Exercise 1 In a Rankine cycle, steam leaves the boiler 4 MPa and 400◦C. The condenser pressure is 10 kPa. Determine the cycle efficiency & Simplified flow diagram for the following cases: a. Basic ideal cycle b. Reheat cycle with 2 turbine stages where the steam at 400 kPa is heated back to 400 ◦C – Also calculate the moisture content at turbine outlet and compare it to case (a) c. Regenerative cycle, where some portion of the steam is extracted at 400 kPa and fed to an open feedwater heater to pre-heat the feedwater to saturated liquid at 400 kPa – Also calculate the weight fraction of steam extracted at 400 kPa d. Actual cycle where the turbine efficiency is 86%, pump efficiency is 80%, Pump outlet pressure 5 Mpa, and the total piping pressure drop between boiler to turbine and pump to boiler is 0.2 Mpa (other pressure drop is considered insignificant) 1
Transcript
Exercise 1
In a Rankine cycle, steam leaves the boiler 4 MPa and 400◦C. The condenser pressure is 10 kPa. Determine the cycle efficiency & Simplified flow diagram for the following cases: a. Basic ideal cycle b. Reheat cycle with 2 turbine stages where the steam at 400 kPa is heated
back to 400 ◦C – Also calculate the moisture content at turbine outlet and compare it to case
(a) c. Regenerative cycle, where some portion of the steam is extracted at 400
kPa and fed to an open feedwater heater to pre-heat the feedwater to saturated liquid at 400 kPa – Also calculate the weight fraction of steam extracted at 400 kPa
d. Actual cycle where the turbine efficiency is 86%, pump efficiency is 80%, Pump outlet pressure 5 Mpa, and the total piping pressure drop between boiler to turbine and pump to boiler is 0.2 Mpa (other pressure drop is considered insignificant)
1
Exercise 1c – Menghitung efisiensi • Dari jawaban sebelumnya:
h5 = 3213.6 h6 = 2685.6 h7 = 2144.1 h1 = 191.8
• kondisi condenser tidak berubah (h1 dan h7 tetap), namun laju alir massa berubah (karena sebagian steam diekstrak sebesar y). Sehingga menghitung Qcond menggunakan: Qcond = (1-y) h1-h7 = (1-0.1654)* -1952.3 kj/kg = -1629.14 kj/kg
• Untuk menghitung efisiensi tinggal menghitung Q boiler = h5-h4
• Karena h5 tetap, yg perlu dicari adalah h4 • Dari kurva T-S terlihat titik 3 pada sat. Liquid
(asumsi, karena tidak mungkin dua fase kavitasi di pompa), dari soal diketahui P3 = 400 kPa. Lalu diperoleh v = 0.001084 m3/kg; h3 = 604.7
• W pompa high pressure (34) dihitung dari : v(P4-P3) = 3.9 kJ/kg
Exercise 1c – Mencari y Koreksi: utk mencari y, lebih mudah menggunakan neraca energi di feed water heater: h6(y) + h2 (1-y) = h3
Diperoleh y = 0.1654
*langkah mencari h2 sama dengan mencari h4 di pompa high pressure Karena neraca energi di turbin yg benar harusnya memperhitungkan daya turbin: Energi masuk = Energi keluar h5 = Wturbine + h6*(y) + h7*(1-y) • Turbine ideal = isentropic, sehingga berlaku:
– Dari s6, pada P = 400 kPa, bisa dicari X6 & h6 – Dari s7, pada P = 10 kPa, bisa dicari X7 & h7
• Lalu dari neraca energi di turbin: Entalpi masuk = Entalpi keluar koreksi lihat diatas h5 = h6(y) + h7(1-y) koreksi lihat diatas
3
Exercise 1d - Turbine
Secara umum langkah pengerjaan mirip dengan contoh 1a. Yang berbeda adalah: – Adanya pressure drop (mengubah kondisi T,P untuk
mencari H/S dari tabel) – Adanya efisiensi isentropis di pompa dan turbin
4
Turbine Pada kondisi inlet 3.8 Mpa/380 °C: h5 = 3169.1 s5 = 6.72 Pada saat kondisi isentropis: s6s = s5 = 6.72 Lalu dicari x6s = (s6s-s6f)/(s6g-s6f) x6s = 0.8098 h6s = hf (1-x6s) + hg (x6s) = 2129.5 Daya turbine dihitung menggunakan efisiensi isentropis: Wt = ηt*(h5-h6s) = 894.1 kJ/kg
Exercise 1d - Pompa
5
Pompa Pada kondisi inlet 10 kPa/42 °C: v = 0.001009 m3/kg Daya pompa: Wp = v (P2-P1)/ηp = 6.3 kJ/kg Wnet = Wt-Wp = 887.8 kJ/kg Qboiler = h4-h3 (langsung didapat dari tabel) = 3041.8 ηcycle = Wnet/Qboiler = 29.2%
Langkah pengerjaan mirip dengan turbine. Di dalam soal tidak disebutkan adanya Pressure drop antara Turbine-Condenser-Pump. Dalam aplikasinya, Lokasi condenser biasanya berada persis di bawah turbine, dan lokasi pompa juga di bawah condenser, sehingga kalaupun ada ∆P biasanya sangat kecil dan bisa diabaikan
Exercise 2
The temperature at the beginning of the compression process of an air-standard Otto cycle with a compression ratio of 8 is 540 °R, the pressure is 1 atm, and the cylinder volume is 0.02 ft3. The maximum temperature during the cycle is 3600 °R. Determine: (a) the temperature and pressure at the end of
each process of the cycle, (b) the thermal efficiency, and (c) the mean effective pressure, in atm 6
Exercise 3
A gas-turbine power plant operating on an ideal Brayton cycle has a pressure ratio of 8. The gas temperature is 300 K at the compressor inlet and 1300 K at the turbine inlet. Utilizing the air-standard assumptions, determine (a) the gas temperature at the exits of the
compressor and the turbine, (b) the back work ratio, and (c) the thermal efficiency.
7
Exercise 4
In an air-standard Brayton cycle the air enters the compressor at 0.1 MPa and 15◦C. The pressure leaving the compressor is 1.0 MPa, and the maximum temperature in the cycle is 1100◦ C. Determine:
a. The pressure and temperature at each point in the cycle. b. The compressor work, turbine work, and cycle efficiency. For each control volume analyzed, the model is ideal gas with constant specific heat at 300 K, and each process is steady state with no kinetic or potential energy changes.
Recalculate answer (a) and (b) if it the compressor efficiency is 80%, turbine efficiency is 85%, and pressure drops between the compressor and turbine are 15 kPa
8
Exercise 4 Pada soal diasumsikan berlaku sifat gas ideal dengan panas spesifik yg konstan. Sehingga untuk menghitung efisiensi langsung menggunakan:
9
Untuk menghitung daya pompa, turbin, dan exchanger diperoleh dari:
Note: nilai h2;h1 tidak dapat diambil dari tabel A-22, karena diasumsikan Cp konstan
Perbandingan korelasi utk Cp konstan dan Cp = f(T)
10
Perbandingan korelasi utk Cp konstan dan Cp = f(T)
