Thermodynamics



  • ef529395-8935-4808-bedb-cb9d534e8d6f-image.png
    ANS A



  • @You-Knowwhere
    Process CBC BCB is done at minimum temperature, while process DAD ADA is done at maximum temperature Process A→b(A \rightarrow b(Ab( isobaric
    VAVB=TATB=800400=2\frac{V_{A}}{V_{B}}=\frac{T_{A}}{T_{B}}=\frac{800}{400}=2VBVA=TBTA=400800=2
    hence VB=2V0V_{B}=2 V_{0}VB=2V0
    B→C( isothermal )B \rightarrow C(\text { isothermal })BC( isothermal )
    C→D(C \rightarrow D(CD( isobaric
    VDVC=TDTC⇒VD=2V0\frac{V_{D}}{V_{C}}=\frac{T_{D}}{T_{C}} \Rightarrow V_{D}=2 V_{0}VCVD=TCTDVD=2V0
    D→(D \rightarrow(D( isothermal WAB=P(VB−VA)=nRDTW_{A B}=P\left(V_{B}-V_{A}\right)=n R D TWAB=P(VBVA)=nRDT
    =nR(−400)=−400nR=n R(-400)=-400 n R=nR(400)=400nR
    WBC=nRT1nVCVB=nR4001nV02V0W_{B C}=n R T 1 n \frac{V_{C}}{V_{B}}=n R 4001 n \frac{V_{0}}{2 V_{0}}WBC=nRT1nVBVC=nR4001n2V0V0
    =−400nR1n2=-400 n R 1 n 2=400nR1n2
    WCD=P(VD−VC)=nR(400)=400nRW_{C D}=P\left(V_{D}-V_{C}\right)=n R(400)=400 n RWCD=P(VDVC)=nR(400)=400nR
    WDA=nRTlnVAVDW_{D A}=n R T \ln \frac{V_{A}}{V_{D}}WDA=nRTlnVDVA
    =nR×800×1n2V0V0=800nR1n2=n R \times 800 \times 1 n \frac{2 V_{0}}{V_{0}}=800 n R 1 n 2=nR×800×1nV02V0=800nR1n2
    ΔW=400nR1n2\Delta W=400 n R 1 n 2ΔW=400nR1n2
    ΔQ=\Delta Q=ΔQ= heat is extracted QCD=nCpΔT=n54R400=1000nRQ_{C D}=n C_{p} \Delta T=n \frac{5}{4} R 400=1000 n RQCD=nCpΔT=n45R400=1000nR
    =800nR1n2=1000nR+800nR1n2=800 n R 1 n 2=1000 n R+800 n R 1 n 2=800nR1n2=1000nR+800nR1n2
    Efficiency =ΔWΔQ=\frac{\Delta W}{\Delta Q}=ΔQΔW
    η=21n25+41n2×100\eta=\frac{21 n 2}{5+41 n 2} \times 100η=5+41n221n2×100%



  • @raj1213 But how Vb = Va and Vc = Vd






Log in to reply
 

Powered by dubbtr | @2020