# Engine 3F Efficiency Analysis part 4

Rev 1: See details at bottom.

Let me summarize what has been covered and what remains in this efficiency analysis:

Component Power percentage of total power Efficiency analysis
Heat lost directly to the environment 40.3w 58.4% Part 1
Thermal shorting: heat conduction directly from the hot end to the cold end 5.9w 8.6% Part 2
Gross engine power: includes net shaft power, plus internal losses 1.6 +/- .2w (part of thermodynamic cycle input heat) Part 3
Thermodynamic cycle input heat 22.8w 33.0%
Regenerator heat & efficiency (part of thermodynamic cycle input heat)
Total 69w 100%

After a lot of thought I came up with an approach that I believe is sound to finish this efficiency analysis. In the process I discovered that my heat bookkeeping method would need to be revised. The changes are shown in the above table. The numbers I’ve computed so far weren’t changed, just how they are grouped.

The gross engine power is now grouped as part of the thermodynamic cycle input. The gross engine power is what remains from the thermodynamic cycle input heat after you’ve paid off all your thermodynamic cycle losses. The regenerator is where you save heat from one cycle to the next, so it also belongs as part of the thermodynamic cycle input.

Although I often think of the regenerator losses in comparison with a perfect regenerator, another way to look at it is comparing it with no regenerator. The more efficient your regenerator, the less heat you need to supply overall for the thermodynamic cycle input. A regenerator should never lose heat, an efficient one saves more heat than an inefficient one.

Here are the assumptions and equations:

(1) After elimination of other avenues for heat loss, the only remaining heat is that which goes into the thermodynamic power cycle. If we call the remaining heat Qavailable then the thermodynamic cycle efficiency is:

Cycle efficiency = gross engine power / Qavailable
Cycle efficiency = 1.6w/22.8w = 7.0%

(2) The heat required in an engine cycle can be computed using the internal energy change for an ideal gas and the PdV work. Air behaves as an ideal gas at the temperatures and pressures we are considering, so this should be a good assumption:

Qrequired = U2 – U1 + PdV work = mCv(T2-T1) ∫12 PdV

Where: U2 – U1 is the internal energy change in joules between two states
m = mass of gas
Cv= heat capacity at constant volume for the gas
T2, T1 = temperatures Kelvin of the two states
12 PdV = PdV work between the two states

This computation is complicated by having three different volumes of gas (hot, cold, and regenerator) that are continually changing mass. I used a spreadsheet to compute these values from my simulation results using the following method:

U2 – U1 = Cv(Σm2T – Σm1T)

The summation includes each of the three masses multiplied by each of the three temperatures. In this case the three temperatures are fixed and the masses in each volume change. Then I sum together all the steps in the simulation where heating takes place (but not cooling) to arrive at a total internal energy increase required of 8.5 j/cycle or 29.8w at 3.5Hz. The internal energy change is the majority of the heating on a low temperature engine such as this one.

The total Qrequired can be made more accurate by including the PdV work that occurs during the part of the cycle where heating occurs. Including the PdV work (from the simulator) increases the total heat required to: 9.3 j/cycle or 32.6w at 3.5Hz.

(3)The heat required minus the heat available is the heat provided by the regenerator. This certainly makes sense and is why we use a regenerator. The regenerator saves heat from one cycle to the next so that the engine requires less heat than it would without the regenerator.

Qrequired – Qavailable = Qregenerator
32.6w – 22.8w = 9.8w

(4) If the regenerator was 0% efficient, then it provides no power. The thermodynamic cycle efficiency would just be the gross engine power divided by the heat required.

Cycle efficiency = gross engine power / Qrequired (0% regeneration)
Cycle efficiency = 1.6w / 32.6w = 4.9% efficiency (without regeneration)

(5) If the regenerator was 100% efficient, then the thermodynamic cycle efficiency would be the Carnot efficiency. This would be a valid assumption if the engine used only reversible processes. Ideal Stirling engines are reversible, you can use them as efficient heat pumps as well as engines. Real Stirling engines are not 100% reversible, they have losses. I think this is a reasonably accurate assumption for the present purpose. Even if a 100% efficient regenerator resulted in 75% of Carnot efficiency, the error introduced in this analysis is 1.7w.

