Exercise set

Thermoelectric Generator and Peltier Cooling Exercises

Worked thermoelectric generator and Peltier cooling exercises for Seebeck voltage, load matching, heat lift, COP, hot-side rejection and release checks.

Branch: Engineering Physics
Content: Exercise set
Updated: Jul 03, 2026
Revision: v1.0.0 · reviewed

These exercises focus on thermoelectric generator and Peltier cooling calculations. The engineering problem is not only to compute voltage, current or heat lift. The release question is whether the installed module has enough temperature difference, electrical load margin, thermal-interface quality, hot-side rejection and validation evidence to support the intended decision.

Use these calculations as screening models. Final design still needs manufacturer curves, temperature-dependent material data, interface-pressure evidence, heat-sink tests, controller limits, condensation checks, converter behavior and environmental qualification.

Release Evidence Notes

A credible thermoelectric release package connects the same thermal boundary through every calculation. The hot source, cold sink, interface resistance, module current, converter load, heat-sink airflow and controller state must describe one installed condition. A module data-sheet rating at a large ideal temperature difference does not prove useful generator power or Peltier cooling in a mounted assembly.

Engineering Boundary Notes

Thermoelectric modules couple heat and electricity. A generator loses useful output when thermal interfaces reduce module temperature difference. A Peltier cooler fails when the hot side cannot reject the cold-side load plus electrical input. A good calculation therefore closes both the electrical and thermal sides.

Common Release Mistakes

using external source and sink temperatures as if they were module face temperatures;
checking cold-side heat lift without hot-side heat rejection;
maximizing generator power while ignoring converter minimum voltage;
treating COP as constant over current and temperature difference;
ignoring condensation, contact pressure, sink fouling, fan failure or controller saturation.

Scenario Map

Scenario	Main calculation	Release decision
Seebeck generation	$V_{oc}=S\Delta T$	Decide whether a converter can start.
Load matching	$P_L=I^2R_L$	Choose power or voltage margin.
Interface derating	$\Delta T_m=\Delta T_{available}-QR_{th}$	Check installed module temperature difference.
Peltier cooling	$Q_h=Q_c+P_e$	Verify hot-side rejection.
Material screening	$ZT=S^2\sigma T/k$	Rank materials without overclaiming device performance.
Release gate	margin, uncertainty and validation evidence	Decide pass, conditional pass or redesign.

Validation Package Checklist

measured source, sink, hot-side and cold-side temperatures;
interface pressure or bond-line evidence;
load voltage, current and converter start behavior;
hot-side heat-sink test at worst credible ambient;
condensation and insulation review for cold surfaces;
uncertainty budget for temperature, voltage, current and heat-flow estimates.

Exercise 1: Seebeck Voltage and Converter Start

A thermoelectric generator has an effective module Seebeck coefficient:

S=0.036\ \text{V/K}

The measured module temperature difference is:

\Delta T=42\ \text{K}

The converter needs $1.1\ \text{V}$ open-circuit before it can start. Compute the open-circuit voltage and decide whether cold start is credible.

Solution

V_{oc}=S\Delta T=0.036(42)=1.51\ \text{V}

The open-circuit voltage exceeds $1.1\ \text{V}$ .

Engineering Comment

The converter can plausibly start at this condition, but the margin is only $0.41\ \text{V}$ . The release review should check the lowest credible source temperature, cold sink warming and converter input threshold tolerance.

Plausibility Check

The voltage is in the low-volt range, which is typical for a compact thermoelectric module under a moderate temperature difference.

Exercise 2: Generator Load Matching

The same module has internal resistance:

R_i=2.8\ \Omega

Use $V_{oc}=1.51\ \text{V}$ and compare $R_L=2.8\ \Omega$ with $R_L=8.0\ \Omega$ .

