Thermodynamics 1: Processes of Ideal Gas, Summaries of Thermodynamics

Download Thermodynamics 1: Processes of Ideal Gas and more Summaries Thermodynamics in PDF only on Docsity! UNIVERSITY OF NUEVA CACERES COLLEGE OF ENGINEERING AND ARCHITECTURE THERMODYNAMICS 1 BY: JOSE ANTONIO E. RODAVIA JR. 43 Week 4: Processes of Ideal Gas Objectives: 1. Describe each of the thermodynamic process. 2. Obtain equations for heat, work, and entropy for each thermodynamic process. 3. Solve problems for working substance undergoing thermodynamic process. XIV. Introduction In solving mechanics problems, we isolate the body under consideration, analyze the external forces acting on it, and then use Newton’s laws to predict its behavior. In thermodynamics, we take a similar approach. We start by identifying the part of the universe we wish to study; it is also known as our system. (We defined a system at the beginning of this chapter as anything whose properties are of interest to us; it can be a single atom or the entire Earth.) Once our system is selected, we determine how the environment, or surroundings, interact with the system. Finally, with the interaction understood, we study the thermal behavior of the system with the help of the laws of thermodynamics. The state of a system can change as a result of its interaction with the environment. The change in a system can be fast or slow and large or small. The manner in which a state of a system can change from an initial state to a final state is called a thermodynamic process. XV. Processes of Ideal Gas  Adiabatic Process - In an adiabatic process, the system is insulated from its environment so that although the state of the system changes, no heat is allowed to enter or leave the system, as seen in Figure 4.1. - When a system expands adiabatically, it must do work against the outside world, and therefore its energy goes down, which is reflected in the lowering of the temperature of the system. An adiabatic expansion leads to a lowering of temperature, and an adiabatic compression leads to an increase of temperature.  Reversible and Irreversible Process - A reversible process is a process in which the system and environment can be restored to exactly the same initial states that they were in before the process occurred, if we go backward along the path of the process. It is a process were any losses are neglected - An irreversible process is what we encounter in reality almost all the time. The system and its environment cannot be restored to their original states at the same time. Figure 4.1: An insulated piston with a hot, compressed gas is released. The piston moves up, the volume expands, and the pressure and temperature decrease. The internal energy goes into work. If the expansion occurs within a time frame in which negligible heat can enter