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8190046)-1-T.(E)(A.)TT1350(E)(M31)TAPRIL EXAMINATIONNATIONAL CERTIFICATEPOWER MACHINES N6(8190046)31 March 2016 (X-Paper)09:00–12:00REQUIREMENTS: Steam Tables (BOE 173)Calculators may be usedThis question paper consists of 7 pages and 1 formula sheet of 6 pages.Copyright reserved

8190046)-2-T1350(E)(M31)TDEPARTMENT OF HIGHER EDUCATION AND TRAININGREPUBLIC OF SOUTH AFRICANATIONAL CERTIFICATEPOWER MACHINES N6TIME: 3 HOURSMARKS: 100NOTE:If you answer more than the required FIVE questions, only the first fivequestions will be marked. All work you do not want to be marked must beclearly crossed out.INSTRUCTIONS AND INFORMATION1.Answer ANY FIVE questions.2.Read ALL the questions carefully.3.Number the answers according to the numbering system used in this questionpaper.4.Questions may be answered in any order, but subsections of questions mustbe kept together.5.ALL formulae used must be written down.6.Show ALL the intermediate steps.7.Questions must be answered in BLUE or BLACK ink8.All sketches and diagrams must be done in pencil in the ANSWER BOOK.9.Write neatly and legibly.

8190046)-3-T1350(E)(M31)TNOTE: Answer ANY FIVE questions in this question paper.QUESTION 1A two-stage, single-acting, reciprocating compressor takes in air at a tempo of0,4 m³/s. The compression index is 1,3. The delivery pressure is 800 kPa. Therotational frequency of the compressor is 550 r/min. The initial pressure andtemperature for the low-pressure cylinder are 104 kPa and 19 C respectively.Conditions for maximum efficiency prevail and free volume is insignificant.Assume R for air as 0, 287 kJ/kg.K.Calculate the following:1.1The intermediate pressure in kPa(3)1.2The volume for the low-pressure and high-pressure cylinder in m³(6)1.3The diameter (D) of the high-pressure cylinder in mm if D L(4)1.4The power required to drive the compressor in kW if the mechanical efficiencyis 88 percent.(7)[20]AND/ORQUESTION 2In a boiler plant consisting of an evaporator, an economiser and a superheater, steamis generated at a pressure of 1 500 kPa and a temperature of 350 C at a rate of5 400 kg/h. The dryness factor of the steam at entry to the super-heater is 0,9.The temperatures of the feed water entering and leaving the economiser are 44,8 Cand 93,5 C respectively. The boiler burns coal at a rate of 600 kg/h.The heat value of the coal is 32 MJ/kg. The atmospheric temperature is 24 C.Air is supplied at a rate of 15 kg of air per kg of coal. The temperature of the flue gas atthe outlet of the economiser is 210 C. The specific heat capacity of the flue gases is1,045 kJ/kg.K.Calculate the following by using steam tables:2.1Efficiency of the plant(3)2.2Equivalent evaporation from and at 100 C per kilogram of fuel(2)2.3Heat to the economiser in kJ/kg(2)2.4Heat to the evaporator in kJ/kg(3)2.5Heat to the superheater in kJ/kg(3)2.6Heat to the stack (chimney) in kJ/kg(2)

8190046)2.7-4-T1350(E)(M31)TDraw up an energy balance in kJ/kg fuel, also as a percentage of the totalheat.(5)[20]AND/ORQUESTION 3A compression-ignition engine working on the dual cycle takes in two thirds of its totalheat supply at constant volume and one third at constant pressure expansion. Thecompression is adiabatic. Heat is lost at a constant volume.The following data is applicable to the engine:Intake conditionsMaximum pressureCompression ratio 101,325 kPa and 16 C5 MPa13 : 1Specific heat capacity at constant volume 0,712 kJ/kg.KSpecific heat capacity at constant pressure 1 kJ/kg.KCalculate the following:3.1The temperature and pressure after compression(5)3.2The temperature after constant volume heat addition(2)3.3The temperature after constant pressure heat addition(5)3.4The volume before expansion in m3(2)3.5The temperature before constant volume heat addition rejection(2)3.6The Air Standard Efficiency (A.S.E) of the cycle.(4)[20]AND/ORQUESTION 4Air enters a nozzle at a pressure of 3 MPa, a temperature of 480 C and a speed of60 m/s. The air leaves the nozzle at a pressure of 600 kPa. The air flow rate throughthe nozzle is 1,2 kg/s. The expansion from the inlet to the throat takes place with anadiabatic efficiency of 90 percent while the adiabatic efficiency from the throat to theoutlet is 85 percent.Assume gamma 1,4; R 0,287 kJ/kg.K and Cp 1,005 kJ/kg.KCalculate the following:4.1The throat area in mm2(10)

