Question 7.13: Calculation of the Time Taken to Boil an Egg An egg with mea......
Calculation of the Time Taken to Boil an Egg
An egg with mean diameter of 40mm and initially at 20°𝐶 is placed in a boiling water pan for 4 minutes and found to be boiled to the consumer’s taste. For how long should a similar egg for the same consumer be boiled when taken from a refrigerator at 5°𝐶. Take the following properties for eggs:
𝑘 = 10𝑤/𝑚°𝐶; 𝜌 = 1200𝑘𝑔/𝑚³; c_{p}\,\, = 2𝑘𝑗/𝑘𝑔°𝐶; and
ℎ (heat transfer coefficient) = 100𝑤/ m^{2^{\circ} } 𝐶.
Use lumped capacitance theory.
Given: r=\frac{40}{2}=20m m=0.02m;\;T_{o}=20^{\circ}c;\;T_{\infty}=100^{\circ}C;\;\tau=4\times60=240s;
k=10w/m^{\circ}C;~~\rho=1200k g/m^{3};~c_{p}=2k j/k g^{\circ}C;h=100w/m^{2^{\circ}}C;
For using the lumped capacitance theory, the required condition 𝐵𝑖 ≪ 0.1 must be valid.
B i={\frac{h\,L}{k}}\,, where L is the characteristic length which is given by,
L=\frac{V}{A_{s}}=\frac{r}{3}=\frac{0.02}{3}m
∴ B i=\frac{h\,L}{k}=\frac{h\times0.02}{k\times3}=\frac{100\times0.02}{10\times3}=0.067
As 𝐵𝑖 ≪ 0.1 , we can use the lumped capacitance system.
The temperature variation with time is given by:
\frac{T(t)-T_{\infty}}{T_{o}-T_{\infty}}=e^{-B i\times F o}
F_{O}=\frac{k}{\rho c_{P}L^{2}}\cdot\tau=\frac{10}{1200\times2\times10^{3}\times\left(\frac{0.02}{3}\right)^{2}}\times240=22.5
B i\times F o=0.067\times22.5=1.5075
{\frac{T(t)-100}{20-100}}=e^{-1.5075}
T(t)=100-80\,e^{-1.5075}=82.3^{\circ}C\simeq82^{\circ}C
Now, let us find 𝜏 when the given data is :T_{o}=5^{\circ}C;\;T_{\infty}=100^{\circ}C and T(t)=82^{\circ}C.
\frac{82-100}{5-100}=e^{-B i\times F o}
F o=\frac{k}{\rho c_{P}L^{2}}\cdot\tau=\frac{10}{1200\times2\times10^{3}\times\left(\frac{0.02}{3}\right)^{2}}\cdot\tau=0.09375\,\,\tau
B i\times F o=0.067 \times 0.09375 \ \tau = 6.281 \times 10^{-3} \tau = 0.00628 \ \tau
0.1895=e^{-0.00628\,\tau}
-0.00628 \ \tau \ \ln e=\ln0.1895
\tau=\frac{ln \ 0.1895}{-0.000628}=264.9~s=4.414\;m i n u t e s