the system, then the process is called adiabatic. Ideally, during an adiabatic process no heat enters or exits the system. UNIVERSITY OF NUEVA CACERES COLLEGE OF ENGINEERING AND ARCHITECTURE THERMODYNAMICS 1 BY: JOSE ANTONIO E. RODAVIA JR. 44  Constant Volume Process - Another term for constant volume process are Isometric, Isochoric, and Isovolumic. - Examples of systems under constant volume process are rigid tank, rigid vessel, and indestructible tank. o Working equations for constant volume process: a. Relationship between pressure and temperature, 𝑷𝟏 𝑻𝟏 = 𝑷𝟐 𝑻𝟐 b. Work, W -The equation for work under constant volume process can be derive by apply the formula for work done by a system. - For non-flow systems 𝑊 = ∫ 𝑃𝑑𝑉 Since the volume is constant and the derivative of a constant is zero so, W = 0 - For steady-flow systems 𝑊 = −𝑉𝑑𝑃 Integrating the pressure, 𝑊 = −𝑉(𝑃 − 𝑃 ) = −∆𝑊 = −𝑚𝑅∆𝑇 c. Heat, Q - For non-flow systems, apply the non-flow energy equation, 𝑄 = ∆𝑈 + 𝑊 The W = 0 for constant volume process so the resulting equation is, 𝑄 = ∆𝑈 = 𝑚𝑐 ∆𝑇 - For steady-flow systems, applying the steady flow energy equation, 𝑄 = ∆𝐻 + 𝑊 The formula for the change in enthalpy is ∆𝐻 = ∆𝑈 + 𝑊 then substitute it to the equation for 𝑄 and 𝑊 = −∆𝑊 , 𝑄 = ∆𝑈 + ∆𝑊 − ∆𝑊 Therefore, 𝑄 = ∆𝑈 = 𝑚𝑐 ∆𝑇 d. Change in Enthalpy, ∆𝐻 ∆𝐻 = 𝑚𝑐 ∆𝑇 e. Change in Internal Energy, ∆𝑈 ∆𝑈 = 𝑚𝑐 ∆𝑇 f. Change in Entropy, ∆𝑆 - Apply the general equation for entropy, ∆𝑆 = ∫ and 𝑑𝑄 = 𝑚𝑐 𝑑𝑇 Then the equation becomes, ∆𝑆 = ∫ Integrate the variable T, ∆𝑆 = 𝒎𝑪𝒗 𝒍𝒏 𝑻𝟐 𝑻𝟏 Figure 4.2: PV diagram for constant volume process UNIVERSITY OF NUEVA CACERES COLLEGE OF ENGINEERING AND ARCHITECTURE THERMODYNAMICS 1 BY: JOSE ANTONIO E. RODAVIA JR. 47  Constant Pressure Process - Another term for constant volume process is Isobaric process. o Working equations for constant pressure process: a. Relationship between volume and temperature, 𝑽𝟏 𝑽𝟐 = 𝑻𝟏 𝑻𝟐 b. Work, W -The equation for work under constant volume process can be derive by apply the formula for work done by a system. - For non-flow systems 𝑊 = ∫ 𝑃𝑑𝑉 Integrating the volume, 𝑊 = 𝑷(𝑉 − 𝑉 ) = ∆𝑊 - For steady-flow systems 𝑊 = −𝑉𝑑𝑃 The work for steady flow systems under constant pressure process is zero, 𝑊 = 0 c. Heat, Q - For non-flow systems, apply the non-flow energy equation, 𝑄 = ∆𝑈 + 𝑊 The formula for the change in enthalpy is ∆𝐻 = ∆𝑈 + 𝑊 𝑄 = ∆𝐻 = 𝑚𝑐 ∆𝑇 - For steady-flow systems, applying the steady flow energy equation, 𝑄 = ∆𝐻 + 𝑊 The work for steady flow systems under constant pressure process is zero, 𝑊 = 0, So, 𝑄 = ∆𝐻 = 𝑚𝑐 ∆𝑇 d. Change in Enthalpy, ∆𝐻 ∆𝐻 = 𝑚𝑐 ∆𝑇 e. Change in Internal Energy, ∆𝑈 ∆𝑈 = 𝑚𝑐 ∆𝑇 f. Change in Entropy, ∆𝑆 - Apply the general equation for entropy, ∆𝑆 = ∫ and 𝑑𝑄 = 𝑚𝑐 𝑑𝑇 Then the equation becomes, ∆𝑆 = ∫ Integrate the variable T, ∆𝑆 = 𝒎𝑪𝒑 𝒍𝒏 𝑻𝟐 𝑻𝟏 Figure 4.3: PV