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Resistivity Conductivity

Reference data and engineering information about resistivity conductivity for hvac systems applications.

resistivityconductivityData Table

Overview

Engineering reference data for Resistivity Conductivity in HVAC systems.

Key Formulas

Electrical Resistance

R=ρ⋅LAR = \rho \cdot \frac{L}{A}

Resistance of a conductor depends on resistivity, length, and cross-section.

Electrical Conductivity

σ=1ρ\sigma = \frac{1}{\rho}

Conductivity is the reciprocal of resistivity.

Temperature Dependence of Resistivity

ρ(T)=ρ0[1+α(T−T0)]\rho(T) = \rho_0 \left[1 + \alpha (T - T_0)\right]

Resistivity increases with temperature for most metals.

Power Dissipation

P=I2R=V2RP = I^2 R = \frac{V^2}{R}

Joule heating in a resistive conductor.

Variables

Symbol Description Unit
ρ\rho Electrical resistivity Ω·m
σ\sigma Electrical conductivity S/m
RR Electrical resistance Ω
LL Conductor length m
AA Cross-sectional area m²
α\alpha Temperature coefficient of resistance 1/°C
T0T_0 Reference temperature (usually 20°C) °C
PP Power dissipated (Joule heating) W

Material Properties Table

Temperature Conversion Factors

Unit Conversions

Electrical resistivity can be expressed in multiple units:

1 Ω⋅m=0.001 kΩ⋅m1 Ω⋅m=100 Ω⋅cm1 Ω⋅m=39.37 Ω⋅inch1 Ω⋅m=3.28 Ω⋅foot\begin{align*} 1\ \Omega\cdot m &= 0.001\ k\Omega\cdot m \\ 1\ \Omega\cdot m &= 100\ \Omega\cdot cm \\ 1\ \Omega\cdot m &= 39.37\ \Omega\cdot inch \\ 1\ \Omega\cdot m &= 3.28\ \Omega\cdot foot \end{align*}

Examples

Example: Aluminum Wire Resistance

Calculate the resistance of an aluminum cable with:

  • Length: L=10 mL = 10\ m
  • Cross-sectional area: A=3 mm2=3×10−6 m2A = 3\ mm^2 = 3 \times 10^{-6}\ m^2
  • Resistivity of aluminum: ρ=2.65×10−8 Ω⋅m\rho = 2.65 \times 10^{-8}\ \Omega\cdot m

Using the resistance formula:

R=ρLA=(2.65×10−8 Ω⋅m)(10 m)3×10−6 m2=0.088 ΩR = \frac{\rho L}{A} = \frac{(2.65 \times 10^{-8}\ \Omega\cdot m)(10\ m)}{3 \times 10^{-6}\ m^2} = 0.088\ \Omega

Example: Resistivity Change with Temperature

For a copper conductor, calculate the resistivity change when temperature increases from 20°C20°C to 50°C50°C:

  • Initial resistivity at 20°C20°C: ρ0=1.724×10−8 Ω⋅m\rho_0 = 1.724 \times 10^{-8}\ \Omega\cdot m
  • Temperature coefficient for copper: α=3.93×10−3 1/°C\alpha = 3.93 \times 10^{-3}\ 1/°C
  • Temperature change: ΔT=50°C−20°C=30°C\Delta T = 50°C - 20°C = 30°C

The change in resistivity is:

Δρ=ρ0αΔT=(1.724×10−8 Ω⋅m)(3.93×10−3 1/°C)(30°C)=2.03×10−9 Ω⋅m\Delta \rho = \rho_0 \alpha \Delta T = (1.724 \times 10^{-8}\ \Omega\cdot m)(3.93 \times 10^{-3}\ 1/°C)(30°C) = 2.03 \times 10^{-9}\ \Omega\cdot m

The new resistivity at 50°C50°C:

ρ50=ρ0+Δρ=1.927×10−8 Ω⋅m\rho_{50} = \rho_0 + \Delta \rho = 1.927 \times 10^{-8}\ \Omega\cdot m

Key Properties and Notes

  1. Conductivity Relationship: Electrical conductivity σ\sigma is the reciprocal of resistivity: σ=1ρ\sigma = \frac{1}{\rho} Conductivity is measured in siemens per meter (S/m), where 1 S/m=1/(Ω⋅m)1\ S/m = 1/(\Omega\cdot m).

  2. Temperature Dependence: For most metals, resistivity increases approximately linearly with temperature over moderate ranges: ρ(T)=ρ0[1+α(T−T0)]\rho(T) = \rho_0[1 + \alpha(T - T_0)] where ρ0\rho_0 is resistivity at reference temperature T0T_0 (typically 20°C20°C).

  3. Material Classification:

    • Conductors: ρ<10−5 Ω⋅m\rho < 10^{-5}\ \Omega\cdot m (e.g., copper, aluminum)
    • Semiconductors: 10−3<ρ<103 Ω⋅m10^{-3} < \rho < 10^{3}\ \Omega\cdot m (e.g., silicon, germanium)
    • Insulators: ρ>108 Ω⋅m\rho > 10^{8}\ \Omega\cdot m (e.g., glass, rubber)

  4. Practical Considerations: Resistivity values depend on:

    • Material purity and alloy composition
    • Temperature and thermal history
    • Mechanical stress and crystal structure
    • Frequency of applied current (for AC applications)

Interactive Charts

References