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18th National Certification Exam Energy Managers & Auditors SEPTEMBER 2017 Paper 2

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Paper 2 Set A with Solutions Regn No: __________________ Name : __________________ (To be written by the candidate) 18th NATIONAL CERTIFICATION EXAMINATION FOR ENERGY MANAGERS & ENERGY AUDITORS September, 2017 PAPER 2: Energy Efficiency in Thermal Utilities Date: 23.09.2017 Timings: 14:00-17:00 HRS Duration: 3 HRS Max. Marks: 150 General instructions: o o o o o Please check that this question paper contains 8 printed pages Please check that this question paper contains 64 questions The question paper is divided into three sections All questions in all three sections are compulsory All parts of a question should be answered at one place Section I: a) b) c) 1. Marks: 50 x 1 = 50 OBJECTIVE TYPE Answer all 50 questions Each question carries one mark Please hatch the appropriate oval in the OMR answer sheet with Black Pen or HB pencil In a Heat exchanger with a high LMTD results in a) higher heat transfer area c) higher u-factor b) lower heat transfer area d) none of the above 2. Heat Loss = Heat gain is the principle of ____ a) Boiler b) Heat Exchanger c) Steam traps d) All of the above 3. Flash steam quantity per kg of condensate depends upon a) condensate pressure only b) condensate pressure and flow c) condensate pressure and flash steam pressure d) none of the above 4. Which of the following waste heat recovery systems is of thermal storage type? _________________________ Bureau of Energy Efficiency 1 Paper 2 Set A with Solutions a) ceramic recuperator c) regenerative burner 5. A boiler trial indicated 2% O2 at boiler exit and 7% O2 at stack. The in-leak air quantity between these two measurement is a) 25% 6. c) 50% d) none of the above b) blow down c) air venting d) deaeration Insulating material made by blending and melting of alumina and silica is known as a) ceramic fibre 8. b) 40% The concentration of solids in a boiler is controlled by a) steam venting 7. b) metallic recuperator d) waste heat boiler b) high alumina brick c) fire brick d) insulating brick Tangential firing is used in which type of boiler: a) CFBC b) Chain Grate c) Spreader Stoker d) Pulverised Fuel 9. _____ is required for the simple estimation of flame temperature of the fuel. a) Ultimate analysis b) Proximate analysis c) Size of the coal d) All of the above 10. Overall heat transfer co-efficient in Heat exchangers depends on ____. a) Conductivity of the wall separating the two fluids b) Convective coefficients of hot and cold fluids c) Fouling coefficients d) All of the above 11. The key property of bio mass fuel which influences storage, handling and transportation a) Calorific Value c) Bulk Density 12. b) Percentage of Ash d) None of the above Which of the following contributes to spluttering of flame at burner tip during combustion of fuel oil? a) ash content c) sulphur content 13. When 100 kg of fuel with 60% carbon is burnt with theoretical air, the mass of CO 2 released will be a) 319 kg 14. b) water content d) humidity of air b) 4400 kg c) 4500 kg d) 220 kg Which of the following heat recovery equipment works on a vapour compression cycle? a) thermo-compressor _________________________ Bureau of Energy Efficiency b) heat wheel 2 Paper 2 Set A with Solutions c) heat pump d) heat pipe 15. Which of the following releases large amount of heat per kg during combustion? a) Hydrogen b) Carbon c) Sulphur d) Nitrogen 16. Auxiliary power consumption for coal preparation will be highest in a; a) stoker fired boiler c) CFBC boiler 17. b) AFBC boiler d) pulverized coal fired boiler In a heat exchanger, for the same heat duty, higher the heat transfer coefficient; a) higher will be the heat transfer area needed b) higher will be the LMTD c) lower will be the heat transfer area needed d) lower will be the LMTD 18. Which of the following increases, when steam is passed through PRV? a) specific volume b) enthalpy of steam c) sensible heat d) saturation temperature 19. Which of the following is not true with respect to improper sizing of coal a) results in poor combustion b) lowers excess air resulting in lesser stack loss c) increases un-burnt in ash d) lowers thermal efficiency 20. When solutions of differing concentrations are separated by a semi-permeable membrane, water from less concentrated solution passes through the membrane to dilute the liquid of high concentration. This is called a) reverse osmosis 21. 