Sje (Paper-Ii) - 2018 Mains Offline Test Series

SJE (PAPER-II) - 2018 MAINS OFFLINE TEST SERIES

MECHANICAL ENGINEERING Full Length MOCK TEST-1 SOLUTIONS

01(a). irreversible process. It is possible to obtain Sol: the values of the gas pressure at  In thermodynamics work done by a system intermediate stages by allowing the partition on its surroundings is defined as an to the move in steps. Then one can plot P interaction whose sole effect, external to versus V and obtain the area under this the system, could be reduced to the raising curve. Yes, the work done by the gas is of a mass through a distance. equal to zero and it is not equal to  P dV.

Free expansion: Situation in which PdV  W =  P dV for a reversible process only.

is finite but work done is zero.  W   P dV for an irreversible process or for  Consider a vessel which is divided into two free expansion. compartments. One compartment contains a gas at a known pressure while the other Paddle Wheel Work: Situation in which  PdV compartment is evacuated. If the partition is is Zero But Work is Done: removed, the gas expands and occupies the  The paddle wheel work process is a process entire container. In this case, the expansion involving friction in which the volume of of the gas is not restrained by an opposing the system does not change at all, and still force (since the other side is vacuum) and work is done on the system. the work done by the gas is equal to zero.  Representation of the process is provided by Such an expansion which is not restrained is a system in which a paddle wheel churns a called a free expansion and it is an fixed mass of fluid as shown in fig.

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 Consider that the system is insulated. Now Electrical work as the paddle wheel runs. Work enters the W = – Eidt for a reversible cell system. It increases the stored energy of the where, E = emf, i = current and t = time. system. The temperature of the fluid  Flow work is a path function increases. There is a corresponding change of state of the system from 1 and 2. But First law of thermodynamics for a steady flow there is no movement of the system energy equation on time basis:

boundary. Hence, PdV is zero, although  V 2   V 2      2    1  Q  W  mh 2   gZ 2   mh1   gZ1  work has been done on the system. Thus, out  2  in  2  PdV does not represent work for this case. Where, Q = rate of heat transfer in kW.

So work may be done on a closed system W = rate of shaft work done by the fluid even though there is no volume change. h = specific enthalpy in kJ/kg ;

System boundary V = velocity of fluid in m/s z = elevation above datum in m ;

 m = mass flow rate into and out of the control work

W volume in kg/s fluid

Fig: Paddle-wheel work COMPRESSOR :

It is a device which converts low pressure fluid to Elastic work: high pressure fluid by giving external work input. W W = – AdL or = –  d V dW 2 h2 (kJ/kg) dm For extension of an elastic rod. dQ dW Where,  = stress,  = strain h   h  ; 1 dm 2 dm dQ A = cross-sectional area of the rod; 1  0 dW dm V = Volume of solid rod, and   (h1  h2 ) h1 (kJ/kg) dm dL = deformation in the rod.  (h 2  h1) kJ/kg Work done in stretching of wire –ve sign indicates work is done on system W = – dL for stretching of a wire (compressor) Where,  = tension.

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THROTTLING PROCESS: Then the first law of thermodynamics  Throttling valves are any type of flow reduces to h2 = h1. restricting devices that cause a significant

pressure drop in the fluid. e.g: capillary 01(b). tubes and valves Sol: Kelvin-Plank Statement:  Unlike turbines they produce a pressure It is impossible to construct a cyclically drop without involving any work. operating device which produces no effect  The pressure drop in the fluid is often other than the extraction of energy as heat accompanied a large drop in temperature. from a single thermal reservoir and  Hence throttling devices are commonly performs an equivalent amount of work. used in refrigeration and air conditioning Clausius Statement: applications. It is impossible to construct a cyclically  Consider the steady –state flow of a fluid working device which produces no effect through a horizontal, insulated pipe in other than the transfer of energy as heat which a porous plug is inserted as shown in from a low temperature body to a high fig. Choose the dotted portion of fig. as the temperature body. control volume and apply the first law of Violation of Clausius statement: thermodynamics. It can be proved that violation of Clausius Porous Plug Insulated statement leads to violation of Kelvin-

Planck statement.

P , T P , T 1 1 2 2 Thermal Reservoir (TH)

Q The flow conditions imply the following: QL H W =Q – Q Steady state: m  m  m and (dE /dt) = 0 W =0 H L 1 2 HE R Horizontal pipe: Z2 = Z1

 QL Insulated: Q = 0 QL

 Thermal Reservoir No shaft work is involved: W s = 0 (TL)

Ignore the changes in KE i.e,V2 = V1

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temperature reservoir and does Work = QH–QL. That is 100% heat is converted to 100% work. Thermal Reservoir ( TH) It is a violation of Kelvin Planck statement.

