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3 years ago

The 10mm diameter rod is made of Kevlar 49. Determine the change in

Engineering

1 answer:

Eddi Din [679]3 years ago

5 0

Answer:

0.815 mm

Explanation:

The rod in made of Kevlar 49, so it has an Young's modulus of

E = 125 GPa

The stiffness of a rod is given by:

k = E * A / L

k = E * π/4 * d^2 / L

So:

k = 125*10^9 * π/4 * 0.01^2 / 0.1 = 98.17 MN/m

Of the pulling forces only one is considered because when you pull on something there is always another force on the other side of equal magnitude and opposite direction to maintain equilibrium.

Hooke's law:

Δl = P/k

Δl = 80*10^3 / 98.17*10^6 = 0.000815 m = 0.815 mm

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Answer:

The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²

Explanation:

Here we have the heat Q given as follows;

Q = 15 × 1075 × (1100 - $t_{A2}$ ) = 10 × 1075 × (850 - 300) = 5912500 J

∴ 1100 - $t_{A2}$ = 1100/3

$t_{A2}$ = 733.33 K

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Where

$\Delta \bar{t}_{a}$ = Arithmetic mean temperature difference

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Hence, plugging in the values, we have;

$\Delta \bar{t}_{a} =\frac{1100+733.33}{2} - \frac{300+850}{2} = 341\tfrac{2}{3} \, K = 341.67 \, K$

Hence, from;

$\dot{Q} = UA\Delta \bar{t}_{a}$ , we have

5912500 = 90 × A × 341.67

$A = \frac{5912500 }{90 \times 341.67} = 192.3 \, m^2$

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3 years ago

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2 years ago

Assume the average fuel flow rate at the peak torque speed (1500 rpm) is 15kg/hr for a sixcylinder four-stroke diesel engine und

sveticcg [70]

Answer:

Q = 8.845 DEGREE

Explanation:

given data:

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2 years ago