Conventional Question Practice Program - IES Masteriesmaster.org/public/archive/2016/IM-1462627832.pdfConventional Question Practice Program ... the process of steam generation. The

Office : F-126, Katwaria Sarai, New Delhi-110016 (Phone : 011-41013406, 7838813406, 9711853908)

Website : www.iesmaster.org E-mail: [email protected]

Conventional Question Practice ProgramDate: 26th March, 2016

1. (b)

2. (a)

3. (c)

4. (b)

5. (a)

6. (c)

7. (b)

8. (a)

9. (c)

10. (b)

11. (d)

12. (c)

13. (b)

14. (d)

15. (a)

16. (b)

17. (c)

18. (d)

19. (b)

20. (c)

21. (d)

22. (b)

23. (d)

24. (b)

25. (c)

26. (b)

27. (d)

28. (a)

29. (b)

30. (d)

31. (a)

32. (a)

33. (a)

34. (b)

35. (b)

36. (d)

37. (b)

38. (a)

39. (a)

40. (c)

41. (c)

42. (c)

43. (c)

44. (d)

45. (b)

46. (d)

47. (b)

48. (c)

49. (d)

50. (a)

51. (c)

52. (d)

53. (b)

54. (d)

55. (b)

56. (b)

57. (c)

58. (c)

59. (d)

60. (a)

61. (a)

62. (c)

63. (d)

64. (d)

65. (a)

66. (c)

67. (d)

68. (a)

69. (c)

70. (a)

71. (c)

72. (a)

73. (a)

74. (a)

75. (b)

76. (b)

77. (b)

78. (a)

79. (b)

80. (d)

81. (a)

82. (b)

83. (b)

84. (d)

85. (a)

86. (d)

87. (c)

88. (b)

89. (a)

90. (a)

91. (a)

92. (b)

93. (c)

94. (b)

95. (b)

96. (d)

97. (a)

98. (a)

99. (c)

100. (b)

101. (a)

102. (b)

103. (d)

104. (d)

105. (a)

106. (a)

107. (a)

108. (b)

109. (c)

110. (a)

111. (b)

112. (c)

113. (a)

114. (d)

115. (c)

116. (a)

117. (d)

118. (d)

119. (a)

120. (a)

ANSWERS

IES M

ASTER

(2) ME (Test-8), Objective Solutions, 26th March 2016

Sol–1: (b)Primary boiler heat transfer surfaces in-clude economiser, evaporator section andsuperheater.

Sol–2: (a)

b = 92%

m = 94%

g = 95%

c = 44%

a = 6%Overall efficiency,

0 = b m g c a(1 )

0 = 0.92 × 0.94 × 0.95 × 0.44× (1 – 0.06)

0 34%

Sol–3: (c)Boiler mountings are fittings for the safetyof the boiler, and for complete control ofthe process of steam generation. Themountings form an integral part of theboiler and are mounted on the body of theboiler itself.Ex- Safety valve, water level indicators,pressure gauge, fusible plug, steam stopvalve, feed check valve, blowoff cock.Boiler accessories are installed to increasethe efficiency of the steam power plants.Ex-Air preheater, Economiser, superheater,feed pump, Injector.

Sol–4: (b)The difference between the temperatureof feed water outlet and saturationtemperature of steam entering the heateris called terminal temperature difference.

Sol–5: (a)I D fans are used to exhaust the productsof combustion from the boiler.Maintaining balanced draft conditions inthe furnace improves boiler operation andprovides energy to move the flue gases atthe velocities needed for good heat transfer.I D fans are subjected to more severeconditions than forced draft fans, because

they must handle larger volumes of gasat high temperatures and containing ashparticles. The physical characteristics ofI D fans must therefore blades are notrecommended for I D fan service.Backward inclined fans are acceptable fornon-abrasive gas service, while radial orradial fans are recommended for abrasiveservice. The higher temperature of gaseshandled by the I D fans sometimes makesit necessary to use water-cooled bearingsto prevent overheating. Inlet dampercontrols or variable speed drives are usedto control I D fan capacity.

Sol–6: (c)It can theoretically be proved that equalincrease in enthalpy of feedwater amongall the heaters and the economizers resultsin maximum efficiency. If n is thenumber of heaters, then rise intemperature in each feed heater and

economizer will be T

n 1 and the total

temperature rise in the feedheater is hT

=

n Tn 1

Effect of number of heaters onincremental gainn hT Incremental gain0 0 –

1 T2 T

2

2 T23 2 T T T=

3 2 6

3 T34 3 T 2 T T=

4 3 12

It is seen from the above table that moreis the number of heaters, more is the risein feedwater temperature and hence moreefficiency. But this also means lessutilization of waste heat from flue gas inthe economizer and hence less boilerefficiency. Morever, incremental gain intemperature decreases with increase innumber of heaters, resulting indiminishing returns while the first costincrease therefore, one has to make

IES M

ASTER


judicious choice for n.In steam turbine plant with very highpressure range, bleeding of steam may benecessary to reduce the volume at thelow pressure end. It is because, as thesteam expands from high initial pressureto low condenser pressure, the volume ofsteam becomes too large due to the lowpressure. Handling such large volume ofsteam would require large flow area,hence bleeding can reduce the volume flowat the low pressure end.

