Question

Problem 2. : A superheated steam at 20 bars and 500^oC is fed to a turbine to generate electricity with a capacity of output of work at 500 kJ/kg. The existing steam is a saturated vapor at 2 bars. The surrounding environment temperature is 25^oC. The outer surface temperature of the turbine is 150^oC 1.) Clearly state your assumptions and draw the process flow diagram with labels ; 2.) Calculate the amount of heat exchanged between the system and surrounding ; 3.) Calculate the internal, external and total entropy generation for the turbine in kJ/kg-K and total entropy generation of the whole processes (turbine + surrounding environment) .

Answer 1

Initial superheated steam enters to the turbine,

T1 = 500 C and P1 = 20 bar

Specific enthalpy of superheated steam h1 = 3468.09 kj/kg

Exit condition of steam : P2 = 2 bar and saturated vapor

By steam table , saturation pressure Ps = 2 bar

Saturation temperature Ts = 120.21 C

Specific enthalpy of saturated vapor, h2 = 2706.2 kJ/kg

1.) assumptions :

a) steady state operation.

b) negligible kinetic and potential energy.

Block diagram for turbine :

2) Applying energy balance for turbine : steady state and open flow system

ΔK. E + ΔP. E + ΔH = Q - Ws

Here negligible change in kinetic energy and potential energy.

ΔK. E = 0 and ΔP. E = 0

ΔH = Q - Ws

Given shaft work Ws = 500 kj/kg

ΔH = h2 - h1 = 2706.2 - 3468.09 = -761.89 kJ/kg

-761.89 = Q - 500 kJ/kg

Q = -261.89  kJ/kg

heat exchange from system and surrounding   Q = -261.89 kJ/kg

3) surrounding environment temperature Tsurr = 25 C = 298 K

Turbine surface temperature, Tsystem = 150 C = 150 + 273 = 423 K

Heat exchange Q = 261.89 kJ/kg

Turbine treated as system.

Heat transfer from higher temperature to lower temperature,

Heat transfer from Tsystem = 150 C to Tsurr = 25 C .

Internal entropy generation means system entropy generation : heat gained by turbine then

Q =-  261.89 kJ/kg

(ΔS)system = Q/Tsystem =- (261.89kj/kg)/423k = - 0.619 kJ/kg.k

External entropy generation means surrounding entropy generation : heat loss the surrounding Q = +261.89 kJ/kg

(ΔS)surr = Q/Tsurr = (261.89kj/kg)/298k = 0.8788 kJ/kg.k

Total entropy generation :

(ΔS)total =( ΔS)surr + ΔS)system = 0.8788 - 0.619 = 0.259 kJ/kg.k

Total entropy generation is always positive for irreversible process.

For reversible process total entropy change is zero.