Automobile Engineering - Solved Problem - Engine Combustion Parameters & Efficiency

----------------------------------------------------------------------------------------------------------

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       
------------------------------------------------------------------------
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.

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

***************************************************************************

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                                       

 

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

Velocity of Metal flow in a Sprue - Metal Casting

VELOCITY OF METAL FLOW IN A SPRUE

Green sand casting is one of the simplest and most popular casting processes. The following figure shows the schematic diagram of a simple green sand mold together with all the components of the gating system. 

The gating system of a sand mold consists of pouring basin, sprue, sprue basin, runner, gates, and runner extension. The design of the gating system mainly involves considerations of fluid flow. To avoid formation of casting defects, it is necessary to control the rate of mold filling. Too fast a metal flow causes air entrapment, porosity and dross formation, and erosion of sand, whereas too slow a flow causes the metal to solidify prematurely, yielding defects such as misrun (an incomplete casting) and cold laps (inhomogeneous or layered casting surface). A well-designed gating system evenly distributes the incoming metal to all parts of the mold without causing turbulence or sand erosion.

A pouring basin is either rectangular or square in cross-section, with a fiat base and a fillet at the base near the sprue entrance. Spherical pouring basins with a curved base cause vortex flow and casting defects. A sprue connects the pouring basin to the runner. A sprue is usually tapered to allow for downward laminar flow, and the sprue basin provides space for the metal to dissipate some energy before changing its direction as it flows into the runners.

EXAMPLE

The figure shows a tapered sprue with the metal head H₁ in the pouring basin, and the total metal head, H₂. The cross-sectional areas at the top and the bottom of the sprue are A₁ and A₂. Determine the metal flow velocity at the top and bottom of the sprue. Also derive a relation between heads H₁, H₂ and cross-sectional areas A₁, A₂ of the sprue.

SOLUTION

From the basic fluid mechanics, using Bernoulli’s Principle, we have

The velocity of fluid (v) at a cross-section with head ‘h’ is given by

Metal head at the top of sprue is H₁ and metal head at the bottom of sprue is H₂.

Similarly, considering the metal flow in the gating system,

The metal flow velocity at the top (v_top) of the sprue is given by

The metal flow velocity at the bottom (v_bottom) of the sprue is given by

We know that the fluid mass crossing the areas A₁ and A₂ per unit time is constant i.e. the metal mass flow rate is constant.

So, the mass flow rate (ṁ) at the top of sprue is equal to that at the bottom of the sprue.

Mass flow rate (ṁ) = velocity × density × Area of cross-section

This yields the fundamental relationship A₁√H₁ = A₂√H₂   .

Forming Processes - EXTRUSION - Numerical Problems with Solutions (solved)

FORMING PROCESSES (Extrusion)

EXAMPLE 3.11 (EXTRUSION)

Estimate the maximum force required for extruding a cylindrical Aluminium billet of 50 mm diameter and 75 mm length to a final diameter of 10 mm. The average tensile yield stress for Aluminium is 170 N/mm². What percent of the total power input will be lost in friction at the start of the operation?

FORMULAE

SOLUTION

The ram force is obviously maximum just after the start of the extrusion process as l is maximum. For the given dimensions and assuming a 45°-dead zone, the value of l at the beginning is given as

Next, the value of µ is determined by solving equation (1) through a trial and error method. Thus, we get µ = 0.15. Now, σ_x|_BB is found out from equation (2) with this value of µ. Hence,

Finally, using equation (3), we have

If the velocity of the ram is V_B mm/sec, then the frictional power loss (using equations (4) and (5)) is given as

Now, the total input power is

Therefore, the power loss in friction is

Forming Processes - DRAWING - Numerical Problems with Solutions (solved)

FORMING PROCESSES (Drawing & Deep Drawing)

EXAMPLE 3.7 (DRAWING)

A steel wire is drawn from an initial diameter of 12.7 mm to a final diameter of 10.2 mm at a speed of 90 m/min. The half-cone angle of the die is 6° and the coefficient of friction at the job-die interface is 0.1. A tensile test on the original steel specimen gives a tensile yield stress 207 N/mm². A similar specimen shows a tensile yield stress of 414 N/mm² at a strain of 0.5. Assuming a linear stress-strain relationship for the material, determine the drawing power and the maximum possible reduction with the same die. No back tension is applied.

