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- 1. BASIC THERMODYNAMICS J2006/3/1 UNIT 3BASIC THERMODYNAMICS OBJECTIVESGeneral Objective : To understand the laws of thermodynamics and its constants.Specific Objectives : At the end of the unit you will be able to: define the definitions of Boyle’s Law, Charles’ Law and Universal Gases Law define and show the application of the specific heat capacity at constant pressure define and apply the specific heat capacity at constant volume
- 2. BASIC THERMODYNAMICS J2006/3/2 INPUT3.0 Definition Of Perfect Gases Did you know, one important type of fluid that has many applications in thermodynamics is the type in which the working temperature of the fluid remains well above the critical temperature of the fluid? In this case, the fluid cannot be liquefied by an isothermal compression, i.e. if it is required to condense the fluid, then cooling of the fluid must first be carried out. In the simple treatment of such fluids, their behavior is likened to that a perfect gas. Although, strictly speaking, a perfect gas is an ideal which can never be realized in practice. The behavior of many ‘permanent’ gases, e.g. hydrogen, oxygen, air etc is very similar to the behavior of a perfect gas to a first approximation. A perfect gas is a collection of particles that: are in constant, random motion, have no intermolecular attractions (which leads to elastic collisions in which no energy is exchanged or lost), are considered to be volume-less points. You are more familiar with the term ‘ideal’ gas. There is actually a distinction between these two terms but for our purposes, you may consider them interchangeable. The principle properties used to define the state of a gaseous system are pressure (P), volume (V) and temperature (T). SI units (Systems International) for these properties are Pascal (Pa) for pressure, m3 for volume (although liters and cm3 are often substituted), and the absolute scale of temperature or Kelvin (K). Two of the laws describing the behavior of a perfect gas are Boyle’s Law and Charles’ Law.
- 3. BASIC THERMODYNAMICS J2006/3/33.1 Boyle’s Law The Boyle’s Law may be stated as follows: Provided the temperature T of a perfect gas remains constant, then volume, V of a given mass of gas is inversely proportional to the pressure P of the gas, i.e. P ∝ 1/V (as shown in Fig. 3.1-1), or P x V = constant if temperature remains constant. P P ∝ 1/V 1/V Figure 3.1-1 Graph P ∝ 1/V If a gas changes from state 1 to state 2 during an isothermal process, then P1 V1 = P2 V2 = constant (3.1) If the process is represented on a graph having axes of pressure P and volume V, the results will be as shown in Fig. 3.1-2. The curve is known as a rectangular hyperbola, having the mathematical equation xy = constant. P P1 1 P2 2 3 P3 V1 V2 V3 V PV = constant Figure 3.1-2 P-V graph for constant temperature
- 4. BASIC THERMODYNAMICS J2006/3/4 Example 3.1 A quantity of a certain perfect gas is heated at a constant temperature from an initial state of 0.22 m3 and 325 kN/m2 to a final state of 170 kN/m2. Calculate the final pressure of the gas. Solution to Example 3.1 From equation P1V1 = P2V2 325 kN/m 2 = ( 0.22 m 3 ) P1 ∴V2 = V1 x 170 kN/m 2 = 0.421 m 3 P2 3.2 Charles’ Law The Charles’s Law may be stated as follows: Provided the pressure P of a given mass of gas remains constant, then the volume V of the gas will be directly proportional to the absolute temperature T of the gas, i.e. V ∝ T, or V = constant x T. Therefore V/T = constant, for constant pressure P. If gas changes from state 1 to state 2 during a constant pressure process, then V1 V2 = = constant (3.2) T1 T2 If the process is represented on a P – V diagram as before, the result will be as shown in Fig. 3.2. P 1 2 0 V V1 V2 Figure 3.2 P-V graph for constant pressure process
