Holooly Plus Logo

Question 7.10: Determination of the Maximum Speed of a Cylindrical Ingot in......

Determination of the Maximum Speed of a Cylindrical Ingot inside a Furnace
A cylinder ingot 10𝑐𝑚 diameter and 30𝑐𝑚 long passes through a heat treatment furnace which is 6m in length. The ingot must reach a temperature of 800°𝐶 before it comes out of the furnace. The furnace gas is at 1250°𝐶 and the ingot initial temperature is 90°𝐶. What is the maximum speed with which the ingot should move in the furnace to attain the required temperature? The combined radiative and convective surface heat transfer coefficient is 100𝑤/ m^{2^{\circ} } 𝐶.
Take: k (steel) = 40𝑤/ m^{2^{\circ} } 𝐶 and 𝛼 (thermal diffusivity of steel) =1.16\times10^{-5}m^{2}/s.

Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.
\rm{d=10c m =0.1m; \ L=30cm=0.3m; \ Length \ of \ the \ furnace=6m;}

 

T_{o}=800^{\circ}C;\;T(t)=800^{\circ}C;\;\;T_{\circ}=90^{\circ}C;

 

{\boldsymbol{v}}_{m a x} of ingot passing through the furnace =?

 

\rm{h=100w/ m^{2^{\circ} } C; \ k(steel)=40w/m^{\circ}C; \ α(steel)=1.16\times 10^{-5}m^{2}/s}

 

Characteristic length,

 

L_{c}=\frac{V}{A_{s}}=\frac{\frac{4}{3}d^{2}L}{[\pi dL + \frac{\pi}{4}d^{2}\times2]}=\frac{d L}{4L+2d}

 

=\frac{0.1\times0.3}{4\times0.3+2\times0.1}=0.02143m

 

B i=\frac{h L_{c}}{k}=\frac{100\times0.02143}{40}=0.0536

 

As 𝐵𝑖 ≪ 0.1 , Then internal thermal resistance of the ingot for conduction heat flow can be neglected.

The time versus temperature relation is given as:

 

\frac{T(t)-T_{\infty}}{T_{o}-T_{o o}}=e^{-B i\times F o}

 

F_{O}\ =\frac{k}{\rho c_{P}L^{2}}\cdot\tau=\frac{\alpha}{L^{2}}\cdot\tau=\frac{1.16\times10^{-5}}{0.02143^{2}}\cdot\tau=0.02526\,\tau

 

B i\times F o=0.0536 \times 0.02526 \ \tau = 1.35410^{-3}\tau=0.001354 \ \tau

 

\frac{\theta}{\theta_{o}}=\frac{T(t)-T_{\infty}}{T_{o}-T_{\infty}}= e^{−𝐵𝑖× 𝐹𝑜}

 

{\frac{\theta}{\theta_{o}}}={\frac{800-90}{1250-90}}=e^{-0.001354\,\tau}

 

0.6121\ =\,e^{-0.001354\,\tau}

 

-0.001354 \ \tau~\ln e=\ln \ 0.6121

 

\tau=\frac{\ln \ 0.6121}{-0.001354}=362.5\;s

 

v^{}_{m a x}  of ingot passing through the furnace,

 

v_{m a x}={\frac{\mathrm{furnace} \ \mathrm{len}\mathrm{g}\mathrm{th}}{\mathrm{ti}\mathrm{me}}}={\frac{6}{362.5}}=0.01655m/s

Related Answered Questions