Thermodynamics - First Stage - PDF

Thermodynamics First stage 2) The change in internal energy for an isothermal process is: ∆𝒖 = 𝟎 Ans. 3) For an isothermal process: 𝒒 = 𝒘 = −𝟐𝟖𝟗. 𝟕 𝐤𝐉/𝐤𝐠 Ans. 4. Adiabatic process: an adiabatic process is one...

Thermodynamics First stage 2) The change in internal energy for an isothermal process is: ∆𝒖 = 𝟎 Ans. 3) For an isothermal process: 𝒒 = 𝒘 = −𝟐𝟖𝟗. 𝟕 𝐤𝐉/𝐤𝐠 Ans. 4. Adiabatic process: an adiabatic process is one in which the system undergoes no heat transfer with the surroundings, but the boundary of the system moves giving displacement work. The arrangement for the adiabatic process is shown in figure below. It consists of a piston cylinder arrangement where the cylinder is insulated from all sides to prevent heat transfer. Since 𝑑𝑄 = 0, therefore 𝑑𝑊 is only due to 𝑑𝑈. The (P-V) diagram for an adiabatic process is shown in figure below. Applying the first law of thermodynamics: 𝛿𝑄 − 𝛿𝑊 = 𝑑𝑈 For an adiabatic process 𝛿𝑄 = 0 Also 𝛿𝑊 = 𝑃𝑑𝑉 and 𝑑𝑈 = 𝑚𝐶𝑣 𝑑𝑇 So 0 − 𝑃𝑑𝑉 = 𝑚𝐶𝑣 𝑑𝑇 𝑃𝑑𝑉 + 𝑚𝐶𝑣 𝑑𝑇 = 0 … … … … … … … (5.40) From the equation of state: 𝑃𝑉 = 𝑚𝑅𝑇 Differentiating both sides, we get: 𝑃𝑑𝑉 + 𝑉𝑑𝑃 = 𝑚𝑅𝑑𝑇 𝑃𝑑𝑉 + 𝑉𝑑𝑃 𝑑𝑇 = … … … … … … … (5.41) 𝑚𝑅 Substituting (5.41) in (5.40), we get: 25 Thermodynamics First stage 𝑚𝐶𝑣 (𝑃𝑑𝑉 + 𝑉𝑑𝑃) 𝑃𝑑𝑉 + =0 … … … … … … … (5.42) 𝑚𝑅 Multiplying both sides by 𝑅, we get: 𝑅𝑃𝑑𝑉 + 𝐶𝑣 (𝑃𝑑𝑉 + 𝑉𝑑𝑃) = 0 … … … … … … … (5.43) Since 𝑅 = 𝐶𝑝 − 𝐶𝑣 , then: (𝐶𝑝 − 𝐶𝑣 )𝑃𝑑𝑉 + 𝐶𝑣 (𝑃𝑑𝑉 + 𝑉𝑑𝑃) = 0 … … … … … … … (5.44) 𝐶𝑝 𝑃𝑑𝑉 − 𝐶𝑣 𝑃𝑑𝑉 + 𝐶𝑣 𝑃𝑑𝑉 + 𝐶𝑣 𝑉𝑑𝑃 = 0 𝐶𝑝 𝑃𝑑𝑉 + 𝐶𝑣 𝑉𝑑𝑃 = 0 … … … … … … … (5.45) Dividing by 𝐶𝑣 𝑃𝑉, we get: 𝐶𝑝 𝑑𝑉 𝑑𝑃 ( ). + =0 … … … … … … … (5.46) 𝐶𝑣 𝑉 𝑃 𝐶𝑝 Since 𝛾 = , then: 𝐶𝑣 𝑑𝑃 𝑑𝑉 +𝛾 =0 … … … … … … … (5.47) 𝑃 𝑉 Integrating, we get: ln 𝑃 + 𝛾 ln 𝑉 = 𝐶 … … … … … … … (5.48) ln(𝑃𝑉 𝛾 ) = 𝐶 𝑃𝑉 𝛾 = 𝐶 … … … … … … … (5.49) For a unit mass: 𝑃𝑣 𝛾 = 𝐶 … … … … … … … (5.50) Now 𝛾 𝛾 𝑃1 𝑉1 = 𝑃2 𝑉2 𝑃2 𝑉1 𝛾 =( ) … … … … … … … (5.51) 𝑃1 𝑉2 For a perfect gas: 𝑃1 𝑉1 𝑃2 𝑉2 = 𝑇1 𝑇2 26 Thermodynamics First stage 𝑇2 𝑃2 𝑉2 𝑉1 𝛾 𝑉2 𝑉1 𝛾−1 =. =( ). =( ) 𝑇1 𝑃1 𝑉1 𝑉2 𝑉1 𝑉2 So 𝑇2 𝑉1 𝛾−1 =( ) … … … … … … … (5.52) 𝑇1 𝑉2 Also 𝛾−1 𝑇2 𝑃2 𝛾 =( ) … … … … … … … (5.53) 𝑇1 𝑃1 The work done is derived as follows: 2 𝑊 = ∫ 𝑃𝑑𝑉 1 𝐶 Since 𝑃𝑉 𝛾 = 𝐶 → 𝑃 = 𝑉 𝛾 , then: 2 2 𝑑𝑉 𝑊 =𝐶∫ = 𝐶 ∫ 𝑉 −𝛾 𝑑𝑉 1 𝑉𝛾 1 2 𝑉 −𝛾+1 𝐶 1−𝛾 1−𝛾 =𝐶| | =[ ] [𝑉2 − 𝑉1 ] −𝛾 + 1 1 1−𝛾 𝛾 𝛾 Since 𝐶 = 𝑃1 𝑉1 = 𝑃2 𝑉2 , then: 𝑃2 𝑉2 − 𝑃1 𝑉1 𝑃1 𝑉1 − 𝑃2 𝑉2 𝑊= = … … … … … … … (5.54) 1−𝛾 𝛾−1 Since 𝑃𝑉 = 𝑚𝑅𝑇, then: 𝑚𝑅(𝑇1 − 𝑇2 ) 𝑊= … … … … … … … (5.55) 𝛾−1 For a unit mass: 𝑃1 𝑣1 − 𝑃2 𝑣2 𝑤= … … … … … … … (5.56) 𝛾−1 or 𝑅(𝑇1 − 𝑇2 ) 𝑤= … … … … … … … (5.57) 𝛾−1 27 Thermodynamics First stage Example (5.6): Air at 1.02 bar and 22°C, initially occupying a cylinder volume of 0.015 m3, is compressed reversibly and adiabatically by a piston to a pressure of 6.8 bar. Calculate: 1) The final temperature. 