LESSON - 33 NUMERICAL PROBLEMS RELATED TO VAPOUR POWER CYCLE

 

Problem 33.1: Dry and saturated steam at pressure 11 bar is supplied to a turbine and expanded isentropically to pressure 0.07 bar. Calculate the following

    (a) Heat rejected, (b) Heat supplied, (c) theoretical thermal efficiency.

Solution:

Given: Dry and saturated steam supplied to a turbine at pressure, p₁ =  11 bar

              Steam expanded isentropically to pressure, p₂ = 0.07 bar.

      From Steam table (Dry saturated steam)

At pressure, p₁ = 11 bar   →   h_g,1= 2779.7 kJ/kg, s_g,1 = 6.550 kJ/kg K

At pressure, p₂= 0.07 bar   →   h_f,3 = 163 kJ/kg,    h_fg =  2409.2 kJ/kg, s _f,3  = 0.559  kJ/kgK,

s_fg = 7.718 kJ/kgK,    v _f,3  = 0.001  m³/kg

(a) Determine heat rejected, q_R

     Formula:   Heat rejected, q_R = h₂ – h_f,₃       

                    Finding unknown, h₂:

Isentropically expansion, therefore   s₂ = s_g,1

                                                                = 6.554

Since for wet steam at point ‘2’,  s₂ = s_f,3+ x₂  s_{fg,(p=0.07 bar)}  

Therefore,          s_f,3+ x₂  s _{fg,(p=0.07 bar)}  = 6.554

                                     0.559 + x₂ (7.718) = 6.550 

                            x₂ = 0.776

Enthalpy for wet steam at point ‘2’,   h₂ = h _f,3 + x₂ h_{fg,(p=0.07 bar)}

 h₂ = 163.4 + 0.776 (2409.2) = 2032.9 kJ/kg

Answer:  Heat rejected, q_R = h₂ – h_f,₃  =  2032.9 - 163 =  1869.9 kJ/kg

(b) Determine heat supplied, q_A

Formula:   Work done during cycle (w) = Heat supplied (q_A)  – heat rejected (q_R)

                                     or  q_A  = w + q_R

Finding unknown, w:

          Net work done during cycle, w = w_t – w_p

Finding unknown, w_t and w_p :

                              Turbine work, w_t =  h_g,1 – h₂  = 2779.7- 2032.9= 746.8 kJ/kg

                              Pump work, w_p = v_f,3 (p_{1 –} p₂) 10²  = 0.001 (11 – 0.07) 10²  = 1.093 kJ/kg         

                              w = w_t – w_p  = 746.8 - 1.093 = 745.7 kJ/kg

Answer:     q_A  = w + q_R = 745.7 +  1869.9 = 2615.6 kJ/kg

(c) Determine theoretical thermal efficiency η_th:

Answer:     = 28.5%

Problem 33.2: A steam turbine receives steam at pressure 20 bar and superheated to 88.6°C. The exhaust pressure is 0.07 bar and the expansion of steam takes place isentropically Using steam table, calculate the following.

(a) Heat rejected (b) Heat supplied, assuming that the feed pump supplies water to the boiler at 20 bar (c) Net work done (d) Work done by the turbine (e) Thermal efficiency (f) Theoretical steam consumption.

Solution.

 

Given: Superheated steam supplied to a turbine at pressure, p₁ =  20 bar  

             Temperature of superheated steam = t_s,1 + 88.6°C.

             From Steam table (Dry saturated steam)

At pressure, p₁ = 20 bar   →   t_s,1 = 212.4°C  

            Therefore, temperature of superheated steam, t_sup,₁ =  2l2.4 + 88.6 = 300°C

            From Steam table (superheated steam)\

At pressure, p₁ = 20 bar  and   t_sup,₁ = 300°C   →   h_sup,1 = 3025 kJ/kg, s_sup,1 = 6.768 kJ/kg 

Given:  Steam expanded isentropically to pressure, p₂ = 0.07 bar.