Sol–7: (b)Stirling: Bent water tubeCochran: Vertical fire tubeLancashire: Horizontal fire tubeBenson: Once through flow

Sol–8: (a)An economiser is the heat exchangerwhich raises the temperature of thefeedwater leaving the highest pressurefeedwater heater to about the saturationtemperature corresponding to the boilerpressure. This is done by the hot fluegases exiting the last superheater orreheater at a temperature varying from370°C to 540°C. By utilizing these gasesin heating feedwater, higher efficiency andbetter economy can be achieved.Fusible plug is a device used to put offfire in the furnace of the boiler when thelevel of water in the boiler falls to anunsafe limit.Superheater is a device used to increasethe temperature of saturated steamwithout raising its pressure.

Sol–9: (c)Steam power plant has efficiencies ofaround 35 to 42%. Gas turbine powerplant has efficiency is the range 28 to33%. Diesel engines also have efficiencyin the range of 35 to 42%. Combined gasturbine (Brayton cycle) and the steamturbine power plant (Rankine cycle)complement each other and thus efficiencycan rise upto 60%. The Brayton cycle hasa high source temperature and rejects

heat at a temperature that is convenientlyused as the energy source for the Rankinecycle plant.

Sol–10: (b)The process of removing non-condensablegases from the steam cycle is calleddeaeration.

Sol–11: (d)The orsat gas analyser gives compositionof O2, CO2 and CO.N2 is found by % N2 = 100 – (% O2 + %CO + % CO2)

Sol–12: (c)Steam drum stores the steam generatedin water tubes and acts as separator forsteam/water mixture. In benson boiler,the boiler pressure is raised to criticalpressure and hence there is no formationof bubbles, hence steam drum is notrequired.

Sol–13: (b)The flame does not leap out of theinspection doors on the walls of the boilerbecause the pressure inside is negative.

Sol–14: (d)The flue gas passes through the stages ofsuperheater, and then economizer. In theair preheaters, the flue gases transfertheir residual heat to the combustion air,during which they are cooled to the exitflue gas temperature of the steamgenerator. For further cleaning, the fluegas is conducted through an electrostaticprecipitator to remove dust.

Sol–15: (a)

Rate of air consumption am = 80kg/s ×11kg/kg of coal = 880 kg/s

Density of air a = 1.2 kg/m3

Pressure developed = 1.2 × 103 × 9.81 Power of FD fan

= 3880 .12 10 9.81

1.2 0.8

= 10.97 MW

IES M

ASTER


Sol–16: (b)Height of hot gas column producingdraught in chimney of height ‘H’,

hg = g

0

TmH 1m 1 T

...(i)

where ‘m’ is mass of air to burn one kgof fuel. Tg is average temperature of hotthe gases in chimney and ‘T0’ is ambienttemperature.The mass flow rate of the gases throughchimney,

gm = 20 g g

m 1 1cm 1 T T T

where c is constant.For maximum discharge,

g

g

dmdT

= 0

This gives,

Tg = 0m 12T

m

From equation (i)

hg = 02TmHm 1

0

m 1T

1

m

= H (2 – 1)= H – Height of chimney

Sol–17: (c)Slight superheating of vapour isdesirable at the suction of compressor, oth-erwise liquid refrigerant would wash awaythe lubricant oil of compressor thus increas-ing wear & tear.

Sol–18: (d)In AJM, as the distance between the nozzletip and the work surface increases, thematerial removal rate first increases thenbecomes stable and after that decreases.

Sol–19: (b)In the 3-2-1 principle of fixture design,pins are located at three perpendicularprimary datum face.

Sol–20: (c)Fixtures reduces setting time. It cannotinterfere with operation.

Sol–21: (d)ECM has the highest material removalrate, while LBM has lowest materialremoval rate with high penetration rates.LBM < EBM < USM < EDM < ECM

Sol–22: (b)USM is mainly used for machining brittlematerials which cannot be machined byECM.While ECM and EDM require commonproperties of material to be machined.

Sol–23: (d)Force fit (interference fit) on H7 - S6hole-shaft pair system.

Sol–24: (b)Density of powder metallurgicalcomponents is higher than the castedproduct. It is due to the force at the timeof compaction and sintering process. Thereis no such force in the casting operation.

Sol–25: (c)Lower limit of hole

= 30.01 + 0.010 × 2= 30.03 mm

Upper limit of hole= 30.5 + 0.015 × 2= 30.08 mm

Dia of hole = 0.080.0330 mm

Sol–26: (b)In trepanning, the cutting tool producesa hole by removing a disc-shaped pieceusually from flat plates. A hole is pro-duced without reducing all the materialremoved to chips, as in case of drilling.Trepanning process can be used to makedisks upto 150 mm in diameter from flatsheet or plate.

