Question

Applying the Entropy Balance: Closed Systems Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston–cylinder assembly from 2 bar, 280 K to 20 bar, 520 K. If the carbon dioxide behaves as an ideal gas, determine the amount of entropy produced, in kJ/K, assuming (a) constant specific heats with c= = 0.939 kJ/kg · K. (b) variable specific heats. Compare the results of parts (a) and (b).

Answer 1

a) the amount of energy produced in kJ/K is 0.73145 kJ/K

b) the amount of energy produced in kJ/K is 0.68975 kJ/K

The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.

Step-by-step explanation:

Draw the T-s diagram.

a)

`C_p` = 0.939 kJ/kg.K , m = 5 kg , T₂ = 520 K , T₁ = 280

R = [8.314 kJ / 44.01 kg.K] , P₂ = 20 bar , P₁ = 2 bar

Δs =
`m[c_p ln((T_2)/(T_1)) - Rln((P_2)/(P_1))]`

Substitute all parameters in the equation

Δs =
`5[(0.939) ln((520)/(280)) - ((8.314)/(44.01))ln((20)/(2))]`

Δs = 5 kg × 0.14629 kJ/kg.K

= 0.73145 kJ/K

b)

Δs =
`m[(s^0(T_2) - s^0(T_1))/(M) - Rln((P_2)/(P_1))]`

Where T₁ = 280 K , s°(T₁) = 211.376 kJ/kmol.K

T₂ = 520 K , s°(T₂) = 236.575 kJ/kmol.K

R = [8.314 kJ / 44.01 kg.K] , M = 44.01 kg.K , P₂ = 20 bar , P₁ = 2 bar

Δs =
`5[(236.575 - 211.376)/(44.01) - ((8.314)/(44.01))ln((20)/(2))]`

= 5 kg (0.13795 kJ/kg.K)

= 0.68975 kJ/K

The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.