Automobile Engineering - Solved Problem - Engine Combustion Parameters & Efficiency

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Question:
An eight cylinder automobile engine of 85.7 mm bore and 82.5 mm stroke with
a compression ratio of 7 is tested at 4000 rpm. on a dynamometer which has a
0.5335 m arm. During a 10 minutes test at a dynamometer scale beam reading
of 400 N, 4.55 kg of gasoline for which the heating value is 46,000 kJ/kg are
burnt, and air at 294 K and 10 × 104 N/m2 is supplied to the carburetor at the
rate of 5.44 kg per minute. Find
a) the brake power (b.p) delivered
b) the brake mean effective pressure (bmep)
c) the brake specific fuel consumption (bsfc)
d) the specific air consumption
e) the brake thermal efficiency
f) the volumetric efficiency
g) the air-fuel ratio (A-F Ratio)

Solution: 2 6 ( ) . 2 2 4000 400 0.5335 89.34 kW 60000 60000 . 89.39 kW ( ) 8( ) ( ) . 2 60000 60000 . 60000 89.39 60000 704.39 kPa 4 4 .0825 (85.7) 4000 4 10 704. a b p T NT b p n bmep Al N bmep Al N b b p x bmep b p lAN bmep                            39 kPa ( ) Fuel consumed in one minute, 0.455 kg/min Fuel consumed in one hour, 0.455 60 kg/hr 27.3 kg/hr 27.3 kg/hr 0.306 kg/b.kWh . 89.34 kW 0.306 kg/b f f f f c w w w w bsfc b p bsfc              .kWh ( ) Air consumption rate, 5.44 kg/min 5.44 60 kg/hr 326.4 kg/hr 326.4 kg/hr 326.4 kg/hr 3.653 kg/b.kWh . 89.34 kW 3.653 kg/b.kWh a a a a d w w w bsac w b p bsac                2 4 3 ( ) Brake Thermal efficiency, . 89.34 kW 89.34 kJ/s 0.455 kg/min 46000 kJ/kg 0.455 46000 kg/s kJ/kg 60 0.256 25.6% ( ) Piston displacement (0.0857) (0.0825) 4.76 10 m 4 b f b b e b p w HV f                     4 3 3 /cycle At 4000 rpm and for four-stroke, eight-cylinder engine, Piston displacement 4.76 10 4000 8 m /min 7.62 m /min 2 Standard conditions of intake: Temperature (T) 298 K,        5 3 5 Pressure ( ) 1 atm 1.01325 10 Pa Volume of air used at intake conditions 5.44 287.1 298 4.593 m /min 1.01325 10 Volumetric efficiency, 4.593 100% 60.3% 7.62 60.3% ( ) A a v v p w RT p g                  ir : Fuel Ratio 5.45 11.98 0.455 A/F Ratio 11.98 a f w w       
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Energy Losses (Heat Balance)

Only a part of the energy supplied to the engine is transformed into useful work whereas
the rest is either wasted or utilized for heating purposes. The main part of the unutilized heat
goes to exhaust gases and to the cooling system. In order to draw a heat balance chart for an
engine, tests should be conducted to give the following information.
(i) Energy supplied to an engine which is known from the heating value of the fuel
consumed.
(ii) Heat converted to useful work.
{Hi) Heat carried away by cooling water.
(iv) Heat carried away by exhaust gases.
(v) Heat unaccounted for (radiation etc.)
It is expected that the heat balance results of CI engine must differ from that of petrol engine
due to much higher compression and expansion ratios in the former. The higher compression
ratio results in lower exhaust gas temperature and also lower flame temperature that in turn
causes lower heat loss to the cylinder walls in CI engines.
The utilization of the fuel’s heat energy is also higher in CI engines because of its higher
compression ratio.
Although the actual value of heat utilization is dependent upon a number of factors like
compression ratio, engine load, fuel injection quantity, timing etc. some average figures for heat

Item S.I. Engine C.I. Engine
Heat converted to useful work (i.p.) 25 to 32% 36 to 45%
Heat carried away by cooling water 33 to 30% 30 to 28%
Heat carried away by exhaust gases 35 to 28% 29 to 20%
Heat unaccounted for 7 to 10% 5 to 7%
Total (= Energy supplied) 100% 100%
balances for both the engines are given below :
If the shaft work (b.p.) is considered instead of useful work, the mechanical losses are to be
accounted for or are generally included in the cooling water heat.

