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Jan 9, 2024

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Carnot Refrigeration Cycle Analysis with Refrigerant-134a*Analysis of a Carnot refrigeration cycle in a closed system. Working fluid: 0.96 kg of Refrigerant-134a. Key points: maximum temperature is 1.2x minimum temperature, net work input is 22 kJ. a) Illustrate the cycle on a Temperature-Entropy (T-s) diagram. b) Calculate the minimum pressure in the cycle, given the refrigerant transitions from saturated vapor to saturated liquid during heat rejectionn 1. Coefficient of Performance (COP) of the cycle: - Formula: COPR = 1 / (TH/TL - 1) - Calculation: COPR = 1 / (1.2 - 1) = 5 2. Heat transfer calculations: - QL (Heat absorbed) = COPR × Win = 5 × 22 kJ = 110 kJ - QH (Heat rejected) = QL + W = 110 kJ + 22 kJ = 132 kJ 3. Determination of TH (high temperature): - Using enthalpy of vaporization (hfg) at TH: hfg @ TH = QH / m = 132 kJ / 0.96 kg = 137.5 kJ/kg - From Refrigerant-134a properties table (A-11): TH ≈ 61.1 °C 4. TL (low temperature) calculation: - TL = TH / 1.2 = 334.1 K / 1.2 ≈ 278.4 K ≈ 5.4 °C 5. Minimum pressure (Pmin) estimation: - Using saturation pressure at TL: Pmin ≈ 354.9 kPa --- Adiabatic Expansion of Saturated Water Vapor in a Piston-Cylinder Device** 2 kg of saturated water vapor at 600 kPa in a piston-cylinder device, expanding adiabatically to 100 kPa, producing 700 kJ work. a) Calculate the entropy change of the water (kJ/kg · K). b) Determine the generated entropy and assess the process's realism. 1. **Entropy Change ( Δ s) of Water**: - Initial State (P1 = 600 kPa, x = 1): u1 (Internal Energy) = 2566.8 kJ/kg, s1 (Entropy) = 6.7593 kJ/kg · K. - Final State Calculation: u2 = u1 - Wout/m = 2216.8 kJ/kg. - Final State (P2 = 100 kPa, u2 = 2216.8 kJ/kg): Saturated mixture, x2 = (u2 - uf) / ufg = 0.8617, s2 = sf + x2 · sfg = 6.5215 kJ/kg · K. - Entropy Change: Δ s = s2 - s1 = -0.238 kJ/kg · K. 2. **Entropy Generation (sgen)**: - Calculation: sgen = Δ s_system = s2 - s1 = -0.238 kJ/kg · K. - Conclusion: This indicates an impossible process as entropy cannot decrease in an adiabatic expansion. --- CO2 in Insulated Piston-Cylinder Device with Paddle-Wheel Work** 0.2 m ³ CO2 at 100 kPa and 17°C. Paddle-wheel work of 200 kJ applied, constant pressure. 1. Determine final temperature. 2. Calculate work produced by the system. 3. Calculate entropy change of CO2. 1. **Final Temperature (T2)**: - Energy Balance: Win - Wb,out = m(u2 - u1) = m(h2 - h1). - Mass (m): m = P1V1 / (R · T1) = 0.5476 kg (CO2 properties used). - Enthalpy at Initial State (h1): h1 = h(T=290K) =9063 kJ/kmol. - Enthalpy at Final State (h2): h2 = h1 + Win / (m/M) = 25133 kJ/kmol. - T2 Calculation: Interpolation from table A-20, T2 ≈ 659.4 K. 2. **Work Produced by System (Wb,out)**: - Wb,out = Win - (m/M) × (u2 - u1) = 200 kJ - (0.5476 kg / 44 kg/kmol) × (19672 - 6651) kJ/kmol. - Wb,out ≈ 37.95 kJ. 3. **Entropy Change of CO2 (Delta S_sys)**: - Delta S_sys = (m/M) × (s0_2 - s0_1) = (0.5476 kg / 44 kg/kmol) × (247.773 - 212.66) kJ/kmol.K. - Delta S_sys ≈ 0.437 kJ/K. **Key Formulas**: - m = P1V1 / (R · T1) - h2 = h1 + Win / (m/M) - Delta S_sys = (m/M) × (s0_2 - s0_1) --- **Problem 4: Mixing of Liquid Water and Superheated Steam in a Piston-Cylinder Device** Liquid water at 200 kPa, 15°C mixed with superheated steam at 200 kPa, 150°C. Water flow rate is 4.3 kg/s. Heat loss rate is 1200 kJ/min to surrounding air at 20°C. Final mixture is at 200 kPa, 80°C. 1. Find mass flow rate of superheated steam. 