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BaLLatris [955]
3 years ago
14

One-dimensional, steady-state conduction with uniform internal energy generation occurs in a plane wall with a thickness of 50 m

m and a constant thermal conductivity of 5 W/m*K. For these condtions, the temperature distribution has the form T(x)=a+bx+cx^2. The surface at x=0 has a temperature of T(0)=To=120 deg. (C) and h=500 W/m^2*K. The surface at x=L is well insulated.(a) Applying an overall energy balance to the wall, calculate the volumetric energy generation rate q.(b) Determine the coefficients a,b,c by applying the boundary conditions to the prescribed temperature distribution. Use the results to calculate and plot the temp. distribution.(c) Consider conditions for whic hthe convection coefficient is halved, but q remains unchanged. Determin the new values for a,b,c and plot the temperature dist. (Hint: T(0) is no longer 120)(d) Under conditions for which q is doubled and the convection coeff. remains unchanged (h=500 W/m^2*K) determine new a,b,c and plot temp distribution. Compare (a),(b),and (c) and discuss the effects of h and q on the distributions.
Engineering
1 answer:
uysha [10]3 years ago
4 0

Answer: please help me tooo

Explanation:

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Given

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To calculate the fraction of atom sites, we make use of the following formula

Nv/N = exp(-Qv/kT)

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Nv/N = exp(-0.55/(8.62E-5 * 600))

Nv/N = 0.000024078672493307

Nv/N = 2.41E-5

b. When T = 298K

Nv/N = exp(-0.55/(8.62E-5 * 298))

Nv/N = 5.026591237904E−10

Nv/N = 5.03E-10 ----- Approximated

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