22. c) softening d) osmosis ______ is used as heat carrier in thermic fluid heaters a) steam b) mineral oil c) water d) all of the above Dissolved CO2 in boiler feed water when left untreated would result in occurrence of ____ in boiler tubes a)creep 23. b) ion exchange b) water side corrosion c)scale d)water hammer In a counter-flow heat exchanger, cold fluid enters at 30 C and leaves at 50 C, whereas the hot fluid enters at 150 C and leaves at 130 C. The LMTD is _________________________ Bureau of Energy Efficiency 3 Paper 2 Set A with Solutions a) 100 C 24. b) 280 C 26. b) fixed carbon b) batch furnace c) arc furnace d) induction furnace In industrial applications the commonly used trap for main steam lines is b) below 5000C 31. b) bottoming cycle d) none of the above Select the odd one among the following b) distillation column c) evaporator d) cooling tower In a boiler, air preheater is installed a) before the economizer c) after economizer b) before superheater d) after ESP A pinch analysis can result in a) reduction in cooling water c) increase in cooling water 32. c) 600-7000C d) 800-9000C In a glass industry waste heat is used for power generation. This type of cogeneration is called a) condenser 30. b) inverted bucket trap d) open bucket trap Fluidized bed combustion takes place in the temperature range of a) topping cycle c) combined cycle 29. d) nitrogen a) continuous furnace a) above 10000C 28. c) sulphur Ceramic fibre gives the maximum energy savings when used in a) thermostatic trap c) thermodynamic trap 27. d) 20 C Which of the following is not measured in the ultimate analysis of a fuel ? a) oxygen 25. c) 0 C b) reduction in steam d) both a & b The main contributor for temporary hardness in Boiler water is ____. a) calcium chloride b) magnesium sulphate c) calcium bicarbonate d) calcium nitrate _________________________ Bureau of Energy Efficiency 4 Paper 2 Set A with Solutions 33. The amount of oxygen required to burn 0.5 kg of Sulphur is a) 1 b) 16 c) 32 d) 0.5 34. _____ is required to accommodate expansion of steam lines. a) prv b) expansion loop c) steam trap d) air vent 35. Steam mains should be run with a falling slope of in the direction of steam flow for effective line condensate drainage 36. 37. a) 50 mm in 30 meters b) 125 mm in 30 meters c) 250 mm in 30 meters d) 350 mm in 30 meters Degasser in water treatment is used to remove a) Dissolved oxygen and other gas b) carbonic acid b) Sulphuric acid d) dissolved CO2 In a boiler Theoretical Air to fuel ratio is 15:1. If 10% excess air is supplied, Flue gas to Fuel ratio will be ____ a) 16:1 b) 17.5:1 c) 16.5:1 d) 17:1 38. Increased Sulphur percentage in furnace oil a)sets lower flue gas temperature limit c) does not add to heat value 39. Which among the following is most viscous fuel? a) furnace oil 40. b) improves viscosity d) forms soot b) HSD c) kerosene d) Light Diesel oil The turbine heat rate is expressed as _________________________ Bureau of Energy Efficiency 5 Paper 2 Set A with Solutions a) kWh/kcal b) kg/kcal c)kcal/kWh d) none of the above 41. Heat loss through openings in furnaces is directly proportional to a) fourth power of furnace temperature b) square of absolute furnace temperature c) absolute furnace temperature d) fourth power of absolute furnace temperature 42. Which of the following is an example of a basic refractory 43. a) Alumino-silicate