Upto the critical radius of insulation:
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Solution
Critical radius is one at which maximum heat transfer exits. Above this critical radius, added insulation decreases heat loss while upto, added insulation increases heat loss.
To minimize the knocking tendency in SI engine, the spark should be located at:
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Solution
The location of spark in case of SI engine should be placed in a close vicinity of exhaust valve to avoid region of final products of gases and hence to minimize the knocking tendency.
If flange thickness, cutting speed and feed are 30 mm, 22 rpm and 0.2 mm/rev,then the machining time for drilling 4 holes of 16 mm diameter on a flange will be ________.
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Solution
Given: Flange thickness = 30mm
cutting speed (vc) = 22 rpm
feed (F) = 0.2mm/rev.
diameter of hole (d) = 16 mm =16 × 10–3m
cutting speed (vc) =πdN
22 = π× 16 × 10–3 × N
N=\(\frac{22}{\pi \times 16\times 10^{-3}}= 437.9 rpm\)
Now,machining time (tm) =\(\frac{\iota }{F\times N}\)
where, F = feed rate, N= speed, l= length to be machined.
tm=\(\frac{(16\times 0.3)+30}{0.2\times 437.9}=\frac{34.8}{87.58}= 0.397min\)
Now,machining time (tm) for 4-holes
= 4 × 0.397 = 1.59 min.
If demand is doubled and ordering cost, unit cost and inventory cost are one halved, then the EOQ will be ______.
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Solution
Given: Demand (d) =2d
ordering cost (Cp) = (Cp/2),EOQ =\(\sqrt{\frac{2dC_{P}}{C_{h}}}\)
holding cost (Ch) = (Ch/2)
New, EOQ(new)=\(\sqrt{\frac{2(2d)(C_{P}/2)}{(C_{h}/2)}}=\sqrt{\frac{4dC_{p}}{C_{h}}}\)
=\(\sqrt[2]{\frac{dC_{p}}{C_{h}}}=\frac{2}{\sqrt{2}}\sqrt{\frac{2dC_{p}}{C_{p}}}=\frac{2}{\sqrt{2}}EOQ=\sqrt{2}EOQ =1.414 times\)
The absolute jet exit velocity from a jet engine is 2800 m/sand the forward flight velocity is 1400 m/s. The propulsive efficiency will be ______.
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Solution
Given:Speed of jet engine = 2800 m/s
forward flight velocity = 1400 m/s
Propulsive efficiency (ηp) =\(\frac{2}{1+\frac{c}{v}}\times 100\)
=\(\frac{2}{1+\frac{2800}{1400}}\times 100=\frac{2\times 1400}{1400+2800}\times 100\)
=\(\frac{2800}{4200}\times 100= 66.67%\)
Two strings are attached to the walls as shown in figure weight is 150N. Then the tensions in the strings will be:______.
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Solution
\(\frac{150}{sin90}=\frac{F_{1}}{sin150}=\frac{F_{2}}{sin120}\)
\(F_{1}=\frac{150sin150}{sin90}= 75N\)
\(F_{2}=\frac{150sin120}{sin90}=129.9 N\)
In a fly wheel, the safe stress is 25.2 MN/m2 and the density is 7g/cm3. Then the maximum peripheral velocity will be_______.
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Solution
Given : Safe stress (σs) = 25.2MN/m2= 25.2 × 106
Density (ρ) = 7g/cm3
Stress induced in flywheel (σs) =ρ r2 ω2
σs= ρv2
where, σs= safe stress, ρ= density, v= tangential velocity.
25.2× 106 =\(\frac{7\times 10^{-3}}{10^{-6}}\times v^{2}\)
v2=\(\frac{25.2\times 10^{6}\times 10^{-6}}{7\times 10^{-3}}\times 3600\)
v= 60 m/s
Which one of the following is correct for ‘climb’?
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Solution
Dislocation introduces imperfections into the structure.The movement of extra half plane in edge dislocation is termed as climb. In edge dislocation Burger vector is perpendicular to the dislocation line.
A metal rod is rigidly fixed at both of its ends. The temperature of the rod is in reased by 1000C. If the coefficient of linear expansion and elastic modulus of metal rod are 10 × 106C° C and 200 G respectively, then the stress produced in the rod will be _______.
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Solution
Given: Increased temperature (Δt) = 100°C
Coefficient of linear thermal expansion (α) = 10 × 10-6
elastic modulus (E) =200 GPa
Stress induced in the rod due to temperature
σt =α D t E
=10 ×10-6× 100 ×200 × 103
σt = 200 mPa.
The materials which shows direction dependent properties are know as:
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Solution
Anisotropic materials are materials whose properties are directionally dependent. Their properties such as modulus of elasticity (E), change with direction alongthe object.