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siniylev [52]
4 years ago
5

Electric heater wires are installed in a solid wall having a thickness of 8 cm and k=2.5 W/m.°C. The right face is exposed to an

environment with h=50 W/m2°C and k'=30°C, while the left face is exposed to h=75 W/m2°C and T[infinity]=50°C. What is the maximum allowable heat generation rate such that the maximum temperature in the solid does not exceed 300°C.

Engineering
2 answers:
Svet_ta [14]4 years ago
6 0

Answer:

2.46 * 10⁵ W/m³

Explanation:

See attached pictures for detailed explanation.

levacccp [35]4 years ago
5 0

Answer:

q^.=2.46*10^5W/m^3

Explanation:

Given\\k=2.5W/m\\h_{1} =75(left)\\h_{2} =50(right)\\T_{1} =50^oC\\T_{2} =30^oC

so

T=-\frac{q^.x^2}{2k} +c_{1}x+ c_{2} \\T=T_{1}  \\at \\x=-0.04\\T=T_{2} \\at\\x=+0.04

dT/dx=-q^.x/k+c_{1} \\T=T_{max} =300\\at\\x=c_{1} \frac{k}{q^.} (1)

h_{1}(T_{1infinity} -T_{1} )=-k\frac{dT}{dx} |_{x=0.04} (2)\\-k\frac{dT}{dx} |_{x=0.04} =h_{2} (T_{2}-T_{2infinity} (3)

300=-\frac{q^.}{2k} [c_{1} \frac{k}{q} ]^2+c_{1} [c_{1} \frac{k}{q} ]+c_{2} (1)

75[50+\frac{q^2}{2k} (0.04)^2+c_{1} (0.04)-c_{2} ]=-k[\frac{+q^2(0.04)}{2k} ](2)

-k[\frac{-q^.(0.04)}{2k} ]=50[\frac{-q^.(0.04)}{2k} +c_{1} (0.04)+c_{2} -30](3)

solving above 3 equations for 3 unknowns c1,c2,q

we get q^.=2.46*10^5W/m^3

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Explanation:

Data provided in the question :

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Explanation:

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The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb

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