2. Calculate rate of entropy generation. **Calculations**: 1. **Mass Flow Rate of Superheated Steam (m˙2)**: - Mass Balance: m˙1 + m˙2 = m˙3. - Energy Balance: m˙1h1 + m˙2h2 = Q˙out + m˙3h3. - Solving for m˙2: m˙2 = (Q˙out - m˙1(h1 - h3)) / (h2 - h3). - Using thermodynamic tables: - h1 ≈ hf at 15°C = 62.98 kJ/kg. - h2 = 2769.1 kJ/kg (superheated steam). - h3 ≈ hf at 80°C = 335.02 kJ/kg. - m˙2 ≈ 0.4806 kg/s. 2. **Rate of Entropy Generation (S˙gen)**: - Entropy Balance: S˙gen = m˙3s3 - m˙1s1 - m˙2s2 + Q˙out/Tsurr. - Using thermodynamic tables: - s1 ≈ sf at 15°C = 0.2245 kJ/kg.K. - s2 = 7.2810 kJ/kg.K (superheated steam). - s3 ≈ sf at 80°C = 1.0756 kJ/kg.K. - S˙gen ≈ 0.746 kW/K. **Key Equations**: - Mass Flow Rate: m˙2 = (Q˙out - m˙1(h1 - h3)) / (h2 - h3). - Entropy Generation Rate: S˙gen = m˙3s3 - m˙1s1 - m˙2s2 + Q˙out/Tsurr. --- **Problem 5: Work Input and Entropy Generation in Water Compression** Water compressed from 100 kPa to 1 MPa in (a) an adiabatic pump (saturated liquid at inlet) and (b) an adiabatic compressor (saturated vapor at inlet). Isentropic efficiency: 85% for both devices. 1. Calculate work input for both pump and compressor. 2. Determine entropy generation for both devices. 1. **Adiabatic Pump (Saturated Liquid Inlet)**: - Initial State (P1 = 100 kPa): h1 = hf = 417.51 kJ/kg, s1 = sf = 1.3028 kJ/kg.K. - Isentropic Efficiency ( η p): h2 = h1 + (h2s - h1) / η p. - h2s Calculation: h2s = h1 + v1(P2 - P1) = 418.45 kJ/kg. - Final Enthalpy (h2): h2 = 418.61 kJ/kg. - Entropy (s2): s2 ≈ 1.3032 kJ/kg.K (interpolation from table A-4). - Work Input (win): win = h2 - h1 = 1.10 kJ/kg. - Entropy Generation (sgen): sgen = s2 - s1 = 0.0004 kJ/kg.K. 2. **Adiabatic Compressor (Saturated Vapor Inlet)**: - Initial State (P1 = 100 kPa): h1 = hg = 2675.0 kJ/kg, s1 = sg = 7.3589 kJ/kg.K. - Isentropic Efficiency ( η p): h2 = h1 + (h2s - h1) / η p. - h2s Calculation: h2s ≈ 3194.48 kJ/kg (interpolation from table A-6). - Final Enthalpy (h2): h2 = 3286.2 kJ/kg. - Entropy (s2): s2 ≈ 7.4971 kJ/kg.K (interpolation from table A-6). - Work Input (win): win = h2 - h1 = 611.2 kJ/kg. - Entropy Generation (sgen): sgen = s2 - s1 = 0.1382 kJ/kg.K. **Key Equations**: - Work Input: win = h2 - h1. - Entropy Generation: sgen = s2 - s1. - Isentropic Efficiency: h2 = h1 + (h2s - h1) / η p. --- **Problem 6: Adiabatic Expansion of Air in a Turbine** Air expanded in an adiabatic turbine. Inlet state: 2800 kPa, 400°C. Outlet pressure: 100 kPa. Isentropic efficiency: 90%. Constant specific heats assumed at 400°C. 1. Calculate outlet temperature of air. 2. Determine work produced by the tu\\\generation. 