Question #129508

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)?

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)?

Expert's answer

Given data are

T_{H} = 32^{0} C + 273 = 305 k

T_{L } = -21^{0} C + 273 = 252 k

V_{1} = 40 cm^{3} , V_{3} = 33 cm^{3}

n = ?

work done in isothermal expansion

w = nRT ln v_{2}/v_{1}

= 2.303 *1 *8.31 * 305 log (40/33)

= 487.313 Joules

in isothermal process change in u = 0

from its law of thermodynamics

dq = change in u + change in w

dq = 0 + w

hence,

heat absorbed q = w

q = 487.313 Joules

Learn more about our help with Assignments: Mechanical Engineering

## Comments

## Leave a comment