Solution

For the matched load:

\displaystyle I=\frac{1.51}{2.8+2.8}=0.270\ \text{A}

V_L=0.270(2.8)=0.756\ \text{V}

P_L=0.270^2(2.8)=0.204\ \text{W}

For the larger load:

\displaystyle I=\frac{1.51}{2.8+8.0}=0.140\ \text{A}

V_L=0.140(8.0)=1.12\ \text{V}

P_L=0.140^2(8.0)=0.157\ \text{W}

Engineering Comment

The matched load gives more power, but the larger load gives converter voltage margin. If the converter cannot boost from $0.756\ \text{V}$ , maximum-power loading is not useful at start.

Plausibility Check

The matched-load power also equals:

\displaystyle \frac{V_{oc}^2}{4R_i}=0.204\ \text{W}

Exercise 3: Interface Temperature-Difference Derating

The external hot-to-cold temperature difference is $90\ \text{K}$ . Heat flow through the stack is $18\ \text{W}$ . Combined interface resistance is:

R_{th,int}=1.4\ \text{K/W}

Compute the module temperature difference.

Solution

The interface temperature drop is:

\Delta T_{int}=QR_{th,int}=18(1.4)=25.2\ \text{K}

Therefore:

\Delta T_m=90-25.2=64.8\ \text{K}

Engineering Comment

The module sees only $72\%$ of the external temperature difference. Generator voltage and Peltier capacity should be calculated from $64.8\ \text{K}$ , not from $90\ \text{K}$ .

Plausibility Check

A thermal contact penalty above $25\ \text{K}$ is large enough to justify a mounting-pressure or interface-material review.

Exercise 4: Heat-Sink Rejection for a Peltier Cooler

A Peltier module must remove:

Q_c=14\ \text{W}

Electrical input is:

P_e=28\ \text{W}

The heat sink has:

R_{th,sink}=1.1\ \text{K/W}

Ambient temperature is $30^\circ\text{C}$ . Compute hot-side heat and heat-sink temperature rise.

Solution

Q_h=Q_c+P_e=14+28=42\ \text{W}

\Delta T_{sink}=Q_hR_{th,sink}=42(1.1)=46.2\ \text{K}

T_h=30+46.2=76.2^\circ\text{C}

Engineering Comment

The cooler has created a hot-side rejection problem. A sink temperature near $76^\circ\text{C}$ may reduce heat lift, stress nearby electronics and force the controller into saturation.

Plausibility Check

The hot side rejects three times the cold-side load, so a large heat-sink rise is expected.

Exercise 5: Peltier COP

Using $Q_c=14\ \text{W}$ and $P_e=28\ \text{W}$ , compute the cooling COP.

Solution

\displaystyle COP=\frac{Q_c}{P_e}=\frac{14}{28}=0.50

Engineering Comment

A COP below one is common for demanding thermoelectric cooling. It means every watt removed from the cold side brings two watts of electrical heat that the hot side must reject.

Plausibility Check

The hot-side heat is $42\ \text{W}$ , consistent with $Q_h=Q_c+P_e$ .

Exercise 6: Current Tradeoff

At $I=3.0\ \text{A}$ , a module lifts $18\ \text{W}$ and consumes $36\ \text{W}$ . At $I=4.5\ \text{A}$ , it lifts $21\ \text{W}$ and consumes $68\ \text{W}$ . Compare incremental cooling and incremental electrical input.

Solution

\Delta Q_c=21-18=3\ \text{W}

\Delta P_e=68-36=32\ \text{W}

Incremental COP is:

\displaystyle COP_{inc}=\frac{3}{32}=0.094

Engineering Comment

The higher current buys little extra cooling and creates a large hot-side load. Unless those $3\ \text{W}$ are essential, the lower current is the better release point.

Plausibility Check

Thermoelectric coolers often lose efficiency at high current because Joule heating rises strongly.

Exercise 7: Material Figure of Merit

A candidate thermoelectric material has:

S=220\ \mu\text{V/K},\quad \sigma=8.5\times10^4\ \text{S/m},\quad k=1.6\ \text{W/(m K)},\quad T=320\ \text{K}

Compute $ZT$ .

