Chapter One Reviews of Thermodynamics Update on 2013/9/13

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1 Chapter One Reviews of Thermodynamics Update on 2013/9/13 (1.1). Thermodynamic system An isolated system is a system that exchanges neither mass nor energy with its environment. An insulated rigid tank filled with air is an example of an isolated system. Something violent may happen inside the system. For example, mixing of gases, internal heat transfer, or even reaction may occur if the system contains fuel and oxygen with proper ratios. A closed system is a system that exchanges energy with its environment. However, it does not exchange mass with its environment. In other words, mass is conserved in the system. A cylinder and piston system is an example of a closed system. Q W An open system is a system that exchanges both energy and mass with its environment. An air compressor is an example of an open system. m i m e Advanced Thermodynamics, ME Dept, NCHU 頁 1

2 A simple system is one in which there is only one way to alter the energy by work. Examples are simple compressible systems, simple elastic systems or simple electric systems. Single phase systems are formed by a single physical structure (solid, liquid or gas) and a single chemical composition. Examples are ice, liquid water or vapor, a water/alcohol mixture, etc. Multiphase systems involve two or more phases either in physical structure or chemical composition. Some examples are water/vapor and water/oil mixtures. Pure substance systems are formed by a single chemical composition. Examples are vapor, oxygen, nitrogen, etc. Mixture systems involve two or more chemical compositions. Some examples are air, air fuel mixture, etc. A system may be composed of either pure substance or mixture. The components of the system may constitute either single phase or multi phases. Single phase Multiphase Pure substance vapor water & vapor Mixture air Air & fuel droplets Advanced Thermodynamics, ME Dept, NCHU 頁 2

3 (1.2). Thermodynamic property Properties are some numerical values used to describe the macroscopic characteristics of the system. Not all properties are independent properties. Some properties can be derived from other properties. Pressure(P), temperature(t), and specific volume(v) are examples of thermodynamic properties. Units of pressure: Pascal, bar, kg/cm 2, mmh 2 O, mmhg, Torr, psi, atm 1 Pascal = 1 N/m 2, 1 kpa = 1000 Pa, 1 MPa = 1000 kpa 1 bar = 100 kpa = 1000 mbar, 1 psi = 1 lbf/in 2, 1 kg/cm 2 = 98 kpa 1 Torr = 1mmHg 1 atm = Pa = psi = 760 mmhg = 760 Torr The properties can be classified in a) directly measurable (pressure, temperature), b) defined by laws of thermodynamics (entropy) and c) defined as combination of other properties (Gibbs function). An extensive property is one that is dependent on the size of a system, e.g. mass and volume. An intensive property is independent of the size of the system. Examples are heat capacity, temperature and pressure. Specific properties are obtained by dividing extensive properties with mass. V H U S v, h, u, s m m m m (1.3). Thermodynamic state A state is made up of several independent properties which can fully define the status of the system. In a simple compressible system, a thermodynamic state can be determined explicitly by two independent properties. The state space is a multi dimensional space constructed by the independent properties of a system. The dimensionality of state space is equal to the number of independent properties of that system. A thermodynamic state is a point in the state space. The state space of a simple compressible system is a two-dimensional space in which a state is represented as a point. Advanced Thermodynamics, ME Dept, NCHU 頁 3

4 P-v diagram or the T-s diagram is often used to represent the thermodynamic state in a simple compressible system. P T V S (1.4). Thermodynamic process When a thermodynamic system changes its state as a result of exchange of energy with other system or the environment, the location of the state in the state space moves, and the locus of all the states that the system goes through make up a process. A process is a curve in the state space. The initial and final conditions of a system are the end states. Infinite number of processes can connected between the same end states. The series of states between end states define the path of the process. The change in the properties when a system undergoes a process between two end states is independent of the path. Thus, the value of a property is independent of the process and depends only on the state of the system. Quantities that depend on the nature of the process are not properties. A process in PV diagram Advanced Thermodynamics, ME Dept, NCHU 頁 4