Sol–27: (d)Grinding process is a finishing process andconsumes high specific energy as the metalremoval is less.

IES M

ASTER


Sol–28: (a)A hob is a surface in which involute rackis wound round the cylinder in a helix,like a worm.Gear Hobbing is gear generation processwhich produces more accurate gears thenmilling because the rotation of hob is likea continuous indexing operation whichcauses the theoritical rack to move alongstraight line.

Sol–29: (b)Plate thickness,

L = 25 mmSpeed of drill,

N = 300 rpmFeed, f = 0.25 mm/revolutionNo. of revolution required to drill the hole

= Lf = 100 revolution

Time taken to drill,

T = 100300 × 60 sec

T 20 sec.

Sol–30: (d)Machinability of steels is improved by theaddition of sulphur, silicon, phosporous etc.

Sol–31: (a)

OB

CA

( )

Shear strain , that the material undergoes

= AB AO OBOC OC OC

= cot tan

Sol–32: (a)

Sol–33: (a)

O

CB

AXd

O´

B´ c

Feed per tooth

BC =2 2

2 2 D DOB OC = d2 2

= d D d

Sol–34: (b)Orthogonal cutting:Cutting edge remains perpendicular to thecutting velocity. Direction of chip flow ve-locity is normal to the cutting edge. Angleof inclination and chip flow velocity are zero.Cutting edge is larger than the width ofcut. Only two mutually perpendicular com-ponents of cutting forces are acting on thetool cutting edge. The maximum chip thick-ness occurs in the middle. The tool pre-pares a surface which is parallel to thework surface.

Sol–35: (b)Sol–36: (d)

ChipTool

FSFH

NS

FV WorkpieceF

N

R

Sol–37: (b)Broaching is a multiple-tooth cutting op-eration with the tool reciprocating. Inbroaching, the machining operation is com-

IES M

ASTER


pleted in a single stroke as the teeth onthe cutting tool, called broach, areat gradually increasing heightcorresponding to the feed per tooth of amilling cutter.

Workpiece

BroachCut

Sol–38: (a)The chip velocity VC is the velocity of thechip relative to the tool and directed alongthe tool face. The shear velocity VS is thevelocity of the chip relative to the workpieceand directed along the shear plane. Thesetwo velocities along with the cutting velocityV would form a closed triangle as shown infigure.

VC

VS

V

From this we can get,

V

sin 90 = S CV V

sin 90 sin

VC =

V sin

cos

VS =

V cos

cos

Sol–39: (a)51 (Manufacturer’s option of nature ofabrasive)A (Type of abrasive) – Al2O3

36 (Grain size) – mediumL (Grade) – MediumT (Structure) – CloseR (Type of bond) – Rubber23 (Manufacturer’s reference)

Sol–40: (c)Sol–41: (c)Sol–42: (c)

Hammer blow is the maximum magnitudeof the unbalanced force along theperpendicular to the line of stroke.

Hammer Blow= 2B b

Sol–43: (c)This problem can be solved without use ofvapour absorption refrigeration system.

Q2 = 100 W

250 K 400 K

R

300 K

Q2 Q3

Q1

1 2

3Using Clausius inequality,

QT < 0

For minimum value of heat required thesystem should be reversible.

QT = 0

32 QQ250 400

= 1Q300 ................(1)

But,Q2 + Q3 = Q1 ................(2)From equation (1) and (2),

3Q 80 W and 1Q 180 W

Alternate Method:For minimum value of heat required, theCOP of vapour absorption system shouldbe maximum.

(COP)max=400 300 250

400 300 250

1 54 = 2Q

Work input

2

3

QQ = 5

4

Q3 = 2Q 45

IES M

ASTER


Sol–44: (d)In exponential smoothing forecast,

Ft = Ft – 1 + (Dt – 1 – Ft – 1)Where, is smoothing constant and isequivalent to 'n' period moving average.

= n

n n 12 =

2n 1

Since, 'n' varies from 2 to (but forextreme case n = 1). Therefore values of varies from 1 to 0.The value of close to 1 has less smoothingeffect and gives higher weightage to recentdata, while value of close to 0 has greatersmoothing effect and d is less responsiveto recent data.

Sol–45: (b)Given, average portion down time = 0.2P = 0.2No. of observation, n = 100For proportion down-time interval attributecontrol P-chart can be usedStandard deviation,

= P 1 P

n

= 0.2 0.8100

= 0.04For 95% confidence levelUpper control limit = P + 2

UCL = 0.2 + 2 × 0.04

UCL 0.28Lower control limit = P – 2

LCL 0.12So, the interval is (0.12, 0.28)

Sol–46: (d)ST = N.T + Allowance

= R.F. × O.T + Allowance= 1.2 × 2 + 0.1 × ST

ST = 2.40.9

ST 2.67 minutes

Sol–47: (b)C-chart are used to measure the no. ofdefects. It follows poisson’s distribution.

