ch20-p051 - 51(a If TH is the temperature of the...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Therefore, () L HL 1 =. fi F dm PT dt T T cT T L §· ¨¸ −+ ©¹ Now, P = 100 × 10 6 W, T L = 0 + 273 = 273 K, T H = 800 + 273 = 1073 K, T i = –40 + 273 = 233 K, T f = 0 + 273 = 273 K, c = 2220 J/kg·K, and L F = 333 × 10 3 J/kg, so ( ) 6 3 100 10 J/s 273 K 1 = 1073 K 273 K 2220 J/kg K 273 K 233 K +333 10 J/kg 82kg/s. dm dt ªº × «» −⋅ × ¬¼ = We note that the engine is now operated between 0°C and 800°C. 51. (a) If T H is the temperature of the high-temperature reservoir and T L is the temperature of the low-temperature reservoir, then the maximum efficiency of the engine is H 800 + 40 K = = = 0.78 or 78%. 800 + 273 K TT T ε (b) The efficiency is defined by = | W |/| Q H |, where W is the work done by the engine and Q H is the heat input. W is positive. Over a complete cycle, Q H = W + | Q L |, where Q L is the heat output, so = W /( W + | Q L |) and | Q L | = W [(1/ ) – 1]. Now = ( T H T L )/ T H , where T H is the temperature of the high-temperature heat reservoir and
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

Ask a homework question - tutors are online