1. **Outlet Temperature (T2)**: - Isentropic Process Exit Temperature (T2s): T2s = T1 * (P2/P1)^((k-1)/k) = (673 K) * (100/2800)^0.395/1.395 = 262 K. - Actual Exit Temperature using Isentropic Efficiency ( η T): T2 = T1 - η T * (T1 - T2s) = 673 K - 0.90 * (673 K - 262 K) = 303.1 K. 2. **Work Produced by Turbine (w˙out)**: - w˙out = cp * (T1 - T2) = 1.013 kJ/kg.K * (673 K - 303.1 K) = 374.71 kJ/kg. 3. **Entropy Generation (sgen)**: - sgen = cp * ln(T2/T1) - R * ln(P2/P1). - Using cp = 1.013 kJ/kg.K and R = 0.287 kJ/kg.K: sgen = 1.013 kJ/kg.K * ln(303.1/673) - 0.287 kJ/kg.K * ln(100/2800) = 0.1482 kJ/kg.K. **Key Equations**: - T2s = T1 * (P2/P1)^((k-1)/k). - w˙out = cp * (T1 - T2). - T2 = T1 - η T * (T1 - T2s). - sgen = cp * ln(T2/T1) - R * ln(P2/P1). Isentropic Compression of Water to 4 MPa** Compressing water from an initial state of 15°C with 80.27% quality to a final pressure of 4 MPa isentropically in a closed system. Calculate the work required (win) in kJ/kg. **Calculations**: 1. **Initial State Properties**: - Temperature (T1): 15 °C. - Quality (x): 80.27% or 0.8027. - From table A-5: - Specific internal energy of liquid (uf): 62.98 kJ/kg. - Specific internal energy of vaporization (ufg): 2332.5 kJ/kg. - Specific entropy of liquid (sf): 0.2245 kJ/kg.K. - Specific entropy of vaporization (sfg): 8.5559 kJ/kg.K. - Initial Internal Energy (u1): \( u1 = uf + x \times ufg \) - Calculation: \( u1 = 62.98 + 0.8027 \times 2332.5 = 1935.28 \) kJ/kg. - Initial Entropy (s1): \( s1 = sf + x \times sfg \) - Calculation: \( s1 = 0.2245 + 0.8027 \times 8.5559 = 7.0922 \) kJ/kg.K. 2. **Final State Properties (Isentropic Compression to P2 = 4 MPa)**: - Final Pressure (P2): 4 MPa. - Entropy remains constant (s2 = s1): 7.0922 kJ/kg.K. - Since s2 > sg (value for saturated vapor at 4 MPa), the state is in the superheated vapor region. - From table A-6: - Final Internal Energy (u2): 3100.3 kJ/kg. 3. **Work Required (win)**: - Work input formula: \( win = u2 - u1 \) - Calculation: \( win = 3100.3 - 1935.28 = 1165.02 \) kJ/kg. **Key Formulas**: - \( u1 = uf + x \times ufg \) - \( s1 = sf + x \times sfg \) - \( win = u2 - u1 \) --- Isentropic Efficiency of an Adiabatic Compressor in a Refrigeration System** An adiabatic compressor compresses saturated R-134a vapor from -2°C to 900 kPa and 50°C. Determine the isentropic efficiency ( η c) of the compressor. **Calculations**: 1. **Isentropic Efficiency Formula**: - η c = wis / wa = (h2s - h1) / (h2 - h1). 2. **Initial State Properties (Saturated Vapor at T1 = -2°C)**: - From table A-11: - Specific enthalpy (h1 = hg): 249.33 kJ/kg. - Specific entropy (s1 = sg): 0.93271 kJ/kg.K. 3. **Final State Properties for Isentropic Compression (P2 = 900 kPa, s2s = s1)**: - Specific entropy remains constant (s2s = s1): 0.93271 kJ/kg.K. - Since s2s > sg, the state is in the superheated vapor region. - From table A-12: - Specific enthalpy (h2s): 274.19 kJ/kg. 4. **Final State Properties for Actual Compression (P2 = 900 kPa, T2 = 50°C)**: - From table A-12 (Superheated vapor region): - Specific enthalpy (h2): 284.79 kJ/kg. 5. **Isentropic Efficiency Calculation**: - η c = (h2s - h1) / (h2 - h1). - Calculation: η c = (274.19 kJ/kg - 249.33 kJ/kg) / (284.79 kJ/kg - 249.33 kJ/kg) = 70.12%. **Key Formulas**: - Isentropic Efficiency: η c = (h2s - h1) / (h2 - h1). --- COP and Heat Rejection Rate of a Refrigerator** A refrigerator removes heat from the food compartment at a rate of 5400 kJ/h and consumes power at a rate of 1 kW. 1. Calculate the Coefficient of Performance (COP) of the refrigerator. 