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9 May, 15:20

Consider a large plane wall of thickness L = 0.3 m, thermal conductivity k = 2.5 W/m · °C, and surface area A = 12 m2. The left side of the wall at x = 0 is subjected to a net heat flux of q0 = 700 W/m2 while the temperature at that surface is measured to be T1 = 80°C. Assuming constant thermal conductivity and no heat generation in the wall, a. Express the differential equation and the boundary conditions for steady onedimensional heat conduction through the wall. b. Obtain a relation for the variation of temperature in the wall by solving the differential equation. c. Evaluate the temperature of the right surface of the wall at x = L.

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  1. 9 May, 16:25
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    a) - k * dT / dx = q_o

    b) T (x) = - 280*x + 80

    c) T (L) = - 4 C

    Explanation:

    Given:

    - large plane wall of thickness L = 0.3 m

    - thermal conductivity k = 2.5 W/m · °C

    - surface area A = 12 m2.

    - left side of the wall at net heat flux q_o = 700 W/m2 @ x = 0

    - temperature at that surface is measured to be T1 = 80°C.

    Find:

    - Express the differential equation and the boundary conditions for steady one dimensional heat conduction through the wall.

    - Obtain a relation for the variation of temperature in the wall by solving the differential equation

    - Evaluate the temperature of the right surface of the wall at x = L.

    Solution:

    - The mathematical formulation of Rate of change of temperature is as follows:

    d^2T / dx^2 = 0

    - Using energy balance:

    E_out = E_in

    -k * dT / dx = q_o

    - Integrate the ODE with respect to x:

    T (x) = - (q_o / k) * x + C

    - Use the boundary conditions, T (0) = T_1 = 80C

    80 = - (q_o / k) * 0 + C

    C = 80 C

    -Hence the Temperature distribution in the wall along the thickness is:

    T (x) = - (q_o / k) * x + 80

    T (x) = - (700/2.5) * x + 80

    T (x) = - 280*x + 80

    - Use the above relation and compute T (L):

    T (L) = - 280*0.3 + 80

    T (L) = - 84 + 80 = - 4 C
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