- 5. BASIC THERMODYNAMICS J2006/3/5 Example 3.2 A quantity of gas at 0.54 m3 and 345 oC undergoes a constant pressure process that causes the volume of the gas to decreases to 0.32 m3. Calculate the temperature of the gas at the end of the process. Solution to Example 3.2 From the question V1 = 0.54 m3 T1 = 345 + 273 K = 618 K V2 = 0.32 m3 V1 V2 = T1 T2 V2 ∴ T2 = T1 x V1 0.32 m 3 = ( 618 K ) 0.54 m 3 = 366 K3.3 Universal Gases Law Charles’ Law gives us the change in volume of a gas with temperature when the pressure remains constant. Boyle’s Law gives us the change in volume of a gas with pressure if the temperature remains constant. The relation which gives the volume of a gas when both temperature and the pressure are changed is stated as equation 3.3 below. PV = constant = R (3.3) T i.e. P1V1 P2V2 (3.4) = T1 T2
- 6. BASIC THERMODYNAMICS J2006/3/6 No gases in practice obey this law rigidly, but many gases tend towards it. An PV imaginary ideal that obeys the law is called a perfect gas, and the equation =R T is called the characteristic equation of state of a perfect gas. The constant, R, is called the gas constant. The unit of R is Nm/kg K or J/kg K. Each perfect gas has a different gas constant. The characteristic equation is usually written PV = RT (3.5) or for m kg, occupying V m3, PV = mRT (3.6) Another form of the characteristic equation can be derived using the kilogram-mole as a unit. The kilogram-mole is defined as a quantity of a gas equivalent to m kg of the gas, where M is the molecular weight of the gas (e.g. since the molecular weight of oxygen is 32, then 1 kg mole of oxygen is equivalent to 32 kg of oxygen). From the definition of the kilogram-mole, for m kg of a gas we have, m = nM (3.7) (where n is the number of moles). Note: Since the standard of mass is the kg, kilogram-mole will be written simply as mole. Substituting for m from equation 3.7 in equation 3.6, PV PV = nMRT or MR = (3.8) nT
- 7. BASIC THERMODYNAMICS J2006/3/7 Now Avogadro’s hypothesis states that the volume of 1 mole of any gas is the same as the volume of 1 mole of any other gas, when the gases are at the same temperature and pressure. Therefore V/n is the same for all gases at the same value of P and T. That is the quantity PV/nT is constant for all gases. This constant is called the universal gas constant, and is given the symbol Ro. i.e. PV (3.9) MR = Ro = or PV = nRo T nT or since MR = Ro then, Ro R= (3.10) M Experiment has shown that the volume of 1 mole of any perfect gas at 1 bar and 1 oC is approximately 22.71 m3. Therefore from equation 3.8 PV 1 x 10 5 x 22.71 R0 = = = 8314.4 J/mole K nT 1 x 273.15 From equation 3.10 the gas constant for any gas can be found when the molecular weight is known, e.g. for oxygen of molecular weight 32, the gas constant is Ro 8314.4 R= = = 259.8 J/kg K M 32 Example 3.3 0.046 m3 of gas are contained in a sealed cylinder at a pressure of 300 kN/m2 and a temperature of 45 oC. The gas is compressed until the pressure reaches 1.27 MN/m2 and the temperature is 83oC. If the gas is assumed to be a perfect gas, determine: a) the mass of gas (kg) b) the final volume of gas (m3) Given: R = 0.29 kJ/kg K
- 8. BASIC THERMODYNAMICS J2006/3/8 Solution to Example 3.3 From the question V1 = 0.046 m3 P1 = 300 kN/m2 T1 = 45 + 273 K = 318 K P2 = 1.27 MN/m2 = 1.27 x 103 kN/m2 T2 = 83 + 273 K = 356 K R = 0.29 kJ/kg K From equation 3.6 PV = mRT P1V1 300 x 0.046 m= = = 0.1496 kg RT1 0.29 x 318 From equation 3.4, the constant volume process i.e. V1 = V2 P1 P2 = T1 T2 P 1.27 x 10 3 T2 = ( T1 ) 2 P = ( 318) 300 = 1346 K 1
- 9. BASIC THERMODYNAMICS J2006/3/9 Activity 3A TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…! 3.1 Study the statements in the table below. Mark the answers as TRUE or FALSE. STATEMENT TRUE or FALSE i. Charles’ Law gives us the change in volume of a gas with temperature when the temperature remains constant. ii. Boyle’s Law gives us the change in volume of a gas with pressure if the pressure remains constant. iii. The characteristic equation of state of a perfect gas is PV = R . T iv. Ro is the symbol for universal gas constant. v. The constant R is called the gas constant. vi. The unit of R is Nm/kg or J/kg. 3.2 0.04 kg of a certain perfect gas occupies a volume of 0.0072 m3 at a pressure 6.76 bar and a temperature of 127 oC. Calculate the molecular weight of the gas (M). When the gas is allowed to expand until the pressure is 2.12 bar the final volume is 0.065 m3. Calculate the final temperature.
