Answer to Question #273275 in Mechanics | Relativity for luca

Question

An ideal Stirling engine using helium as the working fluid operates between temperature limits of 300 and 2000 K and pressure limits of 150 kPa and 3 MPa. Assuming the mass of the helium used in the cycle is 0.12 kg, determine (a) the thermal efficiency of the cycle, (b) the amount of heat transfer in the regenerator, and (c) the work output per cycle.

Answer 1

Draw the T-s diagram of the Stirling cycle as shown below.

(a)

Obtain the characteristic gas constant and specific heat at constant volume for helium gas at 300 K from the table.

Characteristic gas constant,

R = 2.0769 kJ/kgK

Specific heat at constant volume,

C_v = 3.1156 kJ/kgK

Calculate the thermal efficiency of ideal Stirling cycle.

$η_{th} = 1 -\frac{T_L}{T_H}$

Here, T_L is the minimum temperature of the cycle and T_H is the maximum temperature of the cycle.

Substitute 300 K for T_L and 2000 K for T_H.

$η_{th} = 1 -\frac{300}{2000} \\ = 0.85 \\ = 85 \; \%$

Therefore, the thermal efficiency of the ideal Stirling cycle is 85 %.

(b)

Calculate the amount of heat transferred in the regenerator.

$Q_{regen} = mC_v(T_1-T_4)$

Here, m is the mass of helium used in the cycle, C_v is the specific heat at constant volume of helium, T₁ is the temperature of the helium gas at the exit of the regenerator (maximum cycle temperature) and T₄ is the temperature of the helium gas at the inlet of the regenerator (minimum cycle temperature).

Substitute 0.12 kg for m, 3.1156 kJ/kgK for C_v, 300 K for T₄ and 2000 K for T₁.

$Q_{regen} = 0.12 \times 3.1156(2000-300) \\ = 635.58 \; kJ$

Therefore, the amount of heat transferred in the regenerator is 635.58 kJ.

(c)

Obtain the relation for the ratio of specific volumes at the end and beginning of the isothermal heat addition using the combined gas law as follows

$\frac{P_1v_1}{T_1} = \frac{P_3v_3}{T_3} \\ \frac{v_3}{v_1} = \frac{T_3P_1}{T_1P_3}$

Here, v₁ is the specific volume of helium at the beginning of isothermal heat addition, v₃ is the specific volume of helium at the end of isochoric heat rejection process, P₁ is the pressure of helium at the beginning of isothermal heat addition, P₃ is the pressure of helium at the end of isochoric heat rejection process, T₁ is the temperature of helium at the beginning of isothermal heat addition and T₃ is the temperature of helium at the end of isochoric heat rejection process,

The process 2 to 3 is at constant volume. Therefore, the above relation becomes.

$\frac{v_2}{v_1} = \frac{T_3P_1}{T_1P_3}$

Here, v₁ is the specific volume of helium at the beginning of isothermal heat addition, v₂ is the specific volume of helium at the end of isothermal heat addition, P₁ is the pressure of helium at the beginning of isothermal heat addition, P₂ is the pressure of helium at the end of isochoric heat rejection process, T₁ is the temperature of helium at the beginning of isothermal heat addition and T₂ is the temperature of helium at the end of isochoric heat rejection process,

Substitute 2000 K for T₁, 300 K for T₃, 3000 kPa for P₃ and 150 kPa for P₁.

$\frac{v_2}{v_1} = \frac{300 \times 3000}{2000 \times 150} =3$

Calculate the change in entropy during the isothermal heat addition.

$s_2-s_1 = R ln(\frac{v_2}{v_1})$

Here, v₁ is the specific volume of helium at the beginning of isothermal heat addition, v₂ is the specific volume of helium at the end of isothermal heat addition and R is the characteristic gas constant of helium.

Substitute 2.0769 for R, 3 for $\frac{v_2}{v_1}$

$s_2-s_1 = 2.0769 ln(3) = 2.2817 \;kJ/kg \cdot K$

Calculate the heat added during the isothermal heat addition process.

$Q_{in} = mT_1(s_2 -s_1)$

Here, m is the mass of helium used in the cycle, T₁ is the temperature of helium at the beginning of isothermal heat addition and (s₂ -s₁) is the change in entropy during the isothermal heat addition.

Substitute 0.12 kg for m, 2000 K for T₁ and 2.2817 kJ/kgK for (s₂-s₁).

$Q_{in} = 0.12 \times 2000 \times 2.2817 \\ = 547.608 \; kJ$

Calculate the net-work output per cycle.

$W_{net} = η_{th}Q_{in}$

Here, Q_in is the heat added during the isothermal heat addition process and η_th is the thermal efficiency of the ideal Stirling cycle.

Substitute 0.85 for η_th and 547.608 kJ for Q_in.

$W_{net} = 0.85 \times 547.608 \\ = 465.467 \; kJ$

Therefore, the net-work output per cycle is 465.467 kJ.