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Thermodynamic Solution - PDF

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Added on 2020-04-21

Thermodynamic Solution - PDF

Added on 2020-04-21

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THERMODYNAMIC SOLUTIONAnswer-1Hot oil th1=200°C`th2=80°CCold watertc1=18°Ctc2=70°Ca) Arithmetic mean temperature difference (AMTD)AMTD=th1+th22−tc1+tc22AMTD=200+802−70+182AMTD=140−44AMTD=96°Cb)Logarithmic temperature difference (LMTD or θm)LMTD=θ1−¿θ2lnθ1θ2¿θ1=th1−tc2=200−70=130°Cθ2=th2−tc1=80−18=62°CLMTD=130−62ln13062LMTD=91.84°Cc) Overall heat transfer rate 200°C18°C80°C70°CTemperature Diagram

Q=nUAθmU= Overall heat transfer coefficient =75 W/m2KA=πDL = 0.141 m^2Q=80∗75∗0.141∗91.84Q=77.7KWd) Mass flow rate of oilQ=moco(th1−th2¿77.7=mo∗2.19∗(200−80)mo=0.296KgsMass flow rate of waterQ=mwcw(tc2−tc1¿77.7=mw∗4.19∗(70−18)mw=0.357KgsAnswer -2

C2H5OH + 3O2 → 2CO2 + 3H2O If 2.5kg of Ethanol is burnt in air stoichiometrically (i.e. to completion), perform a mass balance to determine:i.The mass of Oxygen(O2) required Molecule wt of ethanol =46 grmsFor 2.5 Kg of ethanol , No. of moles = 2500/46 = 55For 1 mole of ethanol = 3 moles of O2So No. of moles of O2 required = 55 * 3 =165Molecule wt of O2 =32 grmsSo For 2.5 Kg of ethanol , O2 required = 165 *32/1000So For 2.5 Kg of ethanol , O2 required = 5.28 Kgii.The mass of Air (23% O2& 77% N2) requiredAs 23 % of O2 = 5.28 * 23 /100 = 1.214 KgAs 77 % of N2 = 5.28 * 77 /100 = 4.06 Kgiii.The mass of Nitrogen (N2) requiredMolecule wt of N2 =28 grmsN2 required = 165 *28/1000N2 required = 4.62 Kgiv.The Air/Fuel ratio

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