5 (1.4.1) Reversible process If a system undergoes a process from the initial state to the final state, and then the system moves back to the initial state in a reversed direction along the same path. If the process does not cause any net effect on either the system or its environment, then this process is said to be a reversible process. A reversible process is an ideal process. It would not happen in real world. Example, A piston and cylinder system undergoes a compression process, and then expands to its original position. Both processes are adiabatic. Discuss that if the processes are reversible or not. P V Example, A piston and cylinder system undergoes a compression process, and then expands to its original position. Both processes are isothermal. Discuss that if the processes are reversible or not. Thermal Reservoir T V Advanced Thermodynamics, ME Dept, NCHU 頁 5

6 (1.4.2). Irreversible process An irreversible process is that when the system goes back to the initial state with the reversed process, some definite changes happen in either the system or the environment after the process is done. Causes of irreversibility: friction, combustion, diffusion, mixing, chemical reaction, free expansion, heat transfer with finite temperature difference. Results of irreversible process: increases of entropy and loss of availability. Example, A weight is pulled up along a slope for a distance of x. The it falls down to its original position. Discuss that if the process is reversible or not. Example, A piston and cylinder system undergoes a constant pressure process by heating the air inside the system to push a weight upwards a height of h, and then lowers down the weight to its original position by cooling the air. Discuss that if the process is reversible or not. h P V Advanced Thermodynamics, ME Dept, NCHU 頁 6

processes, such that the start of the next

and the end of the last process coincides the

A cycle can be conducted continuously to either

Some well known cycles include Carnot cycle, Otto

7 (1.5). Thermodynamic cycle A system undergoes several processes, such that the start of the next process coincide the end of the previous process, and the end of the last process coincides the start of the first process. All the processes make up a cycle. A cycle is a closed loop in the state space. A cycle can be conducted continuously to either convert heat into work or reversibly. Some well known cycles include Carnot cycle, Otto cycle, Diesel cycle, Rankine cycle, Brayton cycle, Stirling cycle, etc. Otto cycle Diesel cycle Brayton cycle Advanced Thermodynamics, ME Dept, NCHU 頁 7

8 (1.6). Work and Heat Work output/input and heat transfer are the two principal ways of energy exchange between a thermodynamic system and its environment. Work transfer is conducted through ordered motion of atoms or molecules. The motion of a piston driven by expansion gas is an example. Both the piston and the gas are in a bulk motion in which all the atoms composing the piston and the molecules contained in the gas move in a homogeneous way. In contrast, heat transfer is carried out by means of random motion of atoms and molecules. The energy contained in ordered motion can be totally converted to random motion. However, the energy contained in random motion can not be totally converted to ordered motion. The inequality of the conversion between heat and work is the basis of the second law of thermodynamics. (1.6.1). Work of reversible process Wrev Fdx YdX, Wrev YdX Y: generalized force, X: generalized displacement However, in an irreversible process, the work conducted is less than that of a reversible process. W YdX W irr rev (1.6.2). Work in a Simple Compressible Closed System Y P, X V, e.g., piston and cylinder system. Reversible work: W rev Irreversible work: Wirr < In general:w PdV PdV PdV Advanced Thermodynamics, ME Dept, NCHU 頁 8

9 The integration of pressure over volume is the projected area of the cure in the PV diagram. Thus, the reversible work of a process is equal to the projected area of this process represented in the PV diagram. The work of a process depends on the path of the process. Work is a path function, not a point function. As a result, work is not a thermodynamic property. For a closed system consisting of ideal gas, the work and the heat transfer during a specific process are given as the following (Fig ). Fig The work conducted by the piston in a steam engine Reversible work of an isothermal process: mrt V2 Wrev PdV dv mrtln V V1 Heat transfer in an isothermal process: 2 QU W mrtln V V Reversible work of an isobaric process W PdV P dv P( V V ) rev Heat transfer in an isobaric process: cp QU W mcv ( T2 T1) P( V2 V1) P( V2 V1) R Advanced Thermodynamics, ME Dept, NCHU 頁 9

