Mechanical engineering

Question

An ideal Stirling cycle cooler, using helium (considered ideal) as its working fluid, has a compression temperature Tc=32°C and an expansion temperature

has a compression temperature Tc=32°C and an expansion temperature

TE=-21°C.

If state 1, at the beginning of the compression, is defined with a volume V1=40

cm3 and a pressure P1=1.8 MPa, and state 2 is defined with a volume V2=33

cm3, what is the heat qE absorbed during the expansion process (in Joules)?

Answer 1

the stirling cycle is an isothermal expansion, then constant volume cooling, then isothermal compression, then constant volume heating

the isothermal expansion step is the transition from state 3 to state 4 in your problem, which have volumes equal to V2 and V1, respectively.

an isothermal expansion has no change in internal energy, so \Delta U = 0 = Q + W, or Q = - W

the work, W, is the integral of -PdV from V2 to V1, and if the process is carried out reversibly the pressure, P, is nRT/V

after integrating, we find that the work is -nRT ln (V1/V2) and the heat flow, qE in your problem, is nRT ln (V1/V2)

you didn't specify the number of moles of helium, but sometimes the heat quantity is expressed in energy per mole and in this case qE per mol = RT ln (V1/V2) = 8.31 J/K.mol (23 K +273 K) ln (40/33) = 473 J/mol