Ch 10.4 Spontaneous Combustion

Background

Spontaneous combustion may occur in a variety of situations when heat produced by a chemical reaction cannot escape from a system fast enough.
In this case, heat may build up and result in ignition.
We consider the case of a stirred chemical reactor where there is no conduction of heat.

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Background

Spontaneous combustion occurs when the temperature inside a body increases at a rate faster than heat is able to escape from its surface.
The higher the temperature, the more heat is produced by the chemical reaction (usually oxidation).
It is important when designing systems that involve heat production to ensure that heat can escape sufficiently easily.

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Internet Research

Hay that is stored with high moisture continues to respire.
In the process of respiration, plant sugars are converted to water and carbon dioxide.
Plant respiration coupled with bacteria and mold activity produces heat.
The increase in moisture from respiration and the heat from respiration and microbial growth can start a chain reaction that eventually causes temperatures to spiral and potentially result in spontaneous combustion.
When temperatures exceed 170 to 190 F the situation is considered critical and it may be time to think about calling the fire department; see weblink.

Rate of Heat Production

The reaction causing the heat production may be one where some organic material intersects with the atmosphere or oxygen, such as woollen fibers in bales.
It is characterised by its exothermicity, that is, the amount of heat produced by the reaction.
It is also characterised by its rate of reaction, which determines the rate at which the heat is produced.

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Rate of Heat conduction

The process by which heat is able to escape is typically conduction.
Thus the surrounding materials and their ability to conduct heat have a substantial effect on the system.

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Arrhenius Law

Rate of reaction (associated with rate of heat gain):

\[ \small{ \begin{Bmatrix} \mathrm{rate~heat~generated~by} \\ \mathrm{reaction~per~unit~volume} \end{Bmatrix} \\ = \rho Q A e^{-E/(RT)} } \]

\( \rho \) = reacting substance density (moles/volume)
\( Q \) = heat of reaction (J/mole)
\( A \) = reaction parameter (1/sec)
\( E \) = activation energy
\( R = 8.314 \) J/(K mole)
\( T = U + 273.15 \) K

Newton's Law

Recall

\[ \small{ \begin{Bmatrix} \mathrm{rate~of} \\ \mathrm{heat~loss} \end{Bmatrix} = hS(T - T_a) } \]

\( h \) = Newton cooling coefficient J/(\( m^2 \) sec K)
\( S \) = surface area \( m^2 \)
\( T \) = temperature of body K
\( T_a \) = ambient temp K

Model Assumptions

Neglect heat conduction by assuming a uniform temperature throughout the reacting material.
Rate of heat created is given by the Arrhenius law.
Heat is lost from surface of reacting material according to Newton's law of cooling.

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Compartment Diagram, Word Equation & ODE

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\[ \begin{aligned} \small{ \begin{Bmatrix} \mathrm{rate~of~change~of} \\ \mathrm{heat~content} \end{Bmatrix} } &= \small{ \begin{Bmatrix} \mathrm{rate~heat~generated} \\ \mathrm{by~reaction} \end{Bmatrix} -\begin{Bmatrix} \mathrm{rate~heat~lost} \\ \mathrm{to~surroundings} \end{Bmatrix}} \\ \\ \rho V c\frac{dT}{dt} &= \rho V Q A e^{-E/(RT)} - hS(T - T_a) \end{aligned} \]

Differential Equation

From the word equations, our differential equation is

\[ \rho V c\frac{dT}{dt} = \rho V Q A e^{-E/(RT)} - hS(T - T_a) \]

Specific heat \( c \) is used to evaluate rate of change of heat in terms of rate of change of temperature (page 227), with

\[ \begin{aligned} \left[\rho V \right] &= \left(\frac{\mathrm{moles}}{V}\right)V = \mathrm{moles} \\ [c] &= \frac{J}{kg~K} \end{aligned} \]

Seminov model for spontaneous combustion.

Scaled Differential Equation

Our differential equation is

\[ \rho V c\frac{dT}{dt} = \rho V Q A e^{-E/(RT)} - hS(T - T_a) \]

We make a change of variable to obtain

\[ \sigma \frac{d \theta}{dt} = \lambda e^{-1/\theta} - (\theta - \theta_a) \]

See next page, with

\[ \theta = \frac{RT}{E}, ~ \theta_a = \frac{RT_a}{E}, ~ \lambda = \frac{\rho A V Q R}{h S E}, ~\sigma = \frac{\rho V c}{h S} \]

Scaled Differential Equation

\[ \begin{aligned} \theta = \frac{RT}{E} &\Rightarrow T = \frac{E}{R}\theta \\ & \Rightarrow \frac{dT}{dt} = \frac{E}{R}\frac{d\theta}{dt} \\ &\Rightarrow \rho V Q A e^{-E/(RT)} = \rho V Q A e^{-1/\theta} \\ &\Rightarrow hS(T - T_a) = hS(E/R)(\theta - \theta_a) \\ \rho V c T'(t) &= \rho V Q A e^{-E/(RT)} - hS(T - T_a) \\ \frac{\rho V c E}{hSR}\frac{d\theta}{dt} &= \frac{\rho V Q A }{hS} e^{1/\theta} - (E/R)(\theta - \theta_a) \\ \sigma \theta'(t) &= \lambda e^{-1/\theta} - (\theta - \theta_a); ~\sigma = \frac{\rho V c}{h S},~ \lambda = \frac{\rho A V Q R}{h S E} \end{aligned} \]

Numerical Method Outline

Ch104Ex1 <- function(T) {
  #T = time length in seconds for [0, T]
  N <- 10000  #N is the number of time nodes
  h <- T/N    #Time step size in seconds

  #System Parameters
  sigma <- 1.0   #reaction speed 
  theta_a <- 0.2 #dimensionless ambient temp 
  L1<-2.84  #reaction efficiency 
  L2<-2.85  #reaction efficiency 
  L3<-2.86  #reaction efficiency 

  #Slope functions for ODEs
  f1<-function(x) {L1*exp(-1/x)-(x-theta_a)}
  f2<-function(x) {L2*exp(-1/x)-(x-theta_a)}
  f3<-function(x) {L3*exp(-1/x)-(x-theta_a)}

Numerical Solution

Ch104Ex1(300)

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Numerical Solution

Ch104Ex1(300)

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Small changes in reaction efficiency \( \lambda \) near 2.85 results in large change in equilibrium temperature.
This corresponds to ignition, or spontaneous combustion.
Increasing \( \lambda \) corresponds to increasing \( V/S \).

Analytic Solution

The book provides a phase-plane analysis of ODE in order to understand behavior of analytic solution and ignition points.
We will omit this part, and leave it to the interested reader.

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