10 Reversible work of a polytropic Process n PV const. 1 Wrev PdV ( PV 1 1PV 2 2 ) n 1 Heat transfer in a polytropic process: n C Pv C C 1n C Pv C, P, T v, dt (1 n) dv v n n1 n R Rv R Rv cv C C 1 n C k n C qdu Pdv (1 n) dv dv( 1) dv dv n n n n R v v k 1 v k 1 v If both k and n are constants, the amount of heat transfer of the process is k n C k n C 1n 1n k n 1 q dv ( v2 v1 ) ( Pv 2 2 Pv 1 1) n k 1 v k 11n k 11n k n 1 k n Pv ( 1 1 v ) [1 ( ) n q Pv Pv ] k 1 n1 k 1 n1 v2 k n Q ( PV 1 1PV 2 2) ( k 1)( n1) Table 1.1: The relationships among k, n, and the amount of heat transfer n& k dv q n k dv 0 q 0 heat rejection n k dv 0 q 0 heat absorption n k dv 0 q 0 adiabatic process n k dv 0 q 0 adiabatic process n k dv 0 q 0 heat absorption n k dv 0 q 0 heat rejection Example:Calculate the reversible work needed to compress 1 kg of air at 100 kpa and 25 to a volume of 0.5 m 3. (1). Isothermal compression. ( kj) (2). Isobaric compression. ( kj, ) (3). Polytropic process with n=1.5. ( kj) (4). Polytropic process with n=1.3. ( kj) Example:Two kilograms of saturated water vapor at 150 is contained in a piston and cylinder assembly. The system is compressed to 10 bars with a polytropic process in which PV 1.5 =constant. Find the amount of heat transfer during this process. Advanced Thermodynamics, ME Dept, NCHU 頁 10

Assignment 1.1 : Calculate the reversible work needed to compress 1 kg of helium at 100 kpa and 25 to a volume of 0.5 m 3. (1). Isothermal compression. (2). Isobaric compression. (3).

Isothermal compression. (2). Isobaric compression. (3). Polytropic process with n=1.5. (4). Polytropic process with n=1.3. (1.6.3). Work in an open system There are two modes of work that can be conducted in an open system.

11 Assignment 1.1 : Calculate the reversible work needed to compress 1 kg of helium at 100 kpa and 25 to a volume of 0.5 m 3. (1). Isothermal compression. (2). Isobaric compression. (3). Polytropic process with n=1.5. (4). Polytropic process with n=1.3. Assignment 1.2 : Calculate the reversible work needed to expand 1 kg of air at 300 kpa and 250 to a volume of 1.5 m 3. (1). Isothermal compression. (2). Isobaric compression. (3). Polytropic process with n=1.5. (4). Polytropic process with n=1.3. (1.6.3). Work in an open system There are two modes of work that can be conducted in an open system. One is conducted by changing the volume of the system, the same way as that happens in a closed system. An example of this mode of work is the suction of air in a piston type compressor while the intake valve is open and the piston is moving down. Two modes of work: volume variation and shaft work In a steady state steady flow system, the volume of the system remains the same. Work can be conducted by rotating a shaft inside the system. The effect of the shaft might be acting force on the working fluid, or varying volume inside the system. This mode of work is called shaft work. Example of shaft work are pump, rotary compressor, turbine, and expander. Advanced Thermodynamics, ME Dept, NCHU 頁 11

12 The shaft work can be calculated as the following. e W e rev m vdp, wrev vdp i i For ideal gas, the shaft work for some reversible processes are as the following. Reversible work of an isothermal process: RT P w ln i rev vdp dprt P Pe P W ln i rev mrt P e Reversible work of an isobaric process w vdp0 rev Reversible work of a polytropic Process n Pv const. n wrev vdp ( Pv 1 1Pv 2 2) n 1 n n W rev m ( Pv 1 1 Pv 2 2) m ( RT1 RT2) n 1 n1 Example:Calculate the reversible work needed to compress air at 100 kpa and 25 to the pressure of 1 MPa. The flow rate of air is 0.1 m 3 /sec. (1). Isothermal compression. (2). Polytropic process with n=1.5. (3). Polytropic process with n=1.3. (4). Adiabatic process. Assignment 1.3 : Calculate the reversible work needed to compress hydrogen at 100 kpa and 25 to the pressure of 10 MPa. The flow rate is 0.5 m 3 /min. (1). Isothermal compression. (2). Adiabatic compression in single stage. (3). Adiabatic compression in two stages with a perfect intercooler in between. Advanced Thermodynamics, ME Dept, NCHU 頁 12

