Siklus Rankine

 

Update 13 April 2020

Boiler→3→Turbin→4→Condensor→1→Pump

        ↖ ⟵⟵⟵⟵2 ⟵⟵⟵⟵⟵↲

No. Aliran	1	2	3	4
P (bar a)	1	100	100	1
T (oC)	45	?	Saturated	?

ms=10  ton/jam × 1000 Kg/ton × 1/3600 jam/s = 2,778 Kg/s
a. Berapa kerja poros TURBIN teoritis yang di hasilkan (WT,th) ? [MW]

        *Cari di steam table smith van ness*
        H3(100 bar = 10000 Kpa) sat vapor, page 718
        [interpolasi]
       {x-x1} / {x2-x1} = {y-y1} /{y2-y1}
       {10000 - 9870} /{10142,1 - 9870} = {y- 2730} / {2725 - 2730}
        y= H3 = 2727,611 Kj/Kg
       S3=S4
       S3(10000 Kpa) *Sat vapor ,page 721*
       [interpolasi]
       {10000 - 9870} / {10142,1 - 9870} = {y- 5,6278} / {5,6111 - 5,6278}
       y = S3 = 5,619 Kj/Kg.K
       Sv (100 Kpa) *Sat vapor, page 718*
       [Interpolasi]
       {100 - 97,76} / {101,33 - 97,76} = {y- 7,3875} / {7,3554 - 7,3875}
       y = Sv = 7,367 Kj/Kg.K
       SL (100 Kpa) *Sat liquid, page 718*
       [Interpolasi]
       {100 - 97,76} / {101,33 - 97,76} = {y- 1,2956} / {1,3069 - 1,2956}
       y = SL = 1,303 Kj/Kg.K

       Subtitusi S3,Sv,SL
       S3 = S4 = x × Sv + (1-x)  SL
       5,619 Kj/Kg.K = x × 7,367 Kj/Kg.K + (1 – x) × 1,303 Kj/Kg.K
       x = 0,712
       Hv ( 1 bar = 100 Kpa) *sat vapor, page 718*
       [Interpolasi]
        {100 - 97,76} / {101,33 - 97,76} = {y- 2674,4} / {2676 - 2674,4}
       y = Hv = 2675,404 Kj/Kg
       HL ( 1 bar = 100 Kpa) *sat liquid, page 718*
       [Interpolasi]
       {100 - 97,76} / {101,33 - 97,76} = {y- 414,8} / {419,1 - 414,8}
       y = HL = 417,498 kj/kg
       H4,Th = x × Hv + (1-x) × HL
       H4,Th = 0,712 × 2675,404 kj/kg + (1-0,712) × 417,498 kj/kg
       H4,Th = 2025,127kj/kg
       WT,Th = ms (H4,Th - H3)
       WT,Th = 2,778 kg/s ( 2025,127 kj/kg - 2727,611 kj/kg)
       WT,Th = -1951,501 kW
       WT,Th = -1,951 MW

b. Berapa Kerja poros turbin actual (WT, act­) yang di hasilkan, asumsi   efisiensi internal turbin 85%?
       ηT = 85%
       ηT = {H3-H4,act} / {H3-H4,th}
       85% = {2727,611 Kj/Kg  -H4,act} / {2727,611 Kj/Kg - 2025,127 Kj}
       H4,act = 2130,4996 Kj/Kg
       WT, act­ = ms × (H4,act – H3)
       WT, act­ = 2,778 kg/s  (2130,4996 kj/kg – 2727,611 kj/kg)
       WT, act­ = -1658,775 Kw
       WT, act­ = -1,658 MW
       Suhu pada aliran 4 yaitu berdasarkan :

c. Berapa Theoritical Steam Rate (TSR)? [kg/kWh]
       TSR = {3600/H3-H4,th}
       TSR = 3600/{2727,611 kj/kg - 2025,127  kj/kg}
       TSR = 5,125 kg/kWh

d. Berapa Actual Steam Rate (ASR)? [kg/kWh]
       ASR = {3600/H3-H4,act}
       ASR = 3600/{2727,611 Kj/Kg - 2130,4996 Kj/Kg}
       ASR = 6,029 kg/kWh

e. Berapa panas yang di buang dari kondensor (Qc)? [MW]
       H1 (1 bar=100kPa, 45^oC)  berdasarkan suhu, sat liquid, page717
       H1 = 188,4 kj/kg
       Qc = ms (H1-H4,act)
       Qc = 2,778   (188,4 kj/kg– 2130,4996 kj/kg )
       Qc = -5395,153 KW
       Qc = -5,395 MW

f. Berapa kerja shaft pompa yang di perlukan untuk (Wp)? [MW],
     asumsi efisiensi internal pompa 60%
     ῡ (1 bar =100 kPa,45^oC) berdasarkan tekanan karena berbicara
      tentang volume sat liquid, page 718
       [Interpolasi]
       {100-97,76/101,33-97,76} = {y-1,043/1,044-1,043}
       y = ῡ = 1,043 cm3/g × 1000 g/kg × 1 m3/1000000cm3
       ῡ = 0,001043 m3/kg
       Ws = ῡ (P1-P2)
       Ws = 0,001043  (100 kPa – 10000 kPa)
       Ws = -10,3257
       ηP = 60%
       Ws,60% = 60%  (-10,3257)
       Ws,60% = -6,1954
       -Ws,60% = 6,1954
   
       Untuk mencari Wp kita cari terlebih dahulu H2
       -Ws,60% = H2-H1
       H2 = H1 + (-Ws,60%)
       H2 = 188,4  + 6,1954
       H2 = 194,5954
       Wp = ms (H1-H2)
       Wp = 2,778 kg/s (194,5954  - 188,4 )
       Wp = 17,2108 KW
       Wp = 0,01721 MW

       Suhu pada aliran 2 berdasarkan :
g. Berapa panas yang di suplai ke dalam boiler (Qb) ?[MW]
       Qb = ms (H3-H2)
       Qb = 2,778 kg/s (2727,611- 194,5954 )
       Qb = 7036,7173 KW
       Qb =7,036 MW

h. Berapa efisiensi siklus?
       ηsiklus = {|WT|-Wp}/Qb
       ηsiklus = {1,658 MW - 0,01721 MW} / {7,036 MW}
       ηsiklus = 0,23 =23 %