Steam enters a throttling valve at 8 000 kPa and 300 degree

Steam enters a throttling valve at 8, 000 kPa and 300 degree C and leaves at pressure of 1, 750 kPa. I) What is the final temperature of the steam? a) 80 degree C b) 180 degree C c) 100 degree C d) 150 degree C e) 206 degree C I) What is the final specific volume of the steam? a) 0.1267 m^3/kg b) 0.2567 m^3/kg c) 0.0356 m^3/kg d) 0.1467 m^3/kg e) 0.1112 m^3/kg

Solution

During a throttling process, W=0, Q=0, Potential and Kinetic Energiy changes are Zero, and the Enthalpy remains constant.

h₁ = h₂

From Steam tables, properties of steam

at 8000 kPa and 300⁰ C,

h₁= 2786.8 kJ/kg . and v₁ = 0.024264 m³/kg

at 1750 kPa

sat temp, T_s2= 205.72⁰ C

h_f2= 878.27 kJ/kg, h_g2 = 2794.1 kJ/kg, v_f2 = 0.0011656 m³/kg, v_g2= 0.11228 m³/kg

1) for throttling process, h₁=h₂   and let x₂be the state of steam after throttling.

h₁ =h_f2 + x₂ (h_g2-h_f2)

2786.8 = 878.27 + x₂ (2794.1 - 878.27)

x₂ = 0.99

That means the steam slightly wet ( almost dry saturated) after the throttling process. The temperature of steam from saturated water condition to saturated dry steam will be at its saturation temperature cocresponding to its pressure. Hence the final temperature of steam is 205.72 ⁰ C. The nearest option is \"e\"

2) The final specific volume of steam is given by

v₂ = v_f2 + x₂ (v_g2 - v_{f2) = 0.0011656 + 0.99 ( 0.11338 - 0.0011656) = 0.1122 m³/kg.}

_{The nearest option is \"e\".}