Thermodynamics Hipolito Sta Maria Solution Manual Chapter 5 !!better!! (2024)

A reversible heat engine operates between a source at 800°C and a sink at 30°C. It receives 1200 kJ/min from the source. Determine: a) Carnot efficiency b) Power output in kW c) Heat rejected to sink d) Entropy change of source and sink per minute.

d) For the source (heat transfer out, negative Q): [ \Delta S_source = \frac-Q_HT_H = \frac-1200 \ kJ/min1073 \ K = -1.118 \ kJ/K·min ] For the sink: [ \Delta S_sink = \frac+Q_LT_L = \frac339303 = +1.119 \ kJ/K·min ] Total entropy change = (1.119 - 1.118 = 0.001 \ kJ/K·min) (≈0 for reversible engine – rounding). thermodynamics hipolito sta maria solution manual chapter 5

[ \dotm(s_e - s_i) = \frac\dotQT_boundary + \dotS_gen ] A reversible heat engine operates between a source

For a closed system undergoing a process from state 1 to 2: d) For the source (heat transfer out, negative

. This chapter builds on the ideal gas processes (isometric, isobaric, isothermal, isentropic, and polytropic) introduced in the previous chapter to analyze how systems perform work in cycles. Key Concepts and Formulas The core of Chapter 5 is the Carnot Cycle

a) (T_H = 800 + 273 = 1073 K), (T_L = 30 + 273 = 303 K) [ \eta_Carnot = 1 - \frac3031073 = 0.7176 \ (71.76%) ]