MAE 320 Thermodynamics HW 4 Assignment

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1 MAE 0 Thermodynamics HW 4 Assignment The homework is de Friday, October 7 th, 06. Each problem is worth the points indicated. Copying of the soltion from any sorce is not acceptable. (). Mltiple choice (0 points) (a). Which statement is inalid regarding a closed system? (A) Energy can transfer across the bondary of the closed system. (B) The mass of the closed system remains constant (C) The olme of the closed system always remains constant. (D) The pressre of the closed system may change with time (b). Which statement is alid regarding a polytropic process? (A) The pressre of the closed system always remains constant dring a polytropic process. (B) The olme of the closed system always remains constant dring a polytropic process. (C) The temperatre of the closed system always remains constant dring a polytropic process. (D) Heat may transfer across the bondary of the closed system dring a polytropic process. (c). Which statement is inalid regarding a specific heat: (A) c and c p are properties. They are intensie properties. (B) c and c p are applied to any sbstance ndergoing any process. (C) c is related to the changes in internal energy and c p to the changes in enthalpy. (D) c is always greater than c p (E) c and c p are temperatre-dependent for any sbstance. (d). An 8 L of piston-cylinder system contains air at 00 kpa and 00 K. Now air is compressed isothermally to a olme of L. The bondary work dring this compression process is closest to (A). kj (B).8 kj (C) -. kj (D) kj (E) 9. kj V W b PV ln 00kPa 8 0 m ln( ). kj V 8 (e) 0.0 kg of orange is cooled from 0 C to 5 C in a refrigerator. The amont of heat transferred from the orange to the refrigerator is closest to (Hint: aerage specific heat of oranges can be obtained from a Table in Appendix ) (A) 7.40 kj (B).5 kj (C) 4.85 kj

2 (D) kj (E).50 kj From Table A-c to find ot C. o o Q U mc ( T T ) 0.kg (.75kJ / kg C) (5 0) C. 5kJ (f) A closed system ndergoes the series of qasi-eqilibrim processes shown here. The bondary work done dring the whole process is closest to (A) (B) (C) (D) (E) 40 kj 40 kj 40 kj 40 kj 680 kj Recall that the bondary work integral is the area nder qasi-eqilibrim processes on a P-V state diagram. Hence, W 00 (0.7-0.) + 00 (0.5-0.) / + 00 ( ) 40 kj.

3 . A piston-cylinder deice contains. kg R-4a at 4 bar and 50 C. Now R-4a is cooled at constant pressre ntil its temperatre drops to 0 C ( points). (a) draw a T- diagram, label the satrated temperatre, the specific olmes and the temperatres of the initial and final states (b) determine the bondary work done dring this expansion process, in kj. (c) calclate the amont of heat that transferred into the piston, in kj (a) Sate : at P 400 kpa, T sat, 8.9 o C from Table A- Since T > T sat,, it is sperheated apor. From Table A-, m /kg, kj/kg Sate : From Table A-, m /kg, 5.97 kj/kg 50 o C 0 o C 8.9 o C T ( o C) (m /kg) (b) A closed system ndergoes a constant pressre process. Therefore, W b P( V V ) mp( ) W b.kg (400kPa)( ) m / kg.95kpa m. 95kJ (c). For a closed system, the energy balance eqation is: Q W U U m( ) b Q Wb + m( ).95kJ +.kg ( ) kj / kg 8. kj The heat transfered into the system is -8.kJ. The negatie sign indicates the direction of the heat transfer is opposite to the assmed heat transfer into system.

4 4. An ideal gas within a piston-cylinder assembly ndergoes a thermodynamic cycle consisting of the following three processes ( points): Process : Compression with PV constant; from P bar, V.6 m to V 0. m ; U U. Process : Constant pressre expansion process to V V. Process : Constant olme. U - U -549 kj. The changes in the kinetic energy and potential energy can be ignored. (a) Sketch the cycle on a P-V diagram (b) Calclate the bondary work and the net heat transfer dring Process, in kj (c) Calclate the bondary work and the net heat transfer for Process, in kj (d) Calclate the bondary work and the net heat transfer for Process, in kj (a) See P-V diagram on the right V 0. (b) W PV ln 00kPa.6m ln( ). 7kJ V.6 Since Q W U 0, Hence Q W. 7kJ (c) Based on the ideal gas eqation: P V PV C P P V / V ) 00kPa(.6 / 0.) 800kPa ( PVC For a constant process: W P ( V V ) P ( V V ) 800kPa (.6 0.) m 0kJ Since Q W U, Hence Q W + U ) ( U For a cycle, U 0, that is, ( U ) + ( U ) + ( U ) 0 Since U 0, Hence, ( U U ) ( U ) 549kJ Q 0kJ + 549kJ 4669kJ (d) For a constant olme process, W 0 Q W U 549kJ 4

