Respuesta :
Answer:
The final specific internal energy of the system is 1509.91 kJ/kg
Explanation:
The parameters given are;
Mass of steam = 1 kg
Initial pressure of saturated steam p₁ = 1000 kPa
Initial volume of steam, = V₁
Final volume of steam = 5 × V₁
Where condition of steam = saturated at 1000 kPa
Initial temperature, T₁ = 179.866 °C = 453.016 K
External pressure = Atmospheric = 60 kPa
Thermodynamic process = Adiabatic expansion
The specific heat ratio for steam = 1.33
Therefore, we have;
[tex]\dfrac{p_1}{p_2} = \left (\dfrac{V_2}{V_1} \right )^k = \left [\dfrac{T_1}{T_2} \right ]^{\dfrac{k}{k-1}}[/tex]
Adding the effect of the atmospheric pressure, we have;
p = 1000 + 60 = 1060
We therefore have;
[tex]\dfrac{1060}{p_2} = \left (\dfrac{5\cdot V_1}{V_1} \right )^{1.33}[/tex]
[tex]P_2= \dfrac{1060}{5^{1.33}} = 124.65 \ kPa[/tex]
[tex]\left [\dfrac{V_2}{V_1} \right ]^k = \left [\dfrac{T_1}{T_2} \right ]^{\dfrac{k}{k-1}}[/tex]
[tex]\left [\dfrac{V_2}{V_1} \right ]^{k-1} = \left \dfrac{T_1}{T_2} \right[/tex]
[tex]5^{0.33} = \left \dfrac{T_1}{T_2} \right[/tex]
T₁/T₂ = 1.70083
T₁ = 1.70083·T₂
T₂ - T₁ = T₂ - 1.70083·T₂
Whereby the temperature of saturation T₁ = 179.866 °C = 453.016 K, we have;
T₂ = 453.016/1.70083 = 266.35 K
ΔU = 3×[tex]c_v[/tex]×(T₂ - T₁)
[tex]c_v[/tex] = cv for steam at 453.016 K = 1.926 + (453.016 -450)/(500-450)*(1.954-1.926) = 1.93 kJ/(kg·K)
cv for steam at 266.35 K = 1.86 kJ/(kg·K)
We use cv given by (1.93 + 1.86)/2 = 1.895 kJ/(kg·K)
ΔU = 3×[tex]c_v[/tex]×(T₂ - T₁) = 3*1.895 *(266.35 -453.016) = -1061.2 kJ/kg
The internal energy for steam = [tex]U_g = h_g -pV_g[/tex]
[tex]h_g[/tex] = 2777.12 kJ/kg
[tex]V_g[/tex] = 0.194349 m³/kg
p = 1000 kPa
[tex]U_{g1}[/tex] = 2777.12 - 0.194349 * 1060 = 2571.11 kJ/kg
The final specific internal energy of the system is therefore, [tex]U_{g1}[/tex] + ΔU = 2571.11 - 1061.2 = 1509.91 kJ/kg.
