← Back
ExamFull examExam paper only

24 06 21sol

Study material for Energy Systems LM, shared by the Studwiz community and reviewed by moderators.

Energy Systems LMFull exam

Document information

What's included in this study material

Study material for Energy Systems LM, shared by the Studwiz community and reviewed by moderators.

Import quality: text was extracted directly from the original document.

Extracted content from the document

Representative passages recognised in different parts of the material. The full extracted text remains available to search, while this compact preview makes the page easier to read.

Page 1

WRITTEN TEST ENERGY SYSTEMS 24 JUNE 2021 Data Values from tables/Mollier charts Data from person code Person code: 12345678 1 2 3 4 5 6 7 8 X 36 Y 260 Problem 1 Data V biogas: 2600 Nm3/h xCH4 0.6 xCO2 0.4 Air excess 36 % Electric efficiency of the engine: eta,el = 38 % Exhaust gas temperature: Tg = 450 °C DTpp in the heat recovery steam generator: 18 °C cp,gas = 1.1 kJ/kgK LHV CH4 = 50 MJ/kg Results 1) 0.6 CH4 + 0.4 CO2 + 2*0.6*1.15*(O2 + 3.76 N2) -> CO2 + 2*0.6 H2O + 2*0.6*0.15 O2 + 2*0.6*1.15*3.76 N2 Combustion products: CO2 1 kmolCO2/kmolbiogas H2O 1.2 kmolH2O/kmolbiogas O2 0.432 kmolO2/kmolbiogas N2 6.14 kmolN2/kmolbiogas Total 8.77 kmolgas/kmolbiogas Molar fractions of combustion products xCO2 0.114 xH2O 0.137 xO2 0.049 xN2 0.700 2) CH4 input: VCH4 = Vbiogas*xCH4 = 1560 Nm3/h CH4 input: mCH4 = VCH4/3600/22.414*MMCH4 = 0.3093 kg/s Fuel energy input: Qfuel = mCH4*LHV = 15.47 MW Pel = Qfuel*eta,el = 5.88 MW 3) MM biogas = 0.6*16 + 0.4*44 = 27.2 kg/kmol Biogas input: Mbiogas = Vbiogas/3600/22.414 = 0.0322 kmol/s Biogas input: mbiogas = Mbiogas*MMbiogas = 0.876 kg/s Air input: Mair = 2*0.6*1.15*4.76 *Mbiogas = 0.2503 kmolair/s MM air = 0.21*32 + 0.79*28 = 28.84 kg/kmol Air input: mair = Mair*MMair = 7.219 kg/s Exhaust gas flow rate: mgas = mbiogas+mair = 8.095 kg/s Tsat steam at 6 bar = 158.8 °C Tgas, stack = Tsteam + DTpp = 176.8 °C DTgas in heat recovery steam generator = 273.2 °C Q recovered = mgas*cp,gas*DTgas = 2433 kW 4) Thermal efficiency: eta,th = Qrecovered / Qfuel = 15.73 % PES = 1/(eta,el/40 + eta,th/90) = 0.111 Problem 2 Pel = 36 MWe eta,el = 0.98 eta,isc = 0.85 eta,ist = 0.90 Tout,c = 300 °C TOT = 526 °C Tamb = 35 °C pamb = 1.01 bar Dp,filter = 1 % Dp,he = 3 % Dp,ex = 1 % cp,a = 1.05 kJ/kgK MMa = 28.85 kg/kmole gamma = 1.38 1) pin,c = pamb * (1-Dp,filter)…

Preview

First page of the document.

First page: 24 06 21sol