An eight cylinder automobile petrol engine of bore 85.7 mm and stroke 82.5 mm
with a compression ratio of 7: 1 was tested on a dynamometer which has an arm of
533.5 mm long. The dynamometer scale reading was 400 N and the speed of the
engine was 4000 RPM. During the 10 minutes run fuel consumption was 4.55 kg.
The heating value of fuel was 45980 kJ/Kg. Quantity of air supplied through the
carburetor was 5.44 kg per minute at a pressure of 0.1 MPa and at a temperature of
293 K. Find the following: (a) The brake power developed, (b) The brake mean
effective pressure, (c) The brake specific fuel and air consumption (d) The brake
thermal efficiency (e) the volumetric efficiency. Gas constant for air is 287.14 J/kg-K.

Selected Math Riddles - Set 1 (with solutions)

1) If 1/2 of 5 is 3, then what is 1/3 of 10?

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2) Does a pound of gold or a pound of feathers weight more?

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3) If you randomly choose one of the following answers to this question, what is your chance of getting it right?

a) 50%

b) 25%

c) 0%

d) 75%

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4) A claustrophobic person gets on a train. The train enters a tunnel just as it is leaving the station. Where is the best place for him to sit?

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5) Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a sports car; behind all of the others is bicycles. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 2, revealing a bicycle. He then says to you, "Do you want to pick door No. 3?"

Is it to your advantage to switch your choice?

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6) A homeless beggar picks up cigarette butts from the ground and can make a cigarette with 4 butts. If he finds 16 cigarette butts, how many cigarettes can he make?

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7) Little Raju is walking home. He has ₹300 he has to bring home to his mom. While he is walking a man stops him and gives him a chance to double his money. The man says "I'll give you ₹600 if you can roll 1 die and get a 4 or above, you can roll 2 dice and get a 5 or 6 on at least one of them, or you can roll 3 dice and get a 6 on at least on die. If you don't I get your ₹300."

What does Raju do to have the best chance of getting home with the money?

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8) Ram and Krishna are in class when their teacher gives Ram the sum of two numbers and Krishna the product of the same two numbers (these numbers are greater than or equal to 2). They must figure out the two numbers.

Ram:           I don't know what the numbers are.

Krishna:      I knew you didn't know the numbers... But neither do I.

Ram:           In that case, I do know the numbers now.

What are the numbers?

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9) What is the next number in the sequence?

1        11      21     1211          111221       312211

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10) There are several books on a bookshelf. If one book is the 4th from the left and 6th from the right, how many books are on the shelf?

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11) There are 100 coins scattered in a dark room. 90 have heads facing up and 10 are facing tails up. You cannot tell which coins are which. How do you sort the coins into two piles that contain the same number of tails up coins?

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12) A rat is placed at the beginning of a maze and must make it to the end. There are four paths at the start that he has an equal chance of taking: path A takes 5 minutes and leads to the end, path B takes 8 minutes and leads to the start, path C takes 3 minutes and leads to the end, and path D takes 2 minutes and leads to the start.

What is the expected amount of time it will take for the rat to finish the maze?

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13) You have a glass of water that looks about half full. How can you tell, only using the glass of water itself, if the glass is half full or not?

(Hint: The glass is a right cylinder)

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14) Does a ton of feathers or a ton of bricks weigh more?

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15) Mr. Lalith has 4 daughters. Each of his daughters has a brother. How many children does Mr. Lalith have?

