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xeze [42]
2 years ago
14

A hollow aluminum sphere, with an electrical heater in the center, is used in tests to determine the thermal conductivity of ins

ulating materials. The inner and outer radii of the sphere are 0.15 and 0.18 m, respectively, and testing is done under steady-state conditions with the inner surface of the aluminum maintained at 250°C. In a particular test, a spherical shell of insulation is cast on the outer surface of the sphere to a thickness of 0.12 m. The system is in a room for which the air temperature is 20°C and the convection coefficient at the outer surface of the insulation is 30 W/m2·K. If 70W are dissipated by the heater under steady-state conditions, what is the thermal conductivity of the insulation?
Engineering
2 answers:
serg [7]2 years ago
8 0

Answer:

Explanation:

This is a questions that can be answered using electrical resistance analogy. The expression for the

Artyom0805 [142]2 years ago
8 0

Answer:

k_{ins}  = 0.0541 W/m. K

Explanation:

Given data:

inner radii of sphere is 0.15

outer radii of sphere is 0.18 m

thickness of given insulation is 0.12m

outer radius of given insulation is 0.18 + 0.12 = 0.30 m

inner temperature of surface os 250 degree C  = 523 k

T_{\infity} = 293 K

convective coffcient is h= 30 W/m^2 -K

For given condition, rate of heat transfer is given as

q = \frac{T_s -T_{\infty}}{R}

where R is thermal resistance

R = R_{cond} + R_{cond, ins} + R_{conv}

R =\frac{1}{4\pi K_a} [\frac{1}{r_i} - \frac{1}{r_o}] + \frac{1}{4\pi K_{ins}} [\frac{1}{r_0} - \frac{1}{r}] + \frac{1}{4\pi r^2 h}

fro aluminum at T = 523 K, Thermal conductivity isK_a = 230 W/m. K

R =\frac{1}{4\pi \times 230} [\frac{1}{0.15} - \frac{1}{0.18}] + \frac{1}{4\pi K_{ins}} [\frac{1}{.18} - \frac{1}{0.30}] + \frac{1}{4\pi 0.30^2 30}

R = 0.02985 + \frac{0.1763}{k_{ins}} K/W

FROM

q = \frac{T_s -T_{\infty}}{R}

70 = \frac{523 - 293}{0.02985 + \frac{0.1763}{k_{ins}}}

solving forR_{ins}

0.02985 + \frac{0.1763}{k_{ins}} = 3.285

\frac{0.1763}{k_{ins}}  = 3.255

k_{ins}  = 0.0541 W/m. K

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djverab [1.8K]

Answer:

a) 159.07 MPa

b) 10.45 MPa

c) 79.535 MPa

Explanation:

Given data :

length of cantilever beam = 1.5m

outer width and height = 100 mm

wall thickness = 8mm

uniform load carried by beam  along entire length= 6.5 kN/m

concentrated force at free end = 4kN

first we  determine these values :

Mmax = ( 6.5 *(1.5) * (1.5/2) + 4 * 1.5 ) = 13312.5 N.m

Vmax = ( 6.5 * (1.5) + 4 ) = 13750 N

A) determine max bending stress

б = \frac{MC}{I}  =  \frac{13312.5 ( 0.112)}{1/12(0.1^4-0.084^4)}  =  159.07 MPa

B) Determine max transverse shear stress

attached below

   ζ = 10.45 MPa

C) Determine max shear stress in the beam

This occurs at the top of the beam or at the centroidal axis

hence max stress in the beam =  159.07 / 2 = 79.535 MPa  

attached below is the remaining solution

6 0
3 years ago
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Answer:

Check the explanation

Explanation:

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3 years ago
Section BreakProblem 08.048 2.value: 5.00 pointsRequired information Problem 08.048.b Compute the expected error. The expected e
Lady_Fox [76]

Answer:

a) 19 or select the closest answer

b) 5%

Explanation:

a)

from the voltage divide rule

V_{in} = V_0 * \frac{R_2}{R_2 + R_f}

\frac{V_0}{V_{in}} = \frac{R_2 + R_f}{R_2} => 1 + \frac{R_f}{R_2} = 20

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b)

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