b) chrome b) Magnesite d)pure alumina Which property of the refractory determines the deformation under stress? a) b) c) d) 44. creep refractoriness under load (rul) porosity pyrometric cone equivalent (PCE) Electro static precipitator in FBC boiler is used for _______ a) Bottom ash removal b) Fly ash removal c) SO2 removal d) CO2 removal 45. Heat rate is likely to be the least for ________ a) back pressure turbine b) condensing turbine c)extraction condensing turbine d) double extraction turbine 46. Which of the following is true of plate heat exchangers a) close approach b) expandable area c) Counter current d) All of the above 47. In FBC boilers fluidization depends largely on --------- 48. 49. i) Particle size ii) Air velocity a) i & iii b) ii & iii iii) Bed Material c) i & ii iv) Size of Boiler d) i,ii,iii,iv When the boiler bed temperature exceeds beyond 950 C, the result is: a) Low steam temperature b) clinker formation c) Melting of lime stones d) Ash carry over Automatic blowdown controls for boilers work by sensing _________________________ Bureau of Energy Efficiency 6 Paper 2 Set A with Solutions a) TDS 50. b) conductivity c) pH d) conductivity and pH c) zero d) none of the above Latent heat of steam at the critical point is a) infinite b) 540 kcal -------- End of Section - I --------- Section - II: SHORT DESCRIPTIVE QUESTIONS Marks: 8 x 5 = 40 (i) Answer all Eight questions (ii) Each question carries Five marks S-1 Ans Explain the working of Thermic fluid heating system and why it is preferred to steam heating in some cases? Thermic Fluid Heaters: At high temperatures, steam requires a corresponding high operating pressure and establishing high temperatures with steam can be very cumbersome and expensive in some cases. In thermic fluid heaters, a special type of oil-synthetic / mineral is used as heat carrier. This fluid can be heated up to 300 oC at atmospheric pressure. In comparison steam would require a pressure of 85 bar to obtain this temperature. .2.5 marks Advantages: High temperature operation at atmospheric pressure Optional temperature level set points No supply or treatment of hot water and hence no heat loss due to condensate flash steam No risk of corrosion Easy to operate Functioning of Thermic Fluid Heaters Heat is transferred to the fluid through radiation. The hot thermic-fluid is circulated to various process equipments such as dryers, heaters, deodouriser etc. where it gives up the heat. The return oil at a temperature 10 to 20 0C less, comes back to the thermic-fluid heater to get heated up again. The circulation is carried out by a thermic fluid circulation pump. The thermic-fluid heater operates between two temperature ranges. Once the upper limit is reached the burner is switched OFF or goes into the low fire mode. In the case of solid fuel fired system the ID fan switches OFF on _________________________ Bureau of Energy Efficiency 7 Paper 2 Set A with Solutions reaching the upper limit. When the temperature reaches the lower limit due to heat transfer in the process, the burners come ON again and in case of solid fuels, the ID fan comes ON again. Since the thermic fluid heaters operate at a high temperature (250 300 0 C), the leaving exhaust gas temperatures are more than the fluid temperature. Hence, the heat loss through the flue gas is a major component of fuel losses. This offers potential for heat recovery if there is a suitable application. The capacity is specified in terms of Lakh kilo calories per hour or Million kilo calories per hour. (Pg no 56 of Book 2 for explanation) .2.5 marks S-2 Ans In a car manufacture company, Propane is used as fuel in heaters for preheating paints. Calculate the Air to Fuel ratio for complete combustion of C3H8 (Propane), if 15% excess air is supplied to the heater. C3H8 + 5 O2 --> 3 CO2 + 4 H2O .1 mark 1 mole of propane requires 5 moles of Oxygen. Molecular weight of Propane is 44 Kg per mole. .1 mark 44 Kg of Propane requires 160 