. (A). (C) 5. (C) 7. ( to 5) 9. (C) 6. (C). (C). (D) 6. (A) 8. (0.6 to 0.66) 50. (D) 6. (C). (A) 5. (C) 7. (A) 9. (C) 5. (D) 6. (C). () 6. (C) 8. (600) 0. (D) 5. (B) 6. (D) 5. (A) 7. (A) 9. (D). (C) 5. (A) 65. (D) 6. (C) 8. (C) 0. (C). (9 to ) 5. (D) 7. (D) 9. (A). (B). (.9 to.) 55. (A) 8. (5.8 to 5.9) 0. (B, C). ( to ). (0. to 0.6) 56. (C) 9. (D). (D). (B) 5. (A) 57. (B) 0. (A). (B). (D) 6. (0.88 to 0.9) 58. (D). (C). (A) 5. (A) 7. (88 to 9) 59. (C). (C). (B) 6. (C) 8. (50 to 5) 60. (D). (A) Carnot engine E d d c Similarly, d d c d d d c d c Hence decreasing has a higher eiciency.. (C). (A) (Note sign indicates only direction) at constant temperature ranserring units to zero ealuated cells ater reaching optimality gies an alternatie solution with the same transportation cost to the problem. I artiicial ariable which does not hae any physical signiicance is in solution then it is ineasible.. () 5. (A) 6. (C) 7. (D) COre. COH 5 Rerigerating eect COre. = k For minimizing mean tardiness EDD rule is used, or minimizing aerage in process inentory S rule is used. 8. (5.86 [5.8 to 5.9]) 9. (D) 0. (A). (C) x 600 00.567 60 x 5.86 In a regeneratie steam cycle, a part o steam is extracted rom the turbine and utilized to heat up condensate. So temperature at which eed water is supplied in boiler is high. As a result aerage heat addition temperature o steam is increased whereas aerage heat rejection temperature is same so oerall gape between aerage heat addition temperature and aerage heat rejection temperature increase, thereore, thermal eiciency o cycle increases. As we hae less amount o steam remaining in steam turbine at lower stages, the work output will decreases. Rerigerant nomenclature is gien as R m n p where m is number o carbon atoms
and. (C). (C). (D) 5. (C) 6. (C) 7. (A) 8. (C) 9. (A) n is number o hydrogen atoms p is number o luorine atoms R can be written as R0 thereore m 0 m and p Number o clorine atom = = C H F CHFCl emperature can be increased by supplying heat or by doing work. Extent o irreersibility o any process is determined by the entropy increase o the unierse. he mixture o air and liquid air is not a pure substance, because the relatie proportions o oxygen and nitrogen dier in gas and liquid phases in equilibrium. Extent o irreersibility o any process is determined by the entropy increase o the unierse. he exergy o a closed system is either positie or zero. (C) cannot be the alue o exergy. 0. (B, C). (D). (B). (A) n n Maximum useul work in a closed system U E E S S V V On rearrangement 0 0 U E S E S V V et Vapour Increase CO 0 0 0. (B) 5. (C) 6. (A) 7. (A) Aboe the Critical emperature Increase the CO Ammonia is toxic in nature Since electrical energy is high grade energy and heat is low grade energy. In 00K is gien to electric heater it would conert whole 00K into work to perorm the task but i 00K is gien to hearth it can not conert 00K to work 00K to work. In the context o question exergy input to electrical heater is more than exergy input to the hearth but exergy output is same. Hence second law eiciency o hearth is more than second law eiciency o electric heater. So rom the aboe reason we can conclude that electrical heater is more irreersible. 8. (600) Loss in aailable energy = Irreersibility 000 000 I 0 S 00 uni 00 000 9. (D) 0. (C) I 600kJ Reheating S Rankine cycle with Reheating 7 06K 7 7K C R - C Q For Carnot Rerigerator Q=.6 k
