89GES-II - Directorate general of Shipping

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89GES-II
Sr. No. 3
EXAMINATION OF MARINE ENGINEER OFFICER
GENERAL ENGINEERING SCIENCE - II
INDIA (2002)
N.B. -
CLASS- III
(Time Allotted 2 Hours)
Total Marks 100
(1) Attempt FOUR questions ONLY.
(2) All questions carry equal marks.
(3) Neatness in handwriting and clarity in expression carries weightage
1. An eight-pole armature is wound with 480 conductors. The magnetic flux and speed are such that the
average e.m.f generated in each conductor is 2.2A. Calculate the terminal voltage on no-load, the
output current on full load and total power generated on full load when the armature is
(a) lap-connected and
(b) wave-connected
2. A shunt generator supplies 100 amperes at a terminal voltage of 200 volts. The prime mover is
developing 23.8K. Watt. Shunt field resistance 50 ohms and armature resistance 0.1 ohms. Find –
(a) the copper losses,
(b) the iron and friction losses.
(c) the commercial efficiency.
3. A non-linear resistance R1 is in parallel with a linear resistance. R2 across a 110 V D.C. supply.
R2 = 220 
R1 has the following volt-amp. Characteristic :Volts ….
0
32
60
100
125
140
Amperes ….
0
0.05
0.1
0.2
0.3
0.4
R1 is bilateral i.e. conducts equally well in either direction. Plot the characteristics for each resistance
and determine the current taken from the supply.
4. Prove that in a single stage ideal reciprocating air compressor, when the gas is compressed
adiabatically - Work done per cycle = n p1V1 [(p2 / p1) n -1 / n – 1 ] Neglect clearance volume.
n-1
1.146kg of air at a pressure of 1 bar and 25OC is compressed to 13 bar, Calculate the work done per
cubic meter of air taken in the cylinder when the air is compressed adiabatically. Take 1 bar =
100kN/m2, n = 1.4.
5. Steam leaves the nozzle of an impulse turbine at a velocity of 840m/s at an angle of 18O to the
direction of movement of the blades. The linear velocity of the blades is 360 m/s and the inlet and exit
angles of the blades are equal. Calculate (a) The entrance and exit angles of the blades.
(b) The magnitude and the direction of the steam at exit. Neglect the friction of the steam across the
blades.
6. What is meant by Calorific Value?
The mass analysis of a fuel is 87% carbon, 11% hydrogen, 2% oxygen. Calculate the volume of air
required in cubic metres at 1.0 bar and 27OC for perfect combustion per kg of fuel. Take the value of
R for air = 0.287 kJ/kg K. Mass analysis of air = 23% Oxygen, 77% Nitrogen. The atomic weight of
hydrogen is 1, carbon 12 and oxygen 16.
89GES-I
Sr. No. 3
EXAMINATION OF MARINE ENGINEER OFFICER
GENERAL ENGINEERING SCIENCE - II
CLASS- III
(Time Allotted 2 Hours)
INDIA (2002)
N.B. -
Total Marks 100
(1) Attempt FOUR questions ONLY.
(2) All questions carry equal marks.
(3) Neatness in handwriting and clarity in expression carries weightage
Answers
Ans 1. (a) With the armature lap-conducted
No. of parallel paths in the armature winding = number poles = 8
No. of conductors/paths = 480/8 = 60
Terminal voltage on load = (e.m.f./conductor)  (No. of conductors/paths)
= 2.2  60 = 132 V.
Output current on full load = (full-load current per conductor)  No. of parallel paths
= 100  8 = 800 A.
Total power generated on full load
= Output current  generated e.m.f.
= 800  132 = 105600 W = 105.6 kW
(b) With the armature wave-connected
No. of parallel path = 2
 No. of conductors/path = 480/2 240
Terminal voltage on no-load = 2.2  240 = 528 V.
Output current on full load = 100  2 = 200 A
Total power generated on full load = 200  528 = 105600 W = 105.6 kW.
Answer for Question No. 2
1918 watts
1.88 kw
84%
Answer For Question No. 3
R2
0.5
R1
0.4
0.3
Amp
0.2
0.1
20
40
Volt
.I1 from graph = 0.22 Amps
60
80
100
120
I2 from graph = 0.5 Amps
Total current = 0.22 0.05 = 0.72
I1
R1
I2
I = I1+I2
Answer for Question No. 4
378.35 kJ
Answer for Question No. 5
(a) 30036’
(b) 73605’
Answer for Question No. 6
11.91 m3 air / kg fuel
R2
V = 110v
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