2. Determine the rate of heat rejection to the outside air. **Calculations**: 1. **Coefficient of Performance (COPR)**: - Definition: COPR = Q ̇ L / Ẇ net,in. - Given: Heat removal rate (Q ̇ L) = 5400 kJ/h, Power consumption ( Ẇ net,in) = 1 kW. - Convert power consumption to the same units as Q ̇ L: 1 kW = 3600 kJ/h. - Calculation: COPR = 5400 kJ/h / (1 kW * 3600 kJ/h / 1 kW) = 1.5. 2. **Rate of Heat Rejection (Q ̇ H)**: - Energy Balance: Q ̇ H = Q ̇ L + Ẇ net,in. - Calculation: Q ̇ H = 5400 kJ/h + 1 kW * 3600 kJ/h / 1 kW = 9000 kJ/h. **Key Formulas**: - COPR = Q ̇ L / Ẇ net,in. - Q ̇ H = Q ̇ L + Ẇ net,in. --- Maximum Power Output of a Heat Engine** A heat engine operates between a source at 450°C and a sink at 20°C. Heat supplied to the engine is 70,000 kJ/min. Determine the maximum power output in kW of this heat engine. **Calculations**: 1. **Carnot Efficiency ( η max)**: - Formula: η max = 1 - (TL/TH). - Convert temperatures to Kelvin: TH = 450°C + 273 = 723 K, TL = 20°C + 273 = 293 K. - Calculation: η max = 1 - (293 K / 723 K) = 0.595. 2. **Maximum Power Output ( Ẇ net,out)**: - Definition: Ẇ net,out = η max × Q ̇ H. - Given: Heat supply rate (Q ̇ H) = 70,000 kJ/min. - Convert heat supply rate to kW: 1 kW = 60 kJ/min. - Calculation: Ẇ net,out = 0.595 × (70,000 kJ/min) × (1 kW / 60 kJ/min) = 694 kW. **Key Formulas**: - η max = 1 - (TL/TH). - Ẇ net,out = η max × Q ̇ H. --- Validity of a Claimed Heat Pump COP** A thermodynamicist claims a heat pump has a COP of 1.7, operating between thermal energy reservoirs at 273 K and 293 K. Assess the validity of this COP claim. **Calculation and Analysis**: 1. **Maximum COP of a Reversible Heat Pump (COP_HP,max)**: - Formula: COP_HP,max = TH / (TH - TL). - Given temperatures: TH = 293 K, TL = 273 K. - Calculation: COP_HP,max = 293 K / (293 K - 273 K) = 14.65. 2. **Assessment of the Claim**: - Claimed COP: 1.7. - Since the claimed COP (1.7) is less than the maximum possible COP (14.65) for the given temperature limits, the claim is valid. **Key Formula**: - COP_HP,max = TH / (TH - TL). **Conclusion**: The thermodynamicist's claim of a heat pump with a COP of 1.7 is valid as it is less than the theoretical maximum COP for the specified operating conditions. --- Minimum Power Requirement of a Heat Pump** A heat pump is used to maintain a house at 20°C. The outdoor air temperature is -3°C, and the house loses heat at a rate of 75,000 kJ/h. Calculate the minimum power required (in kW) to operate the heat pump. **Calculations**: 1. **COP of a Reversible Heat Pump (COP_HP,max)**: - Formula: COP_HP,max = TH / (TH - TL). - Convert temperatures to Kelvin: TH = 20°C + 273 = 293 K, TL = -3°C + 273 = 270 K. - Calculation: COP_HP,max = 293 K / (293 K - 270 K) = 12.74. 