5 5. As shown in the figre, a piston-cylinder assembly whose piston is resting on a stop contains 800 L of helim gas, initially at 00 kpa and 5 o C. The pressre of helim needed for the piston to start to moe is 50 kpa. (0 points). (a) Calclate the mass of helim in the cylinder (b) Assming a constant specific heat of helim dring this process, determine the heat gain or loss needed for this piston to start moing (State ), in kj. (a). R can be obtained from Table A- consider he is idea gas: P V (00 kpa)(0.8 m ) m RT (.0769 kpa m /kg K)(98 K) 0.9 kg Base on the state eqation of ideal gas: PVmRT T T P / P ) 98K(50 /00) 745K ( (b). For a closed system, the energy balance eqation is: Q Wb U m( ) The specific heat of helim can be obtained from Table A- The bondary work is zero. Q m o ) mc ( T T ) 0.9kg (.56kJ / kg C)(745 98) K kJ ( 5

6 6. Steam at 75 kpa and 8 percent qality is contained in a spring-loaded piston-cylinder deice with an initial olme of m. Steam is now heated ntil the pressre and olme rise to 5 kpa and 5 m. Dring this process, the pressre increases linearly with change in the olme (6 points) (a) Calclate the mass of steam contained in the cylinder (b) Determine the internal energy at the initial and final states. (c) Calclate the work done dring the process, in kj (d) Determine the heat transfer dring the process, in kj Assmptions The cylinder is stationary and ths the kinetic and potential energy changes are zero. There are no work interactions inoled other than the bondary work. The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is qasieqilibrim. Analysis We take the contents of the cylinder as the system. This is closed system since no mass enters or leaes. The energy balance for this stationary closed system can be expressed as (a). The initial state is satrated mixtre at 75 kpa. The specific olme and internal energy at this state are (Table A-5), f + x (0.08)( ) fg The mass of water is V m m. kg 0.78 m /kg 0.78 m /kg (b). the initial internal energy: f + x fg (0.08)(.8) 55.0 kj/kg The final specific olme is V m m m. kg The final state is now fixed. Since f << g, its is a mixtre of two phases. x g f f /kg The internal energy at this specific olme and 5 kpa pressre is + x 50.47kJ / kg kJ kg f g / kj/kg (c). Since this is a linear process, the work done is eqal to the area nder the process line -: P + P (75 + 5)kPa kj W b,ot Area ( V V) (5 )m 450 kj kpa m 6

7 (d) For a closed system: Q in W b,ot Q in U m( W b,ot + m( ) Sbstitting into energy balance eqation gies Qin b,ot ) (since KE PE 0) W + m( ) 450 kj + (. kg)( ) kj/kg,750 kj P 5 kpa 75 kpa 7. Describe the operating principle and the major components between any two ot of the following types of refrigerators inclding Compressor refrigerators, Peltier refrigerators, Solar refrigerators, Thermal mass refrigerators, Magnetic refrigerators and Acostic refrigerators. Yo mst inclde one that ses a refrigerant, and another that does not se any refrigerant. Please describe the history of the deelopment of refrigerants sed for gas-compressor refrigerators, and the state-of-the-art refrigerant. The essay shold cite at least three references, mst be typed in single-space, times new roman font, point, no shorter than one page (5 points). 7

Chapter 4 ENERGY ANALYSIS OF CLOSED SYSTEMS

4- Chapter 4 ENERGY ANALYSIS OF CLOSED SYSTEMS Mog Bondary ork 4-C It represents the bondary work for qasi-eqilibrim processes. 4-C Yes. 4-C The area nder the process cre, and ths the bondary work done,