Chapter 19, Problem 54P, part task bI think there is an error in part task b. It is not clear how to find the final expression for T_max.After it has been made clear that T_max = T_3, there is a description (See blue square in the attached picture part task b) which describes that an expression must substitute T_1, but the written expression applies to T_3?In the calculation after this, there is a sudden jump to the final expression for T_max. But it's not clear how I'm going to get to the final expression? (The orange square in the attached picture of part task b)I have includet pictures from: The task descriptions. Expert sulution to part task a, where we are given the expression, that is needet to solve part task b. And expert sulution to part task b 60. Gasoline engines operate ap-proximately on the Otto cycle,consisting of two adiabaticand two constant-volume seg-ments. Figure 19.25 showsthe Otto cycle for a particularengine. (a) If the gas in theengine has specific-heat ratioy, find the engine's efficiency,assuming all processes arereversible. (b) Find the maxi-mum temperature in terms ofFIGURE 19.25 Problem 60the minimum temperature Tmin. (c) How does the efficiency com-pare with that of a Carnot engine operating between the sametemperature extremes?3p₂PP2-Combustion324AdiabaticexpansionAdiabaticcompressionV₁VExhaust/intake!s (a) Expert SolutionTo determineThe efficiency of the engine.Answer to Problem 54PThe efficiency of the engine is 1-5¹-7.Explanation of SolutionGiven info: The given gasoline engine has two adiabatic and two constant volume segments. The specific heat ratio isy. All processes are assumed to be reversible.Write the expression for the temperature of the hot reservoir from the figure 19.24.Th= T3 - T₂Here,Th is the temperature of the hot reservoirT3 is the temperature at point 3T2 is the temperature at point 23pz (= V₁)nRSubstituteTh=e3P2 (V₁) P2 (V₁)nRnR-=Here,P2 is the pressure at point 2V₁ is the volume at point 1n is the amount of substanceR is the universal gas constantWrite the expression for the pressure of point 1 and 2 from the adiabatic compression.P1 = P2 (²) ²Here,y is the specific heat ratioWrite the expression for the pressure of point 4 and 3 from the adiabatic expansion.P4 =3p2 (1)Write the expression for the temperature of the cold reservoir from the figure 19.24.Tc = T4-T₁Here,Te is the temperature of the cold reservoirT4 is the temperature at point 4T₁ is the temperature at point 1YSubstituteP4V₁nRP₁V₁Rfor T4, 3p2 (3) for p₁, P₁ for T₁ and p2 (3) for p in the above equation.3p2 (3) V₁ P2()'V₁Te =nRnRe = 1-Substitute=Write the expression for efficiency of the engine.= 1.TeTh2p₂ V₁5nRHere,e is the efficiency of the engine2p₂V₁5nR2p2V15/nR2p2V15nR= 1-5¹-y2p₂ V₁5/nRTmaxfor T3 and(b) Expert Solution=Conclusion:Therefore, the efficiency of the engine is 1 - 5¹-1.To determineThe maximum temperature in terms of the minimum temperature.P₂(+V₁)nRAnswer to Problem 54PThe maximum temperature in terms of the minimum temperature is Tmaxfor T₂ in the above equation.Explanation of SolutionWrite the expression for minimum temperature at the point 1.Tmin = T₁P2V15YnRSubstitute2p₂ V₁for Te and for Th in the above equation.5nRT3Substitute for T₁ in the above equation to find the minimum temperature at point 1.P2V157nRTminWrite the expression for maximum temperature at the point 3.T₁ =max3p2V15nR==3.57-¹ Tmin.3p2V1for T₁ in the above equation to find the minimum temperature at point 1.5nR3.5-¹ TminConclusion:Therefore, the maximum temperature in terms of the minimum temperature is Tmax=3.5-¹ min.

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