8190046)-5-4.2The exit area in mm24.3The Mach numberT1350(E)(M31)T(8)(2)[20]AND/ORQUESTION 5A ship is fitted with a turbo-charged two-stroke oil engine, with five single-actingcylinders.The following data is applicable to the engine:Cylinder diameterStroke length per cylinderEngine speedFuel consumptionHeat value of fuelAir consumptionBrake mean effective pressureMechanical efficiencyAtmospheric pressureAtmospheric temperatureGas constant for air 760 mm1 500 mm112 r/min1 050 kg/h44 200 kJ/kg22 kg/kg fuel738 kPa86,5%101,325 kPa16 C0,287 kJ/kg.KCalculate the following:5.1The brake power in kW(4)5.2The indicated power in kW(2)5.3The brake thermal efficiency(3)5.4The torque on the drive shaft in kN.m(3)5.5The volume of air induced per minute(3)5.6The swept volume per minute(3)5.7The volumetric efficiency(2)[20]AND/ORQUESTION 6A two-stage, velocity compound steam turbine has blading designed for an axialdischarge of the steam from the second row of moving blades. All the moving bladesare 30 . The blade circumferential speed, the nozzle angle and the fixed blade anglesare designed for a nozzle discharge velocity of 500 m/s.The velocity coefficient for all blades is 0,9.

8190046)6.16.2-6-T1350(E)(M31)TUse a length of 3 cm for the blade circumferential speed and constructvelocity diagrams for the turbine in the ANSWER BOOK. Indicate the lengthsof ALL the lines as well as the magnitude of the angles on the diagrams andcalculate the scale.(10)Determine the following from the velocity diagrams:6.2.1The blade circumferential speed in m/s(2)6.2.2The power developed per kg of steam per second in kW(3)6.2.3The diagram efficiency(3)6.2.4The axial thrust in kN(2)[20]AND/ORQUESTION 7A refrigeration plant uses methyl chloride and operates between pressure boundariesof 215 kPa and 672 kPa.The following information was obtained:Specific enthalpy of dry saturated vapour at condenser pressure 479 kJ/kgSpecific enthalpy of dry saturated vapour at evaporator pressure 462, 8 kJ/kgSpecific enthalpy of saturated liquid at condenser pressure 110, 2 kJ/kgSpecific enthalpy of saturated liquid at evaporator pressure 51, 6 kJ/kgSpecific volume of dry saturated vapour at evaporator pressure 0,168 m³/kgTemperature of the refrigerant in condenser 31 ºCTemperature of the refrigerant in evaporator -6 ºCThe methyl chloride enters the compressor as a wet vapour, the condenser as a drysaturated vapour and it leaves the condenser as a saturated liquid with no undercooling.The actual coefficient of performance is 90,23 percent of the ideal coefficient ofperformance.The compressor bore size is 0,1285 m; it has a stroke length of 1,2 times the diameterof the piston, a volumetric efficiency of 90 percent and it rotates at 420 r/min.Calculate the following:7.1The ideal coefficient of performance and the actual coefficient of performance7.2The specific enthalpy of the refrigerant at the entrance to the compressor in(3)

8190046)7.37.47.5-7-T1350(E)(M31)TkJ/kg and the dryness factor(5)The swept volume of the compressor in m³ per stroke, the swept volume andthe effective swept volume in m³/s(5)The specific volume of the refrigerant at the entrance to the compressor inm³/kg and the mass flow rate in kg per minute(4)The compressor power in kJ/s and the power required to drive thecompressor in kW if the mechanical efficiency is 80%TOTAL:(3)[20]100

(11040012)-1-T1350(E)(M31)TPOWER MACHINES N6FORMULA SHEETAny other applicable formula may also be used.ENGLISHGENERALAFRIKAANSPaVa mRTaR Cp Cv CpCvPV cPV kPV n cPV n kPV cPV kT2 V1 T1 V2 n 1 P 2 P1 n 1n U m . Cv . TQ U WdQ U Av PV s m Cv . ln 2 Cp . ln 2 P1V1 s m . Cv . lnP2P1 s m . Cp . lnV2V1 s m . R . lnP1P2Q m . Cp . TQ m . Cv . TSsu S g Cp . lnTsuTsS fg S g S fS S f xS fghsu hg Cp (tsu t s )Copyright reservedPlease turn over

(11040012)-2-ENGLISHhws h f xh fgVws xVgT1350(E)(M31)TGENERALAFRIKAANSn 1(hsu 1941)Vsu nPsur hns h f xh fgVs VcVcVs 4Vns xVgd2 LP2 P1 P3rps xPx 1P1Different formulae forVerskillende formules virwork done (Wd)arbeid verrig (Av) P V P1 V1 lnV2V1 P1V1 P2V 2n 1 P1V1 P2V 2 1 m . Cp . Txn P1Ven 1n 1 Px 1 xn 1 P 1 n 1 xn mRT1 (rps ) n 1 n 1 Copyright reservedPlease turn over

