Steam flows through a turbine as shown in the figure. The walls of the turbine are not insulated so that there is heat transfer through the walls as shown. The properties of steam are as given below:

P = 3 MPa, T = 500°C (h = 3456.48 kJ/kg, s = 7.2337 kJ/kg-K)

P = 0 .1013 MPa, T = 100°C (h_{f }= 491.1 kJ/kg, h_{g} = 2676 kJ /kg)

The power output from the turbine in kW is:

This question was previously asked in

BHEL ET Mechanical Held on May 2019

Option 3 :

90.03

BHEL Engineer Trainee Electrical 2019 Paper

1377

240 Questions
240 Marks
150 Mins

__Concept__

The steady flow energy equation for an open system is:

\(\left( {{h_1} +\ q} \right) = \left( {{h_2} +w} \right)\)

__Calculation:__

__Given:__

P_{1 } = 3 MPa, T_{1 } = 500°C, h_{1} = 3456.48 kJ/kg, s_{1} = 7.2337 kJ/kg-K

P_{2} = 0.1013 MPa, T_{2} = 100°C, x_{2 }= 0.97

\({h_f}_2 = 491.1~\frac{{kJ}}{{kg}}{\rm{}},~{h_g}_2 = 2676~\frac{{kJ}}{{kg}}\)

\(\dot m = 500~\frac{{kg}}{{hr}} = \frac{{500}}{{3600}} = 0.139~\frac{{kg}}{{sec}}\)

q = -200 kJ/kg

**Enthalpy at point 2 is:**

\({h_2} = {h_f}_2 + {x_2}\left( {{h_g}_2 - {h_f}_2} \right) = 491.1 + 0.97 × \left( {2676 - 491.1} \right) = 2610.45~\frac{{kJ}}{{kg}}\)

Applying steady flow energy equation,

\(\left( {{h_1} + q} \right) =\left( {{h_2} + w} \right){\rm{}}\)

\(\left( {3456.48 - 200} \right) =\ 2610.45 + w\)

**w = 646.03 kJ/kg**

Now to convert work done in kW, multiply with ṁ

ẇ = w × ṁ = 646.03 × 0.139 = **89.79 kW ≃ 90.03 kW**