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frez [133]
2 years ago
12

As part of a heat treatment process, cylindrical, 304 stainless steel rods of 100-mm diameter are cooled from an initial tempera

ture of 500 C by suspending them in an oil bath at 30 C. If a convection coefficient of 500 W/m2 K is maintained by circulation of the oil, how long does it take for the centerline of a rod to reach a temperature of 50 C, at which point it is withdrawn from the bath
Engineering
1 answer:
saveliy_v [14]2 years ago
5 0

Answer:

Explanation:

Given that:

diameter = 100 mm

initial temperature = 500 ° C

Conventional coefficient = 500 W/m^2 K

length  = 1 m

We obtain the following data from the tables A-1;

For the stainless steel of the rod \overline T = 548 \ K

\rho = 7900 \ kg/m^3

K = 19.0 \ W/mk \\ \\ C_p = 545 \ J/kg.K

\alpha = 4.40 \times 10^{-6} \ m^2/s \\ \\  B_i = \dfrac{h(\rho/4)}{K} \\ \\  =0.657

Here, we can't apply the lumped capacitance method, since Bi > 0.1

\theta_o = \dfrac{T_o-T_{\infty}}{T_i -T_\infty}} \\ \\ \theta_o = \dfrac{50-30}{500 -30}} \\ \\ \theta_o = 0.0426\\

0.0426 = c_1 \ exp (- E^2_1 F_o_)\\ \\ \\  0.0426 = 1.1382 \ exp (-10.9287)^2 \ f_o   \\ \\ = f_o = \dfrac{In(0.0374)}{0.863} \\ \\ f_o = 3.81

t_f = \dfrac{f_o r^2}{\alpha} \\ \\ t_f = \dfrac{3.81 \times (0.05)^2}{4.40 \times 10^{-6}} \\ \\  t_f= 2162.5 \\ \\ t_f = 36 mins

However, on a single rod, the energy extracted is:

\theta = pcv (T_i - T_{\infty} )(1 - \dfrac{2 \theta}{c} J_1 (\zeta) )  \\ \\ = 7900 \\times 546 \times 0.007854 \times (500 -300) (1 - \dfrac{2 \times 0.0426}{1.3643}) \\ \\  \theta = 1.54 \times 10^7 \ J

Hence, for centerline temperature at 50 °C;

The surface temperature is:

T(r_o,t) = T_{\infty} +(T_1 -T_{\infty}) \theta_o \ J_o(\zeta_1) \\ \\ = 30 + (500-30) \times 0.0426 \times 0.5386 \\ \\ \mathbf{T(r_o,t) = 41.69 ^0 \ C}

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