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3 years ago

7

An aluminium alloy tube has a length of 750 mm at a temperature of 223°C. What will be its length at 23°C if its coefficient of

linear expansion is 25.5 x 10^-6/C?

1 answer:

uranmaximum [27]3 years ago

3 0

Answer:

Final length= 746.175 mm

Explanation:

Given that Length of aluminium at 223 C is 750 mm.As we know that when temperature of material increases or decreases then dimensions of material also increases or decreases respectively with temperature.

Here temperature of aluminium decreases so the final length of aluminium decreases .

As we know that

$\Delta L=L\alpha\Delta T$

Now by putting the values

$\Delta L=750\times \25.5\times 10^{-6}\times 200$

ΔL=3.82 mm

So final length =750-3.82 mm

Final length= 746.175 mm

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Answer:

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A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16

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Answer:

the net work per cycle $\mathbf{W_{net} = 0.777593696}$ Btu per cycle

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Explanation:

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length of the stroke = 3.4 in

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The cylindrical volume $V_2 = 0.16 V_1$

the crankshaft N rotates at a speed of 2400 RPM.

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From the given information; the change in volume is:

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the mass in air ( lb) can be determined by using the formula:

$m = \dfrac{P_1V_1}{RT}$

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R = 53.3533 ft.lbf/lb.R°

$m = \dfrac{2088 \ lb/ft^2 \times 0.02518 \ ft^3}{53.3533 \ ft .lbf/lb.^0R \times 519 .67 ^0 R}$

m = 0.0018962 lb

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$v_{r1} =158.58$

$u_1 = 88.62 Btu/lb$

At state of volume 2; the relative volume can be determined as:

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$v_{r4} = v_{r3} \times \dfrac{V_1}{V_2}$

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$W_{net} = Heat \ supplied - Heat \ rejected$

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$W_{net} = m(u_3-u_2- u_4 + u_1)$

$W_{net} = m(1098-184.7- 591.84 + 88.62)$

$W_{net} = 0.0018962 \times (1098-184.7- 591.84 + 88.62)$

$W_{net} = 0.0018962 \times (410.08)$

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$W = 4 \times N' \times W_{net$

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$N' = \dfrac{2400}{2}$

N' = 1200 cycles/min

N' = 1200 cycles/60 seconds

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W = 4 × 20 cycles/sec × 0.777593696

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