From Steam table (Dry saturated steam)

At pressure, p₂ = 0.07 bar   →   h_f,3 = 163 kJ/kg,  h_fg =  2409.2 kJ/kg, s _f,3  = 0.552  kJ/kgK, s_fg = 7.722 kJ/kgK, v _f,3  = 0.001  m³/kg

From Steam table (Dry saturated steam)

At pressure, p₂ = 0.07 bar   →   h_f,3 = 163 kJ/kg,    h_fg =  2409.2 kJ/kg, s _f,3  = 0.552  kJ/kgK, s_fg = 7.722 kJ/kgK,    v _f,3  = 0.001  m³/kg

(a) Determine heat rejected, q_R

     Formula:   Heat rejected, q_R = h₂ – h_f,₃       

                    Finding unknown, h₂:

  Isentropically expansion, therefore   s₂ = s_sup,1

                                                                = 6.768

 Since for wet steam at point ‘2’,  s₂ = s_f,3+ x₂  s_{fg,(p=0.07 bar)}  

 Therefore,          s_f,3+ x₂  s_{fg,(p=0.07 bar)}  = 6.768

                                  0.552 + x₂ (7.722) = 6.768 

                                   x₂ = 0.8047

 Enthalpy for wet steam at point ‘2’,   h₂ = h _f,3 + x₂h_{fg,(p=0.07 bar)}

 h₂ = 163.4 + 0.8047 (2409.2) = 2101.52 kJ/kg

Answer:  Heat rejected, q_R = h₂ – h_f,₃  =  2101.52  - 163 =  1938.52 kJ/kg

(d) Determined work done by the turbine, w_t  :

       Formula:   Turbine work, w_t  =  h_sup,1 – h₂

     Answer:   Turbine work, w_t  =  h_sup,1 – h₂ = 3025 - 2101.52 = 923.48 kJ/kg

(c) Determine net work done, w :

         Formula:   Net work done during cycle, w = w_t – w_p

                                                   Finding unknown, w_p :

                                                   Pump work, w_p = v_f,3 (p_{1 –} p₂) 10²

                                                                              = 0.001 (20 – 0.07) 10²  = 1.993 kJ/kg         

     Answer:  Net work done during cycle, w = w_t – w_p

                                                                            = [923.48 – 1.993] = 921.487 kJ/kg

(b) Determine heat supplied, q_A :

               Formula:   Net work done during cycle (w) = Heat supplied (q_A)  – heat rejected (q_R)

                          or     q_A  = w + q_R

     Answer:             q_A  = w + q_R = 921.487 + 1938.52 = 2860.0 kJ/kg

(e) Thermal efficiency, η_th:

      Answer:   Thermal efficiency,    η_th     = 32.21%

(f) Theoretical steam consumption:

     Answer:   Theoretical steam consumption =  = 3.91 kg/kWh

Problem 33.3: A steam turbine receives superheated steam at a pressure of 17 bar and having a degree of superheat of 110^oC. The exhaust pressure is 0.07 bar and the expansion of steam takes place isentropically. Calculate (a) The heat rejected, (b) the heat supplied, (c) net work done, and (d) thermal efficiency. (Neglect pump work)

 Solution:

Given: Superheated steam supplied to a turbine at pressure, p₁ =  17 bar  

             Temperature of superheated steam = t_s,1 + 110°C.

              Pump work, w_p =  0

            From Steam table (Dry saturated steam)

At pressure, p₄ = p₁' = p₁ = 17 bar   →   t_s,1 = 212.4°C

           Therefore, temperature of superheated steam, t_sup,₁ =  2l2.4 + 110 = 322.4°C

            From Steam table (superheated steam)

At pressure, p₁ = 17 bar  and   t_sup,₁ = 322.4°C   →   h_sup,1 = 3083.4 kJ/kg,      s_sup,1 = 6.939 kJ/kg K

Given:  Steam expanded isentropically to pressure, p₂ = 0.07 bar.

From Steam table (Dry saturated steam)

At pressure, p₂ = p₃ = 0.07 bar   →   h_f,3 = 163.4 kJ/kg,    h_fg =  2409.2 kJ/kg,  s _f,3  = 0.559  kJ/kgK, s_fg = 7.718 kJ/kgK,   v _f,3  = 0.001  m³/kg

(a) Determine heat rejected, q_R

Formula:   Heat rejected, q_R = h₂ – h_f,₃       

Finding unknown, h₂:

  Isentropically expansion, therefore   s₂ = s_sup,1

                                                                = 6.939

Since for wet steam at point ‘2’,  s₂ = s_f,3+ x₂  s _{fg,(p=0.07 bar)}  

Therefore,          s_f,3+ x₂  s _{fg,(p=0.07 bar)}  = 6.939

                                  0.559 + x₂(7.718) = 6.939

                                   x₂ = 0.827

Enthalpy for wet steam at point ‘2’,   h₂ = h _f,3 + x₂h_{fg,(p=0.07 bar)}

 h₂ = 163.4 + 0.827 (2409.2) = 2155.8 kJ/kg

 Answer:   Heat rejected, q_R = h₂ – h_f,₃  =  2155.8  - 163.4 =  1992.4 kJ/kg

(b) Determine heat supplied, q_A

Formula:   q_A= h_sup,1 – h_f,3

    Answer:       q_A  = h_sup,1 – h_f,3 = 3083.4 – 163.4 = 2920 kJ/kg

(c) Determine net work done, w :

