Prove that the “cycle” consisting of only two adiabatic steps (reversible expansion and then compression) will have no efficiency since no work can be produced. Will it violate the 2nd law of the thermodynamics? Clear explain your answer.
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Prove that the “cycle” consisting of only two adiabatic steps (reversible expansion and
then compression) will have no efficiency since no work can be produced. Will it violate
the 2nd law of the
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- For the Carnot cycle of Figure 4.12, what is the entropy change of the hot reservoir, the cold reservoir, and the universe? Figure 4.11 The four processes of the Carnot cycle. The working substance is assumed to be an ideal gas whose thermodynamic path MNOP is represented in Figure 4.12. Figure 4.12 The total work done by the gas in the Carnot cycle is shown and given by the area enclosed by the loop MNOPM.Consider an ideal gas Joule cycle, also called the Brayton cycle, shown below. Find the formula for efficiency of the engine using this cycle in terms of P1 , P2 and .(a) infinitesimal amount of heat is added reversibly to a system. By combining the first and second laws, show that dU=TdSdW. (b) When heat is added to an ideal gas, its temperature and volume change from T1 and V1 to T2 and V2 . Show that the entropy change of n moles of the gas is given by S=CnvlnT2T1nRlnV2V1 .
- Two hundred grams of water at 0 is brought into contact with a heat reservoir at 80 . After thermal equilibrium is reached, what is the temperature of the water? Of the reservoir? How much heat has been transferred in the process? What is the entropy change of the water? Of the reservoir? What is the entropy change of the universe?Two hundred grams of water at 0 is brought into contact into thermal equilibrium successively with reservoirs at 20 , 40 , 60 , and 80 . (a) What is the entropy change of the water? (b) Of the reservoir? (c) What is the entropy change of the universe?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%.
- Check Your Understanding A 50-g copper piece at a temperature of 20 is placed into a large insulated vat of water in 100 . (a) What is the entropy change of the copper piece when it reaches thermal equilibrium with the water? (b) What is the entropy change of the water? (c) What is the entropy change of the universe?A 0.50-kg piece of aluminum at 250 is dropped into 1.0 kg of water at 20 . After equilibrium is reached, what is the net entropy change of the system?In an isochoric process, heat is added to 10 mol of monoatomic ideal gas whose temperature increases from 273 to 373 K. What is the entropy change of the gas?
- Suppose that the temperature of the water in the previous problem is raised by fist bringing it to thermal equilibrium with a reservoir at a temperature of 40 and then with a reservoir at 80 . Calculate the entropy changes of (a) each reservoir, (b) of the water, and (c) of the universe.Check Your Understanding A quantity of heat Q is absorbed from a reservoir at a temperature Th by a cooler reservoir at a temperature Tc . What is the entropy change of the hot reservoir, the cold reservoir, and the universe?Does the entropy increase for a Carnot engine for each cycle?