Kg of Oxygen. Theoretical air required for combustion = 160 / 0.23 = 695.6 Kg/hr .1.5 marks Excess air supplied is 15 %. Actual air supplied for combustion is = 695.6 * 1.15 = 800 Kg/hr of air Air to Fuel ratio = 800 / 44 = 18.18 or 18.2 .1.5 marks S-3 To meet a process plant s heat and power requirements, high pressure steam at 70 MT/hr passes through a back pressure steam turbine, for power generation and the exhaust _________________________ Bureau of Energy Efficiency 8 Paper 2 Set A with Solutions steam is sent for thermal process requirements in the plant. With the following operating data, calculate the heat to power ratio in kW/kW if the turbine and generator efficiencies are 90% and 92% respectively. Steam Inlet conditions Pressure = 50 bar Temperature = 530 C Enthalpy of steam = 3515 kJ/kg Ans S-4 Power generated Heat input to process Heat to power ratio Steam outlet conditions Pressure = 10 bar Temperature = 280 C Enthalpy = 3020 kJ/kg Enthalpy of water = 782 kJ/kg = 70 * 1000 * (3515 - 3020) * 0.90 * 0.92 / 3600 = 7969.5 kW .2 marks = 70 * 1000 * (3020-782) / 3600 = 43517 kW .2 marks = 43517 / 7969.5 = 5.46 kW/kW .1 mark a) Explain why de-superheating is done after pressure reduction in PRVs? b) Why is correction factor required for estimation of LMTD? Ans a) A reduction in steam pressure through a pressure reducing valve (PRV) is an isenthalpic process. Saturated steam when reduced to a lower pressure results in super heated steam. Since process requires only saturated steam, de-superheating is often required, to compensate for superheat gained in PRV application due to isenthalpic expansion. .2.5 marks b) In multi pass shell and tube heat exchangers, the flow pattern is a mixture of cocurrent and counter current flow, as the two streams flow through the exchanger in the same direction on same passes and in the opposite on others. For these reasons, the mean temperature differences is not equal to the logarithmic mean. However it is convenient to retain the LMTD by introducing a correction factor, F which is appropriately termed as the LMTD correction factor. .2.5 marks S-5 Ans What happens to steam properties such as saturation temperature, enthalpy of saturated water, latent heat of steam, enthalpy of steam and specific volume of steam, if the steam pressure is increased? a) Saturation Temperature increases b) Enthalpy of saturated water Increases. c) Latent heat of steam decreases _________________________ Bureau of Energy Efficiency 9 Paper 2 Set A with Solutions d) Enthalpy of steam increases e) Specific Volume decreases .5 marks ( each point carries one mark respectively) S-6 Ans Two identical biomass fired boilers of capacity 10 TPH are operated in a chemical industry. They each have a full load efficiency of 80%. The part load efficiencies at 70% and 40% load are 75% and 68% respectively. For meeting 14 TPH requirement of steam, would you prefer to run both the boilers at 7 TPH capacity or one at full load capacity and other at 40% capacity. Estimate the % savings in the preferred case. Fuel energy required when both the boilers are run at 7 TPH load Governing equation to be used : (Fuel reqd) = [(Qty stm) * (Enth steam Enth feedwater)] / [( GCV Fuel) * (Effy boiler)] (Fuel reqd) = [(Qty stm) * ( h) / (Effy boiler)] Where : h = is same in both cases = [Enth steam Enth feed-water) / ( GCV Fuel)] Fuel required = (2*7* h) / (0.75) = 18.67 h .1 mark Fuel energy required when one boiler is running at full load and the other at part load of 40% Fuel required = [(10* h) / (0.8)] + [(4* h) / (0.68)]= 18.38 h .2 marks The case where one boiler is running at full load and the other at part load of 40% is preferred .1 marks % savings = [(18.67 h) (18.38 h)] * 100 / (18.67 h) = 1.55% .1 mark S-7 Ans A reheating furnace is operating with deteriorated wall insulation. The existing average outer surface temperature of the furnace (of area = 100 m2) with surrounding ambient air temperature of 40 