10-cm-thick brass plates (k = 110 W/m‧K, rho= 8530 kg/m3 , cp = 380 J/kg‧K, α= 40 x 10-6 m2 /s) are cooled in an impinging jet of air at 15

Question

10-cm-thick brass plates (k = 110 W/m‧K, rho= 8530 kg/m3 , cp = 380 J/kg‧K, α= 40 x 10-6 m2 /s) are cooled in an impinging jet of air at 15 o C and convective heat transfer coefficient of 220 W/m2 ‧K. The plate is initially at a uniform temperature of 600 o C and the bottom is insulated. Determine the temperature of plane that is 5 cm from the top surface after 3 minutes of cooling.

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Gerda 5 months 2021-08-05T21:54:05+00:00 1 Answers 4 views 0

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    2021-08-05T21:55:37+00:00

    Answer:

    T=599.9°C

    Explanation:

    To solve this problem you can use the Newton’s law of cooling

    T=T_a+(T_0-T_a)e^{-kt}\\k=\frac{\alpha A}{m c_e}

    Ta: room temperature

    T0: initial temperature

    α: convective heat transfer coefficient

    m: mass

    c: specific heat

    A: area

    Hence, by replacing we have

    V=(0.05m)^3=1.25*10^{-4}m^3\\m=V\rho=(1.25*10^{-4}m^3)(8530\frac{kg}{m^3})=1.06kg\\\\A=(0.5m)^2=0.25m^2\\\\k=\frac{(40*10^{-6}\frac{m^2}{s})(0.25m^2)}{(380\frac{J}{kgK})(1.06kg)}=2.48*10^{-8}s^{-1}\\\\T=(15\°C)+(600\°C-15\°C)e^{-(2.48*10^{-8}s^{-1})(180s)}=599.9\°C

    hope this helps!!

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