Reference ID: MET-B685 | Process Engineering Reference Sheets Calculation Guide
Introduction & Context
The one-dimensional steady-state heat-flow calculation through a plane wall is the simplest—and most frequently used—analogy for transport phenomena in process engineering. It couples three core ideas that re-appear in momentum, heat and mass transfer:
A driving force (temperature difference, pressure difference, concentration difference).
A geometric resistance (thickness ÷ area).
A material property that converts the gradient into a flux (thermal conductivity, viscosity, diffusivity).
Because the same algebraic skeleton is reused for Fourier’s law, Ohm’s law, Fick’s law and Darcy’s law, mastering the wall analogy gives engineers a mental template for scaling equipment, estimating losses, checking insulation thickness, or scoping energy balances in reactors, heat exchangers, dryers, building envelopes and pipelines.
Methodology & Formulas
Convert thickness to base units
\( L = \dfrac{t_{\text{mm}}}{1000} \)
Compute cross-sectional area
\( A = H \cdot W \)
Define the driving force
\( \Delta T = T_{\text{in}} - T_{\text{out}} \)
High-speed compressible flow: density changes make the highway itself expand/contract (choked flow).
When any of these occur, park the analogy and solve the full transport equations.
Worked Example – Heat Loss Through a Plaster Wall
A small process building is kept at 22 °C while the outside winter air sits at 2 °C. One of the long side walls is a monolithic plaster panel 2 m high, 3 m wide and 120 mm thick. We need to know the total conductive heat loss (in watts) and the heat-flux density (W m-2) so that the HVAC engineer can size the trim heater.
Knowns
kplaster = 0.45 W m-1 K-1
Height = 2.0 m
Width = 3.0 m
Thickness = 0.120 m (120 mm)
Tindoor = 22 °C
Toutdoor = 2 °C
Step-by-step calculation
Compute the wall area: \( A = 2.0 \times 3.0 = 6.0 \; \text{m}^2 \)
Temperature difference: \( \Delta T = 22 - 2 = 20 \; \text{K} \)