diagram for constant pressure process UNIVERSITY OF NUEVA CACERES COLLEGE OF ENGINEERING AND ARCHITECTURE THERMODYNAMICS 1 BY: JOSE ANTONIO E. RODAVIA JR. 48 Example 4.2 A certain gas, with cp = 0.529 Btu/lb-oR and R = 96.2 ft-lbf/lbm-oR, expands from 5 cu ft and 80oF to 15 cu ft while the pressure remains constant at 15.5 psia. Compute (a) 𝑇 , (b) ∆𝐻, (c) ∆𝑈, and (d) ∆𝑆 Given: 𝑉 = 5 𝑓𝑡 𝑉 = 15 𝑓𝑡 𝑃 = 15.5 𝑝𝑠𝑖𝑎 𝑡 = 80℉ 𝑅 = 96.2 ° 𝑐 = 0.529 𝐵𝑡𝑢 𝑙𝑏 − °𝑅 Required: a. 𝑇 b. ∆𝐻 c. ∆𝑈 d. ∆𝑆 Solution (a) For 𝑇 1. Apply Charles’s law to determine the final temperature, 𝑉 𝑉 = 𝑇 𝑇 2. Substitute the given values and make sure you convert pressure and temperature to absolute. 5 𝑓𝑡 15 𝑓𝑡 = (80 + 460)°𝑅 𝑇 3. Solve for 𝑇 𝑻𝟐 = 𝟏𝟔𝟐𝟎°𝑹 𝒕𝟐 = 𝟏𝟏𝟔𝟎 ℉ (b) For ∆𝐻 1. Apply the ideal gas equation of state to solve for the mass of air 𝑃𝑉 = 𝑚𝑅𝑇 2. Substitute the given values, 15.5 𝑙𝑏 𝑖𝑛 144𝑖𝑛 1𝑓𝑡 (5𝑓𝑡 ) = 𝑚 96.2 𝑓𝑡 − 𝑙𝑏 𝑙𝑏 − °𝑅 [(80 + 460)°𝑅] 3. Solve for m 𝑚 = 0.2148 𝑙𝑏 4. The formula for ∆𝐻 is, ∆𝐻 = 𝑚𝑐 ∆𝑇 5. Substitute the given values, ∆𝐻 = (0.2148 𝑙𝑏 ) 0.529 𝐵𝑡𝑢 𝑙𝑏 − °𝑅 [(1160 − 80)°𝑅] 6. Solve for ∆𝐻 ∆𝑯 = 𝟏𝟐𝟐. 𝟕𝟐 𝑩𝒕𝒖 (c) For ∆𝑈 1. Solve for 𝑐 using its relation to 𝑐 and 𝑅 UNIVERSITY OF NUEVA CACERES COLLEGE OF ENGINEERING AND ARCHITECTURE THERMODYNAMICS 1 BY: JOSE ANTONIO E. RODAVIA JR. 49 𝑐 = 𝑐 − 𝑅 2. Substitute the given values, 𝑐 = 0.529 𝐵𝑡𝑢 𝑙𝑏 − °𝑅 − 96.2 𝑓𝑡 − 𝑙𝑏 𝑙𝑏 − °𝑅 1 𝐵𝑡𝑢 778 𝑓𝑡 − 𝑙𝑏 3. Solve for 𝑐 𝑐 = 0.405 𝐵𝑡𝑢 𝑙𝑏 − °𝑅 4. The equation for ∆𝑈 under constant volume process is, ∆𝑈 = 𝑚𝑐 ∆𝑇 5. Substitute the given values, ∆𝑈 = (0.2148 𝑙𝑏 ) 0.405 𝐵𝑡𝑢 𝑙𝑏 − °𝑅 [(1160 − 80)°𝑅] 6. Solve for ∆𝑈 ∆𝑼 = 𝟗𝟑. 𝟗𝟓 𝑩𝒕𝒖 (d) For ∆𝑆 1. Apply the equation for ∆𝑆 under constant pressure process ∆𝑆 = 𝑚𝐶 𝑙 𝑇 𝑇 2. Substitute the given values, ∆𝑆 = (0.2148 𝑙𝑏 ) 0.529 𝐵𝑡𝑢 𝑙𝑏 − °𝑅 𝒍𝒏 (1160 + 460)°𝑅 (80 + 460)°𝑅 3. Solve for ∆𝑆 ∆𝑺 = 𝟎. 𝟏𝟐𝟒𝟖 𝑩𝒕𝒖 °𝑹 Learning Activity 4.2 1. A perfect gas has a value of R = 319.2 J/kg-K and k = 1.26. If 120 kJ are added to 2.27 kg of this gas at constant pressure when the initial temperature is 32.2 oC, find (a) T2, (b) ΔH, (c) ΔU, and (d) work for a nonflow process. 2. Assume 5 lb of an ideal gas with R = 38.7 ft-lb/lb-oR and k = 1.668 have 300 Btu of heat added during a reversible constant pressure change of state. The initial temperature is 80oF. Determine (a) final temperature is 80oF. Determine (a) the final temperature, (b) ΔU, ΔH, and ΔS, (c) W 3. For a constant pressure system whose mass is 80 lb, 1 hp-min is required to raise the temperature by 1oF. Determine the specific heat for the system, Btu/lb-oF UNIVERSITY OF NUEVA CACERES COLLEGE OF ENGINEERING AND ARCHITECTURE THERMODYNAMICS 1 BY: JOSE ANTONIO E. RODAVIA JR. 52 d. ∆𝑈 and ∆𝐻 e. ∆𝑆 Solution (a) For 𝑊 1. The equation for the non-flow work under isothermal process is, 𝑊 = 𝑃𝑉𝑙𝑛 𝑃 𝑃 = 𝑚𝑅𝑇𝑙𝑛 𝑃 𝑃 2. Substitute the given values, 𝑊 = (8 𝑙𝑏 ) 