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16) A tomato plant is 3 meters long. The bottom foot of the plant doesn't grow any tomatoes but the rest of the plant grows a tomato every 5 inches. How many vegetables can grow off of the plant?

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17) Ajay has missed an excessive number of days of school, so he must meet with Principal Rajeev. Mr. Rajeev asks him "Why on Earth have you missed so many days?"

Ajay replies "There just isn't enough time for school. I need 8 hours of sleep a day, which adds up to about 122 days a year. Weekends off is 104 days a year. Summer vacation is about 60 days. If I spend about an hour on each meal, that's 3 hours a day or 45 days a year. I need at least 2 hours of exercise and relaxation time each day to stay physically and mentally fit, adding another 30 days.

Add all of that up and you get about 361 days. That only leaves 4 days for school."

The principal knows Ajay is full of it, but can't figure out why. Where is Ajay going wrong?

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18) A deaf and mute man goes to a railway station. Tickets for the train are 50 rupees each. The man goes to the ticket booth and hands the man inside just a hundred rupees. The man in the booth hands him two tickets.

How did the man in the booth know to give him two tickets without even looking at him?

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19) Two cities Delhi and Chennai are exactly 2000 km apart. Praveen leaves City Delhi driving at 50 km/hr and Avinash leaves City Chennai a half an hour later driving 90 km/hr. Who will be closer to City Chennai when they meet?

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20) How long is the answer to this question.

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Non-Traditional Manufacturing Processes (NTMP) Assignment - ElectroChemical Machining (ECM) - MRR

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Question: Prove that Volumetric Material Removal Rate (MRR) for an alloy with ‘n’ elements by Electrochemical Machining is given by: 1 MRR 100 n i i i i I F x M      where, xi is weight percent composition of ith element in the alloy, νi is the valency of ith element, Mi is the atomic weight of ith element in the alloy, ρ is the density of the material, F is the Faraday’s constant, I is the current in the machining process. (Hint: Use Faraday’s Laws) Solution: Electrochemical Machining is governed by Faraday’s laws. The first law states that the amount of electrochemical dissolution is proportional to amount of charge passed through the electrochemical cell, which may be expressed as: m  Q The second law states that the amount of material deposited or dissolved proportional to the Electrochemical Equivalence (ECE) of the material. Thus, m ECE A    m QA ItA m ItA F         where, F = Faraday’s constant Now, MRR m t MRR IA F      Now for passing a current of I for a time t, the mass of material dissolved for any element ‘i’ is given by 100 a i i m x    where Γa is the total volume of alloy dissolved. 100 i i i i i i i i a i i i i m QM F Q F m M Q F x M           1 1 1 1 1 1 100 100 100 Now, 100 100 n T i i n n a i i i i T a i i i i a n i i i i a n i i a i i n i i i i Q It Q Q It F x F x M M It F x M MRR t It F x MRR M t t MRR I F x M                                       

