Question

1 kg of saturated steam at 1000 kPa is in a piston-cylinder and the massless cylinder is held in place by pins. The pins are removed and the system suddenly and adiabatically expands to 5x its original volume before the piston hits a pair of upper pins. The expansion takes place against an atmosphere is 60 kPa. What is the final specific internal energy of the system

Answer 1

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;

$\dfrac{p_1}{p_2} = \left (\dfrac{V_2}{V_1} \right )^k = \left [\dfrac{T_1}{T_2} \right ]^{\dfrac{k}{k-1}}$

Adding the effect of the atmospheric pressure, we have;

p = 1000 + 60 = 1060

We therefore have;

$\dfrac{1060}{p_2} = \left (\dfrac{5\cdot V_1}{V_1} \right )^{1.33}$

$P_2= \dfrac{1060}{5^{1.33}} = 124.65 \ kPa$

$\left [\dfrac{V_2}{V_1} \right ]^k = \left [\dfrac{T_1}{T_2} \right ]^{\dfrac{k}{k-1}}$

$\left [\dfrac{V_2}{V_1} \right ]^{k-1} = \left \dfrac{T_1}{T_2} \right$

$5^{0.33} = \left \dfrac{T_1}{T_2} \right$

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×$c_v$×(T₂ - T₁)

$c_v$ = 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×$c_v$×(T₂ - T₁) = 3*1.895 *(266.35 -453.016) = -1061.2 kJ/kg

The internal energy for steam = $U_g = h_g -pV_g$

$h_g$ = 2777.12 kJ/kg

$V_g$ = 0.194349 m³/kg

p = 1000 kPa

$U_{g1}$ = 2777.12 - 0.194349 * 1060 = 2571.11 kJ/kg

The final specific internal energy of the system is therefore, $U_{g1}$ + ΔU = 2571.11 - 1061.2 = 1509.91 kJ/kg.