2. **Minimum Power Input ( Ẇ net,in)**: - Formula: Ẇ net,in = Q ̇ H / COP_HP. - Given: Heat loss rate (Q ̇ H) = 75,000 kJ/h. - Convert heat loss rate to kW: 1 kW = 3600 kJ/h. - Calculation: Ẇ net,in = (75,000 kJ/h) * (1 kW / 3600 kJ/h) / 12.74 = 1.635 kW. **Key Formulas**: - COP_HP,max = TH / (TH - TL). - Ẇ net,in = Q ̇ H / COP_HP. **Conclusion**: The minimum power required to operate the heat pump under these conditions is approximately 1.635 kW.

Processes: - Isobaric process: P = constant -> W = mP(v2 - v1) - Isochoric process: v = constant -> W = integral of P - Isothermal process: T = constant -> W = PV ln(v2/v1). PV = mRT = c0 rho P = c/V - Adiabatic process: heat transfer Q = 0 (for closed systems) or Q with dot above = 0 (for control volumes) - Polytropic process: PV^n = constant (V: volume) Boundary work in a polytropic process PV^n = C, P = CV^-n W = (P2V2 - P1V1) / (1-n) W = P1V1 ln(v2/v1) = P2V2 ln(v2/v1) = mRT ln(v2/v1) (n = 1) (T in Kelvin) Solid and liquid: Specific heats: c = cp = cv Change in specific internal energy or specific enthalpy: h2 - h1 is approximately equal to u2 - u1 which is approximately equal to c(T2 - T1) Incompressible substance: Delta u is approximately equal to c_avg(T2 - T1) [kJ/kg] Delta h is approximately equal to Delta u + v Delta P is approximately equal to c_avg Delta T + v Delta P [kJ/kg] For Solids: Delta h is approximately equal to Delta u which is approximately equal to c_avg Delta T. For liquids, two special cases are commonly encountered: 1. Constant-pressure processes, as in heaters Delta P = 0: Delta h is approximately equal to Delta u which is approximately equal to c_avg Delta T 2. Constant-temperature processes, as in pumps Delta T = 0: Delta h = v Delta P h rho Delta T is approximately equal to hf rho Delta T + vf rho T(P - Psat at T) "Quality: x = Mass of vapor / Mass mix Specific volume: v = vf + xvfg = (1 – x)vf + xvfg, where vfg = vg – vf Specific internal energy: u = uf + xu(fg) = (1 – x)uf + xu(fg), where u(fg) = ug – uf Specific enthalpy: h = hf + xh(fg) = (1 – x)hf + xh(fg), where h(fg) = hg – hf Specific entropy: s = sf + xs(fg) = (1 – x)sf + xs(fg), where s(fg) = sg – sf Compressed liquid (when tables are not available) v is approximately equal to vf at T u is approximately equal to uf at T h is approximately equal to hf at T s is approximately equal to sf at T Ideal gas EOS: Pv = RT or PV = mRT (R: gas constant; V: volume; v: specific volume) (T in Kelvin) Specific heats: cp = cv + R [kJ/kgK], k = cp/cv, cp = kR / (k-1), cv = R / (k-1), cp = cv + R [kJ/ kmol-K] Specific enthalpy and specific internal energy: h = u + Pv = u + RT/v; u = cvT, h = cpT Change in specific internal energy (assuming constant cv) u2 – u1 = cv0(T2 – T1) [kJ/kg] Change in specific enthalpy (assuming constant cp0) h2 – h1 = cp0(T2 – T1) [kJ/kg] Compressibility factor: Z = Pv / RT; reduced pressure and temperature: Pr = P / Pc, Tr = T / Tc (T in Kelvin) Z = actual volume / ideal volume. Z = 1 for ideal