- 1.
- This HW in one PDF (letter size)
- 2.
- This HW in one PDF (legal size)

The PDE to solve is the parabolic PDE

\[ \frac{1}{\alpha }\frac{\partial T}{\partial t}=\frac{\partial ^{2}T}{\partial x^{2}}\]

There is no source term. We consider only convective heat loss in the ﬁrst part of the problem, then for the second part, add the radiative heat loss.

Before we solve this problem, it is a good idea to write down all the units. This table summarizes this

term | name | dimensional | SI units | Imperial units | SI convert |

\(\alpha =\frac{k}{\left ( \rho \right ) \left ( c_{p}\right ) }\) | diﬀusivity | \(\frac{L^{2}}{T}\) | \(\frac{\text{meter}^{2}}{\sec }\) | \(\frac{\text{ft}^{2}}{\sec }\) | \(0.3048\) |

\(k\) | conductivity | \(\frac{ML}{T^{3}\Theta }\) | \(\frac{\text{Watt}}{\text{meter-Kelvin}}\) | \(\frac{\text{Btu}}{\text{hr-ft-F}^{o}}\) | \(1.73\) |

\(\rho \) | mass density | \(\frac{M}{L^{3}}\) | \(\frac{\text{kg}}{\text{meter}^{3}}\) | \(\frac{\text{lbm}}{\text{ft}^{3}}\) | \(16.019\) |

\(c_{p}\) | speciﬁc heat | \(\frac{L^{2}M}{T^{2}\Theta }\) | \(\frac{\text{joule}}{\text{kg-Kelvin}}\) | \(\frac{\text{Btu}}{\text{lbm-F}^{o}}\) | \(4188\) |

\(\sigma \) | Stefan-Boltzmann | \(\frac{\text{Watt}}{\text{meter}^{2}\text{ Kevin}^{4}}\left ( 5.670367\times 10^{-8}\right ) \) | \(\frac{\text{Btu}}{\text{hr-ft}^{2}\text{-R}^{4}}\left ( 0.171\times 10^{-8}\right ) \) | ||

And to convert \(F^{o}\) to \(C^{o}\) us \(C^{o}=\left ( F^{o}-32\right ) \frac{5}{9}\). To convert from \(F^{o}\) to absolute \(R^{o}\left ( \text{Rankine}\right ) \) add \(459.67\). Hence \(R^{o}=F^{o}+459.67\).

The Matlab program nma_HW_7_problem_1.m solves both parts. Summary of result

time to cool to \(1000\) F (minutes) | |

convective only loss | \(20.18\) |

convective and radiative loss | \(17.26\ \left ( 85.5\%\text{ of above time}\right ) \) |

The following shows the plot for the ﬁrst part (convective heat loss only).

The plot below shows the proﬁle of the heat loss across the whole section. Time is increasing going down in this ﬁgure. From initial time \(t=0\) to \(t=25\) minutes.

Now radiative heat loss was added. It was found that steel will cool faster, by about \(85.5\) percent of the time it took without radiative heat.

A plot is given below which compare both cases on same plot to make it more clear to see the diﬀerence.

In this problem, the independent variable is the concentration of alcohol, called \(c\left ( x,t\right ) \). The initial conditions is that \(c\left ( x,0\right ) =0.02\) and the boundary conditions is the left side (where \(x=0)\) is \(c\left ( L,t\right ) =0\) and on the right side (where \(x=20\) cm), then \(c(R,t)=0.1\).

This is a parabolic pde. It is solved using pdepe. The program nma_HW7_problem_2.m solves this and the following plot shows the result. It took

\[ \fbox{$486$\qquad seconds}\]

For the concentration at \(x=16\) cm to reach \(7.5\%\). The ﬁrst plot below show all proﬁles on same plot, then each one on a separate plot