Condenser    

         Control Volume:  Condenser

      Working fluid : Steam (Water)

            Assumptions: 

            Constant Pressure

            SSSF Process and  ΔKE =0, ΔPE = 0 ; w =0                    state 3 = sat. liquid

Analysis:

              1^st law SSSF Proc.         q + h₂ = w + h₃

                                                                    q_C =  h₃ – h₂

       state 2 >> mixture@ P₂ ,   h₂

      state 3  sat. liq. à h₃ ~ h_f  @ P₃

 Pump

      Control Volume:  Pump

      Working fluid : liq. water

               Assumptions:

           Isentropic Process (ideal)

           SSSF Process  and ΔKE =0, ΔPE = 0 ; q = 0

           Compressed liq. >> v = v₃= constant

           Constant temperature process

Analysis:

1^st law SSSF Proc. q + h₃ = w + h₄

           w_p =  h₃ – h₄     =  v(P₃-P₄)

state 3 >> v₃ ~ v_f @ T₃

(or use approximation for liq. h ~ h_{f @T} + v_f (P - P_sat )

and in this case T = constant 

then   h₃ – h₄     =  v_f (P₃ - P₄)

  Boiler

              Assumptions:

• Constant Pressure Process
• SSSF Process 
•  ΔKE =0, ΔPE = 0 ; w = 0 

Assignment 2

Parametric Study of the Rankine Cycle

1 Analysis of the problem: Find the thermal efficiency of the cycle (assume isentropic efficiency of the turbine to be 92%)

2 Tablulation calculation in EXCEL, varying the PARAMETER

3 Plot graph show the relation of the study parameter vs. the cycle thermal efficiency

4 Show on your group website

5 Submit Report and Present in ClassWed 15 September 2010   

Case 3 The effect of the CONDENSER PRESSURE

Group 10

CONDENSER PRESSURE Begin  5  kPa

CONDENSER PRESSURE End  50  kPa

Boiler pressure  25 MPa

Steam temperature 400 ^oC

การคำนวณหาค่า 

1. Control Volume:  Turbine

Assumptions : SSSF Process  and ΔKE =0, ΔPE = 0 ;

                  q = 0, s₂ = s₁

                  1^st law SSSF Proc.        q + h₁ = w + h₂

                                                                               w_T =  h₁ – h₂

Working fluid : Steam (Water)

          state 1    >>   h₁ =  h @ T₁ and P₁

          steam table..     h₁ = 2578.7  kJ/kg

          state 2   >>  P₂ , P₂ = 7.5 kPa ,  s₂ = s₁ =5.14 kJ/kg.K  >> x= (s2-sf)/sfg

                   s_f=0.6492

                  s_fg= 7.4996

                      x=0.5988

           h₂ = h_f + x₂ h_fg        

           h₂ = 1739.063  kJ/kg

          w_T =  h₁ – h₂     =  912.649  kJ/kg

h_2s=1541.199 kJ/kg

2. Control Volume:  Condenser

Assumptions:

SSSF Process  and ΔKE =0, ΔPE = 0 ; w = 0

1^st law SSSF Proc.      q + h₂ = w + h₃

                                                     q_C =  h₃ – h₂

Working fluid : Steam (Water)

state 2 = mixture@ P₂ ,   h₂ already known

state 3  Saturated liq. >> h₃ = h_f @ P₃  = 191.81  kJ/kg

     q_C =  191.81   – 1739.063  =  -1547.253

3. Control Volume:  Pump

Assumptions:SSSF Process  and ΔKE =0, ΔPE = 0 ; q = 0,

Compressed liq. >> v = v₃= constant

Isentropic Process, s₃ = s₄

1^st law SSSF Proc.      q + h₃ = w + h₄

                                     w_p =  h₃ – h₄     =  v(P₄-P₃)

Working fluid : liq. water at state 3 >> v₃ ~ v_f @ P₃

          w_p =  v(P₄-P₃)  = (0.00101 m³/kg)(25,000 – 10) kPa 

              = 25.2399  kJ/kg

          h₄ =   h₃ + w_p =   191.81  + 25.2399  =  217.0499  kJ/kg

4. Control Volume:  Boiler

Assumptions:SSSF Process  and ΔKE =0, ΔPE = 0 ; w = 0

Solution:

1^st law SSSF Proc.      q + h₄ = w + h₁

                q_B =  h₁ – h₄   =  (2578.7  - 217.0499  ) kJ/kg 

              =  2361.65  kJ/kg

         w_net =  w_out – w_in  = w_turb – w_pump 

              =   (954.5  -25.2399) kJ/kg  = 929.26 kJ/kg

      h_th   =  w_net/q_in     =  929.26 /2361.65 

              = 0.393 (or 39.3%)   answer