Peltier Current and Heat Reversal

The Peltier effect is a thermoelectric phenomenon where a current passing through a junction of two different conductors drives a heat flow. In a thermoelectric cooler, this reversible heat transport can be used to extract heat from a cold reservoir, provided it overcomes the inherent parasitic effects: Joule heating (resistive losses) and thermal conduction. By reversing the direction of the electrical current, the reversible heat pumping also reverses direction, whereas the irreversible Joule heating strictly adds thermal energy regardless of current direction.

Worked Example

Consider a simplified thermoelectric leg with a resistance of 0.02 Ω and a thermal conductance of 0.005 W/K. Let the Seebeck coefficient be 0.0002 V/K. We want to calculate the heat flows when the cold side is at 300 K, the temperature gap is 0 K (both sides are at 300 K), and a current of 2 A is applied.

Calculate the reversible heat (Peltier heat): Q_(reversible) = S · T_c · I
Calculate the cold-side Joule heating. We assume half of the total Joule heat goes to the cold side: Q_(Joule,cold) = 0.5 · I² · R
Calculate the conductive heat leak: Q_(cond) = K · Δ T
The net cooling heat extracted from the cold side is: Q_c = Q_(reversible) - Q_(Joule,cold) - Q_(cond)
The total electrical power input is the sum of Joule heating and the work done against the Seebeck voltage: P = I² · R + S · Δ T · I
The hot side rejects heat: Q_h = Q_c + P

I = 2 A, Tc = 300 K, dT = 0 K, S = 0.0002 V/K R = 0.02 Ω, K = 0.005 W/K Q_reversible = 0.0002 * 300 * 2 = 0.12 W Q_Joule,cold = 0.5 * 2^2 * 0.02 = 0.04 W Q_cond = 0.005 * 0 = 0.0 W Q_c = 0.12 - 0.04 - 0.0 = 0.08 W P = (2^2 * 0.02) + (0.0002 * 0 * 2) = 0.08 W Q_h = 0.08 + 0.08 = 0.16 W COP = Q_c / P = 0.08 / 0.08 = 1.0

In this scenario, a net cooling of 0.08 W is achieved at the cold junction, demonstrating active thermoelectric cooling. If the current is reversed to -2 A, the reversible heat becomes -0.12 W, meaning heat is pumped into the cold side, combining with the 0.04 W of Joule heating to actively heat the cold side with -0.16 W of "cooling".

Interactive Experiment

You can explore these relationships in the Peltier Current and Heat Reversal module. The simulation allows you to adjust the current, temperatures, and Seebeck coefficient to observe the resulting heat flows.

Set the Current to 2 A, with Cold temperature at 300 K and Temperature gap at 0 K. Observe the positive cooling heat indicating active cooling.
Change the Current to -2 A. Notice that the reversible heat arrow flips direction, and the cooling heat becomes negative (net heating), but the Joule heat arrow remains pointing into the cold side.
Increase the Temperature gap to 20 K (with current back at 2 A). See how the conductive leak increases, eventually overwhelming the reversible heat pumping and resulting in a loss of net cooling.
Set the Seebeck coefficient to 0. Observe that all reversible heat pumping ceases, leaving only Joule heating and thermal conduction.

Limitations

This model makes several simplifying assumptions. It uses a constant-property model, meaning the Seebeck coefficient, electrical resistance, and thermal conductance are assumed to be independent of temperature. In real thermoelectric materials, these properties vary significantly with temperature.

Additionally, the model assumes that exactly half of the Joule heat generated within the leg flows to the cold junction and half to the hot junction. The calculation is a 1D idealization of a single leg and does not account for contact resistances, thermal losses to the environment, or the performance of a complete multi-couple device.

References

The governing equations for a thermoelectric leg and the separation of reversible Peltier effects from irreversible Joule heating and conduction are based on standard solid-state physics curricula. For a detailed derivation of these transport equations, see the University of Illinois PHYS 524 Lecture Notes on Thermoelectrics.

Peltier Current and Heat Reversal