January 2017

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Engineering Drawing is the one of the technical subject and is quite differ with the other types of drawing. It is totally based on the accurate measurement and scale system.
The Engineering Drawing would be tough subject to the students who are newer in the drawing sector. This becomes slightly tough due to having low imagination power, i.e thinking power is required to solve the problem of Engineering drawing. Mostly in the Engineering Drawing, we analyze the given condition and imagine the whole body structure from the given least tips. The problem of Engineering drawing would be concerned with the Conversion of three dimension to two dimension and vice versa. According to the principle of to show the maximum possible view of an object the engineering drawing uses many technique to prepare drawing. This would be start from drawing basic feature of two dimension such as point, line, triangle, rectangle and so on, similarly basic feature of three dimension such as tetrahedron, cube, sphere etc. Then it goes simpler to complex. Intersection of line, line and plane’s characteristics, surface development, cross section, orthographic drawing, interaction curve drawing etc. Similarly on three dimensional drawing production such as the isometric drawing, oblique drawing, perspective drawing etc.  The main usefull drawing under  engineering drawing are Machine drawing, Building drawing, circuit drawing (Electronics drawing) etc. These drawing are used for various field such as Designing and Development (Machine design, civil engineering design, PCB board design etc)
Now a development of some software like AutoCAD removes the much tedious traditional task (hand drawing)  to convenient form. 

Engineering  Drawing II is a portion of basic Engineering drawing.

Here are the some material for Engineering Drawing II. You can click the link provided to study/download the available material for Engineering Drawing.
1.EngineeringDrawing II Tutorial Sheet- By Department of Mechanical Engineering Pulchowk Campus
2. EngineeringDrawing II Tutorial Solution-By Narayan Nyaupane
3. EngineeringDrawing II Tutorial Solution – By Bibek Tamang
4.Imagineand Visualize in 3 D Engineering Drawing II Tutorial Solution – By Bibek Tamang
5. Symbolsused in Drawing
6.READING AND INTERPRETING CONSTRUCTION DRAWINGS- By J.N. Ramaswamy, Ph.D., P.E
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Here is a Collection of some important Lecture notes, Numerical solution, Old question and Reference book on Basic Electronics Engineering  for IOE BE I/II for targeting with the examination. You can click these item to study/download.
References Books
1.Electronics Devices and Amplifier Circuits Second Edition - by Steven T Karris
2.Electronics Devices and  Circuits Second Edition -by JIMMIE J. CATHEY 
3. Electronics Devices and  Circuits Theory - by ROBERT BOYLESTAD &LOUIS NASHELSKY
Old Question Collection
1. Six Old question set of IOE exam 2069 to 2071 Bhadra and Magh - by Edunepal.info
Lecture notes and Numerical Solution
1. Bipolar junction transistor's lecture notes and numerical solution- by Introductory Electronics Notes The University of Michigan-Dearborn
3. Some Numerical on flipflop circuit- by Introductory Electronics Notes The University of Michigan-Dearborn
2. Lecture 17 The Bipolar Junction Transistor -by spring 2007
4. Zener diode lecture notes - by Introductory Electronics Notes The University of Michigan-Dearborn
5. Some Numerical on Aplifier circuit- by Introductory Electronics Notes The University of Michigan-Dearborn
6. Some Numerical on low frequency Aplifier circuit- by Introductory Electronics Notes The University of Michigan-Dearborn
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Collection of top five Lecture notes and Numerical Solution on - FUNDAMENTALS OF THERMODYNAMICS AND HEAT TRANSFER - For All First year of IOE

  1. Numerical Solution Mannual on Engineering Thermodynamics  - by Prof. T.T Al Shemmari
  2. Lecture Note on Chapter-1  - by Sujan Shrestha (Kantipur Engineering College)
  3. CHAPTER 3-PROPERTIES OF COMMON SUBSTANCE- by Sujan Shrestha (Kantipur Engineering College)
  4. Lecture note Mannual on FUNDAMENTALS OF THERMODYNAMICS AND HEAT TRANSFER - by PULCHOWK CAMPUS DEPARTMENT OF MECHANICAL ENGINEERING 
  5. Thermodynamics_notes - by GUPSHUPSTUDY
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Tutorial #1

