − T_c,in.
• If ΔT₁ = ΔT₂ (counterflow with C_h = C_c), ΔT_lm = ΔT₁.
• Multipass or cross-flow: Q = U·A·F·ΔT_lm,counterflow.

ε = Q/Q_max, Q_max = C_min·(T_h,in − T_c,in), NTU = U·A/C_min, C_r = C_min/C_max

ε_parallel = [1 − exp(−NTU·(1 + C_r))]/(1 + C_r)

ε_counter = [1 − exp(−NTU·(1 − C_r))]/[1 − C_r·exp(−NTU·(1 − C_r))]; if C_r = 1, ε = NTU/(1 + NTU)

ε = 1 − exp(−NTU)

• Any arrangement when C_r = 0 (condenser, evaporator).

Worked examples

Example 1 (standard): sizing an oil cooler (LMTD method). Given: hot oil, ṁ_h = 2 kg/s, c_ph = 2.2 kJ/kg·K, cooled from 150 °C to 90 °C by water, ṁ_c = 1.5 kg/s, c_pc = 4.18 kJ/kg·K, entering at 25 °C. U = 400 W/m²·K. Find the area for counterflow and for parallel flow.

1. Q = ṁ_h·c_ph·ΔT_h = 2 × 2200 × (150 − 90) = 264 000 W
2. T_c,out = T_c,in + Q/(ṁ_c·c_pc) = 25 + 264 000/(1.5 × 4180) = 25 + 42.1 = 67.1 °C
3. Counterflow: ΔT₁ = 150 − 67.1 = 82.9 K, ΔT₂ = 90 − 25 = 65 K
4. ΔT_lm = (82.9 − 65)/ln(82.9/65) = 73.6 K
5. A = Q/(U·ΔT_lm) = 264 000/(400 × 73.6) = 8.97 m²
6. Parallel flow: ΔT₁ = 150 − 25 = 125 K, ΔT₂ = 90 − 67.1 = 22.9 K, ΔT_lm = (125 − 22.9)/ln(125/22.9) = 60.2 K
7. A = 264 000/(400 × 60.2) = 10.97 m² Answer: counterflow A ≈ 8.97 m²; parallel flow A ≈ 10.97 m² — counterflow needs about 18% less area for the same duty.

Example 2 (GATE level): rating a counterflow exchanger (ε–NTU method). Given: hot water ṁ_h = 0.5 kg/s enters at 120 °C; cold water ṁ_c = 1.0 kg/s enters at 20 °C; take c_p = 4.0 kJ/kg·K for both. Counterflow, U = 500 W/m²·K, A = 8 m². Find Q and both outlet temperatures.

1. C_h = 0.5 × 4000 = 2000 W/K, C_c = 1.0 × 4000 = 4000 W/K, so C_min = 2000 W/K, C_r = 0.5
2. NTU = U·A/C_min = 500 × 8/2000 = 2.0
3. ε = [1 − exp(−2 × 0.5)]/[1 − 0.5·exp(−2 × 0.5)] = (1 − 0.3679)/(1 − 0.1839) = 0.6321/0.8161 = 0.7746
4. Q_max = C_min·(120 − 20) = 2000 × 100 = 200 000 W
5. Q = ε·Q_max = 0.7746 × 200 000 = 154 900 W
6. T_h,out = 120 − 154 900/2000 = 42.5 °C; T_c,out = 20 + 154 900/4000 = 58.7 °C
7. Check with LMTD: ΔT₁ = 120 − 58.7 = 61.3 K, ΔT₂ = 42.5 − 20 = 22.5 K, ΔT_lm = 38.7 K, U·A·ΔT_lm = 500 × 8 × 38.7 = 154 900 W ✓ Answer: Q ≈ 154.9 kW; T_h,out ≈ 42.5 °C; T_c,out ≈ 58.7 °C

Common mistakes

• Pairing the wrong terminal temperatures in ΔT₁ and ΔT₂ — parallel and counterflow pair them differently.
• Using the arithmetic mean temperature difference; it overestimates ΔT and undersizes the exchanger.
• Writing 0/0 when ΔT₁ = ΔT₂ in counterflow; the LMTD is then simply that common ΔT.
• Using C_max instead of C_min in Q_max or NTU.
• Mixing kJ and J (c_p in kJ/kg·K with U in W/m²·K).
• Forgetting that a condensing or boiling fluid makes C_r = 0, so flow arrangement no longer matters.
• Ignoring fouling, or thinking a higher U from cleaning changes the LMTD rather than Q or A.

For GATE PI

Expect LMTD numericals for parallel and counterflow (often with an energy balance first to find a missing outlet temperature), ε–NTU rating problems, condenser problems with C_r = 0, and conceptual questions comparing parallel flow and counterflow or the effect of fouling on U. Practise recognising when to use each method, and the special cases ΔT₁ = ΔT₂ and C_r = 0.

Quick check

1. In counterflow with equal heat capacity rates, the end temperature differences are both 40 K. What is the LMTD?
2. Which flow arrangement allows the cold-fluid outlet to be hotter than the hot-fluid outlet?
3. Define NTU.
4. A condenser has NTU = 1. What is its effectiveness?
5. Does fouling increase or decrease U?

Answers: 1. 40 K 2. Counterflow 3. NTU = U·A/C_min 4. ε = 1 − e⁻¹ ≈ 0.632 5. Decreases it