Unit D-2: List of Subjects

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1 ES312 Energy Transfer Fundamentals Unit D: The Second Law of Thermodynamics ROAD MAP... D-1: From First to Second Law D-2: Second Law of Thermodynamics Unit D-2: List of Subjects Second Law of Thermodynamics Perpetual-Motion Machine (PMM) Reversible and Irreversibile Processes

PAGE 1 of 7 Second Law of Thermodynamics (1) A heat engine that violates the Kelvin-Planck statement of the second law of thermodynamics The schematic of refrigerator The schematic of heat pump THE

The Kelvin-Planck statement of the second law of thermodynamics states: It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of

As a power plant to operate, the working fluid must exchange heat with the environment (sink) as the furnace (source).

2 PAGE 1 of 7 Second Law of Thermodynamics (1) A heat engine that violates the Kelvin-Planck statement of the second law of thermodynamics The schematic of refrigerator The schematic of heat pump THE SECOND LAW: KELVIN-PLANCK STATEMENT (HEAT ENGINES) A heat engine must exchange heat with a low-temperature sink as well as a high-temperature source to keep operating. The Kelvin-Planck statement of the second law of thermodynamics states: It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work. No heat engine can have a thermal efficiency of 100% (not due to friction or other dissipative effects). As a power plant to operate, the working fluid must exchange heat with the environment (sink) as the furnace (source). No heat engine can have 100% efficiency: this is a limitation that applies to both the idealized and the actual heat engines. REFRIGERATORS AND HEAT PUMPS The heat is transferred in the direction of decreasing temperature, and the reverse process cannot occur without a device (refrigerators and heat pumps). Refrigerators and heat pumps are (like heat engines), cyclic devices. The efficiency of refrigerators and heat pumps can be expressed in terms of the COP (Coefficient Of Performance): Desired output Q Q 1 L L COPR (COP of Refrigerator) Required output Wnet,in QH QL QH QL 1 COP Desired output Q Q 1 (COP of Heat Pump) H H HP Required output Wnet,in QH QL 1QL QH COP COP 1 HP R

PAGE 2 of 7 Second Law of Thermodynamics (2) A refrigerator that violates the Clausius statement of the second law Proof that the violation of the Kelvin-Planck statement leads to the violation of

3 PAGE 2 of 7 Second Law of Thermodynamics (2) A refrigerator that violates the Clausius statement of the second law Proof that the violation of the Kelvin-Planck statement leads to the violation of the Clausius statement THE SECOND LAW: CLAUSIUS STATEMENT (REFRIGERATORS OR HEAT PUMPS) A refrigerator (or heat pump) cannot operate unless its compressor is driven by an external power source, such as a motor. The Clausius statement of the second law of thermodynamics states: It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body to a higher-temperature body. The net effect on the surroundings involves the consumption of some energy in the form of work, in addition to the transfer of heat from a colder body to a warmer one. That is, it leaves a trace in the surroundings. EQUIVALENCE BETWEEN KELVIN-PLANCK AND CLAUSIUS STATEMENTS The Kelvin-Planck and Clausius statements are equivalent: any device that violates the Keivin- Planck statement also violates the Clausius statement, and vice versa. Consider the heat-engine-refrigerator combination, operating between the same two reservoirs. The heat engine is assumed to have, in violation of the Kelvin-Planck statement, a thermal efficiency of 100%, and therefore it converts all the heat QH it receives to work W. This work is supplied to a refrigerator that removes heat in the amount of QL from the lowtemperature reservoir and rejects heat in the amount of QL + QH to the high-temperature reservoir. During this process, the high-temperature reservoir receives a net amount of heat QL. Thus, the combination of these two devices can be viewed as a refrigerator. This combined refrigerator is clearly a violation of the Clausius statement. It can also be shown in a similar manner that a violation of the Clausius statement leads to the violation of the Kelvin-Planck statement. The Clausius and the Kelvin-Planck statements are two equivalent expressions of the second law of thermodynamics.

PAGE 3 of 7 Perpetual-Motion Machine (PMM) A perpetual-motion machine that violates the first law of thermodynamics (PMM1) A perpetual-motion machine that violates the second law of thermodynamics

A device that violates the first law of thermodynamics (by creating energy) is called a perpetualmotion machine of the first kind (PMM1).

It is proposed to heat the steam by resistance heaters placed inside the boiler, instead of by the external energy supplied.

The rest of the electrical energy is to be supplied to the electric network as the net work output.

4 PAGE 3 of 7 Perpetual-Motion Machine (PMM) A perpetual-motion machine that violates the first law of thermodynamics (PMM1) A perpetual-motion machine that violates the second law of thermodynamics (PMM2) PERPETUAL MOTION MACHINES (PMMS) A device that violates either first or second law of thermodynamics is called a perpetual-motion machine (PMM). A device that violates the first law of thermodynamics (by creating energy) is called a perpetualmotion machine of the first kind (PMM1). A device that violates the second law of thermodynamics is called a perpetual-motion machine of the second kind (PMM2). PMM1 EXAMPLE Consider a steam power plant. It is proposed to heat the steam by resistance heaters placed inside the boiler, instead of by the external energy supplied. Part of the electricity generated by the plant is to be used to power the resistors as well as the pump. The rest of the electrical energy is to be supplied to the electric network as the net work output. The inventor claims that once the system is started, this power plant will produce electricity indefinitely without requiring any energy input from outside. Violation of the first law: the system (enclosed by the shaded area) is continuously supplying energy to the outside ( Qout Wnet,out ) without receiving any energy. PMM2 EXAMPLE Consider a modified steam power plant. The inventor suggested getting rid of the condenser and sending the steam to the pump as soon as it leaves the turbine. This way, all the heat transferred to the steam in the boiler will be converted to work, and thus the power plant will have a theoretical efficiency of 100%. Violation of the second law: this system is on a cycle and generates a net amount of work while exchanging heat with a single reservoir (source) only (violates Kelvin-Planck statement).