Sol–48: (c)Given,Demand,

D = 1000 units/yearOrdering cost,

Co = ` 100/unit–yearCarrying cost,

Cc = ` 100/unit–yearStock out cost,

Cb = ` 400Optimum level of stock out condition,

Q* = o b

c b c

2C D CC C C

Q* = 2 100 1000 400100 400 100

*Q 40 units.

Sol–49: (d)Lead timeDemand

80100120140

Probability

0.200.250.300.25

Cumulative Prob.Or service level

0.200.450.751.0

Expected value of lead time demand,

x = 80 0.20 100 0.25 120 0.30 140 0.251.0

= 112

Reorder level

= 1.25 × x

= 1.25 × 112

= 140

IES M

ASTER


ROL

140x = 112

From the above table probability,

P(x < 140) = 1

Hence, service level is 100%.Sol–50: (a)

MAD =500 600 680 600800 600 900 600 680 170

4 4

BIAS = 500 600 680 600800 600 900 600

4

= 100 80 200 3004

= 480 1204

Sol–51: (c)A, B & C analysis, which is a method forselective inventory management usesannual usage value of items to classifythem into A, B or C classes.Annual Usage value = Annualconsumption of item × item cost.A class items account for 70% of annualusage value.B class items account for 20% of annualusage value.C class items account for 10% of annualusage value.

Sol–52: (d)For short term forecasts, time seriesmethods like the simple moving average,weighted moving average and exponentialsmoothening are used. Regression analysisis used for mid term forecasts, such asthose required for sales planning, productionplanning, inventory budgeting and so on.Long range forecasts rely more onqualitative techniques like Delphi andmarket research.

Sol–53: (b)Sol–54: (d)

At EOQ,annual ordering cost = annual carrying costAnd annual carrying cost = 4 × 500 = Rs.2000.

Sol–55: (b)Payback period: It refers to the period oftime required to recoup the funds expendedin an investment, or to reach the break-even point. The time value of money is nottaken into account. Payback periodintuitively measures how long somethingtakes to “pay for itself”.Net present value: NPV of a time seriesof cash flows, both incoming and outgoing,is defined as the sum of present values ofthe individual cash flows of the same entity.It measures the excess or shortfall of cashflow in present value terms, above the costof funds. Each inflow/outlfow is discountedback to its present value. Therefore, NPVis the sum of all terms,

NPV =

tt

R1 i

t = time of cash flowi = discount rateRt = net cash flow, at timeInternal rate of return: It is a rate ofreturn to measure and compare theprofitability of investments. It is also calleddiscounted cash flow rate of return. It canalso be defined as the discount rate at whichthe present value of all future cash flow inequal to the initial investment.

Sol–56: (b)Material usage variance is the differencebetween the standard quantity of materialsthat should have been used for the numberof units actually produced and the actualquantity of material used, valued at thestandard cost per unit of the material.Standard material cost is Rs. 20 @ Rs. 10per kg, hence the standard quantity of

material = 20 2 kg=10

IES M

ASTER


Actual material used = 1.8 kgHence, difference = 2 – 1.8 = 0.2 kgStandard rate = Rs. 10 per kg Variance = 0.2 × 10 = Rs. 2A favourable material usage variancesuggests efficient utilization of material.Hence the answer is Rs. 2/- favourable,since the actual material used is less.Material usage variance = (Actual quantity– Standard quantity) × standard price

Sol–57: (c)Inventory Turnover Ratio (ITR)

=

Cos of goods consumed/soldduring the period

Average inventory hold during theperiod

Thus, ITR indicates the number of times theaverage inventory is consumed andreplenished. A high value of ITR indicatesthat the material in question is a fast movingone. A low ITR indicates over-investment andlooking up of working capital on undesirableitems.Raw material inventory turnover ratio

=Material consumed

Average raw material inventory

Sol–58: (c)To produce ultra low temperature, theevaporator temperature will be ultra lowand hence the corresponding pressure willalso be low. This increases the pressureratio of vapour compression cycle andhence this results in decrease in volumetricefficiency of the compressor. To avoid this,cascade refrigeration system is usedinstead of single stage vapour-compressionrefrigeration system.

Sol–59: (d)The curve showing power consumption ofcompressor and evaporator temperature,

Wc

B A

X

Te=Tmin Te=Tamb

Power consumed by compressor,

= rm h

As evaporator temperature reduces mrreduces due to reduced volumetricefficiency of compressor and h increase.Hence rm and h are contradictory.

AX-is pull down period. The design pointof any refrigeration system is in betweenX and B for stability reasons.

Sol–60: (a)Points regarding vapour absorption (VAS)refrigeration system. VA systems operates on low grade

energy (heat) whereas vapourcompression (VCS) refrigerationsystems operates on high grade energy.