QL Hence violation of Clausius leads to violation QH of Kelvin Planck statement.

HE R W = 0 01(c). Sol: Given: QH QL Hot reservoir temperature, T1 = 600 K Thermal Reservoir at TL Cold reservoir temperature, T3 = 300 K

Heat supplied, Q1 = 500 kJ Let us assume a refrigerator is operating To find: between reservoir temperatures TL & TH and (i) Temperature at which heat is rejected transfers QL amount of heat from low by engine E1 temperature reservoir (TL) to high temperature (ii) Work done by engine E1 reservoir (TH) without any work input, Hence it (iii) Work done by engine E2 is a false refrigerator and violation of Clausius (iv) Heat rejected by engine E2 statement. (v) Efficiency of the engines. Let an engine operate in the reverse direction 600K between the same reservoirs TH and TL. The 500KJ engine absorbs QH amount of heat from high E 1 W1 temperature reservoir and rejects QL amount of Q heat to low temperature reservoir and does 2

T2 work = QH – QL. The engine doesn't violates Q2 any law. Its a true engine and truth of Kelvin E2 W2 Planck statement. Q3 Let the engine and the refrigerator be 300K considered as a single self acting device which absorbs QH – QL amount of heat from high temperature reservoir and rejects no heat to low

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T2 T2 01(d). E1  1 =1 T1 600 Sol: Given:

T3 300 41.87 o E2  1 = 1 Cp  2.093  J / C T2 T2 t 100 3 Initial volume , V1 = 2000cm 3 (i) The temperature at which heat is rejected by Final volume , V2 = 2400cm o Temperature , T1 = 0 C engineE1 and received by engine E2 is T2 o Temperature , T2 = 100 C Given E1  E2 To find: T  600 300 2 (i) Magnitude of heat interaction  T2 = 300 2 K = 424.3 K (ii) Increase in internal energy 300 2 1 Assumption: E1  E2 1 1  0.3 600 2  Process is reversible

(ii) Work done by engine, E1 (i) Q = CpdT

100 W1 = E1 Q1 Q  C dT  p = 0.3 500 = 150 kJ 0

100  41.87  = 2.093  dT (iii) Work done by engine, E  2 0  t 100  W = 2 E2 Q2 = (2.093  100) + 41.87ln2 = 0.3 350 = 105 kJ Q = 238.322 J

(iv) Heat rejected by engine, E2 (ii) Work done,

Q3 = Q2 – W2 = 350 – 105 = 245 kJ W = PdV

= P(V2 – V1) (v) The efficiency of the engines = 101.3  (2400 – 2000)  10-3 = 40.52 J

(iii) Q = dU + W dU = 238.322 – 40.52 = 197.802 J

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02(a). cooling and lubrication Sol : lubrication requirements. Higher requirements. Lower rate of wear and tear. Four-Stroke Engine Two-Stroke Engine rate of wear and tear. The thermodynamic The thermodynamic Four-stroke engines Two-stroke engines cycle is completed in cycle is completed in have valves and valve have no valves but only four strokes of the two strokes of the actuating mechanisms ports (some two-stroke piston or in two piston or in one for opening and engines are fitted with revolutions of the revolution of the closing of the intake conventional exhaust crankshaft. Thus, one crankshaft. Thus one and exhaust valves. valve or reed valve). power stroke is power stroke is Because of Because of light weight obtained in every two obtained in each comparatively higher and simplicity due to revolutions of the revolution of the weight and the absence of valve crankshaft. crankshaft. complicated valve actuating mechanism, Because of the above, Because of the above, mechanism, the initial initial cost of the turning moment is not turning moment is cost of the engine is engine is less. so uniform and hence more uniform and more. a heavier flywheel is hence a lighter Volumetric efficiency Volumetric efficiency needed. flywheel can be used. is more due to more is low due to lesser Again, because of one Because of one power time for induction. time for induction. power stroke for two stroke for every Thermal efficiency is Thermal efficiency is revolutions, power revolution, power higher; part load lower; part load produced for same produced for same size efficiency is better. efficiency is poor. size of engine is less, of engine is twice, or or for the same power for the same power the Used where efficiency Used where low cost, the engine is heavier engine is lighter and is important, viz., in compactness and light and bulkier. more compact. cars, buses, trucks, weight are important, tractors, industrial viz., in mopeds, Because of one power Because of one power engines, aero planes, scooters, motorcycles stroke in two stroke in one revolution power generation etc hand sprayers etc. revolutions lesser greater cooling and

