Question 7.11: Determining the Time Required to Cool a Mild Steel Sphere, t......
Determining the Time Required to Cool a Mild Steel Sphere, the Instantaneous Heat Transfer Rate, and the Total Energy Transfer
A 15𝑚𝑚 diameter mild steel sphere (𝑘 = 4.2𝑤/𝑚°C is exposed to cooling air flow at 20° 𝑐 resulting in the convective, coefficient h\,=\,120w/{m}^{2^{\circ}}C. Determine the following:
(i) Time required to cool the sphere from 550°𝐶 to 90°𝐶 .
(ii) Instantaneous heat transfer rate 2 minutes after the start of cooling.
(iii) Total energy transferred from the sphere during the first 2 minutes.
For mild steel take: 𝜌 = 7850𝑘𝑔/𝑚³; c_{p} = 475𝐽/𝑘𝑔°𝐶; and 𝛼 = 0.045 𝑚²/ℎ
Given: r=\frac{15}{2}=7.5m m=0.0075m;\;k=42w/{m^{2}}^{\circ}C;\; T_{\infty}=20^{\circ}C; \ T_{o}=550^{\circ}C;\;
T(t)=90^{\circ}C;\;h=120w/m^{2^{\circ}}C;
(i) Time required to cool the sphere from 550°𝐶 to 90°𝐶, 𝜏 = ?
The characteristic length, L_c is given by,
L_{c}=\frac{r}{3}=\frac{0.0075}{3}=0.0025m
Biot number,
B i=\frac{h L_{c}}{k}=\frac{120\times0.0025}{42}=0.007143
Since, 𝐵𝑖 ≪ 0.1 , so we can use the lumped capacitance theory to solve this problem.
Fourier Number,
F_{O}=\frac{k}{\rho c_{P}L^{2}}\cdot\tau=\frac{\alpha}{L_{c}^{2}}\cdot\tau
\alpha=0.045m^{2}/h=\frac{0.045}{3600}=1.25\times10^{-5}m^{2}/s
F_{O}=\frac{1.25\times10^{-5}}{(0.0025)^{2}}\cdot\tau=2\tau
B i\times F o=0.007143\times2\tau
The temperature variation with time is given by:
\frac{\theta}{\theta_{o}}=\frac{T(t)-T_{\infty}}{T_{o}-T_{\infty}}= e^{−𝐵𝑖×𝐹𝑜}
= \frac{90-20}{550-20}=e^{-0.014286\,\tau}
0.132=e^{-0.014286\,\tau}
-0.014286 \ \tau \ \ln\,e= \ln\, \ 0.132
\tau=\frac{\ln \ 0.132}{-0.014286}=141.7\;s
(ii) Instantaneous heat transfer rate 2 minutes (0.0333h) after the start of cooling,
q^{\prime}(\tau) =?
q^{\prime}(\tau)=h A_{s}\theta_{o}e^{-B i\times F o}
B i\times F o=0.014286\times2\times60=1.7143
q^{\prime}(\tau)=120\times4\pi\times(0.0075)^{2}(550-20)e^{-1.7143}=8.1w
(iii) Total energy transferred from the sphere during the first 2 minutes, (0.0333h)
𝑄(𝑡) = ?
Q(t)=h A_{s}\theta_{o}\bigl[1-e^{-B i\times F o}\bigr]{\frac{\tau}{B i\times F o}}
=120\times4\pi\times(0.0075)^{2}(550-20)[1-e^{-1.7143}]\times{\frac{120}{1.7143}}=2580.2 \ J
Or
F_{O}=\frac{k}{\rho c_{P}L_{c}^{2}}\cdot\tau=\frac{k}{7850\times475\times0.0025^{2}}\times120=206
B i\times F o\ =0.007143\times206\ =1.471
Q(t)=120\times4\pi\times(0.0075)^{2}(550-20)[1-e^{-1.471}]\times{\frac{120}{1.471}}=2825\,J