We fire up our Sterling engine using isopropyl alcohol of 70% and 91% proof. We

Question

We fire up our Sterling engine using isopropyl alcohol of 70% and 91% proof. We have conversion of chemical energy to mechanical energy at its best. Use the IR camera to estimate the efficiency (%) of this conversion in both cases. Discuss the results of this demonstration as well as the differences between the two liquids . Discuss what would increase this efficiency.

I just WONDER how to increase the efficiency? FOR VOULME ? RESSURE ? TEMPERATURE

since we get Isopropyl alcohol 70% the efficiency at 0.625; the hot temperature at 75 the low at 38

the 91% efficiency at 0.5 ; the hot temperature at 80 the low at 30

Qin-Qout+Work. Robert Stirling is the father of the Sterling engine one of the most "successful" inventions of the last century. The Sterling engine's PV is as follows: Two isothermal and two isochoric (constant volume) processes (see fig.2). Let's describe these processes in detail. State 1 to 2: At state 1 the working fluid is at a maximum volume, minimum temperature, and minimum pressure. From state 1 to state 2 the power piston compresses the working fluid while heat is transferred out of the system which keeps the working fluid at a constant low temperature. When the engine is in state 2 the working fluid is in a compressed state (high pressure and low volume), but remains at the same temperature as state 1. The work that was required to compress the volume is provided by stored energy in the engine's flywheel (while the flywheel is turning) State 2 to 3: At state 2 the working fluid is ata minimum volume, minimum temperature, and medium pressure. Between states 2 and state 3 the volume is held constant while heat is added by the hot section to increase temperature. State p4 (3 O 3 to 4: At state 3 the working fluid has achieved Figure 2: The Sterling PV diagram

Explanation / Answer

Efficiency = 1 - Low temperature/High temperature

Thus for low = 38 & high = 75, efficiency = 1 - 38/75

= 0.5

Similarly, for low = 30, high = 80, we have,

Efficiency = 1 - 30/80 = 0.625

For increasing efficiency, the value of the fraction

Low temperature/High temperature

should be decreased which can be achieved by decreasing the low temperature or increasing the high temperature. Hence increase of efficiency is determined by temperature.

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