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02(b). T3 1931.77 T4  1  0.4  809.76 K Sol: re  8.79

(i) P2 57.48 QS P    2.741bar

2 3 3 4  1.4

re  8.79  p=c 4 Heat supplied (qs) = cp(T3 – T2)

Pressure 2 Q R v=c 4 = 1 (1931.77 – 946.946) 1 1 Temperature = 1122.01 kJ/kg Volume Entropy Heat Rejected (qr) = cv (T4 – T1)

T1 = 25C = 298 K = 0.714 (809.76 – 298) = 365.39 kJ/kg P = 1.005 bar = 100.5 kPa RT 0.287 298 1 v  1   0.851 m3 / kg 1 P 100.5 v1 1 rk = 18 = v 2 v1 3 vs = (v1 – v2) = v   0.803 m / kg Cut off = 6.5% of expansion stroke 1 18 6.5 6.5 (iii) Thermal efficiency v  v    (v  v )   (18v  v ) 3 2 100 4 3 100 2 3  q   r  th  1 100  q s   v1    18 & v1  v4  v  365.39   2   1 100  67.43  1122.01 v3 = 1.105 v2 + v2 = 2.04 v2

v3   2.04  rc  2.04 02(c). v2 Sol: Difference between heat pump and v4 rk 18 re     8.79 refrigeration cycle – v3 rc 2.04 A refrigerator is a device which, operating in (ii) Pressures and temperatures at the end of a cycle, maintains a body at a temperature each process  1.4 lower than the temperature of surroundings. P2 = P1  (rk) = 1.005(18) =57.48 bar Performance parameter in a refrigerator T  T r  1  298 180.4  946.946K 2 1 k cycle called the coefficient of performance, T v 3  3 [P = c] abbreviated to COP, which is defined as T v 2 2 Desired effect Q COP =  2 T3 = 2.04  946.946 = 1931.77 K workinput W

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Q Cooling Capacity = 1 TR ∴ [COP]ref  2 …… (1) Q1  Q 2 Bell-Coleman (or) Joule cycle : A heat pump is a device which, operating in T 2 a cycle, maintains a body, say B at a 3 temperature higher than the temperature of 1 the surroundings. By virtue of the 4 temperature difference, there will be heat s leakage Q1 from the body to the surroundings. The body will be maintained p Heat Rejected p=c 3 2 at the constant temperature t1, if heat is Compression   Pv = c Pv = c discharged into the body at the same rate at Expansion which heat leaks out of the body. 4 p=c 1 Heat absorbed

The COP is defined as v

Q COP = 1 Work done during the cycle per kg of air W = Heat rejected – heat absorbed Q1 ∴ [ COP]H.P = ….. (2) = cp (T2 – T3) – cp (T1 – T4) Q1  Q2

Expression for COP in terms of pressure ratio. Relation between COPhp and COPR Heat absorbed in the Refrigerator From Equations (1) and (2), it is found that C.O.P = Net work required [COP]H.P = [COP]ref +1 c T  T  The Cop of heat pump is greater than the  p 1 4 cp T2  T3  cp T1  T4  COP is a refrigerator by unity. T  T   1 4 T2  T3  T1  T4  02(d).  T  Sol:  1  T4  1 o  T4  Temperature at end of heat absorption,T1 = 0 C   T   T  o  2   1  Temperature at end of heat rejection,T3 = 30 C T3  1  T4  1  T3   T4  Pressure ratio P2/P1 = 4 Pressure in the cooler is 4bar.

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For the isentropic compression process 1-2, T  P  1 0.4 3   3   (4) 1.4 1   T4  P4  T  p   2   2  T  p  T 1  1  3 1.48 T Similarly, for the isentropic expansion 4 303 process 3-4 T4  = 204.729 K 1.48 1  T3  p3  NRE (kW) = m N.R.EkJ /kg    T4  p4  NRE = m Cp (T1  T4 ) Since, p2 = p3 and p1 = p4 , the equations 3.5167 = m 1.005  (273  204.729) would be :  m  0.05125 kg/s T T T T 2  3 ; 2  1 T1 T4 T3 T4 We get, (ii) Volume flow rate at inlet of Compressor: T 1 Ideal gas equation is valid for all state COP  4  points. T3  T4  T3   1 T4    P1 V1 = m RT1 1 1   400 V1 = 0.05125  0.287 273 V1  1  1   3  p3   p      2  1 V1 = 0.01 m /s  p   p   4   1  (iii) Volume flow rate at exit of turbine 1  P4 V = m RT4  1 4`   r    1 p  100 V4`= 0.05125 0.287 204.729 p p2 3  3 Where, rp = pressure ratio    V4` = 0.03011 m /s p1 p4

