Peltier cooling inside an enclosure: when it works

Short answer: a thermoelectric (Peltier) cooler can cool a sealed cabinet below loft temperature only if its hot-side heat leaves the cooled space. The power supply can be inside, provided its losses count toward the cooling load. This example puts the supply outside to reduce that load. At an assumed coefficient of performance (COP) of 0.5, the hot side rejects 3 W for every 1 W removed from the cabinet. That ratio changes with operating conditions; verify the selected cooler's curves rather than treating 3 W as a universal sizing rule. If ventilation lowers the loft temperature enough, passive cabinet cooling may meet the electronics' limits without a heat pump.

Why 40 C loft air cools nothing below 40 C

Air at 40 C cannot cool anything that sits below 40 C. Blowing loft air harder over the cabinet only speeds the slide toward loft temperature. A filtered fan trades a dust problem for a cabinet that tracks the attic. The carrying rule: air holds about 1.2 kJ of heat per m³ for each 1 K, so 35 CFM (1 m³/min) of moving air transports roughly 20 W per K of rise. A 30 W load on a 30 CFM filtered fan settles about 2 K above the loft: 42 C in a 40 C attic, past the 40 C rating most electronics carry unless a datasheet allows more.

A sealed box has one heat path left: conduction through the walls. Heat through a stainless wall follows k x area x temperature difference, with k = 3.7 W/m²K for stainless. A 20 x 16 x 8 in box presents about 0.78 m² (1,215 in²) of surface. The stainless wall-mount lines are built to the size you set, from 6 x 4 x 4 in to 60 x 36 x 16 in outside. At 30 W of internal heat that is a 10 K rise over ambient: a 50 C box in a 40 C loft.

Two cheap moves shrink the problem before any heat pump enters. First, exhaust the loft: air exchanged with the outdoors holds the loft near outdoor temperature. At a 30 C loft the same sealed box carries 30 W at about 40 C — at the limit — while a 25 C loft holds it at 35 C with margin. Second, put extra fans on separate openings: two fans in series, one pushing and one pulling, move no more air than one. The sealed-versus-ventilated crossover for a garage or loft box is worked through in the outdoor enclosure guide.

What a thermoelectric cooler actually does

A Peltier module is a solid-state heat pump: DC current moves heat from its cold plate to its hot plate. Cold plate inside the box, finned sink and fan outside. The module pumps heat uphill — from a 35 C cabinet into 40 C loft air, which no fan can do. Three numbers size the job.

COP is heat moved divided by electrical input. A 2022 systematic review of thermoelectric devices puts refrigerator COP at 0.3-0.7, well below the 2-4 a small vapor-compression unit reaches. At the 15-25 K lift a through-wall cooler works across, plan on 0.3-0.6. That efficiency gap is the price of a solid-state pump, paid in input watts.

Rejection is the number that kills undersized installs. The hot side sheds the pumped heat plus the input power: pumped x (1 + 1/COP). At COP 0.5 that is 3 W rejected per 1 W moved; at COP 0.3 it is 4.33 W per 1 W. Size the sink and fan from the selected cooler's actual operating point and worst-case ambient temperature. Rounding that lower-COP case down to 4x would understate the heat rejection.

The headline rating is not the operating capacity. Qmax is specified at zero temperature difference under stated test conditions. Actual pumped heat depends on hot-side temperature, temperature lift and input current. Do not infer a usable capacity by drawing a straight line between Qmax and maximum temperature difference, or by assuming a fixed watts-per-ampere rule. Pick the module from the manufacturer's performance curves at the real operating point, then account for heat-exchanger temperature drops and internal fan losses.

Cold side, wet box: any surface below the internal dew point collects condensate, and a chilled sealed box is a rainmaker for its own boards. A fresh desiccant pack after every opening keeps the internal air near what you sealed in. Put a number on it with the humidity checks in the NEMA enclosure types guide.

The hot side must reject heat outside; the supply need not

Putting both the hot and cold sides in the same closed volume is the fridge-door trap. The cold plate removes heat from the box air while the hot plate returns that heat plus the module's electrical input. At COP 0.5 a module pumping 10 W draws 20 W. If its supply is also inside and 90% efficient, the supply adds about 2.2 W of loss. The box gains about 22.2 W overall; circulating that heat internally does not cool the cabinet.

An internal power supply is a different case: with the hot side correctly rejecting heat outside, only the supply's losses join the internal heat load. Include those losses at the actual module load, along with internal fans and controls, and check the supply's temperature rating and installation instructions. This worked case puts the supply outside in a suitable enclosure and brings DC through sealed glands; the cutouts and glands guide covers the entry hardware. A compressor system follows the same energy balance, with its evaporator absorbing heat inside and its condenser rejecting heat outside; the entire refrigerant loop does not sit outside.

