Special Topic 3 of the Registered Chemical Engineer Professional Examination: Chemical Thermodynamics Processes
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This post was last edited by zhanghp30 on 2019-9-29 at 14:21. 1. Professional knowledge: 1) Problems addressed by thermodynamics: analysis of the feasibility of processes, efficient use of energy； Thermodynamic properties at equilibrium, property prediction, and calculation ； Phase equilibrium and chemical equilibrium calculations in physical and chemical change processes ； 2) The most fundamental principle of thermodynamics: It is impossible to create a perpetual motion machine (First Law of Thermodynamics) ； It is not possible to completely restore a natural process (Second Law of Thermodynamics) ； Absolute zero cannot be attained (Third Law of Thermodynamics). Furthermore, there is another law that people don’t learn in university, though some teachers might mention it; it is the zeroth law of thermodynamics: if two objects both have the same temperature as a third object, then those two objects must also have the same temperature. 3) Characteristics and limitations of thermodynamics: Thermodynamics deals only with the initial and final states, not with the intermediate paths ； When the conclusions of thermodynamics are derived rigorously without any additional conditions, the conclusions are universal. 4) Classification of thermodynamic properties: Extensive properties: They are additive. Such as volume V, mass m, entropy S, enthalpy H, Gibbs free energy G, etc. Strength property: It does not possess additivity. Such as temperature T, pressure P, viscosity u, density p, specific heat r, etc. 5) Calculation of entropy changes for different processes ； Follow the textbook exactly, and pay attention to the applicable conditions of the formulas. 6) Calculation of latent heat: deta H = deta U + P*(Vg - Vl) = H_saturated vapor - H_saturated liquid ； Please note that the latent heat of vaporization for the same substance gradually decreases as the temperature rises, and it becomes zero at the critical point! The heat for a thermodynamic reversible process can be determined using Hess’s law. 7) Definitions of differential and integral heat of solution. There might be case questions! 8*）Continuous flow processes: At constant pressure, Q+W=delta H+delta U^2/2+delta gZ; in a stable process, Q+W=delta H. At constant temperature, Q+W=delta U+delta U^2/2+delta gZ; in a stable process, Q+W=delta U. Note that these two equations are often combined with Maxwell’s relations to create problems of high difficulty, so it is essential to master them! They are key topics in past exams, such as the isenthalpic throttling process and adiabatic processes! 9*) Ideal work, lost work, and exergy: Ideal work: Wid = T0*delta S – delta H. Lost work: Ws = Wid – Ws = T0*delta S – Q = T0*delta S_t. Exergy: Ex = (H – H0) – T0*(S – S0) = –Wid. 10) Fugacity and activity: Pay attention to the definitions of these formulas! 11*）Heat pumps, cooling: Review in detail according to the textbook; it is essential to master this! 【Question Prediction】 1*) Among the following thermodynamic functions, which ones are extensive properties? ( ) A Pressure B Volume C Viscosity D Enthalpy 【Explanation】 BD 2*) Among the following thermodynamic functions, which ones are state functions? ( ) A Heat B Volume C Entropy D Work 【Explanation】 BD; note that heat and work are related to processes and are not thermodynamic state functions! ! ! 3*）For a certain mass of ideal gas undergoing some reversible process, what can possibly happen at the same time is ( ) A Absorbing heat, heating up, and doing positive work on the surroundings B Absorbing heat, cooling down, and doing negative work on the surroundings C Absorbing heat, heating up, and doing negative work on the surroundings D Absorbing heat, cooling down, and doing positive work on the surroundings 【Explanation】ACD; this can be determined by drawing diagrams. 4) When a certain amount of nitrogen undergoes adiabatic free expansion, which of the following statements is correct? （ ） A deta H=0, deta U=0 B deta H>0, detaU>0 C deta H=0, deta U=0 D deta H