(Electric Machine)

QN1         A 30 cm long circular iron rod is bent into circular ring and 600 turns of windings are wound on it. The diameter of the rod is 20mm and relative permeability of the iron is 4000. A time varying current i = 5 sin 314.16t is passed through the winding. Calculate the inductance of the coil and average value of emf induced in the coil.                                               [1.89H, 1890V]

QN.2     For the Magnetic circuit shown below, calculate the Amp-turn (NI) required establishing a flux of 0.75 wb in the central limb. Given that mr = 4000 for iron core.
[40.4 ´ 103 Amp-turn]
 



       
                            

20 cm
 
 



N
 

I
 
                                                                
 





                                                                
 









QN.3        Calculate the net magnetic flux in the core of the following magnetic circuit and show the
            direction of magnetic flux in the core. Given that cross sectional area of the core is 25 sq.cm
            and   mr = 4000.                                                                                     [23.56 mwb, clockwise ]
 





5 cm
 
       










                                                                            


QN4         A circular iron core has a cross sectional area of 5 sq. cm and mean length of 15 cm. It has two coils A and B with 100 turns and 500 turns respectively. The current in the coil A is changed from zero to 10amp in 0.1 sec. Calculate the emf induced in the coil B. Given that the relative permeability of the core is 3000.                                                                                      [62.5 V]


QN5    An iron ring of mean length of 1.2m and cross-sectional area of 0.005 m2 is wound with a coil of 900 turns. If a current of 2 amp in the coil produces a flux density of 2T in the iron ring, calculate:
i)                             The mmf                                              
ii)                           Total flux in the core
iii)                         Magnetic field strength
iv)                         Relative permeability of the core.                  [ 1800AT, 6mwb, 15 AT/m, 637 ]

QN6.   An iron ring has a mean length of 1.5m and cross-sectional area of 0.01 m2. It has a radial air gap of 4mm. The ring is wound with 250 turns. What dc current would be needed in the coil to produce a flux of 0.8 weber in the air gap? Assume that µr = 400 and leakage factor is 1.25.                                                [2.46 Amp]

QN7    An uneven ring shaped core (as shown below) has µr = 100 and flux density in the larger section is 0.75T. If the current through the coil is 500mA determine the number of turns in the coil.                                                                                                                           [5669]
QN8                                                                     A magnetic circuit shown below has cast iron core whose dimensions are given below:


Length (ab + cd) = 50 cm    Cross sectional area of path (ab + cd) = 25 sq.cm
Length (ad) = 20 cm           Cross sectional area of path (ad) = 12.5 sq.cm
Length (dea) = 50 cm           Cross sectional area of path (dea) = 25 sq.cm

Determine the current ‘I’ required to produce a magnetic flux of 0.75 mWb in the central limb.
Given that: Number turns in the coil = 500 and   µr = 2000.                              [ 0.22 A ]

Tutorial #2

(Electric Machine –I)


QN1         A 50 kVA, 50 Hz single phase transformer has 500 turns on primary winding and 100 turns on secondary winding. The primary winding is supplied by 3000V, 50 Hz ac voltage with full resistive load connected on secondary side. Calculate:
i)                                   Emf induced in secondary winding
ii)                                 Primary and secondary windings currents
iii)                               Maximum flux in the core.
Assume that it is an ideal transformer.                       [Ans:   600V, 16.66A, 83.33A, 27.02mwb]

QN3         A step up transformer supplies a current of 5 amp to the load at 200 V. The power factor of the load is 0.8 lagging. Given that R1 = 0.5 ohm, X1 = 1 ohm, R2 = 2 ohm, X2 = 4 ohm, k = 2,
   R0 = 450ohms and X0 = 250 ohms. Calculate the magnitude and phase of V1 and I1 with V2 as reference phasor and calculate the input power factor.
        [V1 = 129.98Ð6.50 V,  I1 = 10.56Ð-37.880  A and input pf = 0.71 lagging ]