. (B) Q CO 7.6 06 7 0.k 00 Heat lost by hot water mc p mc p hot water 60 60 60 60 0 0C K S S sys+ S gen S S = Heat gained by cold water surr cold water S 0, surr sys gen perect insulation mc ln mc ln i i.8 ln.8 ln 0 5. J / K. ( [ to ]) Johnson's rule says to identiy the shortest processing time. I it is at the irst work center, place (the entire) job as early as possible. I it is at second work centre, place the job as late as possible. Eliminate that job rom urther consideration and apply the decision rule to the remaining jobs. Following the algorithm, we get B E D A C he sequential times are as ollows. B E D A C Baking 6 7 5 8 Decoration 9 Job Baking Decoration In Out In Out B 0 E 0 D 0 7 7 A 7 7 0 C 0 So, the minimum total elapsed time or optimum sequence in hours is.. (B). (D) Applying st law o thermodynamics or closed system Q U Q mc d 0.9 0.77 55 0.6kJ Actual work =.6kJ Reersible work == change in aailable energy U U S S 0 here, S S mc ln Reersible work 0.9 0.77 55 0 7 0 0.9.6..kJ State- 500 C 600ka State- 90ka 7 55 0.77 ln 7 0 For highest possible elocity at nozzle exit, system, must be reersible. S gen 0 S S C p ln R ln 90 5.9 ln.07 ln 77 600 S S S S gen sur system gen 90 0 5.9 ln.07 ln 0 77 600 6.7K 89.7C Now steady low energy equation between & is h h C 5. (A) V 000 C V 000 V 5.9 500 89.7C 000 V 06 m / s
By steady low energy equation or unit mass Q = heat transerred per unit mass. = work done per unit mass 5 bar bar Q dq ds (i) O 0. 0.86 (m /kg) ds dh dp Q h h dp (ii) From equation (i) & (ii) 6. (C) V dp V Now dp = Area under - cure projected on axis 0.86 0. 5 00 kj / kg V V 0 0 7000 j / kg 7 kj / kg 7 kj / kg kj / kg (by compressor) dq du + d dq 0 d 0 du 0 mcd 0 constant (insulated) (ree expansion) V S C ln R ln V Air S R ln V constant V V V h Q h R ln 0.87 ln 0.98 kj / kg. K Membrane 7. (. [ to 5]) From st law o thermodynamics Q U F.B.D. o piston A mg A o 0 9.8 o A i A where, Inside air pressure i o o i Outside air pressure Outside air pressure A 0. Soling, we get 90.5a i ia oa mg ork done by gas in raising piston by 0. m 90.5 0. 0. 76.7J Q U U 00 76.7.J 8. (0.65[0.6 0.66]) 9. (C) For imum work min 00 Brayton Cycle 0.7 07 min 00 Carnot 0.7 07 Now Brayton Cycle Carnot 0.7 0.65 0.7 For imum work engine should be reersible Q 0 600 Q 0 500 00 Q 0K 600 0 80K II law 0. (D) actual 00 II law 00 8% 80 Gien: dq 0.8 kj / s dt d 0. kj / s dt 500K HE Q 00K 600 kj/s
. (C) otal heat transer in 0 min dq 0 60 kj 0.8 0 60 0kJ dt otal work done in 0 min d 0 60 0. 0 60 6kJ dt Applying irst law o thermodynamics or closed system. dq d du mc i 0 6 60 0.77 5 6.C ME ork done per cycle stroke olume V V V V. (0 [9 to ]) m i.e. B I F I I d I 0 0.8 I 00 k I B I F 00 0 80k Brake power 80 pth 0% Heat supplied 00. (.06 [.9 to.]) Speciic humidity 0.6 because air is saturated].7 0.6 00.7 t is negatie.7ka 0.006 kg o water apor/kg o dry air otal mass o water apour = 0.006 00 =.06 kg. (0.5 [0. to 0.6]) ' ' h h h x h h h u p Now, h c p s mass o dry air c ( p is small or liquid phase) Similarly h ' c 5. (A) c.5 55 x 0.5 h 0 g eaporator 80 08 80 5.5K 57.7C 0.. 6. (0.9 [0.88 to 0.9]) 0.6 atm 0.0 0.6 0.5.065 0.0 0.6 0.6.065 a.065. ka 0.6 Rs Ra a atm artial pressure o air a a s atm 0.87 5 7 0.87 08 0.5. 98. 0.9 m / kg d. a 7. (90.5 [88 to 9]) From data gien h 9.5 kj / kg hs 89. kj / kg h 07. kj / kg compressor s c h s h h h actual