In VA systems compressor is replacedwith generator, absorber and a pump.

Pump is required to circulates solution(Liquid) in various components sorequire very small or negligible work.

VA systems are used where largewaste heat is available

Solar refrigeration systems are basedon VA cycle.

Sol–61: (a)In electrolux refrigerator, NH3 used asrefrigerant and H2O used as absorbent.Hence, ammonia is absorbed in water.Hydrogen remains always in gaseous formso no question of evaporation.

Sol–62: (c)The ozone which is situated in stratosphereabsorbs harmful ultra voilet radiation andprotect us from these radiation. Thechlorine atom which is present in

IES M

ASTER


refrigerant attacks ozone and thus deplectsits thickness.

Sol–63: (d)Generator : In vapour absorption system,heat is given in generator and vapour ofrefrigerant (ammonia) remove from strongsolution of aqua-Ammonia.

Analyser : Analyser is attached to thegenerator has buffle plates, some watervapour particles escape with ammoniavapour leaving the generator. These watervapour fall on buffle plates and gettransformatic into water which fall backto generator. Hence dehydration takeplace.Rectifier : Water tracks in ammoniavapour leaving the analyser removescompleting and produced dry ammoniavapour.Receiver: It stores high pressure liquidammonia coming out from condenser.

Sol–64: (d)

p

4 4 1

233

h

Through undercooling, refrigeratingeffect increases while there is no effect onpower required.

COP =refrigerating effect

power required

thus, COP increasesSol–65: (a)

T

s

T =T2 33 2

14

s =s4 3 s =s1 2

COP of the refrigeration system workingon reversed Carnot cycle

(COP)R = 1

2 1

TT T

where T1 = lower temperatureand T2 = higher temperatureThus, COP of the reversed Carnot cycledecreases on(i) increasing the higher temperature and(ii) decreasing the lower temperature

Sol–66: (c)Flash chamber increases the refrigeratingeffect. The improvement in COP dependson the choice of intermediate pressure, theflash chamber pressure for given condenserand evaporator pressure. There exists anintermediate pressure pi at which the CoPis maximum.

Sol–67: (d)Sol–68: (a)Sol–69: (c)

(COP)R = (COP)HP–1 = 4 –1 = 3

(COP)R =heat extracted 3=power input

heat extracted = 3 × power input =3 × 3 = = 9 kW

= 9 × 60 = 540 kJ/minSol–70: (a)Sol–71: (c)

h1 = 180 KJ/Kgh2 = 210 KJ/Kgh3 = 80 KJ/Kg = h4

p

A

3 2

1

h

COP =refrigerating effectcompressor work

= 1 4

2 1

h hh h

=

180 80 100= 3.33210 180 30

IES M

ASTER


Sol–72: (a)For liquid vapour regenerative heatexchanger,

1 1h h = 3 3h h

= 4 4h h

1 1h h = 4 4h h

1 4h h = 1 4h h

p

h

3 2

14

3

4 1

Thus, for system A, heat extracted = 1 4h h

For system B, heat extracted = 1 4h h ,Since the vapour is not being superheatedin the evaporator and is rather beingsuperheated in the liquid vapourregenerative heat exchanger.Hence, since other conditions i.e. powerrequirement for the compresor is same, so

(COP)A = (COP)B

Sol–73: (a)The pressure in capillary tube decreasesdue to two factors:(i) The refrigerant has to overcomefrictional resistance offered by tube walls(ii) The liquid refrigerant flashes intomixture of liquid and vapour as its pressurereduces. The density of vapour is less thanthat of the liquid. Hence, the averagedensity of refrigerant decreases as it flowsin the tube. The mass flow rate and thetube diameter and hence area beingconstant, the velocity of refrigerantincreases as m VA . The increase invelocity or acceleration of the refrigerantalso requires pressure drop.

Sol–74: (a)Automatic expansion valve maintainsconstant evaporator pressure regardless ofthe load on the evaporator. It is used with

dry expansion evaporators where the loadis relatively constant.

Sol–75: (b)Sol–76: (b)

Copper and copper-bearing materials areattached by ammonia, so they are not usedin ammonia refrigeration piping system.

Sol–77: (b)(COP)VA < (COP)SE < (COP)VC

Sol–78: (a)Heat generated by resister,

= I2 R= 102 × 23= 2.3 kW

Since, tank is well insulated and heattransfer to system is boundaryphenomenon. So,

Q 0

+ –Since, electricity is flown by electronmovement which crosses the boundary.Hence, these electron movement i.e.electricity does work on system i.e.negative work.

W 2.3 kW

Q = U W

U = – W

U 2.3 kW

Sol–79: (b)For a reversible process centropy change

S = QT

2

V1

1

V

P

V2

P1

P2

IES M

ASTER


We know that for reversible isothermalprocess.

