An isentropic steam turbine processes 5.5 kg/s of steam at 3 MPa, which is exhausted at 50 kPa and 100°C. Five percent of this flow is diverted for feedwater heating at 500 kPa. Determine the power produced by this turbine. Use steam tables.

The answer is 1823.9

Explanation:

Solution

Given that:

m = 5.5 kg/s

= m₁ = m₂ = m₃

The work carried out by the energy balance is given as follows:

m₁h₁ = m₂h₂ +m₃h₃ + w

Now,

By applying the steam table we have that<

p₃ = 50 kPa

T₃ = 100°C

Which is

h₃ = 2682.4 kJ/KJ

s₃ = 7.6953 kJ/kgK

Since it is an isentropic process:

Then,

p₂ =  500 kPa

s₂=s₃ = 7.6953 kJ/kgK

which is

h₂ =3207.21 kJ/KgK

p₁ = 3HP0

s₁ = s₂=s₃ = 7.6953 kJ/kgK

h₁ =3854.85 kJ/kg

Thus,

Since 5 % of this flow diverted to p₂ =  500 kPa

Then

w =m (h₁-0.05 h₂ -0.95 )h₃

5.5(3854.85 - 0.05 * 3207.21  - 0.95 * 2682.4)

5.5( 3854.83 * 3207.21 - 0.95 * 2682.4)

5.5 ( 123363249.32 -0.95 * 2682.4)

w=1823.9