Continuing with the analysis of where the input power (heat) to the engine goes, the next item to consider is the conduction loss often referred to as thermal shorting.
A Stirling engine of gamma configuration, such as the one being considered, uses a displacer to shuttle the operating gas back and forth between the hot end and the cold end. In this engine both the displacer cylinder and the displacer itself are constructed from stainless steel tubes. These tubes conduct heat lengthwise between the hot and cold ends. The illustration below shows the hot end gas as orange near the bottom and the cold end gas as blue near the top.
To compute these losses I would like to have reasonably accurate temperature measurements for the displacer cylinder and displacer. The only temperature I measure directly is the cold end of the displacer cylinder (90F, 32.2C). I don’t directly measure the hot end of the displacer cylinder, just the air temperature nearby (450F, 232C). I have no measure of the gas temperatures inside the displacer volume or of the displacer itself, so I have to make some estimates.
Based on my engine power computations, which I’ll go into further in the next part, I believe that Th for the operating gas is much lower than the measured external temperature and Tc is higher. These temperature differences are also necessary for heat transfer to take place. For this computation I use the values shown in the table below. I use 17 w/m-degK as the average thermal conductivity of stainless steel in the computations.
The rate of conduction heat transfer for the one-dimensional case is:
Qdot = -kA(dT/dx)
Qdot = heat transfer rate in watts
k = thermal conductivity of the material (w/m-degK)
A = cross sectional area of the material (m^2)
(dT/dx) = rate of change of temperature per unit length (degK/m)
This analysis assumes dT/dx is constant along the length of the tubes.
For the computations, I have reduced the conduction length of the tubes from the full length by one-half the stroke length on each end. I think this best represents the points where the temperatures were measured or computed.
|Displacer cylinder||2.0 inch dia x .083 wall x 9.5 inch conduction length. (50.8mm dia x 2.1mm wall x 241.3mm long)|| Th = 450F (232C)
Tc = 90F (32.2C)
Delta T = 360F (200C)
|Displacer||1.75 inch dia x .035 wall x 7.5 inch conduction length. (44.5mm dia x .89mm wall x 190.5mm long)|| TH = 389.3F (198.5C)
TC = 150F (65.5C)
Delta T = 239.3F (133C)
Total conduction losses for the two components are 5.9 watts.
I’d like to make a few comments on the displacer design. The displacer was made relatively long for two reasons:
The longer length reduces dT/dx in the conduction equation to reduce power lost through thermal conduction. The long length was also used to provide more surface area to act as a more efficient regenerator for the operating gas moving back and forth through the annular space. This type of regenerator is not as efficient as a fine mesh screen, but is advantageous from the simplicity standpoint. I’ll look at the efficiency of the regenerator in another part of this analysis.
You may have also noticed the displacer cylinder has a .083” wall, much thicker than it needs to be. This was chosen to reduce the cross sectional area of the annular space the gas travels through to increase the velocity and improve heat transfer. It also reduces the dead volume. A more ideal choice would have been a tube with the same inside diameter and a thinner wall. For simplicity I chose a standard tube size and accepted the penalty (2.9w).
Stainless steel was used for the displacer cylinder and displacer to benefit from the 3.5x lower heat conductivity compared with ordinary steels. If ordinary steel had been used instead of stainless steel, the conduction losses would increase from 5.9 to 20.7 watts. If aluminum was used the losses would be above 35 watts.
So far we’ve accounted for 40.3 watts going directly to the environment and 5.9 watts that does no useful work before going back to the environment through the cold end of the engine. 46.2 watts total.
Part 3 will start to look at the power that goes into doing useful work.