T1 = 273 K ,

T3 = 273 + 30 = 303 K

1 0.4 T2  P2   1.4    = (4) T1  P1 

T2 = 1.48  273 = 405.676 K

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03(a). (iv) Specific gravity (S) or Relative density: Sol: It is the ratio of the mass density of any (i) Mass Density (): It is the mass of the matter to the mass density of a standard fluid matter occupied in unit volume at a standard   (Water). S     temperature and pressure. It is denoted by . water water M  Unit: No unit as it is a ratio of similar   V properties. Dimensional formula of : ML 3 3 Unit : kg / m ………S.I system Matter Specific Gravity Air 0.0012

(ii) Specific Volume (vs) Water 1.0  Volume occupied by unit mass of Wood 0.6 fluid Mercury 13.6

1 3 vs  (m / kg)  Compressibility (β) and Bulk Modulus of  It is the reciprocal of mass density. elasticity (K) Compressibility is the inverse of bulk (iii) Specific Weight ( or w or g): Weight of modulus. the matter per unit volume. 1   K W Mg     g dp dp V V K =  = dV / V d /  Dimensional formula of :

FL–3 = ML–2T–2 Bulk modulus of elasticity is also known as Units: N/m3 Coefficient of compressibility.  It is not absolute quantity and varies

from place to place as ‘g’ changes from place to place.

2 gpoles ≃ 9.83 m/s

2 gequator ≃ 9.78 m/s

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Neutral equilibrium: If a floating object is 03(b). tilted from its equilibrium position by small Sol: displacement and if it remains in the original Stable equilibrium: If a floating object is position after removal of the tilting moment tilted from its equilibrium position by small then the object is said to be in neutral displacement and if it comes to the original equilibrium. For floating bodies metacentre position after removal of the tilting moment must coincide centre of gravity for neutral then the object is said to be in stable equilibrium. equilibrium. For floating bodies metacentre

must be above centre of gravity for stable

equilibrium. Fb G M

F b mg Restoring M moment No Restoring or overturning moment G Unstable: If the metacentre lies below its mg centre gravity then the body is said to be in

an unstable state of equilibrium. Unstable equilibrium: If a floating object is Neutral: If for a body the metacentre tilted from its equilibrium position by small coincides with the centre of gravity of the displacement and if it does not come to the body, then the body will be in a neutral state original position after removal of the tilting of equilibrium moment then the object is said to be in

unstable equilibrium. For floating bodies 03(c). metacentre must be below centre of gravity Sol: Impeller Diameter, D = 80 cm for unstable equilibrium. 2 Discharge, Q = 1 m3/s

F Over turning b Head, H = 80 m moment M Speed, N = 1000rpm G Fb Width at outlet, B2 = 8cm mg Leakage = 3% of Discharge ,  = 80% M hyd Mechanical loss = 10 kW

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Assumptions : Vf 2 5.12 tan 2   1. The flow is steady u2  Vw2 41.89  23.42 2. The fluid is water and it is incompressible   15.5o 3. The blades are infinitesimally thin (ii) Power required: 4. The flow is everywhere tangent to the (V u )  runner blades. Power required = w2 2 gQ  loss  g    (i) Blade angle at outlet = (23.42 41.89 1.03 1000) + 10 103

3 Qth = 1  1.03 = 1.03 m / s = 1020.495 kW Peripheral velocity at outlet is given as 1010.495   100  99% mech 1020.495 D2 N 0.81000 u2 =   41.89m / s 60 60 (iii) Overall efficiency: Discharge, Q2 = D2B2Vf 2

overall = mano  mech 1.03 =   0.8  0.08  Vf2 = 0.80  0.99 = 0.792 = 79.2 %  Vf2 = 5.12 m/s

u2 03(d). Vw2 Sol: Cavitation is the phenomenon which exists  2 in the flowing liquids when the liquid Vf2 V2 Vr2 pressure in the flow system drops below the vapour pressure at some location. Due to this, some of the liquid flashes into vapour forming vapour bubbles (called cavitation bubbles). The vapour bubbles collapse as they are swept away from the low pressure Manometric efficiency of pump is regions, generating highly destructive, extremely high pressure waves. gH m gH m mano =  Vw  V u 2   u Cavitation occurs on the exit side of the w 2 2 mano 2 turbine runner or at inlet to the draft tube in 9.8180  V =  23.42m/s w2 0.841.89 case of turbine and tip regions of impellers or suction side in case of the pumps.