This post was last edited by zhanghp30 on 2015-11-13 at 10:23. 2. Case knowledge: 1) Water at atmospheric pressure can be supercooled below 0°C before freezing; water at -5°C freezes more easily under disturbances. This process can be approximated as adiabatic. Determine the entropy change for this process. The heat of fusion of ice is 334.4 J/g, and the specific heat of water between 0°C and -5°C is 4.22 J/g·K. 【Analysis】Taking 1 g of water as the basis, let the solidification rate be X. Then, 1 g of water at -5°C becomes (1-X) g of water at 0°C plus X g of ice at 0°C. Thus, 1*4.22*(0 + 5) + X*(-334.4) = 0, which gives X = 0.0631. The change in entropy is given by ΔS = ΔS1 + ΔS2 = m*Cp*Ln(T2/T1) + X*ΔHfus/T. Substituting the values, we get ΔS = 1*4.22*Ln(273.15/268.15) + 0.0631*(-334.4)/273.15 = 7.13*10^(-4) J. 2) 1 kg of air is adiabatically and reversibly compressed in a compressor from 0.1034 MPa and 299.7 K to 0.517 MPa. It is assumed that the process occurs under steady-state flow conditions, with any changes in potential energy being negligible. Assuming air is an ideal gas with Cv=0.716 KJ/KG/K and Cp=1.005 KJ/KG/K, what is the compression work? 【Analysis】From the flow equation, dηH + 1/2 dηU² + dηgZ = Q + Ws; with Q = 0 and steady state, it follows that Ws = dηH = mCp*(T2 – T1). An adiabatic reversible process is an isentropic process, and thus T2/T1 = (P2/P1)^(k–1)/k, where k = Cp/Cv = 1.4036. Therefore, T2 = 299.7 * (0.517/0.1034)^(1.4036–1)/1.4036 = 476.1 K. Then, Ws = 1 * 1.005 * (476.1 – 299.7) = 177.28 KJ/kg. 3) When superheated steam at 1.5 MPa and 773 K is used to drive a turbine, with the exhaust pressure at 0.07 MPa, the work output by this turbine is equivalent to 85% of the work produced in an adiabatic reversible process. Heat is dissipated into the surrounding atmosphere at 293.15 K, resulting in a heat loss of 79.4 KJ/kg. Determine the work loss for this process. (Initial state: 1.5 MPa, 773 K; enthalpy H1 is 3473.1 KJ/KG, entropy S1 is 7.5698 KJ/KG/K.) [Explanation] An adiabatic process can be considered; thus S2 = S1 = 7.5698. The enthalpy at this point is H2 = 2680 (an isentrope line will be provided in the exam!) Thus, Ws = ΔH = (2680 – 3473.1) = -793.1 KJ/KG; therefore, the actual value of Ws’ is 0.85 * Ws = 0.85 * (-793.1) = -674.1 KJ/KG. The actual value of H2 is H1 + Q + Ws = 3473.1 – 79.4 – 674.1 = 2719.6 KJ/KG. Checking again, the entropy at 0.07 MPa and 2719.6 KJ/KG is 7.6375; hence, the work lost is Wl = T0 * ΔS – Q = 293.15 * (7.6375 – 7.5698) + 79.4 = 99.2 KJ/KG
This post was last edited by zhanghp30 on 2020-1-16 at 10:39. 4) There is a ammonia refrigeration unit with a cooling capacity of 4.186*10^5 KJ/h; its evaporation temperature is -26°C, and the condensation temperature is 20°C. Assuming that the compressor operates in an adiabatically reversible manner, determine the flow rate of the refrigerant, the work required by the compressor, the load on the condenser, and the coefficient of performance of the refrigeration cycle. (-26°C: H1=1430 KJ/KG, S1=6 KJ/KG/K; 20°C: H2=290 KJ/KG, S2=1.3 KJ/KG/K) 【Analysis】 The refrigerant flow rate G = Q/q = 418600/(H1–H2) = 418600/1430–290 = 367.2 kG/h. The work required by the compressor is Ws = G*W = 367.2*(H3–H1) = –367.3*(1680–1430) = –25.5 Kw. For an adiabatic reversible process, H3 can be determined using the isentropic relation; the value is 1680. A diagram will be provided in the exam! ) The cooling capacity of the condenser is Q = G*(H2 – H1) = 367.2*(290 – 1430) = -5.1*10^5 KJ/h. The coefficient of performance for cooling is S = q0 / (-Ws) = (1430 – 290) / (1680 – 1430) = 4.56. 