This Excel 2003 spreadsheet stirling-cycle-ideal-v-02 lets you input the basic values for a Stirling cycle engine and calculates various quantities such as pressures, work in, work out, heating, cooling, and ideal efficiency.
Don’t use this spreadsheet with the idea that your engine will get close to the values it shows unless you have an engine that truly approximates the ideal Stirling cycle. Most Stirling engines that use a crankshaft to drive the power piston and displacer including the engines I’ve built so far cannot possibly put out more than about 77% of the work per cycle of an ideal Stirling engine. This is assuming piston, displacer, crankshaft and all related mechanisms are frictionless.
A large source of uncertainty when using this spreadsheet is in the minimum and maximum temperatures. The spreadsheet assumes these are the cold and hot temperatures of the working gas in the engine. These temperatures are difficult to measure in an operating engine. I usually measure the hot and cold temperatures as close to the gas as I can. Typically I measure the temperature of a metal plate that directly interfaces with the working gas or the water temperature adjacent to the metal plate. The temperature spread between the hot and cold gas may be only 50% of what I measure or even less depending on the heat transfer. In the future I’ll be working more on the heat transfer efficiency of my engines.
If I use the numbers for my large engine (engine 3d) in the spreadsheet and use the temperatures I actually measure on the metal plates, then I am getting about 11% of the net work out predicted by the spreadsheet (so far). Even if I were to use more realistic hot and cold temperatures, I might still be seeing only 25% of the spreadsheet-predicted work output. Where is the missing work?
Part of the work is lost to friction, something not covered by this spreadsheet at all. In another post I’ve talked about measuring the friction losses on your engine with a flywheel spin-down test. Other friction losses are due to pushing the operating gas around in the engine, especially through the regenerator. Any gas leaks in the engine will also reduce engine power.
This spreadsheet also does not take into account dead space. Any space that is not swept by the displacer or piston is dead space. This includes the space where the regenerator material is placed and also includes internal clearance space above and below the displacer. As a rough approximation the ratio of live volume to total volume (live + dead) will be the ratio of work put out by the engine compared with the ideal output. If your live and dead volumes are equal, then you’ll at best see about 50% of the work out predicted by the spreadsheet. This is after you’ve lost 77% to your engine not following the ideal cycle. My engine 3d has approximately 80 in^3 minimum volume and 41 in^3 dead space including the regenerator. That gives me 77% x (80/121)= 51%.
The idea Stirling cycle also assumes that the compression and expansion of the gas is an isothermal (constant temperature) process. In practice it is normally not possible to closely approximate the isothermal process. A slow turning engine typically can come closer than a high-rpm engine. If no heat is transferred during the compression or expansion cycles, it is an isentropic process. Real engines operate somewhere between isothermal and isentropic and can be simulated by using a polytropic process. I plan to add this feature to a future simulator.
With all these inaccuracies you may be wondering whether this spreadsheet is useful at all. The real value of this spreadsheet is to put an absolute upper bound on a Stirling engine’s work per cycle and efficiency. This spreadsheet also shows the importance of a regenerator, especially in low temperature engines (although it doesn’t assume any volume is required for it).
Using the spreadsheet
I’ve included brief instructions with the spreadsheet. If you aren’t familiar with the Stirling cycle, I suggest this primer as a good starting point.
Notice that this is a cycle spreadsheet so the output is work not power. If instead you’d like to know power then you need to compute the power based on the number of cycles per second. For example, at 10 hz (600 RPM) a joule output per cycle is equivalent to 10 watts.