Answer :
Answer:
a) T₁ = 446.25° C
b) w = 879 KJ/kg
Explanation:
First we consider the initial state of steam:
Pressure of steam = P₁ = 70 bar = 7 MPa
Temperature of steam = T₁ = ?
a)
Now, after passing through throttling valve, the steam properties become:
Pressure = P₂ = atmospheric pressure = 1 bar = 100 KPa
Temperature = T₂ = 400⁰C
From saturated water mixture table, the saturation temperature at 100 KPa is 99.61°C.
Therefore, T₂ >> T(saturation) and the water is in super heated state.
So, from super heated table:
Specific Enthalpy at state 2 = h₂ = 3278.6 KJ/kg
Now, we know that throttling is an isenthalpic process. Therefore:
h₁ = h₂ = 3278.6 KJ/kg
Now, we check hg at state 1 from table
hg₁ = 2772.6 KJ/kg
Therefore, h₁ > hg₁ and steam is in super heated state. So, from super heated table, by interpolation, at h = 3278.6 KJ/kg, and P = 7 MPa, we get:
T₁ = 446.25° C
b)
From super heated table at state 1, we find,the specific entropy and specific enthalpy, linear interpolation:
s₁ = 6.6214 KJ/kg.k
h₁ = 3278.6 KJ/kg
Now, consider state 2 at exit of turbine. Since the process is adiabatic and reversible. Therefore, it will be isentropic:
s₂ = s₁ = 6.6214 KJ/kg.k
P₂ = 100 KPa
At this pressure, sg = 7.3589 KJ/kg > s₂. Hence, the steam is in liquid-vapor mixture state.
Quality of Steam = x = (s₁ - sf)/sfg = (6.6214 - 1.3028)/6.0562
x = 0.878 = 87.8%
Now, the specific enthalpy will be:
h₂ = hf + x(hfg) = 417.51 KJ/kg + (0.878)(2257.5 KJ/kg)
h₂ = 2399.59 KJ/kg
Now, the turbine work done per kg of steam is given by:
w = h₁ - h₂ = 3278.6 KJ/kg - 2399.59 KJ/kg
w = 879 KJ/kg