QN4         A 230V/ 2300V single-phase transformer is excited by 230V ac voltage. The equivalent     resistance and reactance referred to primary side are 0.1 ohm and 0.4 ohm respectively.
                Given that R0 = 500 ohms and X0 = 200 ohms.     The load impedance is (400 + j600) ohm.
                Calculate: a) Primary current   b) Secondary terminal voltage   c) Input power factor
                [I1 = 30 A V2 = 2075.4V, and input pf = 0.52 lagging ]

QN5        A 25 kVA, 6600V/ 250V single phase transformer has the following parameters:
                R1 = 8 ohm, X1 = 15 ohm, R2 = 0.02 ohm, X2 = 0.05 ohm.
                Calculate the full load voltage regulation at a power factor  a) 0.8 lag  b) unity   c) 0.8 lead.
                [ 2.7 %,  1.3 %  and  -0.782 %  ]

QN6        A 20 kVA, 250V/2500V, 50Hz single phase transformer gave the following test results:

               No-load test ( on L.V. side ) :       250V,  1.4 A,    105 watts
               Short circuit test (on H.V. side):  120V,     8 A ,    320 watts

  Calculate the equivalent circuit parameters referred to primary side and draw the equivalent
  circuit.          [ Ro = 595.2 ohm,  Xo = 187.26 ohm, R01 = 0.05 ohm and  X01 = 0.14 ohm  ]

QN7     A single-phase 50 Hz transformer has 100 turns on primary and 400 turns on secondary winding. The net cross-section area of the core is 250 cm2. If the primary winding is connected to a 230 V, 50 Hz supply, determine (a) the emf induced in the secondary winding and (b) the maximum and rms value of the flux density in the core.                                    [920 V, 0.414 Wb/m2, 0.293 Wb/m2]


QN8         The no-load current of a transformer is 15 A at a p.f. of 0.2 when connected to a 460 V, 50 Hz power supply. If the primary winding has 550 turns, calculate :(a) the magnetizing and working component of no-load current.  (b) iron loss (c) maximum and rms value of flux in the core.
                                  [14.7 A, 3 A, 1380 W, 3.77 mWb, 2.66 mWb]

QN9         A 2000V/400V, 50 Hz, single phase transformer draws 2 A at a p.f. of 0.2 lagging when it has no-load. Calculate the primary current and p.f. When secondary current is 200 A at a p.f. of 0.8 lagging. Assume the voltage drop in the winding to be neglected.                               [41.52 A, 0.78]

QN10.      A 100 KVA, 1100/230V, 50 Hz transformer has an HV winding resistance of 0.1Ω and a leakage reactance of 0.4 Ω. The low voltage winding has a resistance of 0.006Ω and a leakage reactance of 0.01 Ω. Find the equivalent winding resistance, reactance and impedance referred to HV and LV side.                            [(0.237+j0.629) Ω, (0.0643+j0.0275) Ω]

QN11       A 50 KVA, 2200/110V transformer when tested gave the following results:
OC test:                400 W              10 A                 110 V
SC test:                             808 W              20.5 A              90   V
Compute all the parameters of the equivalent circuit referred to HV and LV sides of the transformer. Draw the equivalent circuits also.
[HV side: 12120Ω, 4723.2Ω, 1.932Ω, 3.946Ω; LV side: 30.30Ω, 11.80Ω, 4.808*10-3Ω, 9.865*10-3Ω]


QN12       A 1000/500V 1-phase transformer draws a current of 2.4 A at no-load with a p.f. of 0.35 lagging . With secondary terminals short circuited by a thick wire, the primary winding is supplied by an ac voltage of 80 V, the transformer draws a current of 25 A and consumes 250 W. Calculate the equivalent circuit parameters referred to secondary side and draw the equivalent circuit.  [595.23Ω, 222.22Ω, 0.1Ω, 0.7925Ω]

QN13       With the secondary short circuited, if 200 V is applied to a 200 KVA, 1-phase, 3300/400 V transformer, the current through primary was the full load value and the input power was 1650 W. Calculate the secondary p. d. and percentage regulation when the secondary load is passing 300 A at 0.707 p.f. lagging with normal primary voltage.                            [388.41 V, 2.896%]
                                                                