Solution

Convert $S$ :

S=220\times10^{-6}\ \text{V/K}

\displaystyle ZT=\frac{S^2\sigma T}{k}

\displaystyle ZT=\frac{(220\times10^{-6})^2(8.5\times10^4)(320)}{1.6}=0.823

Engineering Comment

The material is useful enough for screening, but $ZT$ is not a device guarantee. Contacts, geometry, interfaces and operating temperature span still decide installed power or cooling.

Plausibility Check

A $ZT$ of order one is plausible for engineering thermoelectric materials.

Exercise 8: Daily Harvested Energy

A generator supplies $0.18\ \text{W}$ for $7\ \text{h}$ each day. Converter efficiency is $72\%$ . Compute delivered daily energy.

Solution

E_{module}=0.18(7)=1.26\ \text{Wh}

E_{delivered}=0.72(1.26)=0.907\ \text{Wh}

Engineering Comment

This can support a low-power sensor, not a continuously transmitting device. The load budget must include sleep current, radio bursts and cold-start reserve.

Plausibility Check

Less than $1\ \text{Wh/day}$ is consistent with sub-watt harvesting over a short hot period.

Exercise 9: Converter Cold-Start Margin

A converter requires $0.80\ \text{V}$ after loading. The generator has $V_{oc}=1.40\ \text{V}$ and $R_i=3.0\ \Omega$ . Startup input resistance is $6.0\ \Omega$ . Compute loaded voltage.

Solution

\displaystyle I=\frac{1.40}{3.0+6.0}=0.1556\ \text{A}

V_L=I(6.0)=0.933\ \text{V}

Engineering Comment

Startup is credible, but the voltage margin is only $0.133\ \text{V}$ . A colder source, higher internal resistance or converter tolerance could prevent start.

Plausibility Check

Because the load is twice the internal resistance, the loaded voltage is two-thirds of open circuit.

Exercise 10: Contact Pressure and Interface Resistance

An interface improves from $1.8\ \text{K/W}$ to $0.9\ \text{K/W}$ after clamp pressure is corrected. Heat flow is $16\ \text{W}$ . Compute recovered module temperature difference.

Solution

Old drop:

\Delta T_1=16(1.8)=28.8\ \text{K}

New drop:

\Delta T_2=16(0.9)=14.4\ \text{K}

Recovered module difference:

28.8-14.4=14.4\ \text{K}

Engineering Comment

If voltage is proportional to temperature difference, this mounting correction can be worth more than changing the module. The release record should include clamp method and torque or compression evidence.

Plausibility Check

Halving thermal resistance halves interface temperature drop at the same heat flow.

Exercise 11: Dew-Point Guard for a Cold Plate

A Peltier cold plate is expected to reach $7^\circ\text{C}$ . The local dew point is $10^\circ\text{C}$ . A $3\ \text{K}$ safety margin above dew point is required. Does the design pass?

Solution

Required minimum cold-plate temperature is:

T_{min}=10+3=13^\circ\text{C}

The predicted plate temperature is:

7^\circ\text{C}<13^\circ\text{C}

Engineering Comment

The design fails the condensation guard. It needs insulation, purge, drainage, dew-point control or a warmer setpoint.

Plausibility Check

Any surface below dew point is already at condensation risk; the specified $3\ \text{K}$ guard makes the failure stronger.

Exercise 12: Hot-Side Feedback on Cooling Capacity

A module can lift $18\ \text{W}$ at the target temperature difference if the hot side is $45^\circ\text{C}$ . Test data show lift drops by $0.35\ \text{W/K}$ above $45^\circ\text{C}$ . The actual hot side is $61^\circ\text{C}$ . Compute corrected lift.

Solution

\Delta T_h=61-45=16\ \text{K}

\Delta Q_c=0.35(16)=5.6\ \text{W}

Q_{c,actual}=18-5.6=12.4\ \text{W}

Engineering Comment

If the cold load is above $12.4\ \text{W}$ , the cooler will not hold the setpoint. Hot-side temperature must be part of the acceptance test.

Plausibility Check

A hotter hot side reducing heat lift is the expected direction.