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Cavitation must be avoided or atleast penetrate into the component and oxygen minimized by selecting proper material such goes out. as stainless steel or alloy steel, by coating  Due to continuous penetration of carbon blades with harder materials, by keeping the atom, the outer envelope will be produced

runner under water or by designing with more iron carbide (Fe3C) phase & turns cavitation free runner. as hard.  The depth up to which form the surface of the 04(a). component has been hardened is known as

Sol: case depth (CD). (i) Carburising:  Since the solubility of carbon is more in  Carburizing is a method of enriching the austenitic state than in ferritic state, fully surface layer of low carbon steel with carbon austenitic state is required for carburizing. in order to produce a hard case.  This can be achieved by heating the steel  This can be carried out by incorporating C above the critical temperature. And diffusion atoms on to the envelope of the L.C.S of carbon is made by holding the heated steel component it will be turned as hard by in contact with carbonaceous material which

forming Fe3C phase in known as carburizing. may be a solid, a liquid or a gas.

O2 Case 1: Heating C+O2 Fe3C coils  Two L.C.S Component with different % of C L.C.S Heating has been carburized, then CD1 > CD2 because C D Jacket if the component contains low carbon content then penetration of external carbon atoms will CO gas be easy more case depth can be achieved.

Case Depth = 0.5 mm/5 hour C C  By heating the component to 8500C 0.1%C temperature CO gas is circulated, in the 0.2%C

heating envelope. CD1 CD2  At that temp carbon monoxide decomposes CD1 > CD2 into carbon and oxygen where carbon will

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Note: steel. The alloying elements having more

 In hardening process to achieve more Dh affinity for nitrogen are aluminum, chromium value, then carbon content should be high but and molybdenum.

in case of case hardening process to achieve H2

to more case depth CD, the carbon content N2 + H2 should be low. Fe3N L.C.S

CD (ii) Nitriding:  Nitriding is the process of enriching the surface of steel with nitrogen by holding for a NH3 gas prolonged period at temperature of ammonia Obtaining more case depth in Nitriding

(NH3). process is difficult because:  In this process the machined and heat treated  The size of the nitrogen atom is large and (hardening by heating to 930C and quenching inert in nature  more ammonia should be 0 in oil, then tempering at 650 to obtain the consumed to obtain more case depth  required properties in core) components are expensive. 0 heated to a temperature of 500 C for between  As it is condition low carbon steel contain 40 to 100 hours (depending on case depth) in low corrosion resistance but after Nitriding a gas tight chamber through which ammonia process they possess extreme corrosion is allowed to circulate. resistance due to inert nature of non nitride  By incorporating Nitrogen atom on the outer phase on the surface.

envelope of L.C.S component it will be  Advantages of this process are termed as hard by forming iron Nitride phase  good surface finish Case Depth = 0.5 mm /hrs.  less distortion and cracks  Ammonia dissociates according to the  good wear resistance following reaction.  used for mass production

2 NH3  3H2 + 2N  better than carburizing

 The atomic nitrogen thus formed diffuses into  Disadvantages of this process are iron forms hard nitrides by combining with  long operational times 100 hours for iron and certain alloying elements present in 0.038mm depth

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 all alloys steels can not be used  Carbon and nitrogen thus formed in atomic  special equipment is needed form diffuse into steel surface.  More oxidation due to prolonged heating  Case Depth = 0.5 mm/10 hrs  This process usually requires 30 to 90 minutes (iii) Cyaniding for completion.  During cyaniding the surface of steel is  After cyaniding, the components are taken out enriched with carbon and nitrogen by and quenched in water or oil followed by low incorporating carbon & Nitrogen atoms temperature tempering. simultaneous on to the outer envelope of the  Advantages are: low carbon steels it will be turned as hard  can be applied to Low carbon and medium by form iron carbide & iron Nitride phases. carbon steels  In this process the components are immersed  bright finish in parts can be obtained

in a liquid bath of 30 % NaCN, 40 % Na2CO3  cracks and distortions are minimized and 30 %NaCl, maintained at a temperature  Most suitable for parts subjected to high of 8000C to 8500C. loads.  Then a measured amount of air is passed  Disadvantages are: through the molten bath.  risk of splitting of poisonous salts  Sodium cyanide reacts with oxygen of the air  unhealthy fumes are formed and is oxidized. The basic reactions in the bath are: 04(b). Sol: 2 NaCN + O2  2 NaCNO (i) TIG or GTAW (Tungsten electrode 2 NaCNO + O2  Na2 Co3 + CO+ 2 N welding or Gas tungsten arc welding):

2 CO  CO2 + C Principle: It is the non-consumable arc Nothing welding in which the electrode is made by (Fe3N+Fe3C) using tungsten as material. In TIG welding when the power supply is given the arc is L.C.S produced between the tip of the tungsten CD electrode and work piece and the heat generated due to this is used for joining of