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Assignment: (Dead Line – 13th March 2015) Question: Prove that Volumetric Material Removal Rate (MRR) for an alloy with ‘n’ elements by Electrochemical Machining is given by: 1 MRR 100 n i i i i I F x M      where, xi is weight percent composition of ith element in the alloy, νi is the valency of ith element, Mi is the atomic weight of ith element in the alloy, ρ is the density of the material, F is the Faraday’s constant, I is the current in the machining process. (Hint: Use Faraday’s Laws) Solution: Volumetric Material removal rate (MRR) is the volume of material removed from the job material per unit time during the machining. In ECM, material removal takes place due to atomic dissolution of work material. Electrochemical dissolution is governed by Faraday’s laws of electrolysis. The first law states that the amount of chemical change i.e., the amount of electrochemical dissolution or deposition is proportional to amount of charge passed through the electrochemical cell that is again the product of current and the time passed, which may be expressed as: m  Q  It …. eqn (1) where m = mass of material dissolved or deposited; and Q = amount of charge passed The second law states that the amount of material deposited or dissolved further depends on Electrochemical Equivalence (ECE) of the material that is again the ratio atomic weight (M) and valency (ν). Thus, m ECE M    …. eqn (2) From equations (1) & (2), we have, m ItM m ItM F      where, F = Faraday’s constant = 96500 coulombs Now, Volume removed / Time taken MRR m m t t ItM MRR F IM t F MRR IM F                     where, I = current, ρ = density of the material Let us assume there are ‘n’ elements in an alloy. The atomic weights are given as Mi (M1, M2,…Mn) with valency during electrochemical dissolution as νi (ν1, ν2,…νn). The weight percentages of different elements are xi (x1, x2,…xn). Now for passing a current of I for a time t, the mass of material dissolved for any element ‘i’ is given by 100 a i i m V x   where Va is the total volume of alloy dissolved. Each element present in the alloy takes a certain amount of charge to dissolve.  /100 100 i i i i i i i i i a i i i a i i i i m QM F Q F m M F V x Q M Q FV x M              The total charge (QT) passed is the algebraic sum of charge passed of individual elements in the alloy. So, we can write, 1 1 1 1 1 1 1 100 100 100 100 Now, Total volume dissolved Total time 100 100 100 n T i i n n a i i i i T a i i i i n i i a i i a n i i i i a n i i a i i n i i i i Q It Q Q It FV x V F x M M It V F x M V It F x M MRR V t It F x MRR V M I t t F x M MRR                                             1 n i i i i I F x M     Thus, Volumetric Material Removal Rate (MRR) for an alloy with ‘n’ elements by Electrochemical Machining is given by: 1 MRR 100 n i i i i I F x M     
Module 9 Non conventional Machining Version 2 ME, IIT Kharagpur Lesson 38 Electro Chemical Machining Version 2 ME, IIT Kharagpur Instructional Objectives (i) Identify electro-chemical machining (ECM) as a particular type of non-tradition processes (ii) Describe the basic working principle of ECM process (iii) Draw schematically the basics of ECM (iv) Draw the tool potential drop (v) Describe material removal mechanism in ECM (vi) Identify the process parameters in ECM (vii) Develop models for material removal rate in ECM (viii) Analyse the dynamics of ECM process (ix) Identify different modules of ECM equipment (x) List four application of ECM (xi) Draw schematics of four such ECM applications 1. Introduction Electrochemical Machining (ECM) is a non-traditional machining (NTM) process belonging to Electrochemical category. ECM is opposite of electrochemical or galvanic coating or deposition process. Thus ECM can be thought of a controlled anodic dissolution at atomic level of the work piece that is electrically conductive by a shaped tool due to flow of high current at relatively