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   →   hg,1= 2779.7 kJ/kg, sg,1 = 6.550 kJ/kg K

At pressure, p2= 0.07 bar   →   hf,3 = 163 kJ/kg,    hfg =  2409.2 kJ/kg, s f,3  = 0.559  kJ/kgK,

sfg = 7.718 kJ/kgK,    v f,3  = 0.001  m3/kg

(a) Determine heat rejected, qR

     Formula:   Heat rejected, qR = h2 – hf,3       

                    Finding unknown, h2:

Isentropically expansion, therefore   s2 = sg,1

                                                                = 6.554

Since for wet steam at point ‘2’,  s2 = sf,3+ x2  sfg,(p=0.07 bar)  

Therefore,          sf,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 hfg,(p=0.07 bar)

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

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

(b) Determine heat supplied, qA

Formula:   Work done during cycle (w) = Heat supplied (qA)  – heat rejected (qR)

                                     or  qA  = w + qR

Finding unknown, w:

          Net work done during cycle, w = wt – wp

Finding unknown, wt and wp :

                              Turbine work, wt =  hg,1 – h2  = 2779.7- 2032.9= 746.8 kJ/kg

                              Pump work, wp = vf,3 (p1 – p2) 102  = 0.001 (11 – 0.07) 102  = 1.093 kJ/kg         

                              w = wt – wp  = 746.8 - 1.093 = 745.7 kJ/kg

Answer:     qA  = w + qR = 745.7 +  1869.9 = 2615.6 kJ/kg

(c) Determine theoretical thermal efficiency ηth:

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 = ts,1 + 88.6°C.

             From Steam table (Dry saturated steam)

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

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

            From Steam table (superheated steam)\

At pressure, p1 = 20 bar  and   tsup,1 = 300°C   →   hsup,1 = 3025 kJ/kg, ssup,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   →   hf,3 = 163 kJ/kg,  hfg =  2409.2 kJ/kg, s f,3  = 0.552  kJ/kgK, sfg = 7.722 kJ/kgK, v f,3  = 0.001  m3/kg

From Steam table (Dry saturated steam)

At pressure, p2 = 0.07 bar   →   hf,3 = 163 kJ/kg,    hfg =  2409.2 kJ/kg, s f,3  = 0.552  kJ/kgK, sfg = 7.722 kJ/kgK,    v f,3  = 0.001  m3/kg

(a) Determine heat rejected, qR

     Formula:   Heat rejected, qR = h2 – hf,3       

                    Finding unknown, h2:

  Isentropically expansion, therefore   s2 = ssup,1

                                                                = 6.768

 Since for wet steam at point ‘2’,  s2 = sf,3+ x2  sfg,(p=0.07 bar)  

 Therefore,          sf,3+ x2  sfg,(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 + x2hfg,(p=0.07 bar)

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

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

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

       Formula:   Turbine work, wt  =  hsup,1 – h2

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

(c) Determine net work done, w :

         Formula:   Net work done during cycle, w = wt – wp

                                                   Finding unknown, wp :

                                                   Pump work, wp = vf,3 (p1 – p2) 102

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

     Answer:  Net work done during cycle, w = wt – wp

                                                                            = [923.48 – 1.993] = 921.487 kJ/kg

(b) Determine heat supplied, qA :

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

                          or     qA  = w + qR

     Answer:             qA  = w + qR = 921.487 + 1938.52 = 2860.0 kJ/kg

(e) Thermal efficiency, ηth:

      Answer:   Thermal efficiency,    ηth     = 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 110oC. 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 = ts,1 + 110°C.

              Pump work, wp =  0

            From Steam table (Dry saturated steam)

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

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

            From Steam table (superheated steam)

At pressure, p1 = 17 bar  and   tsup,1 = 322.4°C   →   hsup,1 = 3083.4 kJ/kg,      ssup,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   →   hf,3 = 163.4 kJ/kg,    hfg =  2409.2 kJ/kg,  s f,3  = 0.559  kJ/kgK, sfg = 7.718 kJ/kgK,   v f,3  = 0.001  m3/kg

(a) Determine heat rejected, qR

Formula:   Heat rejected, qR = h2 – hf,3       

Finding unknown, h2:

  Isentropically expansion, therefore   s2 = ssup,1

                                                                = 6.939

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

Therefore,          sf,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 + x2hfg,(p=0.07 bar)

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

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

(b) Determine heat supplied, qA

Formula:   qA= hsup,1 – hf,3

    Answer:       qA  = hsup,1 – hf,3 = 3083.4 – 163.4 = 2920 kJ/kg

(c) Determine net work done, w :

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

    Answer:  Net work done during cycle, w = (qA)  – (qR)

                                                                          = 2920 – 1992.4 = 927.6 kJ/kg

(d) Thermal efficiency, ηth:

      Answer:     Thermal efficiency, ηth = 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, wp =  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 = wR - wMR

                               where,  wR and  wMR  work done during Rankine and modified Rankine cycle, respectively

Finding unknown, wMR ;

                              Work done during modified Rankine cycle, wMR = (hg,1 – h2') + v3' (P2' – P3')

Finding unknown, h2′  and  v3′  ;

‘1-2′’ Isentropically expansion, therefore   s2' = sg,1

                                                                            = 6.337 

Since for wet steam at point ‘2′’,  s2'  =sf, 2' + x2' sfg, (at p = 3 bar)

Therefore,          s2'  =sf, 2' + x2' sfg, (at p = 3 bar) = 6.337

                                      1.672 + x2' (5.319) = 6.337

                                         x2' = 0.877

Enthalpy for wet steam at point ‘2′’, h2' = hf, 2' + x2' hfg, (at p = 3 bar)

h2' = 561.4 + 0.877 (2163.2) = 2458.5 kJ/kg

Sp. volume for wet steam at point ‘2′’,  v2' = x2' vg,2'

                                                                  = 0.877 (0.6055)

                                                                   = 0.531 m3/kg

Sp. volume for wet steam at point ‘3′’,  v3' = v2' = 0.531 m3/kg

           Therefore, wMR = (hg,1 – h2') + v3' (P2' – P3')

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

                                       = 434.3 kJ/kg

Finding unknown, wR ;

                              Work done during rankine cycle,  wR = hg,1 – h2

Finding unknown, h2  ;

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

                                                                          = 6.337

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

Therefore,          sf,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 + x2hfg,(p=1.2 bar)

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

          Therefore,  wR = hg,1 – h2

                                   = 2797.2 – 2320 = 477.2 kJ/kg

 Answer:  Loss in work due to incomplete expansion = wR - wMR = 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%

Last modified: Wednesday, 11 September 2013, 5:09 AM