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the plates. When the welding is continuing the inert gas will be continuously supplied the Inert gas mainly the argon will be to the weld zone through the cover provided supplied continuously to the weld zone so around the electrode for protecting the weld that the weld pool is protected from pool from atmospheric contamination. atmospheric contamination. In addition the The equipment required for the TIG cooling is supplied through a small pipe welding: which is wrapped around the electrode for Welding torch, continuous electrode wound cooling of the electrode. on the spool, welding power source( DC The equipment required for the TIG generator), Inert gas cylinder, pressure welding: regulator for regulating inert gas supply etc. Welding torch, tungsten electrode, coolant Applications: This process is mainly pump, coolant, welding power source( DC developed for joining of highly reactive generator), filler rod material, Inert gas metals like aluminum, magnesium etc. It cylinder, pressure regulator for regulating also can be used for joining of stainless inert gas supply, etc. steels, heat resistant steels, refractory Applications: This process is mainly materials etc. having thickness greater than developed for joining of highly reactive 5mm in single pass welding operation. Also metals like aluminum, magnesium etc. It the cost of the GMAW is less than the TIG also can be used for joining of stainless welding. steels, heat resistant steels, refractory materials etc. 04(c). Sol: The commonly produced defects in casting (ii) Gas metal arc welding process : It is the are consumable electrode arc welding in which 1. Blow holes : The air or gas bubbles the continuously consumable electrode present in the casting is called blow or which is wound on to the spool is fed by blow holes. The blow hole present on using servo-controls so that the arc the surface of casting is called scar and produced between the tip of the electrode blow hole present very near to the and work piece will generate heat and this surface of casting is called blister. This heat is used for melting of the plates and is due to joining. When the welding is taking place,

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(i) Low porosity property of molding casting process. The chills will provide sand: this can be eliminated by the directional solidification so that it is increasing porosity property using possible to ensure that the solidification selecting large size sand particles, will takes as per the requirement and reducing ramming force, reducing eliminates the shrinkage cavity defect. clay content, providing vent holes 3. Dross : The presence of inclusions in etc. the casting is called dross. This is due to (ii) Aspiration effect during poring of improper separation of impurities molten metal: by selecting the present in the molten metal. Eliminated shape of the sprue as parabolic by using strainer, skim-bob, providing tapered sprue or straight tapered step between the axis of the runner and sprue. ingate etc. (iii) Partial flow of molten metal 4. Sand inclusion : The sand particles during pouring of molten metal : present in the casting is called the sand full flow of molten metal can be inclusion or simply inclusion. This is ensured by placing thin metallic mainly due to sand erosion taking place plate made by using same material during filling of molten metal during as that of the casting so that when casting. The sand erosion can be metal is collecting over the plate the eliminated by ensuring laminar flow of thin plate will get melted and the molten metal in the gating system, molten metal produced by plate and avoiding aspiration effect, using bottom molten metal collected over the plate gating system for filling of loose sand can be flowing through the gating molds, using pads during casting of system which ensures the full flow sharp edged castings etc. of molten metal through the gating 5. Misrun : The method of non filling of system. molten into the projected portion of the 2. Shrinkage cavity : The void produced complex shape of the casting. This is in the casting due to non availability of mainly occurring due to the molten metal during solidification. This solidification has been started before can be eliminated by using chills in the complete filling of the casting cavity.

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This can be eliminated by reducing the Center less Grinding: Used for grinding pouring time of molten into the cavity, circular job without fixing the job between increasing the degree of superheat of centers, but is held against the face of molten metal. Grinding Wheel (GW) by combination of supporting rest and a regulating wheel. So it 04(d). does not require center holes, drivers and Sol: Cylindrical Grinding: The cylindrical other fixtures for holding the work piece. grinding is also called as roll grinding During the process, the work piece is process in which the work piece is mounted supported on work rest plate and the between the centers or in the chuck of regulating wheel holds the work piece against center lathe and rotating continuously. The the horizontal force of action controlling its grinding wheel is rotating and reciprocating size and imparting necessary rotational and the work piece to remove the material from longitudinal feed. Thickness of work rest work piece. The grinding wheel will be blade is usually less than the diameter of work given an independent power to rotate at a piece, for smaller jobs and max thickness is high speed suitable for grinding operation. around 20 mm. The slope of edge is 1/20 to The work which is normally held between 80. the centers is rotated at a much lower speed The speed of regulating wheel in direction opposite to that of the grinding = 15 to 60 m/min. wheel. The table assembly which houses the The speed G.W= 1800 m/min. centers can be reciprocated to provide the Note: The work piece is usually ground necessary traverse feed of the work piece with its centre above the line of centers of past the grinding wheel by using hydraulic the wheels by about half the diameter of arrangements. The infeed is provided by the work piece, the max. value = 12 mm, movement of the grinding wheel head into Regulating wheel the work piece. Typical grinding allowance Work Workpiece  piece kept are 0.1 to 0.3 mm, beyond which the Wheel grinding operation becomes too expensive.