0.06855 𝐵𝑡𝑢 𝑙𝑏 − °𝑅 (88 + 460)𝑙𝑛 80 𝑝𝑠𝑖𝑎 (5 + 14.7)𝑝𝑠𝑖𝑎 3. Solve for 𝑊 𝑊 = 421.15 𝐵𝑡𝑢 (b) For 𝑊 1. The equation for the non-flow work under isothermal process is, 𝑊 = 𝑃𝑉𝑙𝑛 𝑃 𝑃 = 𝑚𝑅𝑇𝑙𝑛 𝑃 𝑃 2. Substitute the given values, 𝑊 = (8 𝑙𝑏 ) 0.06855 𝐵𝑡𝑢 𝑙𝑏 − °𝑅 (88 + 460)𝑙𝑛 80 𝑝𝑠𝑖𝑎 (5 + 14.7)𝑝𝑠𝑖𝑎 3. Solve for 𝑊 𝑊 = 421.15 𝐵𝑡𝑢 (c) For 𝑄 The heat and work are equal for isothermal process, 𝑄 = 𝑊 = 421.15 𝐵𝑡𝑢 (d) For ∆𝑈 and ∆𝐻 ∆𝑈 and ∆𝐻 are both functions of temperature difference and the temperature difference under isothermal process is zero. Therefore, ∆𝑈 = 0 ∆𝐻 = 0 (e) For ∆𝑆 1. Apply the equation for ∆𝑆 under constant temperature process ∆𝑆 = 𝑚𝑅𝑙𝑛 𝑃 𝑃 2. Substitute the given values, ∆𝑆 = (8 𝑙𝑏 ) 0.06855 𝐵𝑡𝑢 𝑙𝑏 − °𝑅 𝑙𝑛 80 𝑝𝑠𝑖𝑎 (5 + 14.7)𝑝𝑠𝑖𝑎 3. Solve for ∆𝑆 ∆𝑺 = 𝟎. 𝟕𝟔𝟖𝟓 𝑩𝒕𝒖 °𝑹 Learning Activity 4.3 UNIVERSITY OF NUEVA CACERES COLLEGE OF ENGINEERING AND ARCHITECTURE THERMODYNAMICS 1 BY: JOSE ANTONIO E. RODAVIA JR. 53 1. During a reversible process there are abstracted 317 kJ/s from 1.134 kg/s of a certain gas while the temperature remains constant at 26.7oC. For this gas, cp = 2.232 and cv = 1.713 kJ/kg-K. The initial pressure is 586 kPa. For both nonflow and steady flow (ΔP = 0, ΔK = 0) process, determine (a) V1, V2 and p2, (b) the work and Q, (c) ΔS and ΔH. 2 Air flows steadily through an engine at constant temperature, 400 K. Find the work per kilogram if the exit pressure is one-third the inlet pressure and the inlet pressure is 207 kPa. Assume that the kinetic and potential energy variation is negligible. 3. Helium at 100 atm, 165 K expands isothermally to 1 atm. For 2 kg, find (a) W for nonlfow and for a steady flow process (ΔP = 0, ΔK = 0), (b) ΔU, (c) Q, and (d) ΔS  Constant Entropy Process (PVk = c) - Another term for constant entropy process is Isentropic process - If a system has adiabatic walls, heat cannot enter or leave the system, and the first cause of possible change of entropy does not occur. A process, taking place without heat entering or exiting, is called an adiabatic process. If a process is quasi-static (also known as reversible), there is no spontaneous increase of entropy, hence, in short, an isentropic process is quasi- static and adiabatic. - In actual practice, quasi-static (reversible) processes do not occur; the concept of reversibility is an idealization, an unachievable limit, but useful for theoretical considerations. o Working equations for constant entropy process: a. Relationship between pressure, volume and temperature,  𝑃 𝑉 = 𝑃 𝑉  =  = b. Work, W -The equation for work under constant temperature process can be derive by apply the formula for work done by a system. - For non-flow systems 𝑊 = ∫ 𝑃𝑑𝑉 Since 𝑃 𝑉 = 𝑃 𝑉 = 𝑃𝑉 = 𝑐 Figure 4.6: PV diagram for constant entropy process UNIVERSITY OF NUEVA CACERES COLLEGE OF ENGINEERING AND ARCHITECTURE THERMODYNAMICS 1 BY: JOSE ANTONIO E. RODAVIA JR. 54 𝑃𝑉 = 𝑐 can be written as 𝑃 = and we substitute it to our working equation for work, 𝑊 = 𝑐 𝑉 𝑑𝑉 Rearranging the expression, 𝑊 = 𝑐 𝑑𝑉 𝑉 Integrate the volume, 𝑊 = 𝑐 𝑉 − 𝑉 1 − 𝑘 Knowing that 𝑐 = 𝑃𝑉 , substitute for c 𝑊 = 𝑃𝑉 𝑉 − 𝑉 1 − 𝑘 Simplify, 𝑊 = 𝑷𝟐𝑽𝟐 𝑷𝟏𝑽𝟏 𝟏 𝒌 - For steady-flow systems 𝑊 = −𝑉𝑑𝑃 Since, 𝑃 𝑉 = 𝑃 𝑉 = 𝑃𝑉 = 𝑐 𝑃𝑉 = 𝑐 can be written as 𝑉 = and we substitute it to our working equation for work, 𝑊 = − 𝑐 𝑃 𝑑𝑃 Rearranging the expression, 𝑊 = −𝑐 𝑑𝑃 𝑃 Integrate the pressure, 𝑊 = −𝑐 𝑃 − 𝑃 𝑘 − 1 𝑘 Knowing that 𝑐 = 𝑃𝑉 , substitute for c 𝑊 = −𝑃 𝑉 𝑃 − 𝑃 𝑘 − 1 𝑘 Simplify, 𝑊 = 𝑘 𝑷𝟐𝑽𝟐 − 𝑷𝟏𝑽𝟏 𝟏 − 𝒌 = 𝑘 𝑊 UNIVERSITY OF NUEVA CACERES COLLEGE OF ENGINEERING AND ARCHITECTURE THERMODYNAMICS 1 BY: JOSE ANTONIO E. RODAVIA JR. 57 𝑊 = 9692.11 𝐵𝑡𝑢 Learning Activity 4.4 1. An adiabatic expansion of air occurs through a nozzle from 828 kPa and 71oC to 138 kPa. The initial kinetic energy is negligible. For an isentropic expansion, compute the specific volume, temperature and speed at the exit section. 2. One pound of an ideal gas undergoes an isentropic process from 95.3 psig and a volume of 0.6 ft3 to a final volume of 3.6 ft3. If cp = 0.124 and cv = 0.093 Btu/lb-oR, what are (a) t2, (b) p2, (c) ΔH and (d) W. 3. A certain ideal gas whose R = 278.6 kJ/kg-K and cp = 1.015 kJ/kg-K expands isentropically from 1517 kPa, 288oC to 965 kPa. For 454 g/s of this gas determine, (a) Wn, (b) V2, (c) ΔU and (d) ΔH  Polytropic Process (PVn = c) - is an internally reversible process. o Working equations for polytropic process: a. Relationship between pressure, volume and temperature,  𝑃 𝑉 = 𝑃 𝑉  =  = b. Work, W -The equation for work under polytropic process can be derive by apply the formula for work done by a system. - For non-flow systems 𝑊 = ∫ 𝑃𝑑𝑉 Since 𝑃 𝑉 = 𝑃 𝑉 = 𝑃𝑉 = 𝑐 𝑃𝑉 = 𝑐 can be written as 𝑃 = and we substitute it to our working equation for work, 𝑊 = 𝑐 𝑉 𝑑𝑉 Rearranging the expression, 𝑊 = 𝑐 𝑑𝑉 𝑉 Integrate the volume, 𝑊 = 𝑐 𝑉 − 𝑉 1 − 𝑛 Knowing that 𝑐 = 𝑃𝑉 , substitute for c Figure 4.7: PV diagram for polytropic process UNIVERSITY OF NUEVA CACERES COLLEGE OF ENGINEERING AND ARCHITECTURE THERMODYNAMICS 1 BY: JOSE ANTONIO E. RODAVIA JR. 58 𝑊 = 𝑃𝑉 𝑉 − 𝑉 1 − 𝑛 Simplify, 𝑊 = 𝑷𝟐𝑽𝟐 𝑷𝟏𝑽𝟏 𝟏 𝒏 - For steady-flow systems 𝑊 = −𝑉𝑑𝑃 Since, 𝑃 𝑉 = 𝑃 𝑉 = 𝑃𝑉 = 𝑐 𝑃𝑉 = 𝑐 can be written as 𝑉 = and we substitute it to our working equation for work, 𝑊 = − 𝑐 𝑃 𝑑𝑃 Rearranging the expression, 𝑊 = −𝑐 𝑑𝑃 𝑃 Integrate the pressure, 𝑊 = −𝑐 𝑃 − 𝑃 𝑛 − 1 𝑛 Knowing that 𝑐 = 𝑃𝑉 , substitute for c 𝑊 = −𝑃 𝑉 𝑃 − 𝑃 𝑛 − 1 𝑛 Simplify, 𝑊 = 𝑛 𝑷𝟐𝑽𝟐 − 𝑷𝟏𝑽𝟏 𝟏 − 𝒏 = 𝑛 𝑊 c. Heat, Q 𝑄 = 𝑚𝑐 ∆𝑇 Where: 𝐶 = 𝐶 d. Change in Enthalpy, ∆𝐻 ∆𝐻 = 𝑚𝑐 ∆𝑇 e. Change in Internal Energy, ∆𝑈 ∆𝑈 = 𝑚𝑐 ∆𝑇 f. Change in Entropy, ∆𝑆 Apply the general equation for entropy, ∆𝑆 = ∫ and 𝑑𝑄 = 𝑚𝑐 𝑑𝑇 Then the equation becomes, ∆𝑆 = ∫ Integrate the variable T,