Thermodynamic Solution - PDF

Added on - 21 Apr 2020

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THERMODYNAMIC SOLUTIONAnswer-1Hot oilth1=200°C`th2=80°CCold watertc1=18°Ctc2=70°Ca) Arithmetic mean temperature difference (AMTD)AMTD=th1+th22tc1+tc22AMTD=200+80270+182AMTD=14044AMTD=96°Cb)Logarithmictemperature difference (LMTD orθm)LMTD=θ1¿θ2lnθ1θ2¿θ1=th1tc2=20070=130°Cθ2=th2tc1=8018=62°CLMTD=13062ln13062LMTD=91.84°Cc) Overall heat transfer rate200°C18°C80°C70°CTemperature Diagram
Q=nUAθmU= Overall heat transfer coefficient=75 W/m2KA=πDL= 0.141 m^2Q=80750.14191.84Q=77.7KWd) Mass flow rate of oilQ=moco(th1th2¿77.7=mo2.19(20080)mo=0.296KgsMass flow rate of waterQ=mwcw(tc2tc1¿77.7=mw4.19(7018)mw=0.357KgsAnswer -2
C2H5OH + 3O2→ 2CO2+ 3H2OIf 2.5kg of Ethanol is burnt inairstoichiometrically (i.e. to completion),perform a mass balance to determine:i.The mass of Oxygen(O2) requiredMolecule 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 O2required = 55 * 3 =165Molecule wt of O2=32 grmsSo For 2.5 Kg of ethanol , O2required = 165 *32/1000So For 2.5 Kg of ethanol , O2required = 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 grmsN2required = 165 *28/1000N2required = 4.62 Kgiv.The Air/Fuel ratio
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