347×10-2 Path2:Q2= nRT In=1×8.315×423×ln 241×10-2 282.13(J) Pah3:g3=nCp(T4-Tc)=1×3.5×8.315×(293-423)=-378333J) The total heat transfer to the gas over the cycle is Qoma=Q+Q2+93=270238+128213-378333=201.18(J) (d)The efficiency of the cycle W 208.35 Qn"+Q22702.38+128250005 (e)T,=293K(coldest); T:=423(hottest); TC=423K The maximum efficiency that a heat engine is the carnot engine's efficiency, that is g=1-=1-293=0.3 423 3. A gasoline internal combustion engine can be approximated by the cycle shown in Fig 4. Assume an ideal diatomic gas and use a compression ratio of 4: 1(V 4V) Assume that P =3P a. Pbl (a)Determine the pressure and temperature of each of the SparktAdiabatic vertex points of the Pl diagram in terms of Pa and Ta.(b) Calculate the efficiency of the cycle Intake Solution Adiabatic For the ideal diatomic gas: F=5, so r=2*\:1.2 Fig 4 a)For the isochoric process a-b: Pb=3Pa T P 3P T=D Ta T=37 P For the adiabatic process b->c: s=Va=4/ P=P→P=P()=3P()=3P()2=0.57P x-=r-7==(1ny-=()23x=227 For the isochoric process c>d: T=TPath 2: 1282.13(J) 2.41 10 3.47 10 ln 1 8.315 423 ln 2 2 2 = × × = = × × × − − B C B V V Q nRT Path 3: ( ) 1 3.5 8.315 (293 423) 3783.33(J) Q3 = nCP TA − TC = × × × − = − The total heat transfer to the gas over the cycle is 2702.38 1282.13 3783.33 201.18(J) Qtotal = Q1 + Q2 + Q3 = + − = (d) The efficiency of the cycle is 0.05 2702.38 1282.59 208.35 1 2 = + = + = = Q Q W Q W total H total ε (e) TA = 293K(coldest); TB = 423(hottest); TC = 423K The maximum efficiency that a heat engine is the carnort engine’s efficiency, that is 0.31 423 293 =1− =1− = H C T T ε 3. A gasoline internal combustion engine can be approximated by the cycle shown in Fig.4. Assume an ideal diatomic gas and use a compression ratio of 4:1 (Vd=4Va). Assume that Pb=3Pa. (a) Determine the pressure and temperature of each of the vertex points of the PV diagram in terms of Pa and Ta. (b) Calculate the efficiency of the cycle. Solution: For the ideal diatomic gas: i=5, so 1.2 1 = + = i i γ (a) For the isochoric process a→b : Pb=3Pa a a a a a a b b a b a b T T P P T P P T P P T T 3 3 = ⇒ = ⋅ = ⋅ = For the adiabatic process b→c : Vc = Vd = 4Va a a d a a c b b b c c c b P P V V P V V PV PV P P ) 0.57 4 1 ( ) 3 ( ) 3 ( 1.2 = ⇒ = = = = γ γ γ γ b a a c b b b c c c T T T V V V T V T T ) 3 2.27 4 1 ( ) ( 1 1 1 1.2 1 = ⇒ = = = γ − γ − γ − − For the isochoric process c→d : Td = Ta P V Fig.4 Pb Va Vd a b c Adiabatic Adiabatic Intake Spark d