Exercise 13: Series Modules for Voltage

Three identical generator modules each produce $0.62\ \text{V}$ open circuit and have $1.5\ \Omega$ internal resistance. They are wired in series. Compute total open-circuit voltage and internal resistance.

Solution

V_{oc,total}=3(0.62)=1.86\ \text{V}

R_{i,total}=3(1.5)=4.5\ \Omega

Engineering Comment

Series wiring helps converter start voltage but also increases source resistance. The load and converter input impedance must be checked at the array level.

Plausibility Check

For series sources, voltages and resistances both add.

Exercise 14: Parallel Modules for Current

Two identical modules each have $V_{oc}=1.2\ \text{V}$ and $R_i=2.0\ \Omega$ . In parallel, estimate equivalent open-circuit voltage and internal resistance.

Solution

The equivalent voltage remains:

V_{oc,eq}=1.2\ \text{V}

The resistance halves:

\displaystyle R_{i,eq}=\frac{2.0}{2}=1.0\ \Omega

Engineering Comment

Parallel wiring improves current capability but does not help a converter that needs higher startup voltage.

Plausibility Check

Parallel identical sources preserve voltage and reduce source resistance.

Exercise 15: Guard Band on Cooling Capacity

A cooler is predicted to lift $16.5\ \text{W}$ . The load is $14.0\ \text{W}$ . Expanded uncertainty on the lift estimate is $1.8\ \text{W}$ . Use a conservative pass rule:

Q_c-U \geq Q_{load}

Solution

Q_c-U=16.5-1.8=14.7\ \text{W}

Since $14.7\geq14.0$ , the calculation passes.

Engineering Comment

The pass is narrow. A fouled heat sink, warmer ambient or worse interface could consume the margin, so the release should remain conditional on hot-side validation.

Plausibility Check

The raw margin is $2.5\ \text{W}$ , and the guarded margin is only $0.7\ \text{W}$ .

Exercise 16: Sink Fouling Sensitivity

A heat sink initially has $R_{th}=0.75\ \text{K/W}$ . Dust fouling increases it by $35\%$ . The hot-side heat is $40\ \text{W}$ . Compute added temperature rise.

Solution

R_{th,new}=0.75(1.35)=1.0125\ \text{K/W}

Added resistance:

\Delta R_{th}=1.0125-0.75=0.2625\ \text{K/W}

Added rise:

\Delta T=40(0.2625)=10.5\ \text{K}

Engineering Comment

A $10.5\ \text{K}$ hot-side penalty can erase Peltier capacity. Maintenance interval and filter condition belong in the release evidence.

Plausibility Check

The added rise is large because the rejected heat is large.

Exercise 17: Waste-Heat Harvester Viability

A remote sensor needs $0.55\ \text{Wh/day}$ . A generator delivers $0.95\ \text{Wh/day}$ before storage losses. Storage and regulator efficiency is $68\%$ . Decide if the energy budget closes.

Solution

E_{usable}=0.95(0.68)=0.646\ \text{Wh/day}

Margin:

0.646-0.55=0.096\ \text{Wh/day}

Engineering Comment

The budget closes with a small margin. The review should test cold days, startup after outage and radio retransmission events.

Plausibility Check

The usable energy is lower than harvested energy because storage and conversion losses are included.

Exercise 18: Thermoelectric Release Gate

A Peltier-cooled detector has:

Q_{load}=9.5\ \text{W},\quad Q_{c,guarded}=10.8\ \text{W}

The hot-side temperature limit is $65^\circ\text{C}$ . Worst-case hot-side test reaches $63^\circ\text{C}$ . The cold plate is $2\ \text{K}$ above the condensation guard. Decide release status.

Solution

Cooling margin:

10.8-9.5=1.3\ \text{W}

Hot-side margin:

65-63=2^\circ\text{C}

Condensation margin is $2\ \text{K}$ .

Engineering Comment

The design passes the numerical gates, but every margin is narrow. Release should be conditional on documented airflow, ambient limit, dew-point control and periodic heat-sink inspection.

Plausibility Check

All margins are positive, so the correct decision is not immediate rejection. The small margins explain why the release is conditional rather than robust.

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