Answer Key THERMODYNAMICS TEST (a) 33. (d) 17. (c) 1. (a) 25. (a) 2. (b) 10. (d) 34. (b) 26. (c) 18. (d) 11. (c) 3. (d) 35. (c) 4. (d) 19.

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1 HERMODYNAMICS ES Answer Key. (a) 9. (a) 7. (c) 5. (a). (d). (b) 0. (d) 8. (d) 6. (c) 4. (b). (d). (c) 9. (b) 7. (c) 5. (c) 4. (d). (a) 0. (b) 8. (b) 6. (b) 5. (b). (d). (a) 9. (a) 7. (b) 6. (a) 4. (d). (d) 0. (d) 8. (d) 7. (c) 8. (d) 5. (a) 6. (d). (a) 4. (a). (d). (b) 9. (b) 40. (d)

2 HERMODYNAMICS ES. (a).(b) () kg 4atm 40 C v m R kg.atm () 4 v R...(i) v m R. v R...(ii) Dividing (i) by (ii) K 7 C 600 ka 00K v 0.5m 00K (Isothermal) v 0.m w dv Isothermal process V C, C/ w w C V dv C n V V V n V n kj ( )ve sign indicates work in done on the system.(d) From st law of thermodynamic 4.(d) W du t [0.8 ( 0.)] mc ( 5) 6.44 C o W du v (Since internal energy for air (ideal gas) is function of temperature alone). W W 0 kj/s kj/h 5.(b) W 0 v c 6. (a) Conservation of mass m m m 4 + m m 7kg/min m m Conservation of energy 7. (c) m m m m C From, Refrigerant -4 (a) - emperature table at 0.4 Ma, Sat. vapour state, h 9.6 kj/kg at 0.9 Ma, 60 C, h 95. kj/kg From steady flow energy equations mh mh w 0.08(9.6). 0.08(95.) w w kw ( )ve sign implies compression 8.(d) From R-4 (a) - emperature able at.6 Ma sat H C 0.88K at 0. Ma sat L 0.09 C 6.9K

3 HERMODYNAMICS ES 9.(a) (CO) max L H.868 (CO) max Heat removed ower input.868 max max.6 kw L 00K max ower output, w mc p ( ) w kw w.76 MW. (c) ds as o f F Cpn R n i E W. (a) s Rn R n 9K 0 C 78K 5 C E W 0. (d) 00K Let the intermediate temperature be. I efficiency of first engine. I II I II K Maximum power output is obtained when gas expands isentropically. For ideal gas K 7.76 C Entropy Balance. (d) S S gen S S gen S S S gen W/K For reversible adiabatic process of ideal gas. s s s.667 for argon gas Cp 0.5 kj/kgk s K Isentropic work produced mc p s ( ) 68. kw

4 4 HERMODYNAMICS ES Actual work produced isentropic wisentropic 4. (d) kw Minimum power input mr n i f (7 + 5) n kw ( )ve sign indicates power input to compressor 5. (a) 0. (b). (a). (d) For isothermal process, C V V constant C d dv C V d dv V V V N C Boiling point Freezing point 00 0 m kg/s i 88 K e 88. K Let the reading on new scale corresponding to 60 C be x, then From entropy balance, i e gen 0 88 K m S S S S m s s gen e i i o ds cv dt e 88. mc n 4.n kw/k Rate of exergy destruction is pipe, 6. (d) 7. (c) 8. (d) 9. (b) I 0 S gen kw Maximum amount of electric power generated/ exergy mgh ,800 kw H U + V dh du + dv + vd dh ds + vd [ for a closed and reversible process, ds du + dv]. (a) 60 0 x x x 80 N At steady state, the internal energy of the resistor and hence its temperature is constant. So, by first law, W he flow of current represents work transfer. At steady state the work is dissipated isothermally into heat transf er to the surroundings. Since the surroundings absorb unit of heat at temperature, S surr W At steady state, S sys 0 W Suniv Ssys Ssurr Here, 00 K i Rt S surr W J 00

5 HERMODYNAMICS ES 5 4. (a) Ssys Ssurr Suniv J/K S sys 00 ( ) 5 kj/k S surr kj/k Ssys Ssurr Suniv 9. (a) Earlier, temperature 7 C 00 K ressure p bar Volume V 0 litres Assuming the volume to remain constant at temperature C 70 K Hence, p p 5. (a) 6. (c) 7. (c) Hence, this is reversible process. Actual dryness fraction x (x ) (x ) where x dryness fraction of steam in seperating calorimeter 0.9 and, x dryness fraction of steam entering throttling calorimeter 0.95 x Work transfer in free expansion is zero, because in free expansion, the pressure against which the gas expands is zero. Hence, work V kw 00 K 0. (d). (d) Efficiency 00 p 70 p.8 bar net work done heat input area enclosed in S diagram heat input Since the chamber is rigid, so volume is constant, HE 00 K W output. (b) Hence work done W 0 he chamber is insulated, So, 0 From first law of thermodynamics, du W du 0 8. (b) s f.6 kj/kjk h fg 800 kj/kg kW 00 s g (entropy of saturated vapour) hfg 800 sf.6 6.kJ/kgK 500. (d) 4. (b) Unavailable work For A B, V For For kj constant B C, V constant C A, constant V constant

6 6 HERMODYNAMICS ES net work done heat input (c) 6. (b) s s h c s s h 0.5 Absolute pressure Gauge pressure + Atmospheric pressure or Absolute pressure bar Initial value V 0.0m Final volume V 0.06m h h c 8. (d) 9. (b) kW 580 Heat rejection W kw Entropy represents the degree of randomness. Increase in entropy means degradation of energy or decrease in available energy. 500 B C Constant pressure 6 Ma 0 a 500 A D Work done W (V V ) F E 4 5 Work done, W area of ABCD 0 kj Heat absorbed 84 kj According to first law of thermodynamics, U W BC AD kJ 84 U 0 Heat rejected, area of ADEF 7. (b) U KJ Change in internal energy U 54 KJ AD AF kj Efficiency W W Maximum efficiency 87 K HE 9 K W 40. (d) For an isothermal process, change in internal energy U 0 From first law of thermodynamics, U W For isothermal process, W 00 kw 9 87 W

Consequences of Second Law of Thermodynamics. Entropy. Clausius Inequity

Consequences of Second Law of Thermodynamics. Entropy. Clausius Inequity onsequences of Second Law of hermodynamics Dr. Md. Zahurul Haq Professor Department of Mechanical Engineering Bangladesh University of Engineering & echnology BUE Dhaka-000, Bangladesh zahurul@me.buet.ac.bd