2) The final volume. 3) The work done. 4) The heat transferred to or from the cylinder walls. Solution: The absolute temperature is: 𝑇1 = 22 + 273 = 295 K To find the mass of the air: 𝑃1 𝑉1 1.02 × 105 × 10−3 × 0.015 𝑃1 𝑉1 = 𝑚𝑅𝑇1 → 𝑚 = = = 0.018 kg 𝑅𝑇1 0.287 × 295 1) The final temperature can be calculated as: 𝛾−1 1.4−1 𝑇2 𝑃2 𝛾 6.8 1.4 =( ) → 𝑇2 = 295 × ( ) 𝑇1 𝑃1 1.02 𝑻𝟐 = 𝟓𝟎𝟕. 𝟐𝟓 𝐊 Ans. 2) The final volume can be calculated as: 1 1 𝑃2 𝑉1 𝛾 𝑃1 𝛾 1.02 1.4 = ( ) → 𝑉2 = 𝑉1 × ( ) → 𝑉2 = 0.015 × ( ) 𝑃1 𝑉2 𝑃2 6.8 𝑽𝟐 = 𝟎. 𝟎𝟎𝟑𝟖𝟕 𝐦𝟑 Ans. 3) The work done is: 𝑚𝑅(𝑇1 − 𝑇2 ) 0.018 × 0.287 × (295 − 507.25) 𝑊= = 𝛾−1 1.4 − 1 𝑾 = −𝟐. 𝟕𝟒𝟏 𝐤𝐉 Ans. 4) The heat transferred for an adiabatic process: 𝑸=𝟎 Ans. 5. Polytropic process: During actual expansion and compression processes of gases, pressure and volume are often related by 𝑃𝑉 𝑛 = 𝐶, where n and C are constants. A process of this kind is called a polytropic process. The (P-V) diagram for such a process is shown below. As mentioned, the general equation for polytropic processes is expressed as: 𝑃𝑉 𝑛 = 𝐶 … … … … … … … (5.58) 28 Thermodynamics First stage From the above equation, we can derive the following equations in the same method as in adiabatic processes: 𝑃2 𝑉1 𝑛 =( ) … … … … … … … (5.59) 𝑃1 𝑉2 𝑇2 𝑉1 𝑛−1 =( ) … … … … … … … (5.60) 𝑇1 𝑉2 𝑛−1 𝑇2 𝑃2 𝑛 =( ) … … … … … … … (5.61) 𝑇1 𝑃1 From equation (5.58), the work done is derived in the same method earlier and expressed as: 𝑃1 𝑉1 − 𝑃2 𝑉2 𝑚𝑅(𝑇1 − 𝑇2 ) 𝑊= = … … … … … … … (5.62) 𝑛−1 𝑛−1 For a unit mass: 𝑃1 𝑣1 − 𝑃2 𝑣2 𝑅(𝑇1 − 𝑇2 ) 𝑤= = … … … … … … … (5.63) 𝑛−1 𝑛−1 The heat transfer for polytropic processes does not equal zero and can be calculated from the following equation: 𝛾−𝑛 𝑄=( )𝑊 … … … … … … … (5.64) 𝛾−1 For a unit mass: 𝛾−𝑛 𝑞=( )𝑤 … … … … … … … (5.65) 𝛾−1 Example (5.7): 1 kg of air at 1.02 bar and 17°C is compressed reversibly according to a law 𝑷𝑽𝟏.𝟑 = 𝑪, to a pressure of 5.5 bar. Calculate the work done on the air and the heat flow to or from the cylinder walls during the compression. Solution: The absolute temperature is: 𝑇1 = 17 + 273 = 290 K First we find the final temperature: 𝑛−1 1.3−1 𝑇2 𝑃2 𝑛 5.5 1.3 =( ) → 𝑇2 = 290 × ( ) = 427.83 K 𝑇1 𝑃1 1.02 The work done is calculated as: 29 Thermodynamics First stage 𝑚𝑅(𝑇1 − 𝑇2 ) 1 × 0.287 × (290 − 427.83) 𝑊= = 𝑛−1 1.3 − 1 𝑾 = −𝟏𝟑𝟏. 𝟖𝟔 𝐤𝐉 Ans. The heat transfer is: 𝛾−𝑛 1.4 − 1.3 𝑄=( )𝑊 = ( ) × −131.86 𝛾−1 1.4 − 1 𝑸 = −𝟑𝟐. 