study problems ch7 - CHAPTER 7 STUDY PROBLEMS THE SECOND...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
CHAPTER 7 STUDY PROBLEMS THE SECOND LAW OF THERMODYNAMICS Heat engines and refrigerators The second law of thermodynamics The reversible and irreversible process The Carnot cycle and its efficiency The thermodynamic and ideal gas temperature scales Real machines and heat transfer with temperature differences FUNDAMENTALS of Thermodynamics Sixth Edition SONNTAG BORGNAKKE VAN WYLEN 61
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
7.1 A window air-conditioner A window air-conditioner is advertised as having a 5000 BTU/h cooling capacity. It requires an electrical power input of 500 W. Find the rate of heat rejected to the outside atmosphere and the coefficient of performance of the unit. Solution: C.V. Air-conditioner. Assume steady state so no storage of energy. The information provided is W . = 500 W and the heat absorbed is Q . L = 5000 BTU/h = 5000 × 1.055 3600 kW = 1.4653 kW The energy equation gives: Q . H = Q . L + W . = 1465.3 + 500 = 1965 W From the definition of the coefficient of performance, Eq.7.2 β REFRIG = Q . L W . = 1465.3 500 = 2.93 H Q W = 500 W L T L T amb REF Remark: Assume the high and low temperatures in the cycle are 45 ° C and 5 ° C then a Carnot refrigerator would have β REFRIG = Q . L W . = T L T H - T L = 278.15 40 = 6.95 62
Background image of page 2
7.2 Influence of T H on the heat engine efficiency Let us examine a heat engine that rejects energy to the ambient, which is at 20 ° C. Assume it burns some fuel and air mixture providing heat addition temperatures of T H = 1500 K, 1000 K and 750 K, respectively. What are the maximum expected thermal efficiencies? Solution: First, to consider the maximum efficiency we must assume the heat engine works as a Carnot heat engine. The efficiency then becomes, Eq.7.5 η Carnot HE = W HE Q H = 1 – T L T H from which we also see that the maximum is obtained with the minimum temperature for the rejection of energy T L . For this we assume ambient so T L = T amb = 20 ° C = 293.15 K and we can compute the three efficiencies T H : 750 1000 1500 η Carnot HE 0.609 0.707 0.805 The efficiency increases with the high temperature approaching the upper limit of 1. Thus an amount of heat transfer at a higher temperature results in more useful work output than the same amount delivered at a lower temperature.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/20/2009 for the course MECHENG MEecheng 3 taught by Professor Borgnakke during the Fall '09 term at University of Michigan.

Page1 / 8

study problems ch7 - CHAPTER 7 STUDY PROBLEMS THE SECOND...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online