Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
17.7k views
in Physics by (53.3k points)

The following data refer to two cylinder locomotive with cranks at ~Oo : Reciprocating mass per cylinder = 300 kg ; Crank radius = 0.3 m ; Driving wheel diameter = 1.8 m ; Distance between cylinder centre tines = 0.65 m ; Distance between the driving wheel central planes = 1.55 m. Determine 

1. the fraction of the reciprocating masses to be balanced, if the hammer blow is not to exceed 46 kN at 96.5 km. p.h. ;

2. the variation in tr active effort ; and

3. The maximum swaying couple.

1 Answer

+1 vote
by (53.4k points)
selected by
 
Best answer

Given: m = 300 kg ;r = 0.3 m ; D = 1.8 m or R = 0.9 m ; a = 0.65 m ; Hammer blow = 46 kN = 46 x 103 N ; v = 96.5 km/h = 26.8 m/s

 1. Fraction of the reciprocating masses to be balanced

Let c = Fraction of the reciprocating masses to be balanced, and B = Magnitude of balancing mass placed at each of the driving wheels at radius b. We know that the mass of the reciprocating parts to be balanced = c.m = 300 c kg

(b)  Position of cranks

The position of planes of the wheels and cylinders is shown in Fig.(a), and the position of cranks is shown in Fig(b). Assuming the plane of wheel A as the reference plane, the data may be tabulated as below:

Plane (1) Mass (m) kg (2) Radius (r) m (3) Cent. force/ω2(m.r) kg-m (4) Distance from plane A (l) m (5) Couple/ω2(m.r.l.) kg-m2 (6
A (R.P.)
B
C
D
B
300 c
300 c
B
b
0.3
0.3
b
B.b
90 c
90 c
B.b
0
0.45

1.1
1.55
0
40.5 c
99 c
1.55 B.b


Now the couple polygon, to some suitable scale, may be drawn with the data given in Table (column 6), as shown in Fig.. The closing side of the polygon (vector c0) represents the balancing couple and is proportional to 1.55 B.b. From the couple polygon,

1.55 B.b = v((40.5c)2 + (99c) 2) = 107c

We know that angular speed, 00=vIR = 26.8/0.9 = 29.8 rad/s

 Hammer blow,

 46 x 103 = B.m2 .b = 69 c (29.8)2 = 61275 c

 c = 46 x 103/61275 = 0.751

2. Variation in tr active effort

 We know that variation in tr active effort 

= ± v2(l-c)m .. 002.r = ±v 2(1-0.751) 300 29.8f 

0.3 = 28 140 N = 28.14 kN

 Maximum swaying couple

 We know the maximum swaying couple

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...