Sizing a cooler for a 30 W cabinet in a 40 C loft

The inputs for this worked case:

• Electronics heat: 30 W, the worked value for a cabinet of this kind — CCTV unit, PIR and lighting boards, DC supplies, stir fan included. Replace it with your metered draw or the sum of the supply nameplates; every number below scales with it.
• Loft air: 40 C, the summer peak this case starts from.
• Target: 35 C inside — 5 K below the loft and 5 K under the 40 C rating most boards carry.
• Box: 20 x 16 x 8 in stainless wall-mount, about 0.78 m² of surface.
• Wall transfer: k = 3.7 W/m²K for stainless.
• COP: 0.5, mid-band in the published 0.3-0.7.

The rule in words: pumped heat = electronics heat + wall gain, with wall gain = k x area x (loft − target); input = pumped / COP; hot-side rejection = pumped + input. The arithmetic:

1. Wall gain = 3.7 x 0.78 x (40 − 35) = 14.4 W.
2. Pumped heat = 30 + 14.4 = 44.4 W.
3. Input = 44.4 / 0.5 = 88.8 W, so a 120 W 12 V supply outside the box runs near 75% loading.
4. Hot-side rejection = 44.4 + 88.8 = 133 W — 3 W rejected per 1 W pumped, or about 454 Btu/hr on sheets that use that unit (1 W = 3.415 Btu/hr).
5. Curve check: a cold plate near 30 C and a hot sink near 55 C would create a 25 K module lift. The illustrative 0.11 K/W hot-side heat sink gives about a 15 K rise when rejecting 133 W. Confirm that the chosen cooler can actually pump at least 44.4 W at those plate temperatures and its intended current, including additional fan or interface losses. Qmax alone cannot establish that capacity; this example does not select a module or guarantee its margin.

The result: a through-wall unit that pumps about 45 W against a 40 C ambient and rejects about 135 W on its outdoor sink. Sizing on the electronics alone — 90 W of rejection for a 30 W load at COP 0.5 — misses the 14 W the walls add. What moves it: hold the box at 30 C and wall gain doubles to 29 W, pumped heat reaches 59 W and rejection 177 W. Under the same assumed COP and wall-transfer coefficient, each extra 1 K of cooling adds 3.7 x 0.78 x 3 = 8.7 W, or about 9 W, of hot-side rejection. A real cooler's COP also changes with temperature lift.

The threshold: above about 100 W of pumped heat, or off-grid on solar, the compressor takes over. At COP 2-4 the same 44.4 W pumps on 11-22 W, 0.3-0.5 kWh a day against the Peltier's 89 W and 2.1 kWh a day. Below that line the Peltier wins on silence, any mounting angle and cycling a setpoint tightly.

Peltier, compressor, fan or nothing: the decision by loft

Read the table top down and stop at the first row that fits. The cheapest cooling that holds the rating is the right one. For the sealed, dusty loft this case describes, the through-wall Peltier with its supply outside is the answer up to roughly 100 W of pumped heat. Past that, or on solar power, spec the small compressor.

One part never belongs in the sealed cabinet: the UPS. Charging lead-acid cells vent hydrogen, which needs free ventilation to the open air. On battery makers' life curves life halves roughly every 10 K above 25 C — at 40 C a battery ages about three times faster than at 25 C. House it in its own ventilated space and run the DC leads up through glands.

What to order for the dusty loft cabinet

• The box: 304 stainless wall-mount in the size the load and cooler flange set — 304 for a dry loft, 316L for coastal salt air. The stainless size finder filters the line by grade and size, and we build custom stainless enclosures to any width, height, depth and cutout from one piece. Manufacturer-rated NEMA 4X-type / IP66; not UL listed.
• The cooler: a through-wall thermoelectric unit on a gasketed flange that matches the cutout drawing, sized per the worked example — about 45 W pumped, 135 W hot-side rejection at a 40 C ambient.
• The supply: a 120 W 12 V unit in its own enclosure outside, DC leads entering through sealed glands.
• The controls: a thermostat cycling the cooler above the 35 C setpoint, plus a humidistat or fresh desiccant against condensation. The enclosure sizing steps in how to size an electrical enclosure cover the wall-gain arithmetic in full.
• The timing: delivered in 10 business days, anywhere in the world; for a custom cutout, counted from your approval of the drawing.

Decide in this order: reduce loft temperature where practical, establish the sealing and equipment-temperature requirements, then calculate the full cooling load. Include wall gain and any internal supply losses. Select the cooler and heat rejection hardware from their performance curves at the actual operating temperatures; the 3 W per watt ratio belongs to this COP 0.5 example.

Catalog dimensions to compare

Compare these catalog dimensions with your component layout. Open each enclosure for its rating, material and usable space, then check those specifications against your application.

Questions