Q = W

= P1V1 ln 2

1

VV

= mRT1 ln1

2

PP

S =1

QT

= mR ln 1

2

PP

1

2

PS mR nP

l

Sol–80: (d)The internal energy of system is two typese.g macroscopic and microscopic. Themacroscopic energy consists of kinetic andpotential energy of molecules. So it ismolecular mass dependent of molecules.The microscopic energy consists of energyat molecular level e.g. molecular rotation,vibration, etc and energy at nuclear levelis also involved. The rotational andvibrational energy also depend uponmolecular structure e.g. inertia ofmolecules depends upon how the atomsare arranged.

Sol–81: (a)If a process is irreversible, then thereshould be finite gradient. The meaning offinite gradient is- The process is occurringacross finite difference. This difference maybe temperature, pressure, concentration,potential etc., from above discussion, it isclear that for reversible process, theremust be equilibrium at all time in allparameters.In question heat transfer due to finitetemperature difference is thermalinequilibrium. Free expansion is pressureinequilibrium. The current flows due toelectrical potential difference i.e. electricalpotential inequilibrium.

Sol–82: (b)1. The assumption in ideal gas model

(PV = RT) is - There is no interactionbetween molecules and volume occupiedby molecules is negligible as comparedto total volume of gas system.

2. At very low pressure and hightemperature, all gases and vapourapproaches to ideal gas behaviour.

3. Since internal energy of ideal gastemperature in function of temperatureand enthalpy is given as

h = u + pV

pV = RT

h = u + RT= f(T) + RT= f(T)

Sol–83: (b)The maximum shear stress theory(sys = 0.5 syt) is the most conservativetheory out of various other theories offailure. It gives the over safe anduneconomical results. While, maximumdistortion energy theory (sys = 0.577 syt)gives the most accurate results (resultsare very close to experimental results). Itgives the safe and economical results.

Sol–84: (d)A. Maximum principal stress theory

2

1

B. Maximum shear stress theory

2

1

IES M

ASTER


C. Maximum octahedral shear stresstheory

2

1

D. Maximum shear strain energy theory

2

1

Sol–85: (a)As per strain energy theory

2 2 21 2 3 1 2 2 3 3 12 2

yf

given 1 = 2 = 3 =

and yf = y

2 23 2 (3 ) 2y

y

3 1 2

Sol–86: (d) According to max. shear stress theory

FOS =

yf2

Absolute max. shear stress

= 1 2 1 2

2802

p p p pMax. , ,2 2 2

= 140100 40 100 40max , ,

2 2 2

= 140100 40 100 40max , ,

2 2 2

= 14070 = 2

Sol–87: (c)Two reversible adiabatics can neverintersect each other, otherwise it will

violate the second law of thermodynamics.Irreversibility I = T0Sgan.

For any process, dSgen = dQdST

Sol–88: (b)

W1 = 21

1

TQ 1T

= 1 1 2

1

Q T TT

W2 = 32

2

TQ 1T

= 312

1 2

TQ T 1T T

because 1 2

1 2

Q QT T

W2 = 2 321

1 2

T TTQT T

=

1 2 3

1

Q T TT

Now, T1 – T2 = T2 – T3

Hence, W1 = W2

T1

T2

T1

HE

Q2

Q1

W1

Q2

HE W2

T3

Q3

Tds = dh – vdPSince pressure is constant, so dp = 0 Tds = dh

dhds = T

Thus, as T increases, dhds will increase.

Sol–89: (a)At steady state, the internal energy ofthe resistor and hence its temperature isconstant. So, by first law,

W = Q

The flow of current represents worktransfer. At steady state the work isdissipated isothermally into heat transferto the surroundings. Since the

IES M

ASTER


surroundings absorb Q unit of heat attemperature T,

surrS = Q WT T

At steady state,

sysS = 0

univS = sys surrWS ST

Here, T = 300 KQ = i2Rt = 0.52 × 100 × 1800

surrS = Q WT T

= 20.5 100 1800

300 = 150 J

univS = sys surrS S

= 0 + 150 = 150 JSol–90: (a)

sysS = 100 × (0.35 – 0.30) = 5 kJ/K

sysS = 75 – 80 = – 5 kJ/K

unirS = sysS + suvvS = 5 – 5 = 0

Hence, this is reversible process.Sol–91: (a)

Actual dryness fraction × = (x1) (x2)where x1 = dryness fraction of steam inseperating calorimeter = 0.9and, x2 = dryness fraction of steamentering throttling calorimeter = 0.95 x = 0.9 × 0.95 = 0.855

Sol–92: (b)Area under T-s diagram represents heattransfer during the reversible process.Since temperature and entropy bothincrease during the process, hence, heattransfer is positive and heat is added orabsorbed.

