In an earlier post I measured 1.05 watts output from my engine 3F using a 69 watt input. This computes to 1.5% efficiency. So where did the other 68 watts go? The following is my analysis of where all the energy went. It won’t be as rigorous as I would like, partly due to lack of instrumentation and partly due to some difficult issues with the thermodynamics of real engines.
The data I’m using for this analysis is not from the video. The measured data for this analysis: 3.5 hz (210 rpm), 71 grams @3 inches torque, Th = 450F, Tc = 90F, halogen bulb power 69w, computed shaft power (from measurements) 1.16w, computed efficiency 1.7%
To start with let’s look at a diagram of the engine.
The halogen light bulb provides heat to the system. The primary goal of the supplied heat is to heat the working gas in the hot end of the displacer cylinder to power the engine. The heat, however, doesn’t just go where you’d like it to go. A large part of the heat from the light bulb dissipates directly to the environment without ever going into the displacer. The thermally conductive displacer and displacer cylinder also conduct heat from the hot end to the cold end of the displacer. Although the engine design uses these surfaces as a regenerator, the conductive losses are neither thermodynamically desirable nor necessary for power production.
Ideally, the regenerator would provide the all the heating and cooling of the working gas except that part which is thermodynamically necessary to generate the power output. This necessary heating is determined by the efficiency of the power cycle used.
The engine itself is not 100% efficient, even after accounting for the thermodynamic inefficiencies. Frictional and other losses in the engine mean that the engine is actually generating more power than you measure at the crankshaft.
In the course of this multi-part analysis I’ll address all of the above issues and do my best to determine where those 69 watts of input heat went.
Heat lost directly to the environment
The first part to start with is the insulated galvanized steel cylinder surrounding most of the displacer cylinder including the hot end. The part of the displacer cylinder that is not inside the insulating can is the cold end plus roughly the displacer stroke length. I measured temperatures over multiple points on the insulated cylinder and then averaged the side, bottom, and top temperatures.
With the average surface temperatures, the ambient temperature and the estimated emissivity of the bright galvanized metal of the insulated cylinder, I used the calculator (This website has some excellent thermal calculator applets) to compute the total power dissipation for natural convection and radiation.
When using the calculator, your choices are to use a rectangular surface that is either horizontal or vertical. You can also select whether to use a the top or bottom surface for a horizontal surface or a single side or both sides in the case of a vertical surface. I used a single side vertical surface for the side of the cylinder with the width equal to the circumference. The top and bottom surfaces were analyzed as squares with the areas equivalent to the circular area.
The values are shown below:
Ambient temperature 63 deg F (17.3 C)
Emissivity of bright galvanized metal 0.28
Altitude 900ft (273m)
|Surface||Dimensions for calculations||Average surface temperature||Computed Surface power dissipation|
|Top||8 inch diameter minus the 2 inch diameter displacer cylinder(174mm x 174mm)||110F (43.3C) Delta T = 26 C||3.3w|
|Side||8 inch diameter x 13 inches long (surface area 330 mm high x 639mm wide)||110.3F (43.5 C) Delta T = 26.2 C||33w|
|Bottom||8 inch diameter (surface area 180mm x 180mm)||133F (56.1 C) Delta T = 38.8 C||5.6w|
The heat loss from the bottom and sides of the insulated cylinder are not necessary for power production. The top surface of the insulated cylinder is more difficult to assess. It connects firmly to the displacer and could act as a cooling fin for the cold end of the displacer or it could be receiving heat through the insulation. Measuring the temperature gradients in the vicinity aren’t conclusive, although some of the heat is definitely coming through the insulation and from the sides of the galvanized steel cylinder. I decide to split it about in half and attribute 1.7w as heat loss. The total heat lost directly to the environment is therefore 33 + 5.6 + 1.7 = 40.3w. The remaining 1.6w of the top surface is considered part of the heat used in producing power.
My next post will go into the conductive losses.