QN14       A 500 KVA, 50 Hz, 6600V/400V, 1- phase transformer have primary and secondary winding resistances are 0.4 Ω and 0.001 Ω respectively. If the iron loss is 3.0 KW, calculate the efficiency at (a) full load (b) half full load.                                                                           [98.64%, 98.438%]

QN15.      A 200 KVA transformer has an efficiency of 98% at full load. If the maximum efficiency occurs at three quarters of full load, calculate the efficiency at half load. Assume p.f. of 0.8 at all loads.                                                                                                                        [97.9%]

QN16       An 11000/230V, 150 KVA, 50 Hz, 1-phase transformer has a core loss of 1.4 KW and full load Cu loss of 1.6 KW. Determine (a) the KVA load for maximum efficiency and the maximum efficiency (b) the efficiency at half full load and full load at 0.8 p.f. lagging.                                                                                                   [140.1312 KVA, 97.566%, 97%, 97.56%]

QN17.      A 600 KVA, 1- phase transformer has an efficiency of 92% both at full load and half load at unity p.f. Determine its efficiency at 60% of full load at 0.8 p.f. lagging.                              [85.9%]

QN18       The primary and secondary of an auto-transformer are 230V and 75V respectively. Calculate the currents indifferent parts of the winding when the load current is 200 A. Also calculate the saving in the use of copper.                                                                         [65.2A, 134.8A, 32.6%]

QN19       If a three phase star/delta, 33 KV/11KV, 50 Hz, transformer is loaded with a delta-connected load of 100 Ω per phase, calculate the primary line current.                                                 [63.5 A]




QN20       A three phase delta/star, 11 KV/400V, 50 Hz, distribution transformer has a star connected balanced load of (4+j6) Ω per phase. Calculate the primary line current.                                [1.16 A]

QN21       A 300 KVA, 11 KV/400V, r/Y, three phase transformer has star connected balanced load of 60 kW at power factor of 0.8 lagging in each phase. Calculate primary line current.                [11.81]

QN22       An 11KV/400V delta/star 3-phase transformer has balanced star connected load of 60 KW at p.f. of 80% lagging per phase. Calculate the primary line current. If the transformer has iron loss of 1.0 KW, calculate the approximate efficiency of the transformer. Given that primary winding resistance and leakage reactance are 25Ω per phase and 30Ω per phase respectively. Secondary winding resistance and leakage reactance are 0.01Ω per phase and 0.02Ω per phase respectively.                                                                                                          [11.81A, 95.92%]
QN23       A 500 KVA, 33/11 KV, 3-phase, 50 Hz delta/star transformer has resistances of 35 Ω per phase at high voltage side and 876 Ω per phases at low voltage side. Calculate the efficiency at full load and one half of full load respectively (a) at unity p.f. (b) at 0.8 p.f. lagging.                    
                                                                                                     [98.54%, 98.2%, 98.35%, 98%]
QN24.      For a transformer with 3-windings shown below, calculate the primary current:
(a)             when the winding ‘B’ is connected to a resistive load of 2 Ω, keeping winding ‘C’ open.
(b)             when the winding ‘C’ is connected to a resistive load of 4 Ω, keeping winding ‘B’ as loaded in case(a)                                                                                                    [25 A, 28.125 A]











Tutorial #3

(Electric Machine –I)

QN1         The data obtained from no-load magnetization test of a dc shunt generator running at 800 rpm is as follow:

   Field current(If in Amp) :   0        1.6              3.2       4.8      6.4        8        9.6       11.2
   Emf ( E in volts )    :   10       148        285      390     460      520     560      590

     The field winding resistance is 60 ohms.

a)                      Draw the OCC and calculate the emf generated by the machine at no-load.
b)                     Find the critical resistance of the machine at 800 rpm.                              [ 550V, 91 ohms]