Formula:   Net work done during cycle, w = Heat supplied (q_A)  – heat rejected (q_R)

    Answer:  Net work done during cycle, w = (q_A)  – (q_R)

                                                                          = 2920 – 1992.4 = 927.6 kJ/kg

(d) Thermal efficiency, η_th:

      Answer:     Thermal efficiency, η_th  = 31.7%

Problem 33.4: A turbine takes dry steam at 20 bar and exhaust at 1.2 bar. The pressure at the release is 3 bar. Find: (a) The theoretical loss of work per kg of steam due to incomplete expansion and (b) The loss in Rankine efficiency due to restricted expansion of steam. (Neglect pump work)

Solution:

 

Given: Dry and saturated steam supplied to a turbine at pressure, p₁ =  20 bar

             Release pressure, p₂'= 3

             Exhaust  pressure, p₂ = 1.2 bar.

            Pump work, w_p =  0

From Steam table (Dry saturated steam)

Pressure (bar)		Enthalpy    (kJ/kg)	Sp. Vol.    (m3/kg)	Entropy    (kJ/kgK)
p1 = 20		hg,1 = 2797.2	v g,1  = 0.0995	s g,1  = 6.337
p2'= 3		hf,2' = 561.4; hfg = 2163.2	vg,2'  =0.6055	sf ,2'  = 1.672; sfg = 5.319
p3 = 1.2		hf,3 = 439.4; hfg = 2244.1		sf,3 = 1.361; sfg = 5.937

(a) Determine the theoretical loss of work per kg of steam due to incomplete expansion:

Formula: Loss in work due to incomplete expansion = w_R - w_MR

                               where,  w_R and  w_MR  work done during Rankine and modified Rankine cycle, respectively

Finding unknown, w_MR ;

                              Work done during modified Rankine cycle, w_MR = (h_g,1 – h_2') + v_3' (P_2' – P_3')

Finding unknown, h_2′  and  v_3′  ;

‘1-2′’ Isentropically expansion, therefore   s_2' = s_g,1

                                                                            = 6.337 

Since for wet steam at point ‘2′’,  s_2'  =s_{f, 2'} + x_2' s_{fg, (at p = 3 bar)}

Therefore,          s_2'  =s_{f, 2'} + x_2' s_{fg, (at p = 3 bar)} = 6.337

                                      1.672 + x_2' (5.319) = 6.337

                                         x_2' = 0.877

Enthalpy for wet steam at point ‘2′’, h_2' = h_{f, 2'} + x_2' h_{fg, (at p = 3 bar)}

h_2' = 561.4 + 0.877 (2163.2) = 2458.5 kJ/kg

Sp. volume for wet steam at point ‘2′’,  v_2' = x_2' v_g,2'

                                                                  = 0.877 (0.6055)

                                                                   = 0.531 m³/kg

Sp. volume for wet steam at point ‘3′’,  v_3' = v_2' = 0.531 m³/kg

           Therefore, w_MR = (h_g,1 – h_2') + v_3' (P_2' – P_3')

                                       = (2797.2 - 2458.5) +  0.531 (3 – 1.2) 10²     

                                       = 434.3 kJ/kg

Finding unknown, w_R ;

                              Work done during rankine cycle,  w_R = h_g,1 – h₂

Finding unknown, h₂  ;

‘1-2’ Isentropically expansion, therefore   s₂ = s_g,1

                                                                          = 6.337

Since for wet steam at point ‘2’,  s₂ = s_f,3+ x₂  s _{fg,(p=1.2 bar)}  

Therefore,          s_f,3+ x₂  s _{fg,(p=0.07 bar)}  = 6.337

                                     1.361 + x₂(5.937) = 6.337

                                     x₂ = 0.838

Enthalpy for wet steam at point ‘2’,   h₂ = h _f,3 + x₂h_{fg,(p=1.2 bar)}

h₂ = 439.4 + 0.838 (2244.1) = 2320 kJ/kg kJ/kg

          Therefore,  w_R = h_g,1 – h₂

                                   = 2797.2 – 2320 = 477.2 kJ/kg

 Answer:  Loss in work due to incomplete expansion = w_R - w_MR = 477.2 - 434.3

 = 42.9 kJ/kg

(b) The loss in Rankine efficiency due to restricted expansion of steam:

Formula:    Loss in Rankine efficiency due to restricted expansion = × 100

Answer:     Loss in Rankine efficiency = × 100

      = 9%