low potential difference through an electrolyte which is quite often water based neutral salt solution. Fig. 1 schematically shows the basic principle of ECM. In ECM, the workpiece is connected to the positive terminal of a low voltage high current DC generator or power source. The tool is shaped and shape of the tool is transferred to the workpiece. The tool is connected to the negative terminal. Machining takes place due to anodic dissolution at atomic level of the work material due to electrochemical reaction. A gap between the tool and the workpiece is required to be maintained to proceed with steady state machining. W O R K T O O L WO R K T O O L electrolyte Fig. 1 Schematic principle of Electro Chemical Machining (ECM) 2. Process During ECM, there will be reactions occurring at the electrodes i.e. at the anode or workpiece and at the cathode or the tool along with within the electrolyte. Let us take an example of machining of low carbon steel which is primarily a ferrous alloy mainly containing iron. For electrochemical machining of steel, generally a neutral salt solution of sodium chloride (NaCl) is taken as the electrolyte. The electrolyte and water undergoes ionic dissociation as shown below as potential difference is applied NaCl ↔ Na+ + Cl- H2O H↔+ + (OH)- Version 2 ME, IIT Kharagpur As the potential difference is applied between the work piece (anode) and the tool (cathode), the positive ions move towards the tool and negative ions move towards the workpiece. Thus the hydrogen ions will take away electrons from the cathode (tool) and from hydrogen gas as: 2H+ + 2e- = H2↑ at cathode Similarly, the iron atoms will come out of the anode (work piece) as: Fe = Fe+ + + 2e- Within the electrolyte iron ions would combine with chloride ions to form iron chloride and similarly sodium ions would combine with hydroxyl ions to form sodium hydroxide Na+ + OH- = NaOH In practice FeCl2 and Fe(OH)2 would form and get precipitated in the form of sludge. In this manner it can be noted that the work piece gets gradually machined and gets precipitated as the sludge. Moreover there is not coating on the tool, only hydrogen gas evolves at the tool or cathode. Fig. 2 depicts the electro-chemical reactions schematically. As the material removal takes place due to atomic level dissociation, the machined surface is of excellent surface finish and stress free. W O R K T O O L Fe++ Fe++ OH−OH−OH−OH− Na+Na+Cl-Cl- H+H+OH−Fe++Fe(OH)2 FeCl2 e e H2 Fig. 2 Schematic representation of electro-chemical reactions The voltage is required to be applied for the electrochemical reaction to proceed at a steady state. That voltage or potential difference is around 2 to 30 V. The applied potential difference, however, also overcomes the following resistances or potential drops. They are: • The electrode potential • The activation over potential • Ohmic potential drop • Concentration over potential • Ohmic resistance of electrolyte Fig. 3 shows the total potential drop in ECM cell. Version 2 ME, IIT Kharagpur Anodic overvoltage Anode potential Activation over potential ohmic potential concentration potential concentration potential Ohmic drop activation overpotentialanode cathode cathodic potential cathodic overpotential Voltage Voltage Fig. 3 Total potential drop in ECM cell 3. Equipment The electrochemical machining system has the following modules: • Power supply • Electrolyte filtration and delivery system • Tool feed system • Working tank Fig. 4 schematically shows an electrochemical drilling unit. Flow control valve Pressure relief valve Pump Filters sludge centrifuge Spent electrolyte Tool Flow meter Pressure gauge Constant feed to the tool Low voltage high current power supply +ve -ve PS O M M O P Fig. 4 Schematic diagram of an electrochemical drilling unit Version 2 ME, IIT Kharagpur 4. Modelling of material removal rate Material removal rate (MRR) is an important characteristic to evaluate efficiency of a