Sol–93: (c)S + L

S + V

L + V VS

LT

s

Triplepoint line

Critical state

Sol–94: (b) U = speed ratio × 2gH

= 0.48 2 9.81 256

= 0.48 16 19.62 = 0.48 × 16 × 4.43

DN60 = 34.022

D = 34.022 60630

= 1.03 m

Sol–95: (b)

(a) Load factor = AverageloadMax. load in given

time period

(b) Capacity factor = Max. load

Plantcapacity(c) Diversity factor

= Sum of individual maximum demands

Simultaneous maximum demands

(d) Plant use factor = Maximum demand

Station capacity

Sol–96: (d)A foreway in a hydel system is providedat the junction of the power channel andthe penstock to store the water temporarilywhich is rejected by the plant when theelectrical load is reduced or to meet theinstantaneous increased demand of waterdue to sudden increase in load. Generalsequence of units in hydel system are

Reservoir Gated opening like sluice gate Canal, tunnel power channel

Power Penstock Intake structure Foreway/surge tank

Tail race River

Sol–97: (a)

Discharge coefficient, 3Q

ND = constant

Q N

IES M

ASTER


and Q D3

Centrifugal pumps can be classified on thebasis of specific speed performance ofdifferent types of pump can be comparedon the basis of specific speed so it formsthe basis for selection of centrifugal pump.

To eliminate cavitation in centifugal pumpThoma’s Cavitation Factor , should beless than crit ical value c. So it can notbe eliminated simply by avoiding sharpbends lowering velocity in suction pipe etc.

Sol–98: (a) Efficiency of centrifugal pump is less than

reciprocating pump In reciprocating pump discharge is

fluctuating and pulsating Reciprocating pumps are suitable for high

heads and low discharges For negative slip to occur, actual discharge

should be more than theoretical discharge,means coefficient of discharge should bemore than 1.

Sol–99: (c)

nd =

2 2i exit

f

2i

V Vh

2gV2g

= 3 .2 .43

= 2.43 = 0.8

Sol–100: (b)The motion equation given in question,

16x 5x 4x = 0

2

2d x dx16 5 4x

dtdt = 0

2

2md x dxc kx

dtdt = 0

The damping ratio,

=c

C C=C 2 km

=5 5

162 4 16

Sol–101: (a)

Due to non-homogeneity of shaft material,the centre of gravity will not lie on axisof rotation and cause deflection duringrotation due to centrifugal force. Butmisalignment of bearing does not deviatethe centre of gravity from axis of rotation.

Sol–102: (b)Sol–103: (d)

When n

2,

then transmissibility ratio

> 1 for all values of damping factor c

CC

This means that the force transmitted tothe foundation through elastic support isgreater than the force applied.

When n

2,

then transmissibility ratio

< 1 for all values of damping factor c

CC

This shows that the force transmittedthrough elastic support is less than the ap-plied force. Thus vibration isolation is pos-

sible only in the range of n

2

Sol–104: (d)

Primary force = 2mr cos

Primary couple = 2mr cos

Secondary force =

2mr cos2n

Secondary couple =

2mr cos2n

An in-line four-cylinder four-stroke enginehas two outer as well as two inner cranksin line. The inner cranks are at 180° tothe outer throws. Thus, the angular posi-tions for the cranks are: for first,

IES M

ASTER


(180 ) for the second, (180 ) for thethird and for the fourth. Primary force =

2mr [cos cos(180 )

cos cos ] 0=180Primary couple =

2 3 1mr cos cos 1802 2

1 cos2

3 cos 0=180 2

Secondary force =

2mr cos2 cos 360 2n cos 360 2

cos2 =

24mr cos2n

Secondary couple =

2 3 1mr cos2 cos 360 22 2

1 3cos cos2360 22 2

= 0

Sol–105: (a)For the transverse vibration of a shaftcarrying several loads, there are twomethods of finding natural frequency of thesystem: Dunkerley’s method and energymethod. Dunkerley ’s method givesapproximate results but is simple. This isused when the diameter of the shaft isuniform. The energy method gives accurateresults but involves heavy calculations incase there are many loads. This is alsoknown as Rayleigh’s method.

Sol–106: (a)The major shortcoming of a fire-tube boileris that definite size and pressurelimitations are inherent in its basic designi.e. the maximum size of the unit andthe maximum operating pressure arelimited. The tensile stress on the drumwall is a function of the drum diameterand the internal pressure given by

= pd2t

where = tensile stressp = gauge pressured = internal diameter of shellt = thickness of wall

Creep mechanism is elongation understress over a period of time, usually atelevated temperatures.

Sol–107: (a)The leaked air in the condenser resultsin increased back pressure on the primemover which means that there is loss ofheat drop and this reduces the output ofthe plant and consequently the thermalefficiency of steam power plant is lowered.