s h h s h h comp 89. 9.5 9.5 0.85 9.5 58.6 97.86 kj / kg Condenser load in / = 97.86 07. = 90.5 kj 8. (50.8 [50 to 5]) s kj kg h h 5C.698ka sat sat sat ;0.5.698.589ka speciic humidity 0.6.589.6 0.5.589 0.6.589 99.70 kg Vap 0.00988 kg dry air h enthalpy o air sts 0 0 0 C p h a c C g p.005 5 0.00988 500.88 5 0 5.5 5.6 50.8 kj / kg dry air Vc / n Vs /.5 V V 8 00 50. (D) V 58cm Now, V V 00 58 58 57cm Now olumetric eiciency, V V 57 78.6% V 000 s Combustion equation can be written as: C H ao.76an CO O CO x 5. (D) y Equating moles or Nitrogen.76a 8 a.00 8N bh O Oxygen.00 b b.0 Carbon x Hydrogen y b.0 Gien equation becomes CH Z x 5x Z 0,0 0 x (0,5) (0,) 9. (C) Vs 000cm V Vc 0.05 Vs 00cm n n V V Z 0 0, 0 5 0 6 5,0,0,0 5 x
Z x 5. (B) 6 5 5 6/5,0 5 0 0, x Z 0 5 0 In case o degenerate solution one o the decision ariable will get zero alue and one o the constraint becomes redundant type howeer it is ully consumed or utilized. Z x 0x Statement: x 6x 8 x x 8 x ; x 0 x, x 6 optimal 0,9 0,8 x x x 8 Z x 0x,6 Z C x C x x C Z x C C y mx c C slope, m C,0 x 9,0 x 6x 8 Slope o the objectie unction 0 Slope o constraint 6 Slope o constraint C C C 0 Fix C 0 C ; C 0 80 6.67 5. (A) C ; 0 C 0 Range o alues o coeicient Cast matrix is to be minimized x 6.67 to 0 5 8 0 5 5 6 5 6 7 0 8 9 9 0 Row ransaction 0 9 6 0 0 9 Column ransaction 9 0 0 0 0 9 0 5. (D) H F F Solution = 5 + + 6 + 7 =9 No. o lines = No. o rows so, solution is optimal are Houses 6 8 00- Supply 00- F 6 00 00 F 8 8 50+ 00 50+ F 5 6 80 0 60 50 Demand 80 60 60 00 600 600 he eect o this on transportation cost is 55. (A) 56. (C) 57. (B) 58. (D) ransport cost increases by `. Aerage in-process inentory = (5 jobs st hour + jobs nd hour + jobs rd hour + jobs 5 th hour + job 5 th hour)/( + + + + 5) 5 5 5 AI. 5 5 ankaj SE 5 km 0 km ankaj
59. (C) 60. (D) 6. (C) 6. (C) otal age o students and principals is 5 5 = 75 Ater excluding the principal's age, the total age o students become = 6 age o principal = 75 6 age o principal = 75 6 = 9 A V S...() A V 9 V S 8 A V S 7...() Equation () () gies V now, A 5 5 Salary o A is 59 5 he ratio o the number o coins is : : or the 50 paise and `.50 coins, respectiely. In terms o monetary alue, the ratio becomes 0.5 : 0.5 :.5 hich equals 0.5 : 0.5 :.5 or :: 9 th o the total alues comes rom 5 paise coins (i.e.) 6600 = `600 is in the orm o 5 paise otal number o 5 paise coins 600 00 0.5 he basic diagram or the gien statements is Females Football players = Ground Males From the aboe basic diagram Choice (a), does not ollow Choice (b), does not ollow Choice (c), ollow Choice (d), not Hence, option (c) is correct. 6. 0.5(A + B + C) = 50% o the work. 6. (D) 65. (D) Means A, B and C can do the ull work in hour. hus, (A + B + C) = 00% From this point it is better to sole through options. Option (C) gies the correct answer based on the ollowing thought process. I c = 50% work per hour, it means C takes hours to complete the work. Consequently, A would take hours and hence do.% work per hour. Since, A + B + C = 00%, this gies as B's hourly work rate = 6.66%. For this option to be correct these numbers should match the second instance and the inormation gien there. According to the second condition: A + B should be equal to 00%. utting A =.% and B = 6.66% we see that he condition is satisied. Hence, this option is correct. 9 In 99, 995 569.05 00 6 In 96, 80.68 00 569.05 Required robability 00 80.68 = 8. = 9% A = Getting a Leap year B = Getting 5 Sunday in a year 5 ; A 00 A C Required probability A B A C B B A A C A A B C 5 7 7 8 Selections