Combine convection and radiation at a surface

A reflective surface can still exchange heat with air

At 60 °C, set emissivity to zero. Radiation transfers no net energy into or out of the ideal reflecting surface, but the model still reports 100 W of outward convection. Low emissivity does not remove other transfer paths.

Convection combines fluid motion with thermal exchange at a surface. Forced convection uses a fan or pump to drive flow; natural convection arises from buoyancy associated with density differences. The model does not solve a fluid-velocity field or predict how a fan changes flow. It uses an assigned convection coefficient to represent that exchange.

Use the coefficient with its assumptions

The convection relation here is Pconv = hA(Ts − Tair). h is the assigned coefficient, 5 W/(m² K); A is exposed area, 0.5 m²; and Ts − Tair is the surface-to-air temperature difference. Air is maintained at 20 °C. A kelvin difference equals the corresponding Celsius difference, so no offset is added to this subtraction.

At 60 °C, Pconv = 5 × 0.5 × (60 − 20) = 100 W outward. At 0 °C, it is −50 W, meaning air supplies energy to the colder surface. A negative value describes direction, not negative thermal energy stored in the material.

Unlike a material conductivity, h depends on flow, geometry, fluid properties and conditions. The fixed value is a teaching input, not a universal constant for air or a measured prediction of natural convection across this entire range. The OpenStax heat-transfer mechanisms reference explains the distinction between convection and conduction.

Add simultaneous transfer paths

Radiation and convection exchange energy with the same surface at the same time. Their signed net rates add: Pout = Prad + Pconv. This differs from successive conduction layers in series, which share one common rate while their temperature drops add.

At 60 °C and ε = 0.8, net radiation is 111.897 W and convection is 100 W. Total outward transfer is 211.897 W. Changing ε to zero removes the radiative term but leaves convection at 100 W because air temperature, area, coefficient and surface temperature are unchanged.

The enclosure and the air happen to have the same 20 °C temperature here, but they represent different boundaries. In a real setting they can differ. A surface exposed to a cold sky and warmer air might receive heat by convection while losing it by radiation. This widget does not simulate that additional boundary variation.

Determine the power needed to hold temperature

The control prescribes surface temperature. Let Phold be thermal power added by an external temperature-control system, positive into the surface. Maintaining the state requires 0 = Phold − Pout, so Phold = Pout.

For the 60 °C, ε = 0.8 case, a controller must supply 211.897 W of thermal power to offset the modeled losses. For the 0 °C case, net radiation is −41.243 W and convection is −50 W, so Pout = −91.243 W. Keeping that cold surface at 0 °C requires removing 91.243 W of thermal power.

The latter is cooling capacity, not electrical power consumption. A refrigerator’s electrical demand depends on its operation and surroundings. Likewise a heater’s electrical input need not equal the thermal power delivered to this area. No device efficiency or coefficient of performance is supplied.

Do not turn an instantaneous balance into a cooling curve

If the controller is removed, temperature can change. Predicting its trajectory requires a heat capacity and possibly internal temperature gradients, while recomputing radiation and convection as temperatures change. The current model supplies none of those dynamics.

The ten-minute energy entry assumes temperatures remain maintained throughout: Qout = Pout × 600 s. At the default settings it is approximately 127.138 kJ outward, replenished by the controller. It is not a claim that the surface loses that much stored energy while holding the same state without an energy source.

At 20 °C, both net mechanisms vanish in this particular model. Emission and absorption still occur at equal nonzero rates when ε is nonzero. At any surface temperature, a fixed ε changes both emission and absorption consistently; changing only one would break the gray-model energy account.

Transfer to a different surface

An area of 0.8 m² has an assigned h = 10 W/(m² K) and is 15 K warmer than its air. Its net outward radiation is separately calculated as 35 W. Find convection, total outward transfer and required thermal holding power. Then consider the same convection conditions with net radiation −50 W instead.

Check the surface-balance transfer

Convection is 10 × 0.8 × 15 = 120 W. With radiation 35 W outward, total loss and required added thermal power are 155 W. If radiation is −50 W, it offsets part of the convective loss, giving 70 W net outward and 70 W required thermal input. The second case needs a different radiative boundary; it is an unfamiliar transfer task, not a setting of the current widget.