C, is recorded to be 120 C. After revamping the refractory, the furnace outer surface temperature reduces to 50 C. If the fuel oil (GCV = 11,000 kcal/kg) cost is Rs. 25,000 per tonne, and efficiency of the furnace is 35%, estimate annual savings for 300 working days per year. Rate of heat loss from furnace surface (existing) S = [10+(Ts Ta) / 20] x ( Ts- Ta) kcals/hr/m2 Heat loss = S x A 120 40 = 100m2 x [10+ ] (120-40) kcals/hr /m2 20 = 112.0 x 103 kcal/hr .1 mark _________________________ Bureau of Energy Efficiency 10 Paper 2 Set A with Solutions Rate of heat loss from furnace surface after revamping insulation refractory 50 40 = 100 [ 10 + ] (50-40) 20 = 10.5 x 103 kcal/hr .1 Energy savings kcal/hr = (112-10.5)x 103 = 101.5x103 kcal/hr Annual energy savings = (112-10.5) x 103 x 300 x 24 = 7.308 x 108 kcal/yr .1 8 7.308 x10 Annual fuel oil savings = = 1.9 x 105 kg of fuel 0.35 x11000 . 1 5 1.9 x10 x 25000 Annual Cost savings = = Rs.4.75 x 106 1000 .1 S8 mark mark mark mark The efficiency of a boiler on GCV basis is 85%. The fuel contains 1.0 % moisture and 12 % hydrogen. The GCV of fuel is 10,500 Kcal/kg. What is the boiler efficiency on the basis of net calorific value? %age of Hydrogen in fuel %age of moisture in fuel NCV = GCV [9 x ------------------------------------ + ------------------------------------] x584 100 100 NCV = 100 12 1.0 10500 [ 9 x ----- + -------- ] x 584 100 100 = 10500 [ 9 x 0.12 + 0.01] x 584 = 10500 - 636.56 = 9863.44 = 9863 kcal / kg .3 marks Boiler efficiency on NCV = = 85 ------ x 10500 9863 90.5% .2 marks -------- End of Section - II --------- _________________________ Bureau of Energy Efficiency 11 Paper 2 Set A with Solutions Section - III: LONG DESCRIPTIVE QUESTIONS Marks: 6 x 10 = 60 (i) Answer all SIX questions (ii) Each question carries Ten marks L-1 In an engineering industry, an electrically heated furnace of efficiency 80%, is used for annealing of the components. The annealing cycle and corresponding energy consumption as follows. S.No 1 2 3 Ans Heat treatment cycle Heat -Up Holding at Cooling Temperature 0 C 30 - 850 850 850 - 60 Time hrs 4 4 4 Power drawn in kW 500 100 20 The electrical energy drawn in each sub-cycle is uniform and plant operates 50 batches per month. The cost of electricity is Rs.8/kWh. The management has decided to replace the electric furnace with FO oil fired furnace with efficiency of 50%. The cost of F.O is Rs.30/kg. Calculate cost savings and payback period of converting from electric to oil fired furnace. Investment for FO fired furnace is Rs.25 lakhs. The GCV of F.O is 10,000 kcal/kg Energy consumption per treatment batch : Heat up time = 500 x 4 Holding time = 100 x 4 Cooling time = 20 x 4 Total energy consumption per batch Monthly energy consumption by electric annealing furnace including losses = 2000kWh; ..1 mark = 400 kWh ..1 mark = 80 kWh ..1 mark = 2480 kWh ..1 mark = 2480 x 50 = 1,24,000 kWh / month .1 mark Actual consumption by the components at electrical furnace efficiency of 80% = (124000 x 0.8) = 99,200 kWh / month .1 mark Eqvt FO required to be supplied to the oil fired furnace at an efficiency of 50% = 99200 x 860 / (10000 x 0.5) = 17062.4 kg / mth .1.5 marks Annual Cost of = [(124000 x 8) (17062.4 x 30) ] x 12 = Rs.57,61,536 / yr .1.5 marks Payback period = 2500000 / 5761536 = 5.2 months .1 mark L-2 Write short notes on any two of the following a. Wet preservation method for boilers b. Reverse osmosis c. Reciprocating engine co-generation system _________________________ Bureau of Energy Efficiency 12 Paper 2 Set A with Solutions Ans a) Wet preservation method for boilers: In the wet method the boiler is filled to the normal level with water at a pH of 10.5 to 11. Hydrazine to the extent of 200 ppm is to be dosed with the water. The unit is to be steamed in service to ensure uniform concentration of boiler water throughout the unit and to eliminate dissolved oxygen from water. Sodium sulphite (Na2SO3), which acts as a