gas Pv = RT or P = RT/v For control volumes: - Ein - Eout = Qin - Qout + Win - Wout - Delta ECV = Qnet,in - Wnet,out - Q = Qnet,in - Qnet,out - W = Wnet,in - Wnet,out With dot notation indicating per unit time: - E dot in - E dot out = Q dot in - Q dot out + W dot in - W dot out - Delta E dot CV = Q dot net,in - W dot net,out - Q dot = Q dot net,in - Q dot net,out - W dot = W dot net,in - W dot net,out Steady-flow process: - E dot in - E dot out = Q dot net,in - W dot net,out = Q dot - W dot (steady state) - m dot (h1 + 1/2 V1^2 + gz1) + Q dot = m dot (h2 + 1/2 V2^2 + gz2) + W dot - Q dot - W dot = m dot (h2 - h1 + 1/2 (V2^2 - V1^2) + g(z2 - z1)) - h2 - h1 + 1/2 (V2^2 - V1^2) + g(z2 - z1) = Q dot / m dot - W dot / m dot Energy, work and heat - Total energy: E = U + KE + PE = mu + 1/2 mV^2 + mgz (V: velocity) - Specific total energy: e = E/m = u + 1/2 V^2 + gz (V: velocity) - Specific internal energy: u = U/m - Specific heat transfer: q = Q/m - Boundary work: W = integral PdV (area under the P-V diagram; V: volume) - Specific work: w = W/m - Spring work: Wspring = integral Fdx = 1/2 K(x2^2 - x1^2) - Shaft power: W dot = T ω (T: torque) Energy balance for a constant-pressure expansion or compression process - General analysis for a closed system undergoing a quasi-equilibrium constant-pressure process. - Q is to the system and W is from the system. - Q - Wother = H2 - H1 [kJ] - Delta U + Wb = Delta H For Closed Systems - Ein - Eout = Qin - Qout + Win - Wout - Delta E sys = Qnet,in - Wnet,out = Q - W - Q = Qnet,in - Qnet,out - W = Wnet,in - Wnet,out Here's the text from the image without special characters, suitable for copying to a cheat sheet: E ﬃ ciency indicates how well an energy conversion or transfer process is accomplished. E ﬃ ciency = Desired output / Required input Combustion equipment e ﬃ ciency eta_comb.equip. = Q_useful / HV Heating value of the fuel: The amount of heat released when a unit amount of fuel at room temperature is completely burned and the combustion products are cooled to the room temperature. E ﬃ ciencies of Mechanical Devices - Mechanical e ﬃ ciency eta_mech = E_mech,out / E_mech,in - Pump eta_pump = E_mech,fluid / W_shaft,in - Turbine eta_turbine = W_shaft,out / Delta E_mech,fluid E ﬃ ciencies of Electrical Devices - Motor eta_motor = W_shaft,out / W_elect,in - Generator eta_generator = W_elect,out / W_shaft,in Energy - Overall e ﬃ ciency of a power plant: eta_overall = eta_comb.equip. * eta_thermal * eta_generator - Combined E ﬃ ciency - Pump-motor: eta_pump-motor = eta_pump * eta_motor - Turbine-generator: eta_turbine-gen = eta_turbine * eta_generator HV stands for Heating Value, Q represents heat, W for work, E for energy, and Delta indicates a change in the respective energy or work quantity. The eta ( η ) symbol denotes e ﬃ ciency, subscripted with the device type or process it's referring to. Steam power plant operating between 5 MPa and 10 kPa. Water is assumed incompressible. Turbine ad pump operate adiabatically and reversibly. **Given Data**: - Turbine Exit Pressure (P4) = 10 kPa. - Turbine Exit Specific