non-traditional machining process. In ECM, material removal takes place due to atomic dissolution of work material. Electrochemical dissolution is governed by Faraday’s laws. The first law states that the amount of electrochemical dissolution or deposition is proportional to amount of charge passed through the electrochemical cell, which may be expressed as: mQ∝, where m = mass of material dissolved or deposited Q = amount of charge passed The second law states that the amount of material deposited or dissolved further depends on Electrochemical Equivalence (ECE) of the material that is again the ratio atomic weigh and valency. Thus of thet νααAECEm Thus ναQAm where F = Faraday’s constant = 96500 coulombs νFItAm=∴ ρνρFIAtmMRR==∴ where I = current ρ= density of the material The engineering materials are quite often alloys rather than element consisting of different elements in a given proportion. Let us assume there are ‘n’ elements in an alloy. The atomic weights are given as A1, A2, ………….., An with valency during electrochemical dissolution as ν1, ν2, …………, νn. The weight percentages of different elements are α1, α2, ………….., αn (in decimal fraction) Now for passing a current of I for a time t, the mass of material dissolved for any element ‘i’ is given by iaimραΓ= where Γa is the total volume of alloy dissolved. Each element present in the alloy takes a certain amount of charge to dissolve. iiiiFAQmν= iiiiAFmQν=⇒ iiiaiAFQνραΓ=⇒ The total charge passed Version 2 ME, IIT Kharagpur Σ==iTQItQ Σ==∴iiiaTAFItQναρΓ Now Σ=Γ=iia IFtMRRναρ.1 5. Dynamics of Electrochemical Machining ECM can be undertaken without any feed to the tool or with a feed to the tool so that a steady machining gap is maintained. Let us first analyse the dynamics with NO FEED to the tool. Fig. 5 schematically shows the machining (ECM) with no feed to the tool and an instantaneous gap between the tool and workpiece of ‘h’. job electrolyte tool dh h Fig. 5 Schematic representation of the ECM process with no feed to the tool Now over a small time period ‘dt’ a current of I is passed through the electrolyte and that leads to a electrochemical dissolution of the material of amount ‘dh’ over an area of S rhVssrhVRVI===∴ then ⎟⎠⎞⎜⎝⎛=s1.rhVsA.F1dtdhxxρν Version 2 ME, IIT Kharagpur rhV.A.F1xxρν= for a given potential difference and alloy hch1.rFVAdtdhxx==ρν where c = constant rFVAxxρν= Σ=iiiArFVcναρ hcdtdh=∴ cdthdh= At t = 0, h = ho and at t = t1 and h = h1 ∫∫=∴1hoht0dtchdh ct2hh2o21=−∴ That is the tool – workpiece gap under zero feed condition grows gradually following a parabolic curve as shown in Fig. 6 h t ho Fig. 6 Variation of tool-workpiece gap under zero feed condition As hcdtdh= Thus dissolution would gradually decrease with increase in gap as the potential drop across the electrolyte would increase Version 2 ME, IIT Kharagpur Now generally in ECM a feed (f) is given to the tool fhcdtdh−=∴ Now if the feed rate is high as compared to rate of dissolution, then after sometime the gap would diminish and may even lead to short circuiting. Under steady state condition the gap is uniform i.e. the approach of the tool is compensated by dissolution of the work material. Thus with respect to the tool, the workpiece is not moving Thus fhc0dtdh=== hcf=∴ or h* = steady state gap = c/f Now under practical ECM condition it is not possible to set exactly the value of h* as the initial gap. Thus it is required to be analysed if the initial gap value would have any effect on progress of the process Now fhcdtdh−= Now chf*hh'h== And ctf*hft't2== dtdh.f1dtdh.c/fc/f'dt'dh2==∴ Thus fhcdtdh−= fc'hcff*h'hc'dt'dhf−=−=⇒ ⎟⎠⎞⎜⎝⎛−=⇒'h'h1f'dt'dhf 'h'h1'dt'dh−=⇒ 'dh'h1'h'dt−=∴ Now integrating between t’ = 0 to t’ = t’ when h’ changes from ho’ to h1’ 'dh'h1'h'dt't0'1h'oh∫∫−=∴ ()()()∫−+∫−−−=∴'1h'oh'1h'oh'h1d'h1'h1d't 1h1hlnhh't'1'o'1'o−−+−= now for different value of ho’, h1’ seems to approach 1 as shown in Fig. 7 Version 2 ME, IIT Kharagpur h1' t' h0= 0 h0= 0.5 1 