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Crank
3 years ago
12

: The interior wall of a furnace is maintained at a temperature of 900 0C. The wall is 60 cm thick, 1 m wide, 1.5 m broad of mat

erial whose thermal conductivity is 0.4 W/m K. The temperature of the outside surface of the wall is 150 0C. Determine the heat through the wall. Also determine thermal resistance to heat flow.
Engineering
1 answer:
Snowcat [4.5K]3 years ago
7 0

Answer:

<em>Heat is lost at the rate of 750 J/s or W</em>

<em>The thermal resistance is 1 K/W</em>

Explanation:

interior temperature T_{2} = 900 °C

wall thickness t = 60 cm = 0.6 m

width = 1 m

breadth = 1.5 m

thermal conductivity k = 0.4 W/m-K

outside temperature T_{1} = 150 °C

heat through the wall = ?

The area of the wall A = w x b = 1 x 1.5 = 1.5 m^2

Temperature difference dt = T_{2} - T_{1} = 900 - 150 = 750 °C

note that dt is also equal to 750 K since to convert from °C to K we'll have to add 273 to both temperature, which will still cancel out when we subtract the two temperatures.

To get the heat that escapes through the wall, we use the equation

Q = Ak\frac{dt}{t}

substituting values, we have

Q = 1.5 x 0.4 x \frac{750}{0.6} = <em>750 J/s or W</em>

Thermal resistance R_{t} = \frac{dt}{Q}

R_{t} = 750/750 =<em> 1 K/W</em>

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Answer:

N = 38546.82 rpm

Explanation:

D_{1} = 150 mm

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A_{2}= \frac{\pi }{4}\times 250^{2}

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The centrifugal force acting on the flywheel is fiven by

F = M ( R_{2} - R_{1} ) x w^{2} ------------(1)

Here F = ( -UTS x A_{1} + UCS x A_{2} )

Since density, \rho = \frac{M}{V}

                        \rho = \frac{M}{A\times t}

                        M = \rho \times A\times tM = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t

                        M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37

                        M = 8252963901

∴ R_{2} - R_{1} = 50 mm

∴ F = 763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}

  F = 33618968.38 N --------(2)

Now comparing (1) and (2)

33618968.38 = 8252963901\times 50\times \omega ^{2}

∴ ω = 4036.61

We know

\omega = \frac{2\pi N}{60}

4036.61 = \frac{2\pi N}{60}

∴ N = 38546.82 rpm

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3 years ago
At an axial load of 22 kN, a 15-mm-thick × 40-mm-wide polyimide polymer bar elongates 4.1 mm while the bar width contracts 0.15
Alenkasestr [34]

Answer:

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The Poisson's ratio is got by using the formula

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The Longitudinal strain is the ratio of the elongation to the original length in the longitudinal direction.

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