𝟗𝟕 𝐤𝐉 Ans. Significance of the First Law The first law of thermodynamics leads directly to the non-flow energy equation and embodies four important concepts, as follows: 1. Heat and work are mutually convertible one into the other as they are both modes of energy transfer. 2. The existence of a type of energy (internal energy) that depends on the thermodynamic state of a system. 3. The possibility of measuring a difference in internal energy between thermodynamic states by making measurements of heat transfer and work. 4. The fact that energy is conserved whenever the thermodynamic state of a closed system changes. Summary Non-flow process is the one in which there is no mass interaction across the system boundaries during the occurrence of the process such as: heating and cooling of a fluid inside a closed container, compression and expansion of a fluid in a piston-cylinder arrangement, etc. For non-flow processes the first law can be written as: 𝑄 − 𝑊 = ∆𝑈 + ∆𝐾𝐸 + ∆𝑃𝐸 For non- flow processes the kinetic and potential energies are very small and can be neglected, so the energy equation becomes: 𝑄 − 𝑊 = 𝑈2 − 𝑈1 𝑞 − 𝑤 = 𝑢2 − 𝑢1 per (kg) where: state (1) refers to the initial state and state (2) refers to the final state. For reversible processes: 2 𝑊 = ∫ 𝑃𝑑𝑉 1 30 Thermodynamics First stage For adiabatic processes (no heat transfer) 𝑄 = 0 For constant volume processes 𝑊 = 0 For constant temperature processes 𝛥𝑈 = 0 The following table contains the governing equations, displacement work equation and heat interaction equation for different non-flow thermodynamic processes: Work Process Governing equations 𝟐 Heat interaction 𝑾 = ∫ 𝑷𝒅𝑽 𝟏 𝑉 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Constant volume 𝑇1 𝑃1 𝑊=0 𝑄 = 𝑚𝐶𝑣 (𝑇2 − 𝑇1 ) (Isochoric) = 𝑇2 𝑃2 𝑃 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Constant pressure 𝑇1 𝑉1 𝑊 = 𝑃(𝑉2 − 𝑉1 ) 𝑄 = 𝑚𝐶𝑝 (𝑇2 − 𝑇1 ) (Isobaric) = 𝑇2 𝑉2 𝑇 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑉2 𝑉2 𝑊 = 𝑃1 𝑉1 ln ( ) 𝑄 = 𝑃1 𝑉1 ln ( ) Constant temperature 𝑃1 𝑉1 = 𝑃2 𝑉2 𝑉1 𝑉1 (Isothermal) 𝑉2 𝑃1 𝑉2 𝑉2 = 𝑊 = 𝑚𝑅𝑇ln ( ) 𝑄 = 𝑚𝑅𝑇ln ( ) 𝑉1 𝑃2 𝑉1 𝑉1 𝑃2 𝑉1 𝛾 ( )=( ) 𝑃1 𝑉1 − 𝑃2 𝑉2 𝑃1 𝑉2 𝑊= 𝑇2 𝑉1 𝛾−1 𝛾−1 Adiabatic ( )=( ) 𝑄=0 𝑇1 𝑉2 𝛾−1 𝑚𝑅(𝑇1 − 𝑇2 ) 𝑇2 𝑃2 𝛾 𝑊= ( )=( ) 𝛾−1 𝑇1 𝑃1 𝑃2 𝑉1 𝑛 ( )=( ) 𝑃1 𝑉1 − 𝑃2 𝑉2 𝑃1 𝑉2 𝑊= 𝛾−𝑛 𝑇2 𝑉1 𝑛−1 𝑛−1 𝑄=( )𝑊 Polytropic ( )=( ) 𝛾−1 𝑇1 𝑉2 𝑚𝑅(𝑇1 − 𝑇2 ) 𝑛−1 𝑊= 𝑇2 𝑃2 𝑛 𝑛−1 ( )=( ) 𝑇1 𝑃1 Note: The heat transferred to the system is positive (+). The heat transferred from the system is negative (-). The work transferred from the system is positive (+). The work transferred to the system is negative (-). 