Assignment on Thermodynamics and Basic Chemistry

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Added on 2020-05-11

Assignment on Thermodynamics and Basic Chemistry

Added on 2020-05-11

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IntroductionThe assignment focuses on thermodynamics and basic chemistry. Especially, the calculations based on the operation cycle of a diesel engine.Part A: Thermodynamics and Engines”Question 1“Thermal efficiency is given as n=1−QoutQ¿=1−mcv(T4−T1)mcP(T3−T2)Let CPCV=γEfficiency¿1−1γ((T4−T1)(T3−T2))all temperatures can be expressed in terms of T3 from:T1=T2rvγ−1 , T2=T3V2V3=T3rc, and T4=T3rvγ−1Therefore, η=1−1γ(T3rvγ−1−T3rcrvγ−1T3−T3rc)η=1−1γ((rcrvγ−1−reγ−1)reγ−1rcrvγ−1)rx−1rcT3T3

Assignment on Thermodynamics and Basic Chemistry_1

η=1−1γ1reγ−1rvγ−1[rc(reγ−1rcγ−1)−reγ−1rc−1]Hence, the thermal efficiency, η=1−1γ1rvγ−1[rcγ−1rc−1]But V3V2=rc=α and V1V2=rv=ΨReplacing these in the thermal efficiency equation yields:η=1−1γ1Ψγ−1[αγ−1α−1]Question 2V1=1949.56litres,V2=140Litres,V3V2=1.86,V3=1.86×140=260.4litres,V4=V1=1949.56litresT4=50°C,T3=T4rvγ−1=50×13.925(¿1.3−1)=110.18°C¿T2=T3V2V3=110.18×11.86=59.24°C, T1=T2rvγ−1=59.2413.925(¿1.3−1)=26.88°C¿P2=P3=150 ̄,P1=P4=P2rvγ−1=15013.925(¿1.3−1)=68.07 ̄¿¿Question 3The specific heat capacity ratio¿CPCV=γ=1.3Swept volume¿V1−V2

Assignment on Thermodynamics and Basic Chemistry_2

¿π4×D2L¿π4×0.962×2.5¿1.80956M3V1−V2=1.80956V1=1.80956+V2=1.80956+140=1949.56LitresCompression ratio Ψ=V1V2=1949.56140=13.925Thermal efficiency, η=1−1γ1Ψγ−1[αγ−1α−1]=1−11.3×113.925(¿1.3−1)[1.861.3−1(1.86−1)]=0.4964=49.64%¿Question 4”Also, thermal efficiency=Heatmassflow×Calorificvalue=80×106 ̇m×45×106 ̇m=80×1060.4964×45×106=3.5813kg/sDiesel consumption per day= ̇3.5813×24×3600=309.424×103Kg“Watt’s GovernorQuestion 1 and 2

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