Chapter 5: The Second Law of Thermodynamics In the precedg two chapters, we have applied the first law of thermodynamics or the conservation of energy prciple, to processes volvg closed and open systems. As poted energy is a conserved property and it can be concluded that a process must satisfy the first law of to occur. However, satisfyg the first law alone does not ensure that the process will take place. For example, a cup of hot coffee left a cooler room eventually cools down. This process satisfies the first law i.e. this energy lost by the coffee is equal to the amount gaed by the surroundg air. Now let s reverse the situation such that the coffee gets even hotter a cooler room as a result of heat transfer from the room air. We know that such a situation will not occur but yet, dog so would not violate the first law. It is clear that processes proceed a certa direction and not the reverse direction. The first law place no restriction on the direction of a process, but satisfyg the first law does not ensure that the process will occur i.e. coffee gettg hotter a cooler room. The adequacy of the first law to identify whether a process can take place is remedied by troducg the second law of thermodynamics. Accordgly, the example where the coffee gets hotter a cool room reverse process, is a violation accordg to the second law. This violation is easily detected with the help of a property called entropy. Therefore, a process will not occur unless it satisfies both the first and second law of thermodynamics. Thermal Energy Reservoir A thermal energy reservoir is a body or system with a relatively large thermal energy capacity (mass specific heat) that can supply or absorb fite amount of heat with undergog any change temperature. A reservoir that supplies energy the form of heat is called a source and one that absorbs energy the form of heat is called the sk. Heat Enges Work can be converted to heat directly and completely, but convertg heat to work required the use of some special device, known as the heat enge. Heat enges differ considerably from one another but can be characterised by the followg. THE SECOND LAW OF THERMODYNAMICS
High temperature source = H Heat Enge W = L Low temperature source. They receive heat form a high temperature source. 2. They convert part of this heat to work (e.g. usually the form of a rotatg shaft). 3. They reject the remag waste heat to a low-temperature sk (e.g. atmosphere, rivers, etc).. They operate on a cycle. The work producg device that best fits to the defition of a heat enge is the steam power plant. A typical steam power plant is shown: THE SECOND LAW OF THERMODYNAMICS 2
2 Boiler Turbe 3 Pump Condenser The net work put of this power plant is given by: W W W Boiler: mh 2 h Turbe: W mh 3 h 2 Condenser: m h 3 h Pump: W mh h m P P P P net m H 2 0 Also, accordg to the first law of thermodynamics, for a cycle: Net work put = Net heat transfer W W W net net Thermal efficiency, η th of a heat enge like a steam power plant: η th = Desired put Required put η th = Net work put Heat th W net W W Example: Heat is transferred to a heat enge form a furnace at a rate of 80 MW. If the rate of waste heat rejection to a nearby rive is 50 MW, determe the net power put and the thermal efficiency for this heat enge. THE SECOND LAW OF THERMODYNAMICS 3
The Second Law of Thermodynamics: Kelv-Plank Statement We have seen earlier that even under ideal condition a heat enge must reject some heat to a low-temperature reservoir order to complete the cycle. That is, no heat enge can convert all the heat it receives to useful work. This limitation on the thermal efficiency of a heat enge forms the basis for the Kelv-Plank Statement of the second law, which is expressed as follow: It is impossible for any device that operates on a cycle to receive heat form a sgle reservoir and produce a net amount of work. That is, a heat enge must exchange heat with a low temperature sk as well as high temperature source to operate. Also, accordg to this statement, no heat enge can have a thermal efficiency of 00%. Example: A steam plant receives heat from a furnace at a rate of 280 GJ/h. Heat losses to the surroundg air from the steam as it passes through the pipes and other components are estimated to be ab 8 GJ/h. If the waste heat is transferred to the coolg water at a rate of 5GJ/h, determe (a) the net power put, (b) thermal efficiency of this power plant. Refrigerators and Heat Pumps We know that heat flows the direction of decreasg temperature, that is, from high temperature mediums to low temperature ones. This heat transfer process occurs nature with requirg any devices. The reverse process, however, cannot occur by itself. The transfer of heat from low temperature medium to high temperature medium requires special devises called refrigerators. The workg fluid used is the refrigeration cycle is called a refrigerant. The most common refrigeration cycle is the vapour-compression refrigeration cycle which volves four ma components: Condenser 3 Compressor Expansion valve 2 Evaporator THE SECOND LAW OF THERMODYNAMICS