5) A heat pump has a power rating of 1 KW; the ambient temperature is 0°C, and the desired heating temperature is 30°C. The coefficient of performance for heating is 80% of that of a reverse Carnot cycle. Determine the heating capacity of this air conditioner as well as the amount of heat absorbed by the heat pump from the environment. [Solution] Qh = Sh * Wn = 0.8 * (Th/Th – Tl) = 0.8 * (303.15/30 – 3) = 8.084 KW. The amount of heat absorbed from the environment is Q = Qh – Wn = 8.084 – 1 = 7.084 KW. 6) A seawater desalination project in China. The raw material is seawater containing 3.5% NaCl, and the product is pure water; both are at a constant temperature of 25°C. It is known that the saturated vapor pressures of seawater and pure water at 25°C are 3098.1 Pa and 3167.2 Pa respectively, while the osmotic pressure of seawater is 3049.9 kPa. Determine the ideal work for this process. 【Analysis】 Ws = T0*delta S – delta H. The principle of seawater desalination is that the change in free energy caused by pressure represents the ideal work for the process; thus, Ws = T0*delta S – delta H = -delta G = -8.314*298.15*Ln(3167.2/3098.1) = -54.68 J/mol. 7*) There is a dry and clean room with a volume of 1000 m3, a temperature of 10°C, and atmospheric pressure. How much heat is required to raise the temperature to 20°C while keeping the pressure constant? If the outdoor temperature is 0°C, and the indoor temperature is reduced from 20°C to 10°C, how much heat is released? Assume air to be an ideal gas, with Cpm = 29.29 J/K/mol. 【Analysis】During heating at constant pressure, n = f(T) = PV/RT. The heat required is Qp = dΔH = f(n)·CPm·dT = [PV/(RT)]·CPm·dT = PV/R·(CPm)·Ln(T2/T1) = 101325·1000/8.3145·29.29·Ln(293.15/283.15) = 1.24·10^7 J. When cooling down, air enters the room; at 20°C, n1 = PV/293.15, and at 10°C, n2 = PV/283.15. The additional amount of air is n3 = n2 – n1 = PV/283.15R – PV/293.15R. The total heat transferred is the sum of these two values: Q = Q1 + Q2 = f(PV/293.15R)·29.29·dT + f(PV/283.15R – PV/293.15R)·29.29·dT = 101325·1000·29.29·[-10/293.15 + (1/283.15 – 1/293.15)·10] = -1.2·10^7 J. 8) There is a pipe carrying hot water at 90°C; due to poor insulation, the water temperature drops to 70°C upon reaching the user’s location. Calculate the work lost during this temperature drop process as well as the heat loss, assuming the atmospheric temperature is 25°C and the specific heat capacity of water is 4.18 J/mol/K. 【Analysis】Using 1 mol of water as a reference, dΔH = Q – Ws, with Ws = 0. Thus, Q = dΔH = 1·4.18·(343 – 363) = -83.6 J. dΔS = Cp·Ln(T2/T1) = 4.18·Ln(343/363) = -0.2369 J/K. According to the second law of thermodynamics, -dΔS + q/T0 + dΔSg = 0. Therefore, dΔSg = dΔS – Q/T0 = -0.2369 + 83.6/298.15 = 0.0436 J/K. WL = T0·dΔSg = 298.15·0.0436 = 13 J. 9) A steam compression refrigeration cycle has a cooling capacity of 4·10^4 KJ/h. The temperature in the evaporator is -10°C, while the inlet temperature of water used in the condenser is 8°C. The amount of water in the cycle is unlimited. Design a minimal set of components for this cycle and calculate the minimum work required for its operation. If air cooling is used, with an indoor temperature of 25°C, what is the minimum work required? 【Analysis】The minimum-power device designed is the reverse Carnot cycle! Thus, Sc = Q0 / (-Ws) = TL/Th – TL; 40000 / (-Ws) = 263.15 – 281.15 – 263.15, so WS = -2736 KJ/h. When cooled by air at a temperature of 25°C, 40000 / (-Ws) = 263.15 – 298.15 – 263.15, giving WS = -5320 KJ/h. 10) It is known that the vapor pressure of ice at -5°C is 0.4 kPa and its relative density is 0.915. Calculate the fugacity of ice at -5°C and 100 MPa. 【Analysis】-5C; at 0.4 KPa, water vapor can be approximated as an ideal gas, so f = 0.4 KPa. Under isothermal conditions, (df/dp)T = V/RT. Considering that the molar volume of ice changes little with pressure, it can be treated as a constant. Thus, Vs = M/p = 18/0.915 = 19.69 cm3/mol. Furthermore, fs = f*e^(Vs(100–0.4)/R(273.15–5)) = 0.4*e^(1.969*10^-5*(1000–0.4)/8.314*10^-5*(273.15–5)) = 0.97 kPa
Amazing, it has the vibe of an experienced teacher
This section has always been the one I find most confusing. Study it properly*~~~