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 = inclination of regulating wheel with

respect to Grinding wheel D C 4 So velocity component is two parts 3 B i) vsin : gives necessary feed 1  

ii) vcos : gives rotational motion to the  2  6 11 1 wheel. C A C 5

05(a). As in the figure link ‘2’, i.e. AB is fixed and Sol: In this type of inversion ‘A’ and ‘B’ are the link BC acts as crank. Length of the link both held fixed and the link BC rotates in a AD is extended on the side of ‘A’ such that an circle around B. The slider C slides back extra link is connected with a cutting tool at and forth along the rotating AD (link ‘1’). ‘E’. Circular path traced by the slider with crank rotation provides information of cutting Applications: tool travel time during the operation. Whit-worth quick return motion mechanism Stroke time is proportional to the distance and rotary engine. traveled by the slider on the circumference of 4 C the circle D Time of cutting Forwardstroke time 3  B Time of return Re turn stroke time 1 2 FT 180  2r 180  2    A RT 180  2r 180  2 Whitworth Quick-Return Mechanism obtuseangleC1BC 11 at B  This mechanism is used in shaper machine acute angleC1BC 11 at B tools for cutting metals. Forward stroke cuts IIIrd Inversion: the metal which takes a little longer time than In this type of inversion B and C are both held return stroke that is idle, thus called a quick- fixed but the slider is allowed to swivel. return mechanism. In this the slider guide link B is made to rotate. This linkage also converts 3 2 rotary motion of the crank into oscillatory C 4 motion of the slider link (tool). A 1 D

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Applications: 05(b). Crank and slotted lever mechanism Sol: Single plate clutch Given that  = 0.225 Crank and slotted lever mechanism d 1 1.25 Here link ‘BC’ is fixed and ‘AB’ is crank which d 2 -3 makes the link ‘AD’ to move in the slotted Pmax = 0.019.8 MPa = 98 10 MPa lever (through the slider). N = 3000 rpm

Power = 33.5 HP = 33.5 0.7457 kW 5 6

1 = 24.832 kW A   B B D A1 A11 Assumptions:

 Uniform wear theory 4 3

  Active on both sides.  C C d1 = outer diameter As in figure, the link ‘CD’ can be made into a d2 = inner sides groove with a slider attached to the crank. P60 P = T  T  0.07904 kNm Because of this crank rotation, the groove 2N experiences rotational oscillation which in turn makes cutting tool into linear oscillations But in uniform wear, max torque

through an extra link connected to it. At r2 T  2 2 Pr2 dr r extreme positions BA  CA and BA  2 We multiply by 2 because it is active on both CA. As in the whit-worth quick return sides. mechanism forward stroke takes more time We know that, C = Pr = P r = P r than return stroke and 1 1 2 2

r1 1 11 T  2 2c rdr Time of cutting obtuse angle A BA atB r  2 Time of return acuteangle A1BA11 atB  r2  r2   1 2  180  2  4 P2r2   =  2  180  2

r1 but  1.25 (where P2 = Pmax) r2

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2T Then maximum fluctuation occurs between  r 3  2 2  r   these angles.  1   4pmax   1  r2     150 sin3 T

2 0.07904 103 150 A C = 6 2 4  0.225  0.098 10 1.25 1 Tmean 500 3 3 r2 1.014 10 m B

r 0.10047m 2  r2 = 100 mm 30 60 120 150 90 180  r1 = 125.58 mm

60 KE  E  T d 05(c).  max  0 Sol: Given that 60  150 sin 3d  50 cos3 60 T = (500 + 150 sin 3) kgf-m  0 0 I = 103 kg-m2, N = 300 rpm = 100 kg-f-m = 981 Nm The turning moment curve could be E 981 repeated for every 120. Cs    0.099% I2 103  102 1 2/ 3  T  Td mean 2  0 3 05(d). Sol: Sensitiveness:  2  3  150 3  A governor is said to be sensitive when it  500  cos3  2  3   0  readily responds to a small change of speed. = 500 kgf-m The movement of the sleeve for a fractional

(a) Power = Tmean × = 500 × 10 kgf – m/s change of speed is the measure of = 15.708 9.81 kW = 154.095 kW sensitivity. N  N 2N  N  (b) (i) When the resisting torque is constant Sensitiveness  2 1  2 1 N N 2  N1  Tmean = Tresistance = 500