Enthalpy (h4) = 2583.9 kJ/kg. - Pump Exit Pressure (P1) = 5 MPa. - Pump Exit Specific Enthalpy (h3) = 4608.1 kJ/kg. - Specific Volume at 10 kPa (vi) = 0.00101 m ³ /kg. **Objective**: Find the ratio of Turbine Work (WTurbine_out) to Pump Work (WPump_in). **Key Equations**: - Turbine Work: WTurbine_out = h3 - h4. - Pump Work: WPump_in = vi * (P1 - P4). - Work Ratio: WTurbine_out / WPump_in. **Solution**: - Turbine Work Calculation: WTurbine_out = 4608.1 kJ/kg - 2583.9 kJ/kg = 2024.2 kJ/kg - Pump Work Calculation: WPump_in = 0.00101 m ³ /kg * (5000 kPa - 10 kPa) = 5.041 kJ/kg. - Work Ratio Calculation: 2024.2 kJ/kg / 5.041 kJ/kg ≈ 402. **Conclusion**: The ratio of 402 indicates the turbine does significantly more work compared to the work input to the pump __ Cylinder with air, heated for 15 minutes at constant pressure. - Objective: To determine entropy change for both constant and variable specific heats. **Assumptions:** - Air is considered as an ideal gas. **Properties:** - Gas constant of air: R = 0.287 kJ/kg.K (from Table A-1). **Analysis:** 1. **Mass Calculation:** - Conditions: Pressure (P) = 120 kPa, Volume (V) = 0.3 m^3, Temperature (T) = 290 K. - Formula for mass (m): m = PV / RT. - Calculated mass: m = 0.4325 kg. 2. **Electrical Work:** - Power = 0.2 kJ/s. - Time = 15 minutes (900 seconds). - Total work (Welec) = 180 kJ. 3. **Energy Balance:** - Change in internal energy (Delta U) plus work due to heat (Wheat) equals change in enthalpy (Delta H) in a constant pressure process. 4. **Entropy Change with Constant cp:** - Average Temperature assumed: 450 K. - Final Temperature (T2): T2 = T1 + Welec / (m * cp_avg). - Entropy Change (Delta S): Delta S = m * cp_avg * ln(T2/T1) - R * ln(P2/P1). 5. **Entropy Change with Variable Specific Heats:** - Enthalpy Calculation: h2 = h1 + Welec / m. - Determine s2 value from air table. - Entropy Change (Delta S): Delta S = m * (s2 - s1). **Key Values:** - Initial Conditions: Volume = 0.3 m^3, Pressure = 120 kPa, Temperature = 17 degrees Celsius. - Use cp, h, s values from relevant tables for calculations. *** Air is preheated by hot exhaust gases in a cross-flow heat exchanger. The rate of heat transfer, the outlet temperature of the air, and the rate of entropy generation are to be determined. Assumptions: 1. Steady operating conditions exist. 2. The heat exchanger is well-insulated, so heat loss to the surroundings is negligible. Heat transfer from the hot fluid equals heat transfer to the cold fluid. 3. Changes in kinetic and potential energies of fluid streams are negligible. 4. Fluid properties are constant. Properties: - Specific heats of air and combustion gases: 1.005 and 1.10 kJ/kg°C, respectively. - Gas constant of air: R = 0.287 kJ/kg.K (Table A-1). Analysis: - System: exhaust pipes as a control volume. - Energy balance (steady-flow system): E_in - E_out = 0. - Rate of net energy transfer = Rate of change in internal, kinetic, potential energies. - E_in = E_out. - mh1 = Q_out + mh2 (since kinetic and potential energies ≈ 0). - Q_out = mc (T_out - T_in). Rate of heat transfer from exhaust