13 (1.6.4). Simple Elastic System The work is carried out by deformation of the system. compression and bending of a beam. Examples are spring Y F, X L Wrev FdL F L, A L Wrev FdL ALd ALd Vd d Young s modulus: Y ( ) T, d Y W d wrev d V Y 2 wrev d Y 2Y Example:The tensile force in a steel wire 1 m long and 1 mm diameter is increased reversibly at constant temperature of 30 from zero to 100 N. Calculate the work done. Assume that Y = 2 x 10 8 kn/m 2. Assignment 1.4 : For a certain rubber band, the restoring force is related to its initial length L 0 and the displacement x as follows: 2 3 x F x N L0 L 0 The initial length of the rubber band is 5 cm. Determine the work required to stretch the band to a final length of 10 cm. (1.6.5). Simple Surface Tension System The work is carried out by varying the surface area of the system. Examples are formation of liquid droplets and bubble generation. Y, surface tension (N/m) X A, area (m 2 ) Wrev da F 2L Advanced Thermodynamics, ME Dept, NCHU 頁 13

14 F Wrev da da 2L Liquid Temperature C Surface tension, γ ( mn/m) Water Water Water Water Ethanol Mercury Formation of liquid droplets from water pool: 4 3 Mass of droplets: m r n 3 3m Number of droplets: n 3 4 r 2 2 3m 3m 3V Total work: Wrev A4r n4r 3 4r r r W 3 Specific work: w (in volume) V r Specific work: W 3 V 3 w (in mass) m mr r Example:An atomizer shoots out water droplets of an average radius of 10-4 cm into the air. Estimate the work required when 1 kg of water is atomized isothermally at 25. The surface tension of water in contact with air at 25 is 7.2x10-4 N/cm. ANS: x 10 2 J/kg. m W P KE PE+ U work = kinetic energy + potential energy + thermal energy 2 V 3m KE m, PE A 2 r m = 1 kg,δp = 250 kpa,ρ = 1000 kg/m 3 Advanced Thermodynamics, ME Dept, NCHU 頁 14

Bubble formation: 2 2 Pi R PoR 22R 4 Pi Po R Work of surface tension: 4 4 2 dws PdV r dr 16 rdr r 2 2 Ws 16 rdr 8 ( R2 R1 ) 2 ( A2 A1) Work to expel outside fluid: 2 dw 4 o P0dV P0 r dr 2 4 3 Wo P0 4

How much work should be done to boost a bubble of this size? Assignment 1.5 : A spherical air bubble with a radius of 2.5 10-6 m is suspended at the bottom of a pool water of 10 meters deep.

15 Bubble formation: 2 2 Pi R PoR 22R 4 Pi Po R Work of surface tension: dws PdV r dr 16 rdr r 2 2 Ws 16 rdr 8 ( R2 R1 ) 2 ( A2 A1) Work to expel outside fluid: 2 dw 4 o P0dV P0 r dr Wo P0 4 r dr P0 R P0( V2 V1) 3 Total work for the bubble growth: W W W 2 ( A A) P( V V ) s o Example:Find the pressure inside a soap bubble of 1 cm in diameter. How much work should be done to boost a bubble of this size? Assignment 1.5 : A spherical air bubble with a radius of m is suspended at the bottom of a pool water of 10 meters deep. The surface tension of water in contact with air at 25 is 7.2x10-4 N/cm. Calculate the pressure of air inside the bubble. Assignment 1.6: An atomizer shoots out water droplets into the air. The distribution of droplet size is shown below. Estimate the work required when 1 kg of water is atomized isothermally at 25. The surface tension of water in contact with air at 25 is 7.2x10-4 N/cm. Advanced Thermodynamics, ME Dept, NCHU 頁 15

16 Advanced Thermodynamics, ME Dept, NCHU 頁 16

Chapter 5: The First Law of Thermodynamics: Closed Systems

Chapter 5: The First Law of Thermodynamics: Closed Systems The first law of thermodynamics can be simply stated as follows: during an interaction between a system and its surroundings, the amount of energy