Heat Exchangers: LMTD and Effectiveness-NTU

A heat exchanger moves heat from a hot fluid to a cold one across a wall. Two methods size and rate them: LMTD when all temperatures are known, and effectiveness–NTU when the outlet temperatures are not.

Types

• Parallel flow: both fluids enter at the same end. The cold outlet can never be hotter than the hot outlet.
• Counter flow: opposite ends. The most effective arrangement; the cold outlet can exceed the hot outlet.
• Cross flow (car radiators) and shell-and-tube exchangers use correction factors.

LMTD method

Q = U A ΔT_lm, with

ΔT_lm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂)

• Parallel flow: ΔT₁ = T_h,in − T_c,in; ΔT₂ = T_h,out − T_c,out.
• Counter flow: ΔT₁ = T_h,in − T_c,out; ΔT₂ = T_h,out − T_c,in.
• If ΔT₁ = ΔT₂ (a balanced counter-flow exchanger), ΔT_lm = ΔT.
• For the same terminal temperatures, counter flow gives a larger LMTD, so it needs less area.

Fouling (deposits on the surfaces) adds resistance: 1/U = 1/h_i + R_f + 1/h_o (thin wall).

Effectiveness–NTU method

• Heat capacity rates: C_h = ṁ_h c_h, C_c = ṁ_c c_c. C_min is the smaller.
• Capacity ratio C = C_min/C_max.
• Q_max = C_min (T_h,in − T_c,in).
• Effectiveness ε = Q/Q_max.
• NTU = UA/C_min.

Boiling and condensation

• Pool boiling curve: natural convection → nucleate boiling (very high h) → critical heat flux → transition → film boiling. Operating beyond the critical heat flux can burn out the surface.
• Condensation: dropwise condensation gives much higher h than filmwise. Nusselt's theory gives filmwise h on vertical plates.

Worked example

Hot oil cools from 150 °C to 90 °C while water heats from 30 °C to 70 °C, in counter flow, with U = 300 W/m²·K and Q = 60 kW.

• ΔT₁ = 150 − 70 = 80; ΔT₂ = 90 − 30 = 60; ΔT_lm = 20/ln(80/60) = 69.5 K.
• A = 60,000/(300 × 69.5) = 2.88 m².

Practice