The Clausius inequality holds good for Aany process Bany cycle Conly reversible process
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- As shown in the figure below, a reversible power cycle receives energy QH by heat transfer from a hot reservoir at TH and rejects energy Qc by heat transfer to a cold reservoir at Tc. Hot reservoir at TH Boundary- R Wa cycle Cold reservoir2c at Te (a) If TH = 1100 K and Tc = 400 K, what is the thermal efficiency? (b) If TH = 500°C, Tc = 20°C, and Wcycle = 1000 kJ, what are QH and Qc, each in kJ? (c) If n = 60% and Tc = 40°F, what is TH, in °F? (d) If ŋ = 40% and TH = 727°C, what is Tc, in °C?As shown in the figure below, a reversible power cycle receives energy QH by heat transfer from a hot reservoir at TH and rejects energy Qc by heat transfer to a cold reservoir at Tc. Hot reservoir at TH Qu Boundary- -Weyele Cold reservoir e at T (a) If TH = 1600 Kand Tc = 400 K, what is the thermal efficiency? (b) If TH = 500°C, Tc = 20°C, and Weycle = 200 kJ, what are QH and Qc, each in kJ? (c) If n = 50% and Tc = 40°F, what is TH, in °F? (d) If n = 40% and TH = 727°C, what is Tc, in °C?Problem 4.01. A carnot refrigerator (carnot cycle heat pump in reverse) operating between Th and Te is used to cool and freeze a bottle of water, volume V, at a temperature To < Th to freezing temperature T (known density Pw, heat capacity cw). (a) Find the work required to cool and freeze the water. (b) Find the change in entropy in the heat baths, and use it to place a limit on the change in entropy of the water (without calculating the entropy change in the water). The C.O.P. of a carnot refrigerator: KR= Qc = W Te Th-Te
- As shown in the figure below, a reversible power cycle receives energy QH by heat transfer from a hot reservoir at TH and rejects energy Qc by heat transfer to a cold reservoir at Tc. Hot reservoir at TH Boundary- Weycle Cold reservoir2c at Te (a) If TH = 600 Kand Tc = 400 K, what is the thermal efficiency? (b) If TH = 500°C, Tc = 20°C, and Weycle = 200 kJ, what are QH and Qc, each in kJ? (c) If ŋ = 50% and Tc = 40°F, what is TH, in °F? (d) If n = 40% and TH = 427°C, what is Tc, in °C?IS meve7STBle (If Te=TE a reversible process ) S,. S; = SS: Creversible adiabatic) the process 2-→ ź P=c = loos h 423 =0.0355 Ig1lg.c 408 If AS >o irreversible process > =0 reversible process for adiabatic o imposi ble process process H.w: 1. Caleuate the change of entropyot dng of aiv espording polytropically in acylindes belind apston from 6.3 bar and 550'e to 1.05 bor The inder of expanston is 1-3 . kg et Carbone dioxide (M=44) is compressed trom I bor d 15Ć until the presure is 8:3 bar and the volume is then o-0oum. calculate the change of entropy. Take : Cp= o-88 5 |2. o 05 University of Anbar College of Engineering Mechanionl Enginerng Depertment 2 Year (Sophomore year) THIERMODYNAMICS L ME2307 Instructor: Asst. Prof. Dr. Saad M. Jalil Page Na. (Sheet No. 6) Q1: A rigid cylinder containing o.o06ni of nitrogen at 1.04 bar, 15c, is heated reversibly until the temperature is 90'c. Calculate the change of entropy and the heat supplied Sketch the processon perfect gas- Ts diagram.…Q.: Define the following: 1. Entropy, 2. Carnot cycle efficiency 3. РМ)». 4. Reversible process 5. (C.O.P.)neat pump
- Discuss the importance of reversible processes. Says in engineering practice. Briefly discuss the major factors that irreversibility.Two reversible cycles are in series, each process doing the same net work, Wcycle. The first cycle receives energy QH by heat transfer from a hot reservoir at 1000°R and energy Q is reinjected by heat transfer to a reservoir at an intermediate temperature, T. The second cycle receives energy Q by heat transfer from the reservoir at temperature T and reinjects the QC energy by heat transfer to the reservoir at a temperature of 400°R. All energy transferred is positive in the direction of the arrow. Determine: a) the intermediate temperature T, in °R, and the thermal efficiency for each of the two cycles; b) the thermal efficiency of a simple reversible cycle operating between the hot and cold reservoirs at 1000°R and 400°C, respectively. Then determine the net work done by the simple cycle, expressed in terms of the net work done by each of the two cycles, Wcycle.GIVEN: Gas HAS CONSTANT MASS IDEAL GAS SPECIFIC HEAT RATIO OF GAS GAS CYCLE раз At 1: P1 At 3: At 2 + 2 = : = 2 BAR T2 = 12: 23: T3 300°K 31: VOLUME is Y = 1.6 ISENTRO Pic REVERSIBLE ISOTHERMAL EXPANSION CONSTANT T1 = 149.8 x 150⁰ K V₁ = √3 ASK: A) ENTROPY B) TOTAL ENTROPY VARIATION FOR V₁ = 0.0005 m m³ PRODUCTION EACH FOR CYCLE WHOLE CY C L F
- As shown in the figure below, a reversible power cycle receives energy QH by heat transfer from a hot reservoir at TH and rejects energy Qc by heat transfer to a cold reservoir at Tc. Hot reservoir at TH Boundary- Cold reservoir 20 at Te (a) If TH = 1100 Kand Tc = 400 K, what is the thermal efficiency? (b) If TH = 500°C, Tc= 20°C, and Weycle = 1800 kJ, what are QH and Qc, each in kJ? (c) If n = 70% and Tc= 40°F, what is TH, in °F? (d) If n = 40% and TH = 1027°C, what is Tc, in °C?H.W. 4.6 A steam power plant has steam entering at 70 bar, 450°C into HP turbine. Steam is extracted at 30 bar and reheated up to 400°C before being expanded in LP turbine upto 0.075 bar. Some portion of steam is bled out during expansion in LP turbine so as to yield saturated liquid at 140°C at the exit of open feed water heater. Considering HP and LP turbine efficiencies of 80% and 85% determine the cycle efficiency. Also give layout and T-s diagram.H.W. 4.6 A steam power plant has steam entering at 70 bar, 450°C into HP turbine. Steam is extracted at 30 bar and reheated up to 400°C before being expanded in LP turbine upto 0.075 bar. Some portion of steam is bled out during expansion in LP turbine so as to yield saturated liquid at 140°C at the exit of open feed water heater. Considering HP and LP turbine efficiencies of 80% and 85% determine the cycle efficiency. Also give layout and T-s diagram. Answers: Thermal efficiency = 38.99% !!