Question

A metal bar with an initial temperature, 𝑇0, in the interval of 30°C ≤ 𝑇0 ≤ 35°C is dropped into a container of boiling water (100°C). The temperature of the metal bar, 𝑇 at any time, 𝑡 satisfies the following Newton’s Law of Cooling model 𝑑𝑇 𝑑𝑡 = −𝑘(𝑇 − 𝑇𝑚) where 𝑇𝑚 is the ambient temperature and 𝑘 is the constant. After 5 seconds, the temperature of the bar, 𝑇1 is in the interval of 40°C ≤ 𝑇1 ≤ 50°C. a. Find the equation that models the temperature of the metal bar, 𝑇 at any time, 𝑡 (choose a value of 𝑇0 and 𝑇1 from the given intervals, respectively). By using an appropriate analytical method, solve the derived model and explain the reason for the selection of the method. b. Compute the temperature of the metal bar after 100 seconds by using the derived model in Part 1(a) with THREE (3) different numerical methods with step size, ℎ = 10 seconds. Select the best numerical method to compute the temperature of the metal bar and justify your answer

Answer 1

a)

$-\frac{dT}{T-T_m}=kdt$

$ln(T_m-T)=kt+c$

$T=T_m-e^{kt+c}=100-e^{kt+c}$

$T(0)=100-e^c=T_0$

$c=ln(100-T_0)$

$ln65\le c\le ln70$

$T(5)=100-e^{c+5k}=T_1$

$k=(ln(100-T_1)-c)/5$

$(ln50-ln70)/5\le k\le (ln60-ln65)/5$

$-0.067\le k\le -0.016$

$100-70e^{-0.016t}\le T\le 100-65e^{-0.067t}$

reason for the selection of the method:

this is method to find exact value of variable

temperature of the metal bar after 100 seconds:

$100-70e^{-0.016\cdot100}\le T(100)\le 100-65e^{-0.067\cdot100}$

$85.87\le T(100)\le99.92$

b)

Euler method 

$T(0)=30$

$T'(t)=0.067(100-T)=f(t,T)$

$T_n=T_{n-1}+10f(t_{n-1},T_{n-1})$

$T_{10}=T(100)=99.8989\degree C$

error = $99.92-99.8989=0.0211$

Runge-Kutta 2 method:

$k_2=10f(t_0+10,T_0+k_1)$

$k_1=10f(t_0,T_0)=(10)f(0,30)=(10)⋅(4.69)=46.9$

$k_2=10f(t_0+10,T_0+k_1)=(10)f(10,76.9)=(10)⋅(1.5477)=15.477$

$T_1=T_0+(k_1+k_2)/2=30+31.1885=61.1885$

$T(10)=61.1885$

........................

$T(100)=99.8078\degree C$

error = $99.92-99.8078=0.1122$

Runge-Kutta 3 method:

$k_1=10f(t_0,T_0)=(10)f(0,30)=(10)⋅(4.69)=46.9$

$k_2=10f(t_0+10/2,T_0+k_1/2)=(10)f(5,53.45)=(10)⋅(3.1188)=31.1885$

$k_3=10f(t_0+10,T_0+2k_2-k_1)=(10)f(10,45.477)=(10)⋅(3.653)=36.5304$

$T_1=T_0+(k_1+4k_2+k_3)/6$

$T_1=30+[46.9+4(31.1885)+(36.5304)]/6=64.6974$

$T(10)=64.6974$

...................................................

$T(100)=99.9255\degree C$

error = $99.9255-99.92=0.0055$

Minimal error is for Runge-Kutta 3 method, so this is the best numerical method to compute the temperature.