- 10. BASIC THERMODYNAMICS J2006/3/10 Feedback To Activity 3A 3.1 i. False ii. False iii. True iv. True v. True vi. False 3.2 From the question, m = 0.04 kg V1 = 0.072 m3 V2 = 0.072 m3 P1 = 6.76 bar = 6.76 x 102 kN/m2 P2 = 2.12 bar = 2.12 x 102 kN/m2 T1 = 127 + 273 K = 400 K From equation 3.6 P1V1 = mRT1 P1V1 6.76 x 10 2 x 0.0072 ∴R = = = 0.3042 kJ/kg K mT1 0.04 x 400 Then from equation 3.10 R R= o M Ro 8.3144 ∴M = = = 27 kg/kmol R 0.3042 i.e. Molecular weight = 27
- 11. BASIC THERMODYNAMICS J2006/3/11 From equation 3.6 P2V2 = mRT2 P2V2 2.12 x 10 2 x 0.065 ∴ T2 = = = 1132.5 K mR 0.04 x 0.3042 i.e. Final temperature = 1132.5 – 273 = 859.5 oC. CONGRATULATIONS, IF YOUR ANSWERS ARE CORRECT YOU CAN PROCEED TO THE NEXT INPUT…..
- 12. BASIC THERMODYNAMICS J2006/3/12 INPUT3.4 Specific Heat Capacity at Constant Volume (Cv) The specific heat capacities of any substance is defined as the amount of heat energy required to raise the unit mass through one degree temperature raise. In thermodynamics, two specified conditions are used, those of constant volume and constant pressure. The two specific heat capacities do not have the same value and it is essential to distinguish them. If 1 kg of a gas is supplied with an amount of heat energy sufficient to raise the temperature of the gas by 1 degree whilst the volume of the gas remains constant, then the amount of heat energy supplied is known as the specific heat capacity at constant volume, and is denoted by Cv. The unit of Cv is J/kg K or kJ/kg K. For a reversible non-flow process at constant volume, we have dQ = mCvdT (3.11) For a perfect gas the values of Cv are constant for any one gas at all pressures and temperatures. Equations (3.11) can then be expanded as follows : Heat flow in a constant volume process, Q12 = mCv(T2 – T1) (3.12) Also, from the non-flow energy equation Q – W = (U2 – U1) mcv(T2 – T1) – 0 = (U2 – U1) ∴ (U2 – U1) = mCv(T2 – T1) (3.13) i.e. dU = Q Note: In a reversible constant volume process, no work energy transfer can take place since the piston will be unable to move i.e. W = 0.
