Heat Exchanger LMTD

Size or rate counter-flow and parallel-flow heat exchangers using the log mean temperature difference and effectiveness-NTU methods. Reports LMTD, heat duty, required area, effectiveness and NTU.


TEMA · Cengel & Ghajar · Incropera

Configuration

Flow arrangement
Solve for

Temperatures & Properties

°C
°C
°C
°C
W/m²K
kg/s
J/kg·K

Results

44.81K
LMTD
44.81K
F × LMTD
40.000 × 10³W
Heat duty Q
1.785
Required area
50.00K
ΔT₁
40.00K
ΔT₂
0.500
Effectiveness ε
0.893
NTU
0.750
Capacity ratio Cr

Temperature Profile

0255075100Length along exchanger (%)0255075100°C
  • Hot
  • Cold

About Heat Exchanger LMTD Calculator

The heat exchanger LMTD calculator sizes or rates a two-stream heat exchanger using the Log Mean Temperature Difference method and cross-checks the result with the effectiveness-NTU method. Enter the four terminal temperatures (hot inlet/outlet and cold inlet/outlet), the overall heat-transfer coefficient U, and either the area to rate an existing unit or a stream flow and specific heat to size a new one.

It reports the two terminal temperature differences, the LMTD, the effective mean temperature difference F x LMTD, the heat duty Q, the required or supplied area A, and -- when stream capacity rates are known -- the effectiveness epsilon, the number of transfer units NTU, and the capacity ratio Cr. The arrangement (counter-flow or parallel-flow) changes how the terminal differences are paired and is fully accounted for.

How It Works

  1. Pick the flow arrangement: counter-flow (streams move in opposite directions) or parallel-flow (same direction).
  2. Enter the hot inlet/outlet and cold inlet/outlet temperatures, and the overall coefficient U in W/(m^2.K).
  3. Choose to solve for area (supply a stream mass flow and cp so the energy balance fixes the duty) or for duty (supply the area A).
  4. The calculator forms dT1 and dT2 for the chosen arrangement, evaluates LMTD = (dT1 - dT2)/ln(dT1/dT2), applies the optional correction factor F, then computes Q = U A F LMTD and the effectiveness-NTU figures.

Worked Example

A counter-flow exchanger cools a hot stream from 100 C to 60 C while heating a cold stream from 20 C to 50 C, with U = 500 W/(m^2.K). The hot stream has m*cp = 1000 W/K. Terminal differences are dT1 = 100 - 50 = 50 K and dT2 = 60 - 20 = 40 K, so LMTD = (50 - 40)/ln(50/40) = 10/0.22314 = 44.81 K. The duty is Q = 1000 x (100 - 60) = 40,000 W, giving required area A = Q/(U x LMTD) = 40000/(500 x 44.81) = 1.785 m^2. With Cmin = 1000 W/K and Cmax = 1333 W/K, Cr = 0.75, NTU = U A/Cmin = 0.893 and effectiveness epsilon = Q/(Cmin x (100 - 20)) = 40000/80000 = 0.50.

Formulas

Terminal temperature differences
counter-flow: dT1 = Th_in - Tc_out, dT2 = Th_out - Tc_in | parallel-flow: dT1 = Th_in - Tc_in, dT2 = Th_out - Tc_out
Log Mean Temperature Difference
LMTD = (dT1 - dT2) / ln(dT1 / dT2) [limit: LMTD = dT1 when dT1 = dT2]
Heat duty (LMTD method)
Q = U * A * F * LMTD
Energy balance per stream
Q = m_h * cp_h * (Th_in - Th_out) = m_c * cp_c * (Tc_out - Tc_in)
Effectiveness-NTU
NTU = U A / Cmin, Cr = Cmin/Cmax | CF: eps = (1 - e^-NTU(1-Cr))/(1 - Cr e^-NTU(1-Cr)) | PF: eps = (1 - e^-NTU(1+Cr))/(1 + Cr)

Standards & References

  • TEMA (Tubular Exchanger Manufacturers Association) standards
  • Cengel & Ghajar, Heat and Mass Transfer
  • Incropera, Fundamentals of Heat and Mass Transfer
  • Kays & London, Compact Heat Exchangers (effectiveness-NTU)

Frequently Asked Questions

What is the difference between the LMTD and effectiveness-NTU methods?

LMTD is convenient when all four terminal temperatures are known (sizing): you compute the mean driving temperature and then the area. Effectiveness-NTU is convenient when outlet temperatures are unknown (rating): you compute effectiveness from NTU and capacity ratio. This tool reports both so you can cross-check.

Why does counter-flow need less area than parallel-flow?

For the same inlet and outlet temperatures, counter-flow produces a larger and more uniform temperature difference along the exchanger, so its LMTD is higher. Because Q = U A F LMTD, a larger LMTD means less area is needed for the same duty. Parallel-flow also cannot raise the cold outlet above the hot outlet.

What happens when the two terminal temperature differences are equal?

When dT1 equals dT2 the formula (dT1 - dT2)/ln(dT1/dT2) becomes 0/0. The analytic limit is simply dT1, so the calculator returns the common terminal difference instead of dividing by zero.

What is the LMTD correction factor F?

F accounts for shell-and-tube or cross-flow arrangements that are neither pure counter-flow nor pure parallel-flow. It is 1 for ideal counter-flow or parallel-flow and less than 1 otherwise, increasing the required area. Read F from TEMA charts for the chosen configuration and enter it here.