QN2         A 4-pole dc shunt generator has wave wound armature. The armature and field winding resistances are 0.2 ohm and 60 ohms respectively. The brush contact drop is 1V per brush. The generator is delivering a power of 3kW at 120V. Calculate :

a)                       Total armature current coming out from the brush.
b)                       Current in each armature conductor
c)                       Generated EMF (E)                                                                     [ 27A, 13.5A, 127.4V ]

QN3         A dc series generator is running at 800 rpm and delivering a power of 6kW to the load at 120V. The armature and field winding resistance are 0.1 ohm and 0.3 ohm respectively. When the load is increased to 9kW, the speed is increased to 1200 rpm. Calculate the new values of armature current and load terminal voltage.                                                [46.85A,  145.27 V ]

QN4         A dc compound generator has to supply a current of 120A at 120V. The shunt field, series field and armature winding resistances are 30 ohm, 0.05 ohm and 0.1 ohm respectively. Calculate the emf generated by the armature in the following two cases :

i)                                   Long shunt connection
ii)                                 Short shunt connection                                                               [ 138.6V,  138.42V ]

QN5         Calculate the resistance of the load which consumes a power of 5 kW from a dc shunt generator whose load characteristic is described by the equation : VL = 250 – 0.5*IL.         [ 11.48 ohms ]

QN6         A short shunt cumulative compound dc generator supplies 7.5kw at 230V. The shunt field, series field and armature resistance are 100, 0.3and 0.4 ohms respectively. Calculate the induced emf and the load resistance.                                                                           [253.8V, 7.05W]

QN7.    The resistance of the field circuit of a shunt excited dc generator is 200W. When the output of the  generator is 100 KW, the terminal voltage is 500V and the generated emf 525V.Calculate (a) the armature resistance and (b) the value of the generated emf when the output is 60kw, if the terminal voltage then is 520V.                                                              [0.123W, 534.56V]

     Contd…….

QN8.        A 6 pole wave-wound dc shunt generator has 1200 armature conductors. The useful flux per pole is 0.02wb, the armature resistance is 0.4W and the speed is 400 rpm. If the shunt resistance is 220W, calculate the maximum current which the generator can deliver to an external load if the terminal voltage is not to fall below 440V.                                                                     [ 98 A ]

QN9   A separately excited dc generator  running at 1200rpm supplies 200A at 125V to a load of constant resistance. What will be the current when the speed is dropped to 1000 rpm if the field current is unaltered? Given that armature resistance =0.04W, total drop at brushes = 2 V.     [1.66.17A]

QN10.   A dc generator is connected to 220V dc mains. The current delivered by the generator to the main is 100A. The armature resistance is 0.1W.The generator is driven at a speed of 400rpm.Calculate:(a) the induced emf (b) the electromagnetic torque (c) the power input and output of the armature when the speed drops to 350 rpm. State weather the machine is generating or motoring. Assume constant flux.
                                                                                                              [230V, 549.08N-m, 23kW, 37.73kW]

QN.11  A dc shunt machine, connected to 250V mains has an armature resistance of 0.12W and the resistance of the field circuit is 100W. Calculate the ratio of the speed as a generator to the speed as a motor, the line current in each case being 80A.                                                                   [1.081]

 QN12.     A 1500 kW, 500V, 16 pole, dc shunt generator runs at 150 rpm.What must be the useful flux per pole if there are 2500 conductors in the armature and the winding is lap connected and full-load armature copper loss is 25kW? Calculate the area of the pole shoe if the air gap flux density has a uniform value of 0.9wb/m2. Neglect charge in speed. Take Rf =55W.

QN13.   A shunt generator delivers 50 KW at 250 V and 400 rpm. The armature resistance is 0.02W and field resistance is 50W . Calculate the speed of the machine when running as a shunt motor and taking 50 KW input at 250 V.                                                                                          [387 rpm]

QN14.      A dc shunt generator has an output of 10 KW at 500 V; the speed being 1000 rpm. The armature circuit resistance is 0.5 and the field resistance is 250. Calculate speed when running as a shunt motor taking 50 KW at 500 V.              



