Subjects
Applied Mathematics for Electrical Engineering - 3130908
Complex Variables and Partial Differential Equations - 3130005
Engineering Graphics and Design - 3110013
Basic Electronics - 3110016
Mathematics-II - 3110015
Basic Civil Engineering - 3110004
Physics Group - II - 3110018
Basic Electrical Engineering - 3110005
Basic Mechanical Engineering - 3110006
Programming for Problem Solving - 3110003
Physics Group - I - 3110011
Mathematics-I - 3110014
English - 3110002
Environmental Science - 3110007
Software Engineering - 2160701
Data Structure - 2130702
Database Management Systems - 2130703
Operating System - 2140702
Advanced Java - 2160707
Compiler Design - 2170701
Data Mining And Business Intelligence - 2170715
Information And Network Security - 2170709
Mobile Computing And Wireless Communication - 2170710
Theory Of Computation - 2160704
Semester
Semester - 1
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Semester - 3
Semester - 4
Semester - 5
Semester - 6
Semester - 7
Semester - 8
Basic Mechanical Engineering
(3110006)
BME-3110006
Summer-2020
Question-3c-OR
BE | Semester-
1
Summer-2020
|
11-06-2020
Q3) (c)
7 Marks
A petrol engine with a stroke length of
200
mm
and diameter of
150
mm
has a clearance volume of
7
×
10
5
mm
3
. If the indicted thermal efficiency is
0
.
30
, find the relation efficiency. If the effective pressure is
5
bar
and engine runs at
1000
rpm
. Find the IP of the engine. Take
γ
γ
=
1
.
4
.
Given data:
L
=
0
.
2
m
d
=
0
.
15
m
V
C
=
7
×
10
5
mm
3
=
7
×
10
-
4
m
3
η
η
ith
=
0
.
30
bar
p
m
=
5
bar
=
500
kPa
N
=
100
rpm
γ
γ
=
1
.
4
Find:
η
η
rel
=
?
I
.
P
.
=
?
Solution:
The relative efficiency is givne by
η
η
η
η
rel
=
η
ith
η
air
×
100
%
But, the air standard efficiency is given by,
η
γ
η
air
=
1
-
1
r
γ
-
1
Where,
π
r
=
compression
ratio
=
V
s
+
V
c
V
c
=
π
4
×
0
.
15
2
×
0
.
2
+
7
×
10
-
4
7
×
10
-
4
=
6
.
046
Now,
η
η
air
=
1
-
1
6
.
046
1
.
4
-
1
=
0
.
5131
Thus,
η
η
rel
=
0
.
30
0
.
51
×
100
%
=
58
.
82
%
Indicated power is given by,
I
.
P
.
=
p
m
LAN
60
×
2
×
no
.
of
cylinder
,
kW
π
I
.
P
.
=
500
×
0
.
2
×
π
4
×
0
.
15
2
×
1000
60
×
2
×
1
I
.
P
.
=
14
.
718
kW
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