A bearing supports a vertical shaft of 140 mm diameter rotating at 9 rad/s.The shaft carries a vertical load of 18 KN.Asssuming uniform pressure distribution and coefficient of friction is 0.06, then the power lost in friction will be ________.
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Solution
The shear force diagram (SFD) for abeam is shown in figure.The bending moment diagram (BMD) is represented by the following:
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Solution
For a cantilever beam with find at one end and loading is uniform distributed load (UDL),
Consider the following statements with regard to PERT and CPM.
1.PERT is event oriented while CPM is activity oriented
2.PERT is probabilistic while CPM is deterministic
3.Livelling and smoothing are the techniques related to resource scheduling in CPM.
Which of the statements given above are correct?
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Solution
(i) PERT is event oriented and CPM is activity oriented
(ii) PERT →Probabilistic
(iii) CPM →Deterministic
If n is the polytrophic index of compression and P2/P1 is the pressure ratio for a three stage compressor with ideal inter cooling expression for total work of three stage is:
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Solution
Total work performed =\(\frac{3n}{n-1}P_{1}P_{2}\left [ \left ( \frac{P_{2}}{P_{1}} \right )^{\frac{n-1}{3n}}-1 \right ]\)
A rectangular strain rosette gives the following reading in as train measurement taste,
ε1= 1000 × 10–6 , ε2= 8000 ×10–6 , ε3 = 1000× 10-6
The direction of major principal strain with respect to gauge length is
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Solution
A fluid flowing through a pipe line undergoes a throttling process from 10 bar to 1 bar in passing through a partially open value. Before throtting, the specific value of the fluid is 0.5 m3/kg and after throttling is 2 m3/kg. Then the change in specific internal energy during throttling process will be________.
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Solution
Given: Pressure(PA) = 10 bar = 10× 105
Pressure(PB) =1 bar = 1× 105
(before) volume of fluid (VA) = 0.5 m3/kg
(after) volume of fluid (VB) = 2 m3/kg
Change in internal energy (ΔU) = PAVA – PBVB
= 106/sup> × 0.5 - 105 × 2
= 5 × 105 - 2 × 105
= 3× 105 J/kg
In a two stage compressor with ideal intercooling for the work requirement to be minimum, the intermediate pressure Pi in terms of condenser and evaporator pressure Pc and Pe respectively is
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Solution
For minimum work requirment,
Intermediate pressure (Pi) =\(\sqrt{P_{c}P_{e}}\)
where, Pc= condenser pressure
Pe= evaporator pressure
A car of 9 KN weight is moving at 90 kmph. After applying brakes, the time required for car to stop will be _________.
If, coefficient of friction between road and tyre is 0.81.
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Solution
Given: Speed of car = 90 km/h = 25 m/s
Coefficient of friction(µ) = 0.81 weight of car (w) = 9 kN = 9 × 103N Mass of car (m) = \(\frac{9\times 10^{3}}{9.8}\) =918.37 kg
Friction force (F) = μ.w = 0.81 × 9000 = 7290 N
Now,using impulse momentum principle,
Ft = m(v – u)
7290 × t = 918.37 (0 – 25)
t = 3.14 seconds
Match List-I with List-II
List-I List-II
A.Laser welding 1.Uniting large area sheets
B.Friction welding 2.Repairing large parts
C.Ultrasonic welding 3.Welding a rod to flatsurface
D.Explosive welding 4.Fabrication of nuclear reactor components
5.Welding very thin mate-rials
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Solution
Laser welding = welding very thin materials
Friction welding = welding a rod to a flat surface
Ultrasonic welding = Fabrication of nuclear reactor components.
Explosive welding = uniting large area shuts
A cube having each side of length a, is constrained in all directions and is heated uniformly so that the temperature is raised to T°C. If a= thermal coefficient of expansion of cube material and E = young’s modular of elasticity, the stress developed in the cube is
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Solution