Diploma-Diploma Mechanical Engineering 4th Sem Theory of Machines(Gujarat Technological University-2013)
Thursday, 29 August 2013 06:36anudouglas
GUJARAT TECHNOLOGICAL UNIVERSITY
DIPLOMA ENGINEERING - SEMESTER–IV • EXAMINATION – SUMMER 2013
Subject Code: 341904 Date: 23-05-2013
Subject Name: Theory of Machines
Time: 10:30 am - 01:00 pm Total Marks: 70
Instructions:
1. Attempt all questions.
2. Make suitable assumptions wherever necessary.
3. Figures to the right indicate full marks.
4. English version is considered to be Authentic.
Q.1 (a) What is kinematic pair? Classify various types of kinematic pairs? Explain them with neat sketches. 07
(b) The crank of reciprocating engine is 200mm long and connecting rod is 800mm long. The crank is at 40 from IDC and engine speed is 160 rpm. Find (I) Velocity (II) Acceleration of Piston (III) Angular Velocity of connecting rod. Solve by Klein’s Construction Method. 07
Q.2 (a) Draw two Inversion of single slider Crank chain & Explain it. 07
(b) What is cam & follower? State types of cam & follower and explain with sketch Roller & Flat face follower. 07
OR
(b) Draw cam profile to move the knife edge follower to give 60 mm lift with SHM. During 120 of cam rotation & it dwells for 40 of cam rotation then I returns with uniform velocity during 120 rotation & then follower remains in rest during remaining period. The axis of follower passes through the axis of cam shaft. Base circle diameter of the cam is 50mm. 07
Q.3 (a) Prove that P/W= Tan (α+Ø) for screw jeck. 07
(b) In a steamer have 8 collars on thrust shaft find power lost in friction when (I) Uniform Pressure (II) Uniform Wear, Which one is better? Why? Use following data.
(1) Outside diameter of shaft = 600mm
(2) Inside diameter of shaft = 300mm
(3) Total Thrust = 120 KN
(4) Coefficient of friction = 0.14
(5) Engine Speed = 120 rpm (07)
OR
Q.3 (a) (1) Differentiate between brake & Dynamometer.
(2) Define friction state its types & write law of fluid friction. 07
(b) Derive expression for friction torque for flat collar bearing assuming uniform pressure. 07
Q.4 (a) Derive the expression of centrifugal tension for power transmission in belt. 07
(b) Find the power for flat belt from the following data:
(1) Speed of belt = 540m/min
(2) Maximum Tension = 3320 N
(3) µ= ૦.4
(4) Ɵ= 3 Radian (07)
OR
Q. 4 (a) (I) Explain Compound Gear train.
(II) Classify mechanical power transmission & state advantages of chain drive. (07)
(b) Find the width of belt from following data:
(1) Permissible Tension = 15 N/mm width
(2) Lap angle of belt = 0.9 π Radian
(3) Coefficient of friction = 0.27
(4) Power transmitted = 15 KW
(5) Speed of belt =1200 m/min (07)
Q.5 (a) (I)State five effect of unbalancing
(II) Compare Fly wheel with governer. 07
(b) Four masses 8k.g , 10k.g , 12k.g & 15k.g are revolving about an axis in same plane at radii of 10 cm, 15cm, 12cm & 9cm respectively. The angular position of masses 10k.g, 12k.g & 15k.g are 45˚, 120˚, 225˚ from the position of 8k.g.Determine the magnitude & direction of the balance mass at a radius of 13 CM for complete dynamic balancing. (07)
OR
Q.5 (a) What is vibration? State causes of vibration & measures to reduce vibrations. (07)
(b) The following information recorded from multi cylinder engine turning moment diagram.
(1) Vertical axis T.M 1mm=650 N.M
(2) Horizontal axis crank angle 1mm=4.5˚The area in mm² around average torque line respectively are -28,+380, -260,+310,-300,+282,-380,+265 & -229.The fluctuation of speed are ±1.8% of average speed.The average speed 400 rpm.Then find mass of flywheel of 0.7m radius.
| Earning: ₹ 6.50/- |
