Termodinamica

6-1

Chapter 6 THE SECOND LAW OF THERMODYNAMICS
The Second Law of Thermodynamics and Thermal Energy Reservoirs 6-1C Water is not a fuel; thus the claim is false. 6-2C Transferring 5 kWh of heat to an electric resistance wire and producing 5 kWh of electricity. 6-3C An electric resistance heater which consumes 5 kWh of electricity and supplies 6 kWh of heat to a room. 6-4C Transferring 5 kWh ofheat to an electric resistance wire and producing 6 kWh of electricity. 6-5C No. Heat cannot flow from a low-temperature medium to a higher temperature medium. 6-6C A thermal-energy reservoir is a body that can supply or absorb finite quantities of heat isothermally. Some examples are the oceans, the lakes, and the atmosphere. 6-7C Yes. Because the temperature of the oven remains constant nomatter how much heat is transferred to the potatoes. 6-8C The surrounding air in the room that houses the TV set. Heat Engines and Thermal Efficiency 6-9C No. Such an engine violates the Kelvin-Planck statement of the second law of thermodynamics. 6-10C Heat engines are cyclic devices that receive heat from a source, convert some of it to work, and reject the rest to a sink. 6-11C Method (b). With theheating element in the water, heat losses to the surrounding air are minimized, and thus the desired heating can be achieved with less electrical energy input. 6-12C No. Because 100% of the work can be converted to heat. 6-13C It is expressed as "No heat engine can exchange heat with a single reservoir, and produce an equivalent amount of work". 6-14C (a) No, (b) Yes. According to the second law,no heat engine can have and efficiency of 100%. 6-15C No. Such an engine violates the Kelvin-Planck statement of the second law of thermodynamics. 6-16C No. The Kelvin-Plank limitation applies only to heat engines; engines that receive heat and convert some of it to work.

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6-17 The power output and thermal efficiency of a power plant are given. The rate of heat rejection is to bedetermined, and the result is to be compared to the actual case in practice. Assumptions 1 The plant operates steadily. 2 Heat losses from the working fluid at the pipes and other components are negligible. Analysis The rate of heat supply to the power plant is determined from the thermal efficiency relation, & W 600 MW & = 1500 MW QH = net,out = Furnace η th 0.4 The rate of heat transfer to the riverwater is determined from the first law relation for a heat engine,
& & & QL = QH − Wnet,out = 1500 − 600 = 900 MW
sink ηth = 40% HE 600 MW

In reality the amount of heat rejected to the river will be lower since part of the heat will be lost to the surrounding air from the working fluid as it passes through the pipes and other components.

6-18 The rates of heat supply and heat rejection of apower plant are given. The power output and the thermal efficiency of this power plant are to be determined. Assumptions 1 The plant operates steadily. 2 Heat losses from the working fluid at the pipes and other components are taken into consideration. Analysis (a) The total heat rejected by this power plant is
& QL = 145 + 8 = 153 GJ/h
Furnace

Then the net power output of the plant becomes& & & Wnet,out = QH − QL = 280 − 153 = 127 GJ/h = 35.3 MW

& Q H = 280 GJ/h
HE

& QL
sink

(b) The thermal efficiency of the plant is determined from its definition,
& Wnet,out 127 GJ/h η th = & = = 0.454 = 45.4% 280 GJ/h QH

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6-19E The power output and thermal efficiency of a car engine are given. The rate of fuel consumption is to be determined. Assumptions The car operatessteadily. Properties The heating value of the fuel is given to be 19,000 Btu/lbm. Analysis This car engine is converting 28% of the chemical energy released during the combustion process into work. The amount of energy input required to produce a power output of 110 hp is determined from the definition of thermal efficiency to be
& Wnet,out 110 hp  2545 Btu/h  &  = 999,598 Btu/h  QH = = ...