Heat exchanger LMTD

Introduction

The log-mean temperature difference method answers the sizing question for any two-fluid heat exchanger: given the four terminal temperatures, how much heat is exchanged and how much surface is needed for a given overall heat transfer coefficient. This calculator computes the LMTD for counterflow and parallel flow, the Bowman correction factor F for a one-shell-pass shell-and-tube unit with an even number of tube passes, the duty from either stream's mass flow and specific heat (or taken directly), the service U after fouling, and the required area. It also reports effectiveness, NTU and capacity ratio, which follow from the temperatures alone and link to the ε-NTU method.

Marine engineers use it for lube-oil and jacket-water coolers, fuel heaters and central coolers; process and HVAC engineers for any shell-and-tube or plate exchanger pre-sizing; students for the standard heat transfer course problems.

Formulas

Heat duty (energy balance on either stream):

Q = ṁ_h c_p,h (T_h,in − T_h,out) = ṁ_c c_p,c (T_c,out − T_c,in)

• ṁ — mass flow, kg/s; c_p — specific heat, J/(kg·K); T — terminal temperatures, °C

Log-mean temperature difference:

LMTD = (ΔT₁ − ΔT₂) / ln(ΔT₁/ΔT₂)       (LMTD = ΔT₁ when ΔT₁ = ΔT₂)
Counterflow: ΔT₁ = T_h,in − T_c,out,  ΔT₂ = T_h,out − T_c,in
Parallel:    ΔT₁ = T_h,in − T_c,in,   ΔT₂ = T_h,out − T_c,out

Correction factor for one shell pass, 2/4/6… tube passes (Bowman, Mueller and Nagle 1940), applied to the counterflow LMTD:

R = (T_h,in − T_h,out) / (T_c,out − T_c,in),   P = (T_c,out − T_c,in) / (T_h,in − T_c,in)
F = √(R²+1) ln[(1−P)/(1−RP)] / { (R−1) ln[ (2 − P(R+1−√(R²+1))) / (2 − P(R+1+√(R²+1))) ] }
R = 1:  F = P√2 / { (1−P) ln[ (2 − P(2−√2)) / (2 − P(2+√2)) ] }

Area and fouling:

1/U_service = 1/U_clean + R_f
A = Q / (U_service × F × LMTD)

• U — overall heat transfer coefficient, W/(m²·K); R_f — total fouling resistance, m²·K/W

Effectiveness, NTU and capacity ratio (from the temperatures alone):

ε = max(ΔT_h, ΔT_c) / (T_h,in − T_c,in)
NTU = U A / C_min = max(ΔT_h, ΔT_c) / (F × LMTD)
C_r = min(ΔT_h, ΔT_c) / max(ΔT_h, ΔT_c)

Assumptions and limitations

• Steady state, no heat loss to the surroundings, constant specific heats and constant U along the exchanger. For large temperature ranges or condensing/boiling, split the exchanger into zones.
• The LMTD method needs all four temperatures. When only inlets are known (rating an existing exchanger), use the ε-NTU method; the ε, NTU and C_r outputs here are the bridge.
• The F factor is the Bowman analytical result for one shell pass with any even number of tube passes. Two or more shell passes in series, cross-flow and plate exchangers have different F curves; for plate exchangers F is close to 1 in pure counterflow arrangement.
• F below about 0.75 means the design sits on the steep part of the curve and small temperature changes cause large area changes; add shell passes or use counterflow.
• Fouling resistances are the TEMA design values; actual fouling depends on velocity, temperature and cleaning schedule. Fouling on each side should be referred to the same area basis as U.
• U values in the help text are order-of-magnitude ranges for pre-sizing; a real design computes U from the film coefficients, wall resistance and fin efficiency.

Worked example

Incropera, Example 11.1. A counterflow concentric-tube cooler takes lubricating oil from 100 °C to 60 °C at 0.1 kg/s (c_p = 2 131 J/(kg·K)); cooling water at 0.2 kg/s enters at 30 °C. U = 38.1 W/(m²·K) based on the 25 mm inner tube.

1. Duty: 0.1 × 2 131 × (100 − 60) = 8 524 W.
2. Water outlet: 30 + 8 524 / (0.2 × 4 178) = 40.2 °C.
3. Terminal differences, counterflow: ΔT₁ = 100 − 40.2 = 59.8 K, ΔT₂ = 60 − 30 = 30 K.
4. LMTD: (59.8 − 30) / ln(59.8/30) = 43.2 K.
5. Area: 8 524 / (38.1 × 43.2) = 5.18 m²; as a 25 mm tube that is 5.18 / (π × 0.025) = 65.9 m, Incropera's answer.
6. Effectiveness: oil has the smaller capacity rate; ε = 40 / 70 = 0.571; NTU = 40 / 43.2 = 0.93; C_r = 10.2 / 40 = 0.255.

In parallel flow the same temperatures give LMTD = 39.75 K and about 9 % more area. In a 1-2 shell-and-tube with the water heated to 60 °C instead (R = 1, P = 0.5), F = 0.80 and the effective difference drops from 40 K to 32 K.

FAQ

Related calculators

• Pipe flow and pressure drop — tube-side and shell-side nozzle losses.
• Pump and fan affinity laws — cooling-water flow changes that move the duty.
• Thermal expansion and restraint stress — tube bundle expansion between the two fluid temperatures.
• Engineering unit converter — Btu/(h·ft²·°F), °F, kJ/(kg·K).