FORMULAE

SOLUTION

To start with, we determine the average tensile yield stress of the material during the process. For the given operation, the strain is

Since the stress-strain relationship is linear, the value of the tensile yield stress at the end of the operation is given by

So, the average yield stress is

From the given data,

For the strain hardening a material, the value of σ_xf, strictly speaking, can go up to the value of the tensile yield stress at the end of the operation (σ_Yf) which may be considerably greater than the average value σ_Y. So, when (σ_xf)_max is put equal to σ_Yf,

Now, from equation (1) with F_b = 0,

Using equations (2) and (3), the drawing power we obtain is

It is interesting to note that the drawing stress is more than the tensile yield stress of the original material. To find out the maximum allowable reduction, we use equation (4) with F_b = 0 and ϕ = 2. Hence,

If we use equation (5), then D_max = 0.652

EXAMPLE 3.8 (DEEP DRAWING)

A cold rolled steel cup with an inside radius 30 mm and a thickness 3 mm is to be drawn from a blank of radius 40 mm. The shear yield stress and the maximum allowable stress of the material can be taken as 210 N/mm² and 600 N/mm², respectively.

(i)                Determine the drawing force, assuming that the coefficient  of friction µ = 0.1 and β = 0.05

(ii)             Determine the minimum possible radius of the cup which can be drawn from the given blank without causing a fracture.

FORMULAE

SOLUTION

(i) We first calculate the blank holding force F_h from the given data as

Next, we find the value of σ_r at r → r_d by using equation (1). Thus,

Now, using equation (2), we get

It should be noted that this is much less than the fracture strength though r_j – r_p = 10 mm = 3.33t, i.e., very close to the limit set by the condition of plastic buckling. From equation (3), the drawing force is given as

(ii) In this case, σ_z = 600 N/mm². From equation (2), we have

Using equation (1) and taking F_h = 52778 N, we get

Again, it is interesting to note that r_j – r_p = 30.8 mm >> 4t, and this goes much beyond the limit set by the plastic buckling condition.

Forming Processes - FORGING - Numerical Problems with Solutions (solved)

FORMING PROCESSES (Forging of Strip & Forging of Disc)

EXAMPLE 3.4 (FORGING OF A STRIP)

A strip of lead with initial dimensions 24 mm × 24 mm × 150 mm is forged between two flat dies to a final size of 6 mm × 96 mm × 150 mm. If the coefficient of friction between the job and the die is 0.25, determine the maximum forging force. The average yield stress of lead in tension is 7 N/mm².

SOLUTION

First, let us determine the shear yield stress K for lead by using following equation. Thus,

To find p, the value of x_s is required. From following equation, we have

Now, from the following equations, the expressions for the pressures p₁ and p₂ (for the non-sticking and the sticking zones, respectively) can be found out. Thus,

Using following equation, the force per unit length we get is

Since the length of the strip is 150 mm, the total forging force is

EXAMPLE 3.5

Solve Example 3.4 when the coefficient of friction µ = 0.08

SOLUTION

Using the following equation, we obtain

Since, x_s is more than l, the entire zone is non-sticking, and, as a result, the expression for the pressure throughout the contact surface is given by following equation. Thus,

So,

The corresponding value of the total forging force is

EXAMPLE 3.6 (DISC FORGING)

A circular disc of lead of radius 150 mm and thickness 50 mm is reduced to a thickness of 25 mm by open die forging. If the coefficient of friction between the jib and the die is 0.25, determine the maximum forging force. The average shear yield stress of lead can be taken as 4 N/mm².

FORMULAE:

SOLUTION

Since the volume remains constant, the final radius of the disc is

R = √2 × 150 mm = 212.1 mm.