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N1 = minimum speed corresponding to full 06(a). load conditions T  Sol: Torsion formula,  ...... (1) N2 = maximum speed corresponding to no- J R T J load conditions    R Stability: As both shafts (solid and hollow) are A governor is said to be stable if it brings subjected to same torque and same the speed of the engine to the required value maximum shear stress, hence ratio J/R must and there is not much hunting. The ball also be same for them. masses occupy a definite position for each

speed of the engine within the working (i) For solid shaft, range. Obviously, the stability and the 3 sensitivity are two opposite characteristics. J  4 2  d  .d    ...... 2 R 32 d 16 Isochronisms:   A governor with a range of speed zero is known as an isochronous governor. This (ii) For hollow shaft, mean that for all positions of the sleeve or J   2  4 4    .Do  Di .  the balls, the governor has the same R 32  Do  equilibrium speed. However, it is not  4    4 2    Do   Do   practical due to the friction at the sleeve 16.Do  3   Hunting: D3 65 A governor is said to be hunt if the speed of  o  ...... 3 16 81 the engine fluctuates continuously above From (1) , (2) and (3), and below the mean speed. This is caused d3 D3 65 by too sensitive governor which changes the  o  16 16 81 fuel supply by a large amount when a small 1/ 3 D  81  change in speed of rotation takes place.  o   . ...... 4 d  65  A governor is said to be isochronous when

the equilibrium speed is constant (i.e., range (iii) Weight of solid shaft, of speed is zero) for Degree of hunting is  more in unstable governors. W  .d2.L g s 4 s

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Weight of hollow shaft,  W  D2  D2 L g 50 kN 40 kN H  o i  H 20 kN/m 4 100 kN.m   2   D2   D2 .L g C D E F G 4  o 3 o  H A    0.5 m 1 m 2 m 1.5 m 0.5 m 0.5 m

5  2 R W  . .D L g RB E H o H 40 kN 9 4 35 kN 35 kN Here, H = s (+ve) (+ve) Thus, A (–ve) C D 2 2/3 B E F G W 5 D  5 81 (–ve) H   o      0.6433 WS 9  d  9 65 –10 kN –15 kN SFD Hence, for given conditions, hollow shaft 32.5 kN.m will be 35.67% lighter than the solid shaft.

2.5 kN.m B (+ve) C (+ve) E F 06(b). A (–ve) G (–ve) (–ve) D Sol: Reactions: –2.5 kN.m –20 kN.m Take Fy = 0, BMD  R  R  20 0.5 40 50 B E –67.5 kN.m Consider a moment at point B, Shear Force: MB = 0 (SF) = 0 kN  G (SF)F = 40 kN RE 4.5 40 550 3100  20 0.50.25  0  RE = 55 kN (SF)E,Right = 40 kN

and RB = 45 kN (SF)E,Left = 40 – RE = –15 kN

(SF)D,Right = –15 kN

(SF)D,Left = –15 + 50 = 35 kN

(SF)B,Right = 35 kN

(SF) Left = 35 – R = –10 kN B, B

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Bending Moment: N2 = 100 N

 0.5  F2 = 1 N2 = 0.25  100 = 25 N BM   20  0.5    2.5 kN.m B 2   T = F2 = 25 N  0.5  Tension in the cord is 25 N BM  20  0.5 1  R 1  Cleft   B  2  = 32.5 kN.m From FBD of block B: BM  32.5 100  67.5 kN.m  Cright N1 + P sin 45 = 250

N1 = 250 – P sin 45 ...... (a) BMD  40 1.5  0.5 R E 1.5  2.5 kN.m P cos 45 = 25 + 2 N1 ……… (b) BM  40 0.5  20 kN.m P E P cos 45 = 25 + 0.3(250– ) 2

P 06(c). ( 1+ 0.3) = 25 + 75 = 100 2 Sol: 1 = 0.25 00  2 P = 100 N 1.3 P A  = 45  P = 108.78 N 150 N B 2 = 0.3  T = 25 N Free body diagram ‘B’ is shown below

N2 06(d). Sol: Consider a thin cylinder of diameter, D A T F2 thickness, t subjected to internal pressure P. Hoop stress: 100 N Consider the cross-section of thin cylinder as shown in figure P sin 45

F2 h B P cos 45

2.N1 = F1 L

250 N

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Bursting force = resisting force Bursting force = resisting force PLD   2L t   h P D 2   .D.t 4   PD     h 2t PD 1    h Longitudinal stress 4t 2 Consider the cross section as shown in figure.

P  l

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