appropriate fans—from very small doorway corner fans to circulate room to room, to various inline fans mounted in round duct to move air longer distances, are effective.

These marketing techniques are the same used by the foil bubble-wrap “insulation” (and even “insulation paint”) people.

If there’s a sucker born every minute, and only a sucker would buy an Amish fireplace that uses 1.5kW, but they are limited to only purchasing two, at what rate does the kWh/year increase over time? And a bonus question: if it’s 66F in my office and my hands are cold, how long will I have to walk on my treadmill at 2 MPH in order to heat myself up as much as I would if I just ran a cheap 1500 space heater from Kmart in my office to warm it up in here? Ok, thanks. Time for lunch.

David, yep, all light or other radiation emitted in a room ends up heating the space unless it passes through a window. I am not sure if I would say that it is adding to the kinetic energy of the air or surroundings, unless you are talking about at the molecular level. In terms of classical thermodynamics, it is just increasing the internal energy of the space (air or solid surfaces). Even the microwaves in the oven end up as heat.

So Roy, what about light? We all know that incandescent bulbs are mostly electric heaters, but assuming no light escapes the room, doesn’t the energy converted to visible light also manifest as heat?

We think of thermal radiation as somehow being different from visible light, but all wavelengths of electromagnetic energy impart kinetic energy to the surroundings, right? It’s just a matter of frequency and energy level. Energy in = energy out.

Bobby, I think that you pretty much nailed it. For an “ideal gas”, which is a good assumption for air at our typical conditions, the internal energy or enthalpy is only a function of temperature. Thus, it doesn’t matter if you add heat or work to the air stream in the duct; the change in temperature is the same regardless of any change in pressure. Some people think that as the air passes through ductwork with a corresponding pressure drop due to friction that the temperature will go up, but that is not true. The pressure will drop, but the density will also drop such that the enthalpy is constant (assuming no heat gains or losses through the duct walls).

You did mention energy associated with the air velocity, and that is a separate consideration from air enthalpy. That is the kinetic energy of the air stream and is proportional to the velocity squared. For typical residential ductwork, that kinetic energy is negligible.

So the bottom line is that any electrical energy to the blower ends up being additional heat load to the air stream regardless of the motor or blower efficiency.

Roy, interesting question. I will take an amateur stab at it. Since we know mechanical energy and heat to be equivalent, the amount of energy in the unit is equivalent. Temperature is a funny thing, it is the “average” heat energy and doesn’t really equate to any real measure except in a theoretical closed system of finite volume. But it is a useful mental concept. My answer would be that in the case of using a fan, more of the energy in the system would be found in mechanical energy than in a purely passive resistance heater. The theoretical temperature would be equal because mechanical energy in a turbulent fluid converts to heat at the molecular level. The measured temperature at a thermometer would not pick up the higher energy level of the moving airstream, because the energy is not equally distributed in the space and has not all converted to heat through mechanical decay. To arrive at the correct energy total, you would need a thermoanemometer to measure both temperature and mechanical energy in the airstream. (But good luck getting an accurate measurement in a turbulent flow!)

My guess is that if you had a perfectly insulated chamber and ran a 400w device in it for an hour (whether a fan or a heater), then after a sufficiently (near infinite?) long time for friction to return the air movement to near stagnant, the measured temperature would be the same.

The motor is converting 300 W to shaft power which goes to the blower and the other 100 W is converted to heat which goes directly to the air stream around the motor. The 300 W of shaft power to the blower raises the pressure and temperature of the air stream. But if you look at the whole blower and motor assembly, does the temperature of the air stream go up more or less than it would for just a 400 W electric resistance heater in the air stream with no air pressure change?