31 Thermodynamics First stage Exercises Problem (5.1): In an air motor cylinder the compressed air has an internal energy of 450 kJ/kg at the beginning of the expansion and an internal energy of 220 kJ/kg after expansion. If the work done by the air during the expansion is 120 kJ/kg, calculate the heat flow to or from the cylinder. Ans. (-110 kJ/kg) Problem (5.2): 2 kg of gas, occupying 0.7 m3, has an initial temperature of 15°C. It was then heated at constant volume until its temperature became 135°C. How much heat was transferred to the gas and what is its final pressure? Take Cv = 0.72 kJ/kg.K and R = 0.29 kJ/kg.K. Ans. (172.3 kJ, 338.1 kPa) Problem (5.3): A mass of air whose pressure, volume and temperature are 275 kPa, 0.09 m3 and 185°C respectively has its state changed at constant pressure until its temperature becomes 15°C. How much heat is transferred from the gas and how much work is done on the gas during the process? Ans. (-32.12 kJ, -9.19 kJ) Problem (5.4): A quantity of air occupies a volume of 0.3 m3 at a pressure of 100 kPa and a temperature of 20°C. The air is compressed isothermally to a pressure of 500 kPa. Draw the (P-V) diagram of the process and determine: 1) The heat received or rejected (stating which) during the compression process. 2) The mass of the air. 3) The final volume of the air. Ans. (-48.3 kJ, 0.388 kg, 0.06 m3) Problem (5.5): 0.05 kg of carbon dioxide (molecular weight 44), occupying a volume of 0.03 m3 at 1.025 bar, is compressed in a perfectly thermally insulated cylinder, until the pressure is 6.15 bar. Calculate the final temperature, the work done on the gas and the heat flow to or from the cylinder walls. Assume carbon dioxide to be a perfect gas and take γ = 1.3. Ans. (490 K, -5.25 kJ, 0 kJ) Problem (5.6): A cylinder contains 0.07 kg of fluid having a pressure of 1 bar, a volume of 0.06 m3 and a specific internal energy of 200 kJ/kg. After a polytropic compression process, the pressure and volume of the gas become 9 bar and 0.0111 m3 respectively and the internal energy becomes 370 kJ/kg. Draw the (P-V) diagram of the process and determine: 1) The amount of work required for the compression. 2) The quantity and direction of heat transferred during the compression process. Ans. (-13.2 kJ, -1.3 kJ) 32 Thermodynamics First stage Problem (5.7): Air at a pressure of 1.06 bar and a temperature of 15°C, is compressed isothermally to 14 bar and is then expanded adiabatically to the original pressure. Draw the (P-V) diagram of the processes then calculate: 1) The final temperature and specific volume of the gas. 2) The net work done. 3) The heat transferred to or from the surroundings. Ans. (137.8 K, 0.361 m3/kg, -105.6 kJ/kg, -213.3 kJ/kg) 33