Workg operation: Refrigerant enters the compressor as a vapour and is compressed to the condenser pressure. It leaves the compressor at a relatively high temperature and cools down and condenses as it flows through the coils of the condenser by rejectg heat to the surroundg. The refrigerant then enters an expansion valve where its pressure and temperature drop drastically due to throttlg effect. The low temperature refrigerant then enters the evaporator, where it evaporates by absorbg heat from the refrigerated space. The cycle is completed as the refrigerant leaves the evaporator and re-enters the compressor. Coefficient of Performance (COP) The efficiency of a refrigerator is expressed terms off COP R. COP R = Desired put Required put COP R = Heat removed at the Evaporator Work done by the compressor Applyg the first law to a cycle: COP R THE SECOND LAW OF THERMODYNAMICS 5
Heat Pumps Another device that transfers heat form a low temperature medium to a high temperature one is a heat pump. Refrigerators and heat pumps operate on a cycle but they differ their objectives. The objective of a refrigerator is to mata the refrigerated space at a low temperature by removg heat from it and dischargg this heat to a high temperature medium. The objective of a heat pump is to mata a heated space at a high temperature. This is done by absorbg heat form a low temperature source and supplyg this heat to the high temperature medium. The measure of performance of a heat pump is also expressed terms of COP HP defed as: COP HP = Desired put Required put COP HP = Heat absorbed by Condenser Work done by the compressor Applyg the first law to a cycle: COP HP Therefore: COP HP COP R Example: The food compartment of a refrigerator is mataed at C by removg heat form it at a rate of 360kJ/m. If the required power put to the refrigerator is 2 kw, determe: (a) the COP of the refrigerator, (b) the rate of heat rejection to the room that houses the refrigerator. Reversible and Irreversible Processes A Reversible process is defed as a process that can be reversed with leavg any trace on the surroundgs. That is, both the system and the surroundgs are returned to their origal state at the end of the reverse process. Processes that are not reversible are called irreversible processes. Reversible processes are merely idealizations of actual processes. All processes occurrg nature are irreversible. Reversible process can be viewed as theoretical limits for the correspondg irreversible ones. THE SECOND LAW OF THERMODYNAMICS 6
Factors that cause a process to be irreversible are called irreversibilities. They clude: Friction Unrestraed expansion Mixg of two gases Electrical resistance Chemical reaction A process is called ternally reversible if no irreversibilities occur with the boundaries of the system durg the process. A process is called externally reversible if no irreversibilites occur side the system boundaries. Totally reversible processes (of just reversible process) are both ternally reversible and externally reversible. The Carnot Cycle Reversible cycles cannot be achieved practise because the irrevrsibilites associated with each process cannot be elimated. Reversible cycles provide upper limits on the performance of real cycles. Reversible cycles also serve as the startg pots the development of actual cycles. The best known reversible cycle is the Carnot cycles. The theoretical heat enge tat operates on the Carnot cycle is called the Carnot Heat Enge. A Carnot cycle is composed of four reversible processes two isothermal and two adiabatic as shown: P 2 T H = const W net 3 T L = const v Process 2 3: Work done by the gas durg expansion process Process 3 : work done on the gas durg compression process THE SECOND LAW OF THERMODYNAMICS 7
The Carnot Prciple Accordg to the second law of thermodynamics, a heat enge cannot operate by exchangg heat with a sgle reservoir (Kelv Plank Statement), and a refrigerator cannot operate with a net work put from an external source. Similarly, accordg to the Carnot prciple, () the efficiency of all irreversible heat enges are is always less than the efficiency of a reversible one operatg between the same reservoirs and, (2) the efficiency of all reversible heat enge operatg between the same two reservoirs are the same. The Carnot Heat Enge The total efficiency of any heat enge, reversible or irreversible, is given by: th L H Where: H = (heat transferred to the heat enge from a high temperature source). L = (heat transferred from the heat enge to a low temperature sk). Thus, the thermal efficiency for a Carnot enge or any reversible heat enge operatg between T H and T L can be expressed as: th T T L H T H and T L must be Kelv!!! This is the highest efficiency a heat enge can have and all other irreversible heat enge operatg between T H and T L has lower efficiencies. Most work producg devices operation today have efficiencies under 0%. The thermal efficiency of actual heat enges can be maximised by: () Supplyg heat to the enge at the highest possible temperature (limited by material strength) (2) Rejectg heat from the enge at the lowest possible temperature (limited by the temperature of coolg medium). Example: A Carnot heat enge receives 500 kj of heat per cycle from a high temperature source at 652 C and rejects heat to a low temperature sk at 30 C. Determe the (a) thermal efficiency (b) amount of heat rejected to the sk (c) net work put from this enge. THE SECOND LAW OF THERMODYNAMICS 8