Sol–108: (b)As load increases, the demand for steamincreases, hence the fuel and air flow ratesincrease. Thus, the combustion productsmass rate of flow increases, whichincreases the convective heat transfercoefficient.Alternatively, as demand for steamincreases with increase in load, fuel andair flow and hence, combustion gas floware increased. The convective heat transfercoefficients (hi and ho) increase both insideand outside the tubes, increasing theoverall heat transfer coefficient (Uo)between gas and steam faster than theincrease in mass flow rate of steam alone.The combustion temperature do notmaterially change with load. Thus, thesteam receives greater heat transfer perunit mass flow rate, and its temperatureincreases with load.

Radiant Superheater

Convective SuperheaterCombined Superheater

Steam flow, percent

Stea

m-o

utle

t tem

pera

ture

IES M

ASTER


Sol–109: (c)Atomization involves mechanicaldisintegration of molten metal into fineparticles by means of a jet of compressed air,inert gases or water.

Sol–110: (a)Sol–111: (b)

The economic lot size depends on the setupcost. If the setup time can be reduced, thelot size can be reduced. The economic lotsize is not more than a direct relationshipbetween the inventory cost and setup cost.The effect of setup cost decreasesexponentially according to the increase inbatch size. Therefore, in order to calculatethe economic lot size, it is supposed that thesetup cost is constant i.e. the setup time isconstant. This traditional hypothesis is basedon a constant setup time i.e. it is not possibleto reduce the setup time. More often thannot, though, setup time can be reduced. Asetup cost decreases, the economic lot sizewould also decrease until, the unit productlot size is reached.

Customer satisfaction depends on variousfactors such as the quality of product,meeting due dates, price, volume andflexibility.

Sol–112: (c)Throttling is a constant enthalpy process,in which h1 = h2. Whether the temperatureand internal energy change in throttlingprocess depends on whether the fluidbehaves as an ideal gas or not. Since theenthalpy of an ideal gas is a function oftemperature alone, hence, T1 = T2 forthrottling process for ideal gas.For an ideal gas, throttling process takesplace at(i) constant enthalpy(ii) constant temperature(iii) constant internal energyThus, for real gas, after throttling changein internal energy is not zero, but is givenby 2 1u u = 1 2p p .A reversible cycle consist of reversible

processes only. Since VCS consists ofirreversible processes, viz. throttling, henceit can not be reversible. For a cycle to bereversible all the processes should bereversible.

Sol–113: (a)T2

Q2W R,HP

Q1T1

(COP)HP = (COP)R + 1

(COP)HP = 2 1R

Q Q; COPW W

where W = Q2 – Q1

Sol–114: (d)The state of a pure substance gets fixedif two independent properties are given.In the two-phase region, pressure andtemperature are dependent on each other.Thus, to fix the state, apart from eithertemperature or pressure, one moreproperty, such as specific volume,enthalpy or steam quality is required.

Sol–115: (c)All the three phases co-exist at the triplepoint.

Sol–116: (a)Sol–117: (d)

Specific speed is defined as the speed of anideal pump geometrically similar to theactual pump, which when running at thisspeed will raise a unit volume, in a unitof time through a unit of head. It is usedas a criteria for the :(i) Selection of shape of the pump curve

(ii) Determination of efficiency of thepump

(iii) Prediction of NPSH requirements(iv) Selection of lowest cost pump for their

application.Sol–118: (d)

In the vapour absorption systems usingwater- lithium bromide, water is used as

IES M

ASTER


refrigerant and a solution of lithiumbromide in water is used as absorbent.Since water is used as refrigerant usingthese systems, it is not possible to providerefrigeration at sub-zero temperatures.Hence, it is used only in applicationsrequiring refrigeration at temperaturesabove 0°C. Hence, these systems are usedfor air conditioning appl ications.In ammonia based vapour absorptionsystem, ammonia is used as refrigerant andwater as absorbent. Further, the Li-Br-water systems operate under very low(high vacuum) pressures, the ammonia-water system is operated at pressures muchhigher than atmosphere.

Sol–119: (a)Direct costs are those for activities or ser-vices that benefit specific projects e.g. sala-ries for project staff and materials requiredfor a particular project. Because these ac-tivities are easily traced to projects andtheir costs are usually charged to projects

on an item-by-item basis. Indirect cost ismore difficult to assess than the direct cost.Indirect costs are those for activities orservices that benefit more than one project.Their precise benefits to a specific projectare often difficult or impossible to trace.For example, it may be difficult to deter-mine precisely how the activities of thedirector of an organization benefit a spe-cific project. The actual practice of divid-ing up or allocating the indirect costs is anapproximate and time-consuming activity.

Sol–120: (a)The distinguishing feature of diamond toolsis their hardness, which is greater than thatof any other material. It is also chemicallyinert and has high thermal conductivity.However, oxidation of diamond starts at about450°C and thereafter, it can even crack. Forthis reason, the diamond tool is kept floodedby the coolant during cutting, and light feedsare used.

* * * * *

Documents

Conventional Question Practice Program - IES Masteriesmaster.org/public/archive/2016/IM-1462627832.pdfConventional Question Practice Program ... the process of steam generation. The