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Elena-2011 [213]
3 years ago
8

The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me

tal sheets. The oven air is at T[infinity],????= 300℃, and the corresponding convection coefficient is ℎ???? = 30 W/ m2 °K. The inner wall absorbs a radiant flux of ????′′????????????= 100 W/ m2 from hotter objects within the oven. The room air at T[infinity],o= 25℃, and overall coefficient for convection and radiation from the outer surface is ℎo = 10 W/ m2 °K. What insulation thickness L is required to maintain the outer wall surface at a safe-to-touch temperature of To = 40 degree C?

Engineering
1 answer:
masha68 [24]3 years ago
8 0

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0 Equation 1

Similarly, the equation for outer node “o” will be

\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0 Equation 2

The conventive thermal resistance in i-node will be

R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w Equation 3

The conventive hermal resistance per unit area is

R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w Equation 4

The conductive thermal resistance per unit area is

R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w Equation 5

Since q_{rad}  is given as 100, T_{o}  is 40 T_ \infty  is 300 T_{\infty, o}  is 25  

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0  Equation 6

\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0

T_{i}-40= \frac {L}{0.05}*150

T_{i}-40=3000L

T_{i}=3000L+40 Equation 7

From equation 6 we can substitute wherever there’s T_{i} with 3000L+40 as seen in equation 7 hence we obtain

\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0

The above can be simplified to be

\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0

\frac {260-3000L}{0.033}=50

-3000L=1.665-260

L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm

Therefore, insulation thickness is 86mm

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Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

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Data Given:

x = 27 , 44 , 32 , 47, 23 , 40, 34, 52

y = 30, 19,  24,  13 , 29,  19,  21,  14

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We are asked to verify the above values manually in this question.

So,

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Yes, it is equal to the given value. Hence, verified.

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∑y = 30 + 19 +  24 + 13 + 29 + 19 +  21 +  14

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Let's verify it:

For this to find,  first we need to square all the value of x individually and then add them together to verify.

∑x^{2} = 27^{2} + 44^{2} + 32^{2} + 47^{2} + 23^{2} + 40^{2} + 34^{2} + 52^{2}

∑x^{2} = 11,887

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Let's verify it:

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Answer:

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the negative sign means reduction

Therefore, there is 30.12% reduction

b) the true stress at fracture is 658.26 Mpa

Explanation:

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Final diameter d_{f} = 10.7

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a) determine The ductility in terms of percent reduction in area;

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% = π/4 [ ( (d_{f} )² - (d_{o} )²) / ( π/4  (d_{o} )²) ]

= ( (d_{f} )² - (d_{o} )²) / (d_{o} )² × 100

we substitute

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the negative sign means reduction

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True stress  \alpha _{T} = \alpha _{E} ( 1 +  E_{E} )

E_{E}  is engineering strain

E_{E}  = dL / Lo

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True stress  \alpha _{T} = 460 (1.431)

True stress  \alpha _{T} = 658.26 Mpa

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Answer:

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Explanation:

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