Question

Steam enters a turbine operating at steady state at 1 MPa, 200 °C and exits at 40 °C with a quality of 83%. Stray heat transfer and kinetic and potential energy effects are negligible. Determine a. The power produced by the turbine, in kJ/kg b. The change in specific entropy from inlet to exit, in kJ/kg-K

Answer 1

Answer:

(a) Work out put=692.83$\frac{KJ}{Kg}$

(b) Change in specific entropy=0.0044$\frac{KJ}{Kg-K}$

Explanation:

Properties of steam at 1 MPa and 200°C

        $h_1=2827.4\frac{KJ}{Kg},s_1=6.69\frac{KJ}{Kg-K}$

We know that if we know only one property in side the dome then we will find the other property by using steam property table.

Given that dryness or quality of steam at the exit of turbine is 0.83 and temperature T=40°C.So from steam table we can find pressure corresponding to saturation temperature 40°C.

Properties of saturated steam at 40°C

      $h_f= 167.5\frac{KJ}{Kg} ,h_g= 2537.4\frac{KJ}{Kg}$

 $s_f= 0.57\frac{KJ}{Kg-K} ,s_g= 8.25\frac{KJ}{Kg-K}$

So the enthalpy of steam at the exit of turbine  

$h_2=h_f+x(h_g-h_f)\frac{KJ}{Kg}$

$h_2=167.5+0.83(2537.4-167.5)\frac{KJ}{Kg}$

$h_2=2134.57\frac{KJ}{Kg}$

$s_2=s_f+x(s_g-s_f)\frac{KJ}{Kg-K}$

$s_2=0.57+0.83(8.25-0.57)\frac{KJ}{Kg-K}$

$s_2=6.6944\frac{KJ}{Kg-K}$

(a)

Work out put =$h_1-h_2$

                      =2827.4-2134.57 $\frac{KJ}{Kg}$

Work out put =692.83 $\frac{KJ}{Kg}$

(b) Change in specific entropy

     $s_2-s_1=6.6944-6.69\frac{KJ}{Kg-K}$

Change in specific entropy =0.0044$\frac{KJ}{Kg-K}$

Answer 2

(a) Power produced by turbine be "663.7707 kJ/kg".

(b) Change in specific entropy be "0.2540903 kJ/kg-K".

Kinetic and Potential energy

According to the question,

At 1 MPa and 200°C, Inlet

h₁ = 2827.3 kJ/kg

s₁ = 6.692 kJ/kg-K

At 40°C, Outlet

$h_f$ = 166.79 kJ/kg, $h_g$ = 2572.5 kJ/kg

$s_f$ = 0.56969 kJ/kg-k, $s_g$ = 8.2521 kJ/kg-K

Quality, x = 83%

Now,

→ h₂ = $h_f$ + x ($h_g -h_f$)

By substituting the values,

       = 166.79 + 0.83 × (2572.5 - 166.79)

       = 2163.5293 kJ/kg

→ s₂ = $s_f$ + x ($s_g -s_f$)

      = 0.56969 + 0.83 × (8.2521 - 0.56969)

      = 6.9460903 kJ/kg-K

Now,

(a) The power developed be:

= h₁ - h₂

= 2827.3 - 2163.5293

= 663.7707 kJ/kg

(b) Change in specific entropy be:

= s₂ - s₁

= 6.9460903 - 6.692

= 0.2540903 kJ/kg-K

Thus the above answer is appropriate.

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