Þ R = 212.1 mm

Now, using equation (1)

The expressions for p₁ and p₂ using equations (2) and (3) are

Substituting the values of K, h, R, µ and r_s in equation (4), the total forging force we obtain is

Forming Processes - ROLLING - Numerical Problems with Solutions (solved)

FORMING PROCESSES

Example 3.1 (ROLLING)

A strip with a cross-section of 150 mm × 6 mm is being rolled with 20% reduction in area, using 400-mm-diameter steel rolls. Before and after rolling, the shear yield stress of the material is 0.35 kN/mm² and 0.4 kN/mm², respectively. Calculate (i) the final strip thickness, (ii) the average shear yield stress during the process, (iii) the angle subtended by the deformation zone at the roll centre, and (iv) the location of the neutral point θ_n. Assume the coefficient of friction to be 0.1

SOLUTION

(i) As no widening is considered during rolling, 20% reduction in the area implies a longitudinal strain of 0.2 with consequent 20% reduction in the thickness. Therefore, the final strip thickness is given as

(ii) The average shear yield stress during the process is taken to be the arithmetic mean of the initial and the final values of the yield stress. So,

(iii) From figure, it is clear that

Substituting the values, we get

(iv) To determine θ_n, first λ_n has to be calculated. Since nothing has been mentioned regarding the forward and the back tension, σ_xi and σ_xt will be assumed zero. Hence,

where

Substituting the values of R, t_f, and θ_i in the expression for λ_i, we get λ_i = 5.99.

Now using this value of λ_i, we find that the value of λ_n is 1.88

θ_n can be expressed as

Hence,

EXAMPLE 3.2 (ROLLING)

Assuming the speed of rolling to be 30 m/min, determine (i) the roll separating force and (ii) the power required in the rolling process describes in Example 3.1

SOLUTION

From above equation, we see that to find out the roll separating force F, we need to know the pressure p (as a function of θ), θ_n and θ_i. In Example 3.1, we have already calculated θ_n and θ_i. For the numerical integration of the above equation, we employ Simpson’s rule, using four divisions after θ_n and eight divisions before θ_n. Thus,

With these intervals, the pressure at different station points can be computed, using following equations.

Next, we compile the following data. After the neutral point:

Station Point	θ (rad)	(mm)		pafter (kN/mm2)
1	0.00000	2.40000	0.000	0.75
2	0.00575	2.40331	0.479	0.788
3	0.0115	2.4132	0.958	0.830
4	0.01725	2.4298	1.432	0.876
5	0.023	2.453	1.88	0.925

Before the neutral point:

Station Point	θ (rad)	(mm)		Pbefore (kN/mm2)
1	0.023	2.453	1.88	0.925
2	0.02981	2.489	2.454	0.887
3	0.03662	2.534	2.997	0.855
4	0.04343	2.589	3.529	0.828
5	0.05024	2.65	4.049	0.804
6	0.05705	2.725	4.555	0.786
7	0.06386	2.81	5.048	0.772
8	0.07067	2.899	5.525	0.759
9	0.0775	3.000	5.99	0.75

Now, from equation (1), the roll separating force per unit width is

(i) Since the width of the strip is 150 mm, the roll separating force is

(ii) The driving torque per unit width for each roll can be computed, as

Taking the values before and after the neutral point and using Simpson’s rule, we find

So, the total driving torque for each roll is 100.8 × 150 N-m = 15210 N-m

Now, the total power required to drive the two rolls is 2 × 15, 120 × ω W, ω being the roll speed in rad/sec. The rolling speed (same as peripheral speed of rolls) is given to be 30 m/min. So,

Thus, the total power (2P_R) = 75.6 kW.

The power required in the rolling process is 75.6 kW

EXAMPLE 3.3

If the diameter of the bearings in Examples 3.1 and 3.2 is 150 mm and the coefficient of friction (µ_b) is 0.005, estimate the required mill power.

SOLUTION

Substituting the values of µ_b, d_b, ω, and F in following equation, we obtain

So, the mill power is given as