*****  End  ******



















Tutorial #4

DC Generator
(Electric Machine –I)



QN1         A 4-pole dc shunt generator has wave wound armature. The armature and field winding resistances are 0.2 ohm and 60 ohms respectively. The brush contact drop is 1V per brush. The generator is delivering a power of 3kW at 120V. Calculate :

d)                       Total armature current coming out from the brush.
e)                       Current in each armature conductor
f)                        Generated EMF (E)                                                                     [ 27A, 13.5A, 127.4V ]

QN2         A dc series generator is running at 800 rpm and delivering a power of 6kW to the load at 120V. The armature and field winding resistance are 0.1 ohm and 0.3 ohm respectively. When the load is increased to 9kW, the speed is increased to 1200 rpm. Calculate the new values of armature current and load terminal voltage.                                                [46.85A,  145.27 V ]

QN3         A dc compound generator has to supply a current of 120A at 120V. The shunt field, series field and armature winding resistances are 30 ohm, 0.05 ohm and 0.1 ohm respectively. Calculate the emf generated by the armature in the following two cases :

iii)                               Long shunt connection
iv)                               Short shunt connection                                                               [ 138.6V,  138.42V ]

QN4         Calculate the resistance of the load which consumes a power of 5 kW from a dc shunt generator whose load characteristic is described by the equation : VL = 250 – 0.5*IL.         [ 11.48 ohms ]

QN5         A short shunt cumulative compound dc generator supplies 7.5kw at 230V. The shunt field, series field and armature resistance are 100, 0.3and 0.4 ohms respectively. Calculate the induced emf and the load resistance.                                                                           [253.8V, 7.05W]

QN6.    The resistance of the field circuit of a shunt excited dc generator is 200W. When the output of the  generator is 100 KW, the terminal voltage is 500V and the generated emf 525V.Calculate (a) the armature resistance and (b) the value of the generated emf when the output is 60kw, if the terminal voltage then is 520V.                                                              [0.123W, 534.56V]


QN7.        A 6 pole wave-wound dc shunt generator has 1200 armature conductors. The useful flux per pole is 0.02wb, the armature resistance is 0.4W and the speed is 400 rpm. If the shunt resistance is 220W, calculate the maximum current which the generator can deliver to an external load if the terminal voltage is not to fall below 440V.                                                                     [ 98 A ]

QN8   A separately excited dc generator  running at 1200rpm supplies 200A at 125V to a load of constant  resistance. What will be the current when the speed is dropped to 1000 rpm if the field current is unaltered? Given that armature resistance =0.04W, total drop at brushes = 2 V.     [1.66.17A]

QN9   A dc generator is connected to 220V dc mains. The current delivered by the generator to the main is 100A. The armature resistance is 0.1W.The generator is driven at a speed of 400rpm.Calculate:(a) the induced emf (b) the electromagnetic torque (c) the power input and output of the armature when the speed drops to 350 rpm. State weather the machine is generating or motoring. Assume constant flux.
                                                                                                              [230V, 549.08N-m, 23kW, 37.73kW]

QN11  A dc shunt machine, connected to 250V mains has an armature resistance of 0.12W and the resistance of the field circuit is 100W. Calculate the ratio of the speed as a generator to the speed as a motor, the line current in each case being 80A.                                                                   [1.081]

 QN12.     A 1500 kW, 500V, 16 pole, dc shunt generator runs at 150 rpm.What must be the useful flux per pole if there are 2500 conductors in the armature and the winding is lap connected and full-load armature copper loss is 25kW? Calculate the area of the pole shoe if the air gap flux density has a uniform value of 0.9wb/m2. Neglect charge in speed. Take Rf =55W.

QN13.   A shunt generator delivers 50 KW at 250 V and 400 rpm. The armature resistance is 0.02W and field resistance is 50W . Calculate the speed of the machine when running as a shunt motor and taking 50 KW input at 250 V.                                                                                          [387 rpm]

QN14.      A dc shunt generator has an output of 10 KW at 500 V; the speed being 1000 rpm. The armature circuit resistance is 0.5 and the field resistance is 250. Calculate speed when running as a shunt motor taking 50 KW at 500 V.              



*****  End  ******

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