Cycle efficiency = Carnot cycle (100% regeneration)
Carnot Efficiency = 1- (Tc/Th) Where Th and Tc are absolute temperatures

On my cycle simulator the 1.6w gross engine power corresponds to Th and Tc temperatures of 205.3 C and 58.8C.

Carnot Efficiency = 1 – (58.8 + 273.15)/(205.3 + 273.15) = 30.6%

Thermodynamic cycle input = gross power out / cycle efficiency
= 1.6w/.306 = 5.2w

If the regenerator were 100% efficient, then Qavailable would need to be 5.2w. Using assumption (3) then:

Qrequired – Qavailable = Qregenerator =
32.6 – 5.2 = 27.4w (100% efficient)

Knowing the heat supplied by a 100% efficient regenerator lets us compute the actual regenerator efficiency:

Qregen(actual) / Qregen(100%) =
9.8/27.4 = 36% regenerator efficiency

Using the above computations the updated heat loss table is now:

Component Power percentage of total power Efficiency analysis
Heat lost directly to the environment 40.3w 58.4% Part 1
Thermal shorting: heat conduction directly from the hot end to the cold end 5.9w 8.6% Part 2
Gross engine power: includes net shaft power, plus internal losses 1.6 +/- .2w (part of thermodynamic cycle input heat) Part 3
Thermodynamic cycle input heat 22.8w 33.0% Part 4
Regenerator heat & efficiency 9.8w, 36% efficient (part of thermodynamic cycle input heat) Part 4
Total 69w 100%

## Conclusion

Although I believe the methods and assumptions I used in this part of the analysis are reasonably sound, the regenerator efficiency seems low. I expected something around 80% and have difficulty believing it is below 50%.

Part of the problem may be caused by the possibly inaccurate 22.8w value arrived at by subtraction from the total. For example, a 5 watt increase in estimating the heat lost directly to the environment (a 12.5% error) would result in a 5 watt higher regenerator output and an efficiency of about 54%.

There is also the possibility that the 36% regenerator efficiency is close to correct and I need to design a better regenerator. Because I believe my analysis is not that far off, I think I have to accept this result.

Overall I was surprised by the amount of heat that goes directly from the heater (the light bulb) into the environment. When I say I’m surprised, I don’t mean by the computed result. When I ran the test I could feel considerable heat coming from the insulated cylinder. This large loss was in spite of my efforts to insulate the hot section.

I’ve also tried propane and kerosene (paraffin) burners. As heat sources they are fine for powering this enginge, but the exhaust gases carry away a significant amount of heat and complicate insulating the engine. Another problem is measuring the power (heat). At the low power level this engine uses, I only burn a few grams an hour. I’d need a laboratory grade scale to measure the input power accurately. My energy loss table would also need another entry for heater efficiency.

Rev 1 details:

Section (2) was replaced above. The original is shown below. Because the value for Qrequired was changed, it required changing other sections, but only for the value change. The essential change was adding the effect of the PdV work to Q required.

(2) The heat required in an engine cycle can be computed using the internal energy change for an ideal gas. Air behaves as an ideal gas at the temperatures and pressures we are considering, so this should be a good assumption:

Qrequired = U2 – U1 = mCv(T2-T1)

Where: U2 – U1 is the internal energy change in joules between two states
m = mass of gas
Cv= heat capacity at constant volume for the gas
T2, T1 = temperatures Kelvin of the two states

This computation is complicated by having three different volumes of gas (hot, cold, and regenerator) that are continually changing mass. I used a spreadsheet to compute these values from my simulation results using the following method:

U2 – U1 = Cv(Σm2T – Σm1T)

The summation includes each of the three masses multiplied by each of the three temperatures. In this case the three temperatures are fixed and the masses in each volume change. Then I sum together all the steps in the simulation where heating takes place and arrive at a total heat required of 8.5 j/cycle or 29.8w at 3.5Hz.