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Heat exchanger duty (LMTD)

Heat transfer

The log-mean-temperature-difference rating equation, and the sharply nonlinear sensitivity of duty to the terminal temperature approach.

Governing equation

Q=UAΔTlm,ΔTlm=ΔT1ΔT2ln(ΔT1/ΔT2)Q = U A\,\Delta T_{\mathrm{lm}},\qquad \Delta T_{\mathrm{lm}} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1/\Delta T_2)}
QQ
Heat duty (W) — an unknown.
ΔT1\Delta T_1
Terminal temperature difference at end 1 (K) — an unknown.
ΔT2\Delta T_2
Terminal temperature difference at end 2 (K).
UU
Overall heat transfer coefficient (W m⁻² K⁻¹).
AA
Heat transfer area (m²).

Derivation

  1. 1
    dQ=U(ThTc)dAdQ = U\,(T_h - T_c)\,dA

    Local rate equation across a differential area, with the driving force the local hot-cold temperature difference ΔT.

  2. 2
    d(ΔT)dA=UΔT(1m˙hch+1m˙ccc)\frac{d(\Delta T)}{dA} = -U\,\Delta T\left(\frac{1}{\dot m_h c_h} + \frac{1}{\dot m_c c_c}\right)

    For constant capacity rates the driving force decays exponentially along the exchanger.

  3. 3
    Q=UAΔT1ΔT2ln(ΔT1/ΔT2)Q = U A\,\frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1/\Delta T_2)}

    Integrating over the area replaces the varying ΔT with its log-mean — exact for constant U and capacity rates.

Jacobian

The matrix Jij=Fi/xjJ_{ij} = \partial F_i / \partial x_j of the residual vector with respect to the unknowns, at the representative operating point. Every value below is computed by the exact forward-mode autodiff engine the solver uses — click a non-zero cell to see the symbolic derivative, each one verified against that engine in our test suite. This is the block a Newton step factorizes.

∂/∂Q∂/∂dT1
Duty energy balance

Shaded cells are non-zero (the sparsity pattern); blanks are structural zeros.

Symbolic derivative
F1Q=\dfrac{\partial F_{1}}{\partial\, \mathrm{Q}} =
11
Evaluated at the operating point: 1

For the residual Duty energy balance with respect to Q. This exact expression is asserted equal to the autodiff value in the test suite.

Operating point

Representative operating point — a real solution of this instance.

Q = 1.637e+5dT1 = 50dT2 = 20U = 500A = 10

References

  • Incropera & DeWitt, Fundamentals of Heat and Mass Transfer, 6th ed., §11.3.
  • Perry's Chemical Engineers' Handbook, 8th ed., §11 (heat-transfer equipment).

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