The heat machine receives 1200 J of heat from high-temperature heat source and re lease the heat of 900 Ĵ to low tenperatuve hent sovrce in each Y ound Que sion : Find the power of ma chine if each cycle takes 0-2 S. Angwer a) 6000 w 67 4500 c) 60 w d) 1500 w F) 240 W 92 180 W

The heat machine receives 1200 J of heat from high-temperature heat source and re lease the heat of 900 Ĵ to low tenperatuve hent sovrce in each Y ound Que sion : Find the power of ma chine if each cycle takes 0-2 S. Angwer a) 6000 w 67 4500 c) 60 w d) 1500 w F) 240 W 92 180 W

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter4: The Second Law Of Thermodynamics

Section: Chapter Questions

Problem 77AP: A heat engine operates between two temperatures such that the working substance of the engine...

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Check your Understanding Show that QhQh=QcQc for the hypothetical engine of Figure 4.10 The second property to be demonstrated is that all reversible engines operating between the same two reservoirs have the same efficiency. To this, stat with the two engines D and E of Figure 4.10 (a), which are operating between two common heat reservoirs at temperatures Th and Tc . First, assume that D is a reversible engine and that E is a hypothetical irreversible engine that has a higher efficiency than D. If both engines perform the same amount of work W per cycle, it follows from Equation 4.2 that QhQh . It then follows from the first law that QcQc . Figure 4.10 (a) Two uncoupled engines D and E working between the same reservoirs. (b) The engines, With D working reverse. Suppose the cycle of D is so that it operates as a refrigerator, and the two engines are coupled such that the work output of E is used to drive D, as shown in Figure 4.10(b). Since QhQh and QcQc , the net result of each cycle is equivalent to a spontaneous transfer of heat from the cold reservoir to the hot reservoir, a process second law does not allow. The original assumption must therefore be wrong, and it is impossible to construct an irreversible engine such that E is more efficient than the reversible engine D. Now it is quite easy to demonstrate that the efficiencies of all reversible engines operating between the same reservoirs are equal. Suppose that D and E are reversible engines. If they are as shown in Figure 4.10(b), the efficiency of E cannot be greater than the efficiency of D, or second law would violated. If both engines are then reversed, the same reasoning implies that the efficiency of D cannot be greater than the efficiency of E. Combining these results leads to the conclusion that all reversible engines working between same two reservoirs have the same efficiency.
Check Your Understanding In Example 4.7, the spontaneous flow of heat from a hot object to a cold object results in a net increase in entropy of the universe. Discuss how this result can be related to an increase in disorder of the system.
The gasoline internal combustion engine operates in a cycle consisting of six parts. Four of these parts involve, among other things, friction, heat exchange through finite temperature differences, and accelerations of the piston; it is irreversible. Nevertheless, it is represented by the ideal reversible Otto cycle, which is illustrated below. The working substance of the cycle is assumed to be air. The six steps of the Otto cycle ale as follows: i. Isobaric intake stroke (OA). A mixture of gasoline and air is drawn into the combustion chamber at atmospheric pressure P0 as the piston expands, increasing the volume of the cylinder from zero to VA . ii. Adiabatic compression stroke (AB). The temperature of the mixture rises as the piston compresses it adiabatically from a volume VA to VB . iii. Ignition at constant volume (BC). The mixture is ignited by a spark. The combustion happens so fast that there is essentially no motion of the piston. During this process, the added heat Q1 causes the pressure to increase from pB to pc at the constant volume VB(=Vc) . iv. Adiabatic expansion (CD). The heated mixture of gasoline and air expands against the piston, increasing the volume from VC to VD . This is called the power stroke, as it is the part of the cycle that delivers most of the power to the crankshaft. v. Constant-volume exhaust (DA). When the exhaust valve opens, some of the combustion products escape. There is almost no movement of the piston during this part of the cycle, so the volume remains constant at VA(=VD) . Most of the available energy is lost here, as represented by the heat exhaust Q2 . vi. Isobaric compression (AO). The exhaust valve remains open, and the compression from VA to zero drives out the remaining combustion products. (a). Using (i)e=W/Q1; (ii)w=Q1Q2; and (iii)Q1=nCv(TCTB),Q2=nCv(TDTA), Show that e=1TDTATCTB. (b). Use the fact that steps (ii) and (iv) are adiabatic to show that e=11r1 where r=VA/VB . The quantity r is called the compression ratio of the engine. (c) In practice, r is kept less than around 7. For larger values, the gasoline-air mixture is compressed to temperatures so high that it explodes before the finely timed spark is delivered. This preignition causes engine knock and loss of power. Show that for r=6 and =1.4 (the value for air), e=0.51 , or an efficiency of 51%. Because of the many irreversible processes, an actual internal combustion engine has an efficiency much less than this ideal value. A typical efficiency for a tuned engine is about 25% to 30%.