PAGE 4 of 7 Reversible and Irreversible Processes A reversible process involves no internal and external irreversibilities 2 examples of reversible processes Irreversible compression and expansion

5 PAGE 4 of 7 Reversible and Irreversible Processes A reversible process involves no internal and external irreversibilities 2 examples of reversible processes Irreversible compression and expansion processes REVERSIBLE PROCESS A reversible process is defined as a process that can be reversed without leaving any trace on the surroundings. That is, both the system and the surroundings are returned to their initial states at the end of the reverse process. This is only possible if the net heat and net work exchange between the system and the surroundings is zero for the combined (original and reverse) process. Reversible processes actually do not occur in nature (merely an ideal model). Process that are not reversible are called irreversible processes. Reversible processes can be viewed as theoretical limits for the corresponding irreversible ones. The concept of reversible processes leads to the definition of the second law efficiency for actual processes. IRREVERSIBILITIES The factors that cause a process to be irreversible are called irreversibilities. Friction Unrestrained expansion Mixing of two fluids Heat transfer across a finite temperature difference Electric resistance Inelastic deformation of solids Chemical reactions A process is called internally reversible if no irreversibilities occur within the boundaries of the system during the process. A process is called externally reversible if no irreversibilities occur outside the system boundaries during the process. A process is called totally reversible ( reversible ) if it involves no irreversibilities within the system or its surroundings.

6 PAGE 5 of 7 EXERCISE D-2-1 (Do-It-Yourself) The food compartment of a refrigerator is maintained at 4 C by removing heat from it at a rate of 360 kj/min. If the required power input to the refrigerator is 2 kw, determine (a) the coefficient of performance of the refrigerator and (b) the rate of heat rejection (in kj/min ) to the room that houses the refrigerator. Solution The power consumption of a refrigerator is given. The COP and the rate of heat rejection are to be determined. Assumptions Steady operating conditions exist. Analysis (a) The coefficient of performance of the refrigerator is: 1 min 360 kj/min 60 sec Q COPR L 3 Wnet,in 2 kw This means that 3 kj of heat is removed from the refrigerated space for each kj of work supplied. (b) The rate at which heat is rejected to the room that houses the refrigerator is determined from the conservation of energy relation for cyclic devices: 60 sec Q Q W 360 kj/min 2 kw 480 kj/min 1 min H L net,in

7 PAGE 6 of 7 EXERCISE D-2-2 (Do-It-Yourself) A heat pump is used to meet the heating requirements of a house and maintain it at 20 C. On a day when the outdoor air temperature drops to 2 C, the house is estimated to lose heat at a rate of 80,000 kj/h. If the heat pump under these condition has a COP of 2.5, determine (a) the power consumed by the heat pump (in kj/h ) and (b) the rate at which heat is absorbed from the cold outdoor air (in kj/h ). Solution The COP of a heat pump is given. The power consumption and the rate of heat absorption are to be determined. Assumptions Steady operating conditions exist. Analysis (a) The power consumed by this heat pump, shown in the figure, is determined from the definition of the coefficient of performance: Q 80,000 kj/h Wnet,in H 32,000 kj/h COPHP 2.5 (b) The house is losing heat at a rate of 80,000 kj/h. If the house is to be maintained at a constant temperature of 20 C, the heat pump must deliver heat to the house at the same rate, that is, at a rate of 80,00 kj/h. Then the rate of heat transfer from the outdoor becomes: QL QH Wnet,in 80, 000 kj/h 32, 000 kj/h 48,000 kj/h

8 PAGE 7 of 7 Water enters an ice machine at 55 F and leaves as ice at 25 F. If the COP of the ice machine is 2.4 during this operation, determine the required power input (in HP ) for an ice production rate of 28 lbm/h. Note: 169 BTU of energy needs to be removed from each lbm of water at 55 F to turn it into ice at 25 F. Solution The COP and the refrigeration rate of an ice machine are given. The power consumption is to be determined. Assumptions The ice machine operates steadily. Unit Conversion 1 HP = 2,545 BTU/h Analysis The cooling load of this ice machine is: Q mq 28 lbm/h 169 BTU/lbm 4, 732 BTU/h L L Using the definition of the coefficient of performance, the power input to the ice machine system is determined to be: Q L 4,732 BTU/h 1 HP Wnet,in COPR 2.4 2,545 BTU/h HP

Engineering Thermodynamics. Chapter 5. The Second Law of Thermodynamics

Engineering Thermodynamics. Chapter 5. The Second Law of Thermodynamics 5.1 Introduction Chapter 5 The Second aw of Thermodynamics The second law of thermodynamics states that processes occur in a certain direction, not in just any direction. Physical processes in nature can