Find the temperature at each of the six positions given in the chart at the inst

Question

Find the temperature at each of the six positions given in the chart at the instant t=150s.

The heat diffusion equation for transient in a one-dimensional slab is: Suppose the heat flux is qx = 0 at x = 0 and the at x = L is exposed to a convection environment with fluid temperature and convection coefficient h . The boundary conditions are: Suppose the slab is initially at uniform T(x,0) = Ti. The dimensionless form of this equation, its initial condition, and conditions are: Figure 5.7 in section 5.4 of the text has a response chart for the solution to this problem. Use the chart to solve the problem: A large 100-mm thick steel plate is initially at temperature Ti = 200 degree C at time t = 0. Both sides of the plate are exposed 50 degree C oil with convection coefficient h = 1500 W/m2.K. The thermal conductivity, capacity, and density of the steel are k = 38 W/m.K , cp = 485 J/kg.K , and rho = 7800 g/m3. Find the temperature at each of the six positions given in the chart at the t = 150 s. Figure 5.7 The transient temperature distribution in a slab at six positions: x/L = 0 is the center, x/L = 1 is one outside boundary.

Explanation / Answer

Fo = alpha*t/L^2 = K*t/(rho*Cp*L^2)

Fo = 38*150/(7800*485*0.1^2)

Fo = 0.15

Bi = h*L/K = 1500*0.1/38

Bi = 4

Ti = 200

Tinf = 50

1)

x/L = 0

(T - Tinf)/(Ti-Tinf) = 1

Therefore

T =200

2)

x/L = 0.2

(T - Tinf)/(Ti-Tinf) = 0.99

Therefore

T = 198.5

3)

x/L = 0.4

(T - Tinf)/(Ti-Tinf) = 0.99

Therefore

T = 198.5

4)

x/L = 0.6

(T - Tinf)/(Ti-Tinf) = 0.95

Therefore

T = 192.5

5)

x/L = 0.8

(T - Tinf)/(Ti-Tinf) = 0.9

Therefore

T = 185

6)

x/L = 1

(T - Tinf)/(Ti-Tinf) = 0.85

Therefore

T = 177.5

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