de-oxygenerator, can also be used as an alternative to hydrazine and the sulphite concentration has to be maintained at 300-400 ppm. Analysis of boiler water should be carried out frequently. If the hydrazine concentration in water happens to drop below 50 ppm, the water in the drum should be lowered to the normal operating level and an appropriate quantity of chemicals should be dosed to bring back 200 the concentration of hydrazine or sodium sulphite. The boiler should be steamed to circulate chemicals to uniform concentration. ..5 marks b) Reverse osmosis: When solutions of differing concentrations are separated by a semipermeable membrane, water from less concentrated solution passes through the membrane to dilute the liquid of high concentration, which is called osmosis. If the solution of high concentration is pressurized, the process is reversed and the water from the solution of high concentration flows to the weaker solution. This is known as reverse osmosis. ..5 marks c) Reciprocating engine co-generation system: Also known as internal combustion (I. C.) engines, these cogeneration systems have high power generation efficiencies in comparison with other prime movers. There are two sources of heat for recovery: exhaust gas at high temperature and engine jacket cooling water system at low temperature. As heat recovery can be quite efficient for smaller systems, these systems are more popular with smaller energy consuming facilities, particularly those having a greater need for electricity than thermal energy and where the quality of heat required is not high, e.g. low pressure steam or hot water. L3 ..5 marks (Consider any two of the above) A fluidized bed boiler generates 24TPH at 22 kg/cm 2(g), out of which, 18 TPH is going to the back pressure turbine and exhausts from it at 5kg/cm2(g) to meet the process steam requirement in the plant. The balance steam from the boiler is passed through a PRDS to supply 10 kg/cm2(g) steam at dry saturated condition to another process. DM water at 1050C is used for de-superheating in PRDS. Given data: a) b) c) d) e) Mechanical Efficiency of steam turbine = 95 % Losses in gear transmission = 4% Efficiency of alternator = 96 % 2 The total heat of steam at turbine inlet condition at 22 kg/cm (g) = 708 kcal/kg The total heat of steam at turbine outlet condition at 5 kg/cm2 (g) = 658 kcal/kg f) The total heat of the steam at PRDS exit at 10kg/cm2(g) = 670 kcal/kg Calculate 1. Power generation from back pressure turbine 2. Steam flow rate going to process at the exit of PRDS ANS 1: Power generation from back pressure turbine: Enthalpy drop across the turbine per kg of inlet steam = (h1-h2) = (708-658)= 50 kcal/kg _________________________ Bureau of Energy Efficiency 13 Paper 2 Set A with Solutions ..1 mark Total steam flow rate through turbine Total enthalpy drop across the turbine = 18,000 kg/hr =18,000*50 = 900000 kcal/hr ..1 mark Mechanical Efficiency of steam turbine Efficiency of alternator % losses in gear transmission Over all efficiency of the turbo alternator = 95% = 96 % = 4% = 0.95 x 0.96 x 0.96 = 0.8755 = 87.55% ..2 marks Energy output of turbine = 900000 x 0.8755 = 787950 ..1 mark Power output of the alternator = 787950/ 860 = 916 kW ..1 mark 2: Steam Flow rate going to process at the exit of PRDS: Flow rate of DM water =m Heat content of steam at exit of PRDS = Heat supplied by water and steam to PRDS (6000 + m) x 670 = (6000 x 708) + (m x 105) m = 403.54 kg/ hr = 403.54 kg/hr ..3 marks Steam flow at outlet of PRDS L4 Ans = 6000 + 403.54 = 6403.54 kg / hr ..1 mark In a chlor-alkali plant, 100 TPD caustic solution at 30% concentration is dried to 55% concentration in a single effect evaporator, where the ratio of steam input to moisture removal is 1.0 kg/kg. It is proposed to be replaced by a triple effect evaporator at an investment cost of Rs. 5 crore, for which the ratio of steam input to moisture removal is 0.45 kg/kg. Steam for the evaporator is