- 13. BASIC THERMODYNAMICS J2006/3/13 The reversible constant volume process is shown on a P-V diagram in Fig. 3.4. P P2 2 P1 1 V V1 = V2 Figure 3.4 P-V diagram for reversible constant volume process Example 3.4 3.4 kg of gas is heated at a constant volume of 0.92 m 3 and temperature 17 oC until the temperature rose to 147 oC. If the gas is assumed to be a perfect gas, determine: c) the heat flow during the process d) the beginning pressure of gas e) the final pressure of gas Given Cv = 0.72 kJ/kg K R = 0.287 kJ/kg K Solution to Example 3.4 From the question m = 3.4 kg V1 = V2 = 0.92 m3 T1 = 17 + 273 K = 290 K T2 = 147 + 273 K = 420 K Cv = 0.72 kJ/kg K R = 0.287 kJ/kg K
- 14. BASIC THERMODYNAMICS J2006/3/14 a) From equation 3.13, Q12 = mCv(T2 – T1) = 3.4 x 0.72(420 – 290) = 318.24 kJ b) From equation 3.6, PV = mRT Hence for state 1, P1V1 = mRT1 mRT1 3.4 kg x 0.287 kJ/kgK x 290 K P1 = = 3 = 307.6 kN/m 2 V1 0.92 m c) For state 2, P2V2 = mRT2 mRT2 3.4 kg x 0.287 kJ/kgK x 420 K P2 = = 3 = 445.5 kN/m 2 V2 0.92 m3.5 Specific Heat Capacity at Constant Pressure (Cp) If 1 kg of a gas is supplied with an amount of heat energy sufficient to raise the temperature of the gas by 1 degree whilst the pressure of the gas remains constant, then the amount of heat energy supplied is known as the specific heat capacity at constant pressure, and is denoted by Cp. The unit of Cp is J/kg K or kJ/kg K. For a reversible non-flow process at constant pressure, we have dQ = mCpdT (3.14) For a perfect gas the values of Cp are constant for any one gas at all pressures and temperatures. Equation (3.14) can then be expanded as follows: Heat flow in a reversible constant pressure process Q = mCp(T2 – T1) (3.15)
- 15. BASIC THERMODYNAMICS J2006/3/153.6 Relationship Between The Specific Heats Let a perfect gas be heated at constant pressure from T1 to T2. With reference to the non-flow equation Q = U2 – U1 + W, and the equation for a perfect gas U2 – U1 = mCv(T2 – T1), hence, Q = mCv(T2 – T1) + W In a constant pressure process, the work done by the fluid is given by the pressure times the change in volume, i.e. W = P(V2 – V1). Then using equation PV = mRT, we have W = mR(T2 – T1) Therefore substituting, Q = mCv(T2 – T1) + mR(T2 – T1) = m(Cv + R)(T2 – T1) But for a constant pressure process from equation 3.15, Q = mCp(T2 – T1) Hence, by equating the two expressions for the heat flow Q, we have mCp(T2 – T1) = m(Cv + R)(T2 – T1) ∴Cp = Cv + R Alternatively, it is usually written as R = Cp - C v 3.163.7 Specific Heat Ratio (γ) The ratio of the specific heat at constant pressure to the specific heat at constant volume is given the symbol γ (gamma), Cp i.e. γ= (3.17) Cv Note that since Cp - Cv= R, from equation 3.16, it is clear that Cp must be greater than Cv for any perfect gas. It follows therefore that the ratio Cp/Cv = γ , is always greater than unity. In general, γ is about 1.4 for diatomic gases such as carbon monoxide (CO), hydrogen (H2), nitrogen (N2), and oxygen (O2). For monatomic gases such as
- 16. BASIC THERMODYNAMICS J2006/3/16 argon (A), and helium (He), γ is about 1.6, and for triatomic gases such as carbon dioxide (CO2), and sulphur dioxide (SO2), γ is about 1.3. For some hydro-carbons the value of γ is quite low (e.g. for ethane (C2H6), γ = 1.22, and for iso-butane (C4H10), γ = 1.11. Some useful relationships between Cp , Cv , R, and γ can be derived. From equation 3.17 Cp - Cv= R Dividing through by Cv Cp R −1 = Cv Cv Cp Therefore using equation 3.17, γ = , then, Cv R γ −1 = Cv R Cv = 3.18 (γ − 1) Also from equation 3.17, Cp = γCv hence substituting in equation 3.18, γR Cp = γ Cv = (γ − 1) γR Cp = 3.19 (γ − 1)