(11040012)-3-ENGLISHT1350(E)(M31)TGENERALDifferent formulae forwork done (Wd)AFRIKAANSVerskillende formulesvir arbeid verrig (Av) area of PV-diagram area van PV-diagram work done first stage work done secondstage . arbeid verrig eerstestadium arbeid verrig tweede stadium .Wd nett Wd t Wd cAvnett Avt AvkWd nett QnettAvnett QnettDifferent formulae forair standard efficiencies (ASE)Verskillende formulesvir lugstandaardrendemente (LSR) 1 1 1 r 1 heat added heat rejectedheat added 1 rv 1rp rc 1[(rp 1) 1 r 1 ( 1)Volume of air taken in Volume of free airSwept volume(rc 1)]warmte toegevoeg warmte afgestaanwarmte toegevoegDifferent volumetricefficiencies, vol rpVerskillende volumetrieserendemente, vol Volume lug ingeneemSlagvolume Volume vrylugSlagvolumeSwept volume1 Vc P2 n 1 1 Vs P1 Copyright reservedPlease turn over

(11040012)-4-ENGLISHGENERALDifferent thermalefficiencies, therm.Wd heat supplied brake therm. therm. c BPm f / s CVms (hs hw)m f CVT2' T1T2 T1 mech. AFRIKAANSVerskillende termieserendemente, term.Av warmte toegevoegIPm f / s CV ind . therm. T1350(E)(M31)T rem term. RDmb / s WW ind . term. IDmb / s WWm s (hs hw)mb WW term. ηt T3 - T4T3 - T4'BPIP k T2' T1T2 T1 meg . RDIDIndicated efficiency ratioIndikateurrendementverhouding ind . therm.ASE ind . term.LSRBrake efficiency ratioRemrendementverhouding brake therm.ASETNBP 2 60T F r rem. term.LSRRD 2 TN60BP Pbrake mean LANERD Prem gem. LANEIP Pind. mean LANEID Pind . gem. LANEISFC mfBSFC mf/hIP/hBPISBV mb / hIDRSBV mb / hRDCOP T1T2 T1KVW T1T2 T1COP REWdKVW VEAvP m . U . ΔVwD m . U . ΔVwFax m . V fFaks. m . V fCopyright reservedPlease turn over

(11040012)-5-ENGLISHT1350(E)(M31)TGENERALηdia. AFRIKAANS2 . U . ΔVwV12 2 Pc P1 1 1 2 Tc T1 1 Cc 2 103 (h1 hc ) C12C2 2 103 (h1 h2 ) C12Cc 2 103 C p (T1 Tc ) C12C2 2 103 C p (T1 T2 ) C12Ac mVcCc h1 hc hc h2 h1 h2h1 hc hc'h2'h1 h2'A2 mV2C2 T1 Tc Tc T2 T1 T2T1 Tc'Tc T2'T1 T2'm s (hs hw )mb 2 257EE m s (hs hw )m f 2 257EV iso. Wd iso.Wd poly. iso. WdQ rank. rank. Copyright reservedAviso.Av poli.AvQPlease turn over

(11040012)-6- carn. 1 T1350(E)(M31)TT2T1h u pVC12gZ1 U1 P1V1 Q 2gZ 2 U 2 P2V2 Copyright reservedC 22 Wd2C12gZ1 U1 P1V1 Q 2gZ 2 U 2 P2V2 C22 Av2

MARKING GUIDELINENATIONAL CERTIFICATEAPRIL EXAMINATIONPOWER MACHINES N5APRIL 2016This marking guideline consists of 8 pages.Copyright reservedPlease turn over

MARKING GUIDELINE-2POWER MACHINES N5T.(E)(A.)TQUESTION 11.1U 1.2U Ci 1.3 DN60 1,96 389860400 m / s(3)0,4 CiU0,44000,41000 m / s(3)Diagram on next page.Copyright reservedPlease turn over

MARKING GUIDELINE-3POWER MACHINES N5T.(E)(A.)T(8)Copyright reservedPlease turn over

MARKING GUIDELINE1.4-4POWER MACHINES N5T.(E)(A.)T1.4.1Cri 645 m/s(1)1.4.2Cre 567,6 m/s(1)1.4.3Ce 330 m/s(1)1.4.4Fc m cfi cfe 100 375 325 6083,333 N(3)[20]QUESTION 22.12.2The efficiency of any boiler is defined by the ratio of heat transferred to feedwaterin converting it to steam divided by the heat suppliedby thecomplete combustion of the fuel.2.2.1EE Total Heat 2.2.2h1 Total Heat Transferre d22572257 EE2257 11,225278,4 kJ / kgTotal Heat100 Calorific Value 125278,4 100 30000184,26 %(6)158 kJ / kgHeat Gained M s h2 h1 h2 Heat LostMf Cv Mf Cv h1Ms1775 30000 0,8426 158180002650,692 kJ / kgAt 2500 kPa the hg for dry steamis 2801kJ / kg . The steamis wet , because hg is greater than h2 .Copyright reserved(4)(6)Please turn over

MARKING GUIDELINE2.2.3-5POWER MACHINES N5 hwetT.(E)(A.)Thf xhfghw hfhfg2650,692 962 1839 0,918 The steam is 91,8% dry. x(4)[20]QUESTION 33.1PV R MRTPVV MT T115 11