generated from an oil fired boiler at an evaporation ratio of 14. Calculate annual fuel savings in TPD. % salt concentration at inlet = 30% % salt concentration at outlet = 55% Input quantity of caustic solution to drier = 100 TPD Amount of bone dry salt at drier inlet = 100x 0.3 = 30 TPD ..1 mark Amount of water at drier inlet = 100 - 30 = 70 TPD ..1 mark Flow rate of salt solution at drier outlet = 30 / 0.55 = 54.5 TPD ..1 mark Amount of water at drier outlet = 54.5 - 30 = 24.5 TPD ..1 mark Amount of water removed = 70 - 24.5 = 45.5 TPD ..1 mark Ratio of steam / moisture for single effect = 1.0 ..1 mark Amount of steam required for single effect = 45.5 TPD Ratio of steam / moisture for triple effect = 0.45 Amount of steam required for triple effect = 45.5 x 0.45 = 20.475 TPD ..1 mark Amount of steam saved by triple effect = 45.5 20.45 = 25.05 TPD ..1 mark Evaporation ratio = 14 Amount of fuel savings = 25.05 / 14 = 1.789 TPD 2 marks _________________________ Bureau of Energy Efficiency 14 Paper 2 Set A with Solutions L-5 Ans L-6 Write short notes on following refractory properties and their significance. a) Porosity b) Bulk density c) Pyrometric cone equivalent d) Thermal conductivity Porosity Low porosity is desirable as it would prevent easy penetration of refractory and also larger number of small pores are preferred over small number of large pores ..2.5 marks Bulk density Increase in bulk density increases its volume stability, heat capacity and resistance to slag penetration ..2.5 marks Pyrometric Cone Equivalent (PCE) temperature at which refractory will deform under its own weight is its softening temperature indicated by PCE . ..2.5 marks Thermal conductivity It is the heat loss per unit area per unit insulation thickness per unit temperature difference W-m2/m C or W-m/ C. Thermal conductivity of materials increases with temperature. So thermal conductivity is always specified at the mean temperature (mean of hot and cold face temperatures) of the insulation material ..2.5 marks Refer Book 2, Page no 156 157 In a Crude Oil refining unit, a counter-flow shell & tube heat exchanger is used to preheat LDO (Light Diesel Oil) flowing at 60 m3/hr at 50 C using steam at 8 kg/cm2g as a heating medium. Steam enters the heat exchanger through a pipe of 6 diameter. Density and Specific heat of LDO is 830 kg/m3 & 0.7 kcal/kg C respectively. Properties of steam at 8 kg/cm2 is given below, Pressure, (kg/cm2g) 8.0 Temperature, ( C) 170 Specific volume (m3/kg) 0.22 Enthalpy (kcal/kg) Water Evaporation Steam 170 490 660 a. Calculate the mass flow rate of Steam, if the maximum permissible velocity in the pipeline is 25 m/sec. b. Temperature of the Fuel oil, after preheating in the heat exchanger Ans Steam Flow rate calculation: Diameter of the pipe = 6 inch = 6 * 2.54 = 15.24 cm _________________________ Bureau of Energy Efficiency 15 Paper 2 Set A with Solutions ..1 mark Area of the pipe 2 = 3.14 * D / 4 = 3.14* (15.24)2 / 4 = 182.3 cm2 = 0.0182 m2 Volume flow rate of steam = Area * Velocity Velocity permissible = 25 m/sec Volumetric flow rate of steam = 0.0182 * 25 = 0.455 m3/s * 3600 = 1638 m3/hr ..2 marks 3 Specific volume of steam = 0.22 m /kg Mass flow rate of steam = 1638 / 0.22 = 7445 kg/hr a. Mass flow rate of steam = 7.445 TPH ..1 mark Temperature of the Fuel oil (LDO) Heat transfer rate of Steam =m*H = 7445 * 490 = 3648050 kcal/hr Heat transfer rate of Steam ..2 marks = heat transfer rate of LDO Heat transfer rate of LDO = 3648050 kcal/hr Heat transfer rate of LDO = m * Cp * (T1 T2) = 60 * 830 * 0.7 * (T 50) = 3648050 / (60 * 830 * 0.7) Delta Temperature = 104.65 C Outlet Temperature of LDO = (104.65 + 50) C ..2 marks _________________________ Bureau of Energy Efficiency 16 Paper 2 Set A with Solutions = 154.65 C b) Outlet temperature of LDO = 154.65 C ..2 marks . . End of Section III .. . _________________________ Bureau of Energy Efficiency 17

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