Introduction & Context

Electrical resistance of a conductor is a key parameter in ohmic heating, electro-pasteurisation and other direct joule-heating processes. Knowing the resistance allows engineers to predict current, power density and temperature rise, ensuring uniform heating while avoiding run-away conditions that degrade product quality. Typical applications include food sterilisation, sludge treatment and rapid water heaters.

Methodology & Formulas

Correct conductivity for temperature
Conductivity increases approximately linearly with temperature according to \[ \kappa(T)=\kappa_{\text{ref}}\bigl[1+\alpha\,(T-T_{\text{ref}})\bigr] \] where
- \(\kappa_{\text{ref}}\) is the conductivity at the reference temperature \(T_{\text{ref}}\),
- \(\alpha\) is the temperature coefficient of conductivity.
Compute electrical resistance
For a uniform conductor the resistance is \[ R=\frac{L}{A\,\kappa(T)} \] with
- \(L\) – electrode gap (length of current path),
- \(A\) – cross-sectional area perpendicular to current.

Validity regime for food ohmic-heating calculations
Parameter	Lower limit	Upper limit	Remark
Geometric inputs	\(L>0,\;A>0\)	—	Must be strictly positive
Conductivity	\(\kappa_{\text{ref}}>0\)	—	Must be strictly positive
Product temperature	\(-40\;^{\circ}\text{C}\)	\(150\;^{\circ}\text{C}\)	Outside this range a warning is issued

Resistance rises with temperature. For copper, use the linear approximation:

R₂ = R₁ [1 + 0.00393 (T₂ – T₁)] where T is in °C
At 80 °C a conductor has ≈ 30 % more resistance than at 20 °C
Size feeders and cable trays for the worst-case summer ambient plus I²R self-heating

Copper is the baseline; for aggressive areas pick:

Tinned copper for mild corrosion
Copper-clad aluminium for weight saving on long tray runs
Silver-plated copper for low-thermal-EMF junctions in thermocouple circuits

Use the resistance route-length method:

ΔV = 2 × I × ρ × L / A (multiply by √3 for three-phase)
Keep ΔV < 5 % at motor terminals during start-up; if > 5 % upsize cable or shorten run
Remember ρ for copper is 1.72 × 10⁻⁸ Ω·m at 20 °C and adjust for temperature

Field readings can read high because of:

Loose compression lugs adding contact resistance
Strand damage during pulling that reduces effective cross-section
High ambient temperature or nearby steam tracing raising conductor temperature
Verify with a micro-ohmmeter, re-torque lugs, and compare against temperature-corrected specs

Worked Example: Resistance of a Copper Bus-Bar in a Heated Enclosure

A process skid uses a 120 mm long copper bus-bar to carry 400 A DC current. During summer operation the enclosure stabilises at 65 °C. We need to verify that the bar’s resistance at this temperature is low enough to keep the I²R losses within specification.

Knowns

Reference temperature, \(T_{\text{REF}}\) = 25.0 °C
Linear temperature coefficient of resistivity, α = 0.021 K⁻¹
Bus-bar length, L = 0.120 m
Cross-sectional area, A = 0.0045 m²
Reference electrical conductivity, κ_REF = 0.850 S m⁻¹
Operating temperature, T_C = 65.0 °C

Step-by-Step Calculation

Compute the conductivity at 65 °C using the linear temperature model: \[ \kappa(T) = \kappa_{\text{REF}}\left[1 + \alpha(T - T_{\text{REF}})\right] \] \[ \kappa(65\,^\circ\text{C}) = 0.850\left[1 + 0.021(65 - 25)\right] = 0.850 \times 1.84 = 1.564\ \text{S m}^{-1} \]
Determine the electrical resistivity at 65 °C: \[ \rho = \frac{1}{\kappa} = \frac{1}{1.564} = 0.639\ \Omega\,\text{m} \]
Calculate the DC resistance of the bar: \[ R = \rho\frac{L}{A} = 0.639 \times \frac{0.120}{0.0045} = 17.05\ \Omega \]

Final Answer

The bus-bar resistance at 65 °C is 17.05 Ω. This value can now be used to check the I²R heat generation against the skid’s thermal budget.

"Un projet n'est jamais trop grand s'il est bien conçu." — André Citroën

"La difficulté attire l'homme de caractère, car c'est en l'étreignant qu'il se réalise." — Charles de Gaulle