Steam is produced at 70 bar and some unknown temperature. A small amount of steam is bled off just before entering a turbine and goes through an adiabatic throttling valve to atmospheric pressure. The temperature of the steam exiting the throttling valve is 400°C. The unthrottled steam is fed into the turbine, where it is adiabatically expanded to atmospheric pressure.
a. What is the temperature of the steam entering the turbine?
b. What is the maximum work per kilogram of steam that can& obtained using the turbine in its present mode of operation?

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

gets less

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