Answer to Question #215803 in Molecular Physics | Thermodynamics for Wavie

Question

0.5 kg of air executes a carnot cycle having a thermal efficiency of 50%. The heat transfer to the air during the isothermal expansion is 40 kJ. At the beginning of the isothermal expansion, the pressure is 7 bar and the volume is 0.12 m3. Determine:

i) The maximum and minimum temperature of the cycle in K.

ii) The volume of air at the end of isothermal expansion.

iii) The work and heat transfer for each of the four processes.

Answer 1

$m=0.5\;kg \\ η_{th}= 50 \; \%$

Heat transferred during isothermal expansion = 40 kJ

$p_1=7 \;bar \\ V_1 = 0.12 \;m^3 \\ c_v= 0.721 \;kJ/kgK \\ c_p= 1.008 \;kJ/kgK$

I) The maximum and minimum temperature of the cycle in K.

$p_1T_1 = mRT \\ 7 \times 10^5 \times 0.12 = 0.5 \times 287 \times T_1$

Maximum temperature

$T_1 = \frac{7 \times 10^5 \times 0.12}{0.5 \times 287} = 585.4 \;K \\ η_{cycle}= \frac{T_1-T_2}{T_1}=0.5 = \frac{584.5-T_2}{585.4}$

Minimum temperature

$T_2 = 585.4-0.5 \times 585.4 = 292.7 \;K$

ii) The volume of air at the end of isothermal expansion.

Heat transferred during isothermal expansion =

$= p_1V_1ln(r) = mRT_1ln(\frac{V_2}{V_1}) = 40 \times 10^3 \\ 0.5 \times 287 \times 585.4 \times ln(\frac{V_2}{0.12}) = 40 \times 10^3 \\ ln(\frac{V_2}{0.12})= \frac{40 \times 10^3}{0.5 \times 287 \times 585.4}= 0.476 \\ V_2= 0.12 \times e^{0.476} = 0.193 \;m^3$

iii) The work and heat transfer for each of the four processes.

Process 1-2 Isothermal expansion 40 kJ

Process 2-3 Adiabatic expansion 0

Process 3-4 Isothermal compression -40 kJ

Process 4-1 Adiabatic compression 0