- 17. BASIC THERMODYNAMICS J2006/3/17 Example 3.5 A certain perfect gas has specific heat as follows Cp = 0.846 kJ/kg K and Cv = 0.657 kJ/kg K Find the gas constant and the molecular weight of the gas. Solution to Example 3.5 From equation 3.16 R = Cp - C v i.e. R = 0.846 – 0.657 = 0.189 kJ/kg K or R = 189 Nm/kg K From equation 3.10 R0 M= R 8314 i.e. M= = 44 189
- 18. BASIC THERMODYNAMICS J2006/3/18 Activity 3B TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…! 3.3 Two kilograms of a gas receive 200 kJ as heat at constant volume process. If the temperature of the gas increases by 100 oC, determine the Cv of the process. 3.4 A perfect gas is contained in a rigid vessel at 3 bar and 315 oC. The gas is then cooled until the pressure falls to 1.5 bar. Calculate the heat rejected per kg of gas. Given: M = 26 kg/kmol and γ = 1.26. 3.5 A mass of 0.18 kg gas is at a temperature of 15 oC and pressure 130 kN/m2. If the gas has a value of Cv = 720 J/kg K, calculate the: i. gas constant ii. molecular weight iii. specific heat at constant pressure iv. specific heat ratio
- 19. BASIC THERMODYNAMICS J2006/3/19 Feedback To Activity 3B 3.3 From the question, m = 2 kg Q = 200 kJ (T2 – T1) = 100 oC = 373 K Q = mCv(T2 – T1) Q 200 Cv = = = 0.268 kJ/kgK m( T2 − T1 ) 2(373) 3.4 From the question, P1 = 3 bar T1 = 315 oC = 588 K P2 = 1.5 bar M = 26 kg/kmol γ = 1.26 From equation 3.10, R 8314 R= o = = 319.8 J/kg K M 26 From equation 3.18, R 319.8 Cv = = = 1230 J/kg K = 1.230 kJ/kg K (γ − 1) 1.26 − 1 During the process, the volume remains constant (i.e. rigid vessel) for the mass of gas present, and from equation 3.4,
- 20. BASIC THERMODYNAMICS J2006/3/20 P1V1 P2V2 = T1 T2 Therefore since V1 = V2, P 1.5 T2 = T1 2 = 588 x = 294 K P1 3 Then from equation 3.12, Heat rejected per kg gas, Q = Cv(T2 – T1) = 1.230(588 – 294) = 361.6 kJ/kg 3.5 From the question m = 0.18 kg T = 15 oC = 288 K V = 0.17 m3 Cv = 720 J/kg K = 0.720 kJ/kg K i. From equation 3.6, PV = mRT PV 130 x 0.17 R= = = 0.426 kJ/kgK mT 0.18 x 288 ii. From equation 3.10, R R= o M R 8.3144 M = o = = 19.52 kg/kmol R 0.426 iii. From equation 3.16, R = Cp - C v Cp = R + Cv = 0.426 + 0.720 = 1.146 kJ/kg K iv. From equation 3.17, C p 1.146 γ = = = 1.59 C v 0.720
- 21. BASIC THERMODYNAMICS J2006/3/21 SELF-ASSESSMENT You are approaching success. Try all the questions in this self-assessment section and check your answers with those given in the Feedback to Self- Assessment on the next page. If you face any problem, discuss it with your lecturer. Good luck. 1. 1 m3 of air at 8 bar and 120 oC is cooled at constant pressure process until the temperature drops to 27 oC. Given R = 0.287 kJ/kg K and Cp = 1.005 kJ/kg K, calculate the: i. mass of air ii. heat rejected in the process iii. volume of the air after cooling. 2. A system undergoes a process in which 42 kJ of heat is rejected. If the pressure is kept constant at 125 kN/m2 while the volume changes from 0.20 m3 to 0.006 m3, determine the work done and the change in internal energy. 3. Heat is supplied to a gas in a rigid container.The mass of the container is 1 kg and the volume of gas is 0.6 m3. 100 kJ is added as heat. If gas has Cv = 0.7186 kJ/kg K during a process, determine the: i. change in temperature ii. change in internal energy
- 22. BASIC THERMODYNAMICS J2006/3/22 Feedback To Self-Assessment Have you tried the questions????? If “YES”, check your answers now. 1. i. m = 7.093 kg ii. Q = 663 kJ iii. V2 = 0.763 m3 2. W = -24.25 kJ (U2 – U1) = -17.75 kJ 3. i. (T2 – T1) = 139.2 K ii. (U2 – U1) = 100 kJ CONGRATULATIONS!!!! …..May success be with you always….

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