Simulation for ho'= 0, 0.5, 1, 2, 3, 4, 5 Fig. 7 Variation in steady state gap with time for different initial gap Thus irrespective of initial gap 1cfh1*hh'h=⇒== fch=∴ or h1.rFVAhcfxxρν== si.FArhV.FAfxxxxρνρν==∴ s/mminMRRsI.FAfxx==∴ρν Thus it seems from the above equation that ECM is self regulating as MRR is equal to feed rate. 6. Applications ECM technique removes material by atomic level dissolution of the same by electrochemical action. Thus the material removal rate or machining is not dependent on the mechanical or physical properties of the work material. It only depends on the atomic weight and valency of the work material and the condition that it should be electrically conductive. Thus ECM can machine any electrically conductive work material irrespective of their hardness, strength or even thermal properties. Moreover Version 2 ME, IIT Kharagpur as ECM leads to atomic level dissolution, the surface finish is excellent with almost stress free machined surface and without any thermal damage. ECM is used for • Die sinking • Profiling and contouring • Trepanning • Grinding • Drilling • Micro-machining Die sinking 3D profiling Work Tool Fig. 8 Different applications of Electro Chemical Machining drilling (drilling) work tool Version 2 ME, IIT Kharagpur trepanning toolwork Fig. 9 Drilling and Trepanning by ECM 7. Process Parameters Power Supply Type direct current Voltage 2 to 35 V Current 50 to 40,000 A Current density 0.1 A/mm2 to 5 A/mm2 Electrolyte Material NaCl and NaNO3 Temperature 20oC – 50oC Flow rate 20 lpm per 100 A current Pressure 0.5 to 20 bar Dilution 100 g/l to 500 g/l Working gap 0.1 mm to 2 mm Overcut 0.2 mm to 3 mm Feed rate 0.5 mm/min to 15 mm/min Electrode material Copper, brass, bronze Surface roughness, Ra 0.2 to 1.5 μm Version 2 ME, IIT Kharagpur Quiz Test 1. For ECM of steel which is used as the electrolyte (a) kerosene (b) NaCl (c) Deionised water (d) HNO3 2. MRR in ECM depends on (a) Hardness of work material (b) atomic weight of work material (c) thermal conductivity of work material (d) ductility of work material 3. ECM cannot be undertaken for (a) steel (b) Nickel based superalloy (c) Al2O3 (d) Titanium alloy 4. Commercial ECM is carried out at a combination of (a) low voltage high current (b) low current low voltage (c) high current high voltage (d) low current low voltage Problems 1. In electrochemical machining of pure iron a material removal rate of 600 mm3/min is required. Estimate current requirement. 2. Composition of a Nickel superalloy is as follows: Ni = 70.0%, Cr = 20.0%, Fe = 5.0% and rest Titanium Calculate rate of dissolution if the area of the tool is 1500 mm2 and a current of 2000 A is being passed through the cell. Assume dissolution to take place at lowest valency of the elements. ANi = 58.71 ρNi = 8.9 νNi = 2 ACr = 51.99 ρCr = 7.19 νCr = 2 AFe = 55.85 ρFe = 7.86 νFe = 2 ATi = 47.9 ρTi = 4.51 νTi = 3 3. In ECM operation of pure iron an equilibrium gap of 2 mm is to be kept. Determine supply voltage, if the total overvoltage is 2.5 V. The resistivity of the electrolyte is 50 Ω-mm and the set feed rate is 0.25 mm/min. Version 2 ME, IIT Kharagpur Answers Answers to Quiz Test 1 – (b) 2 – (b) 3 – (c) 4 – (a) Solution to Prob. 1 νFAItmmMRR.=== ρνρΓFAItmMRR.===∴ MRR = 600 mm3/min = 600/60 mm3/s = 10 mm3/s = 10x10-3cc/s 2x8.7x96500xI5610x103=∴− As AFe = 56 νFe = 2 F = 96500 coulomb ρ = 7.8 gm/cc 562x8.7x10x10x96500I3−=∴ I = 268.8 A Answer Solution of Problem 2 Now, Σ=iialloy1ραρ TiTiFeFeCrCrNiNi1ραραραρα+++= cc/gm07.851.405.086.705.019.72.09.87.01=+++= Now Σ==iiiAFItmMRRναρρ ⎭⎬⎫⎩⎨⎧+++=9.473x05.085.552x05.099.512x2.071.582x75.0x07.8x965001000 = 0.0356 cc/sec = 2.14 cc/min = 2140 mm3/min Version 2 ME, IIT Kharagpur min/mm43.115002140AreaMRRndissolutioofRate===∴ answer Solution to Prob. 3 fc*h= where FeFeFerFVAcνρ= ()2x50x10x8.7x9650085.55x5.2VC3−−= ()7.13475.2V−= ()6025.0x13475.2Vfc2*h−=== 615.55.2V2−= .Volt73.8V=∴ Answer Version 2 ME, IIT Kharagpur