gases: - Q = me (Cp_gas) (T_gas_out - T_gas_in) = (2.2 kg/s)(1.1 kJ/kg°C)(180°C-95°C) = 205.7 kW. Mass flow rate of air: - m_air = (P_air V_air) / (R T_air) = (95 kPa)(1.6 m3/s) / (0.287 kPa m3/kg K)(293 K) = 1.808 kg/s. Outlet temperature of air: - Q = m_air Cp_air (T_out - T_in), - T_out = T_in + Q / (m_air Cp_air) = 20°C + 205.7 kW / [(1.808 kg/s)(1.005 kJ/kg°C)] = 133.2°C. Entropy generation rate: - S_gen = m_exhaust Cp ln(T2/T1) + m_air Cp ln(T4/T3). - S_gen = (2.2 kg/s)(1.1 kJ/kg K) ln[(95+273)K/(180+273)K] + (1.808 kg/s)(1.005 kJ/kg K) ln[(133.2+273)K/(20+273)K]. - S_gen ≈ 0.091 kW/K." . **Acceleration:** - 1 m/s ² = 100 cm/s ² | 1 m/s ² = 3.2808 ft/s ² - 1 ft/s ² = 0.3048 m/s ² **Area:** - 1 m ² = 10^4 cm ² = 10^6 mm ² = 10^-6 km ² | 1 m ² = 1550 in ² = 10.764 ft ² - 1 ft ² = 144 in ² = 0.09290304 m ² **Density:** - 1 g/cm ³ = 1 kg/L = 1000 kg/m ³ | 1 g/cm ³ = 62.428 lbm/ ft ³ = 0.036127 lbm/in ³ - 1 lbm/in ³ = 1728 lbm/ft ³ - 1 kg/m ³ = 0.062428 lbm/ft ³ **Energy, heat, work, internal energy, enthalpy:** - 1 kJ = 1000 J = 1000 N·m = 1 kPa·m ³ | 1 kJ = 0.94782 Btu - 1 kJ/kg = 1000 m ² /s ² | 1 Btu = 1.055056 kJ - 1 kWh = 3600 kJ = 5.40395 psia·ft ³ = 778.169 lbf·ft **Force:** - 1 N = 1 kg·m/s ² = 10^5 dyne | 1 N = 0.22481 lbf - 1 kgf = 9.80665 N | 1 lbf = 32.174 lbm·ft/s ² = 4.44822 N **Length:** - 1 m = 100 cm = 1000 mm = 10^6 μ m | 1 m = 39.370 in = 3.2808 ft = 1.0926 yd - 1 km = 1000 m | 1 ft = 12 in = 0.3048 m - 1 mile = 5280 ft = 1.6093 km | 1 in = 2.54 cm **Mass:** - 1 kg = 1000 g | 1 kg = 2.2046226 lbm - 1 metric ton = 1000 kg | 1 lbm = 0.45359237 kg - 1 ounce = 28.3495 g - 1 slug = 32.174 lbm = 14.5939 kg - 1 short ton = 2000 lbm = 907.1847 kg **Power, heat transfer rate:** - 1 W = 1 J/s | 1 kW = 3412.14 Btu/h - 1 kW = 1000 W = 1.341 hp = 737.56 lbf·ft/s - 1 hp = 745.7 W | 1 hp = 550 lbf·ft/s = 0.7068 Btu/s - 1 boiler hp = 33,475 Btu - 1 Btu/h = 1.055056 kJ/h - 1 ton of refrigeration = 200 Btu/min **Pressure:** - 1 Pa = 1 N/m ² | 1 Pa = 1.4504 × 10^ − 4 psia - 1 kPa = 10^3 Pa = 10^ − 3 MPa = 0.020886 lbf/ft ² - 1 atm = 101.325 kPa = 1.01325 bars | 1 psi = 144 lbf/ ft ² = 6.894757 kPa - 1 atm = 14.696 psia = 29.92 in Hg at 30°F - 1 in Hg = 3.387 kPa - 1 mm Hg = 0.1333 kPa **Specific heat:** - 1 kJ/kg·°C = 1 kJ/kg·K = 1 J/g·°C | 1 Btu/lbm·°F = 4.1868 kJ/kg·°C - 1 Btu/lbmol·R = 4.1868 kJ/kmol·K - 1 kJ/kg·°C = 0.23885 Btu/lbm·°F **Specific volume:** - 1 m ³ /kg = 1000 L/kg = 1000 cm ³ /g | 1 m ³ /kg = 16.02 ft ³ /lbm - 1 ft ³ /lbm = 0.062428 m ³ /kg **Temperature:** - T(K) = T(°C) + 273.15 | T(R) = T(°F) + 459.67 = 1.8T(K) - Δ T(K) = Δ T(°C) | T(°F) = 1.8T(°C) + 32 - Δ T(°F) = Δ T(R) = 1.8 Δ T(K) **Velocity:** - 1 m/s = 3.60 km/h | 1 m/s = 3.2808 ft/s = 2.237 mi/h - 1 mi/h = 1.46667 ft/s - 1 mi/h = 1.6093 km/h **Volume:** - 1 m ³ = 1000 L = 10^6 cm ³ (cc) | 1 m ³ = 6.1024 × 10^4 in ³ = 35.315 ft ³ - 1 U.S. gallon = 231 in ³ = 3.7854 L **Volume flow rate** -1 m ³ /s = 60,000 L/min = 10 ⁶ cm ³ /s | 1 m ³ /s = 15,850 gal/min (gpm) = 35.315 ft ³ /s = 2,118.9 ft ³ /min (cfm)

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