Homework_7 thermodynamics

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1 Student name : Shyanne Smith UIN: Date : 11/03/2023 Homework Assignment No. 7: Solution I. Indicate whether the following statements are true or false 1. Exergy is not conserved in an actual process. A: True 2. A closed system exergy balance consists of three major contributions: exergy change, exergy transfer, and exergy destruction. A: True 3. Exergy is destroyed by irreversibilities. A: True 4. The units of exergy are the same as those of energy. A: True Same because enthalkpy change is also a component of exergy transfer process

2 5. Exergy transfer accompanying heat transfer is a function of the environment temperature and the temperature of the boundary where heat transfer occurs. A: True 6. To define exergy, we think of two systems: a system of interest and an exergy reference environment. A: True 7. When a closed system is at the dead state, it is in thermal and mechanical equilibrium with the exergy reference environment, and the values of the system’s energy and thermomechanical exergy are each zero. B: False Thermodynamic energy is not zero because at the dead state there is still chemical exergy left. 8. The energy of an isolated system must remain constant, but its exergy can only increase. B: False Since entropy is produced in all actual processes, the only processes that can occur are those for which the entropy of the isolated system increases; this is known as the increase of entropy principle. 9. Exergy destruction is proportional to entropy production. A: True 10. Like entropy, exergy is produced by action of irreversibilities. B: False II. Checking understanding 1. When the exergy of a system increases, its state moves ____away_______the dead state. 2. The following term reduces to zero within the steady-state form of the closed system exergy rate balance. a) The time rate of the change of exergy within the closed system b) The time rate of exergy transfer by work

3 c) The time rate of exergy transfer by heat d) The time rate of exergy destruction 3. Which of the following are true statements about a system at the dead state? a) It is in thermal equilibrium with the environment. b) It is in mechanical equilibrium with the environment. c) Thevalueofthethermomechanicalcontributiontoexergyiszero. d) All of the above. 4. For a closed system, as the exergy reference environment temperature____increases______, the exergy destruction rate decreases. III. Refrigerant 134a initially at -36 °C fills a rigid vessel. The refrigerant is heated until the temperature becomes 25 °C and the pressure is 1 bar. There is no work during the process. For the refrigerant, determine the heat transfer per unit mass and the change in specific exergy, each in kJ/kg. Let T0 = 20°C, p0 = 0.1 MPa and ignore the effects of motion and gravity. Solution (use problem-solving techniques for this problem , KNWN, FIND, … ) KNOWN: T1= -36°C T2=25°C P2= 1bar ambient temperature=20°C ambient pressure= 0.1MPa FIND:heat transfer per unit mass and change in specific exergy SCHEMATIC/DATA: given in problem ENGINEERING MODEL: From the properties table of Refrigerant 134a at T2=25 ∘ C and P2=1 bar specific internal energy is u2=250.61 kJkg specific volume is v2=0.23783 m3kg specific entropy is s2=1.0976 kJkg ⋅ K ANALYSIS: Now, the volume and mass of refrigerant in the vessel remains constant so, specific volume remains constant initial specific volume is v1=v2=0.23783 m3kg From the properties table of Refrigerant 134a at T1= − 36 ∘ C: the specific volume of saturated liquid is vf=0.0007111 m3kg the specific volume of saturated vapor is vg=0.29740 m3kg the specific internal energy of the saturated liquid is uf=4.987 kJkg

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4 the specific internal energy of the saturated vapor is ug=209.68 kJkg The specific entropy of saturated liquid is sf=0.02137 kJkg ⋅ K The specific entropy of saturated vapor is sg=0.96319 kJkg ⋅ K here vf<v1<vg so, initially vessel contains a liquid-vapor mixture let the initial quality is =x1 initial specific volume is given by v1=vf+x1(vg-vf) 0.23783=0.0007111+x1 * (0.29740 – 0.0007111) x1= 0.79922 the initial internal energy is given by u1=uf+x1(ug−uf)= 4.987+0.79922×(209.68−4.987) u1=168.5817 kJ/kg the initial specific entropy is given by s1=sf+x1(sg−sf)= 0.02137+0.79922×(0.96319−0.02137) s1=0.77409 kJ/kg ⋅ K let the heat transfer per unit mass is =q From the first law of thermodynamics u2−u1=q−w 2 50.61−168.5817=q−0 q=82.0283 kJ/kg The change in specific exergy is given by Δ e= (u2-u1)+P0(v2-v1)-T0(s2-s1)= (250.61 – 168.5817)+0.1*10^3 * (0.23783 – 0.23783) – 293.15 * (1.0976 – 0.77409) Δe= -12.8087 kJ/kg IV. Figure below provides steady state data for the outer wall of a dwelling on a day when the indoor temperature is maintained at 25°C and the outdoor temperature is 35°C. The heat transfer rate through the wall is 1000 W. Determine, in W, the rate of exergy destruction (a) within the wall and (b) within the enlarged system shown on the figure by the dashed line. Let T 0 = 35°C. Solution (use problem- solving techniques for this problem, KNWN, FIND, …)

5 KNOWN: indoor temperature is maintained at 25°C outdoor temperature is 35°C The heat transfer rate through the wall is 1000 W FIND: Determine, in W, the rate of exergy destruction (a) within the wall and (b) within the enlarged system shown on the figure by the dashed line. SCHEMATIC/GIVEN DATA: ENGINEERING MODEL: Let T 0 = 35°C. The heat transfer rate through the wall is 1000 W ANALYSIS: A) Rate of entropy generation within the wall: Δ s= Δ Q/T1 – Δ Q/T2 = 1000(1/300 – 1/306) = 0.065 W/K = Δ s State of exergy destruction= Δ E= Δ s*T0= 0.065*308= 20.13

6 Δ E= 20.13 W b) within the enlarged system: Δ s= Δ Q/T1 – Δ Q/T2 = 1000(1/298 – 1/308) = 0.1089 W/K = Δ s State of exergy destruction: Δ E= Δ s * T0 = 0.1089 * 308 = 33.56 Δ E= 33.56 W

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