Ch 10.4 Spontaneous Combustion

What do you give to a sick lemon?

Lemon aid!

What do you call a man with a rubber toe?

Roberto!

• 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.

• Hay that is stored with high moisture continues to respire.
• 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 heat from respiration and microbial growth can start a chain reaction that can lead to spontaneous combustion.

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 rate of reaction, which determines the rate at which the heat is produced.

• Heat typically escapes via conduction.
• Thus the surrounding materials and their ability to conduct heat have a substantial effect on the system.

The main components of this process are shown below.

• The compartment will be the container of material.
  • This could be a hay-bale, drum, reaction chamber, etc.
• Heat input is via an exothermic reaction.
• Heat output is via heat conduction.

From the compartment diagram and assumptions, we can state our word equation for the balance law as follows:

\[ \small{ \begin{aligned} \begin{Bmatrix} \mathrm{rate~of~change~of} \\ \mathrm{heat~content} \end{Bmatrix} &= \begin{Bmatrix} \mathrm{rate~heat~generated} \\ \mathrm{by~reaction} \end{Bmatrix} -\begin{Bmatrix} \mathrm{rate~heat~lost} \\ \mathrm{to~surroundings} \end{Bmatrix} \end{aligned} } \]

• We consider simplest case: container is a stirred chemical reactor where there is no conduction of heat.
  
  • Thus we neglect internal heat conduction by assuming uniform temperature throughout 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.
  • Thus conduction is limited to surface heat exchange.

• Rate of reaction (gives rate of heat gain) shown below.
• Like Fourier's law, this formula was observed experimentally.

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

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

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

Rate of heat generated:

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

\( \left[\rho Q A \right] = \left(\frac{moles}{V}\right)\left(\frac{J}{moles}\right)\left(\frac{1}{sec}\right) = \frac{J}{sec\cdot V} \)

\( \left[\frac{E}{RT} \right] = \frac{\frac{J}{mole}}{\left(\frac{J}{K \cdot mole}\right)K} = \frac{\frac{J}{mole}}{\frac{J}{mole}} =1 \)

• \( \rho \) = reacting substance density; \( Q \) = heat of reaction
• \( A \) = reaction parameter; \( E \) = activation energy
• \( R \) = universal gas constant; \( T \) = temperature in Kelvin

Recall from Ch9.2:

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

Parameters:

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

From Ch9.2, the fundamental equation relating rate of change of heat to rate of change of temperature is

\[ Q = cm \frac{dT}{dt}= \rho Vc \frac{dT}{dt} \]

Parameters:

• \( c \) = specific heat of reacting substance, in J/(kg K).
• \( m \) = mass of reacting substance
• \( \rho \) = density
• \( V \) = volume

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

• 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

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

We have \[ \small{ \begin{aligned} \theta = \frac{RT}{E} &\Rightarrow T = \frac{E}{R}\theta \\ & \Rightarrow \frac{dT}{dt} = \frac{E}{R}\frac{d\theta}{dt} \\ &\Rightarrow \rho V Q A e^{-E/(RT)} = \rho V Q A e^{-1/\theta} \\ &\Rightarrow hS(T - T_a) = hS(E/R)(\theta - \theta_a) \end{aligned} } \]

Then \[ \small{ \begin{aligned} \rho V c\frac{dT}{dt} &= \rho V Q A e^{-E/(RT)} - hS(T - T_a) \\ \sigma \frac{d \theta}{dt} &= \lambda e^{-1/\theta} - (\theta - \theta_a) \end{aligned} } \]

\[ \small{ \begin{aligned} \rho V c\frac{dT}{dt} &= \rho V Q A e^{-E/(RT)} - hS(T - T_a) \\ \\ \sigma \frac{d\theta}{dt} &= \lambda e^{-1/\theta} - (\theta - \theta_a) \\ \\ \theta &= \frac{RT}{E}, \, \sigma = \frac{\rho V c}{h S}, \, \lambda = \frac{\rho A V Q R}{h S E} \end{aligned} } \]

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)}

• Analytic solution difficult to find.
  
• Can do equilibrium analysis of ODE to investigate behavior of solution and ignition points.
  

• Left graph: Temp on \( x \)-axis; rate of change of heat on \( y \)-axis.
• https://www.desmos.com/calculator/dvjsbbpk2x

• Left graph: Temp on \( x \)-axis; rate of change of heat on \( y \)-axis

• Left graph: Temp on \( x \)-axis; rate of change of heat on \( y \)-axis

• Left graph: Temp on \( x \)-axis; rate of change of heat on \( y \)-axis
• https://www.desmos.com/calculator/dvjsbbpk2x

• Left graph: Temp on \( x \)-axis; rate of change of heat on \( y \)-axis
• https://www.desmos.com/calculator/dvjsbbpk2x

• Left graph: Temp on \( x \)-axis; rate of change of heat on \( y \)-axis
• https://www.desmos.com/calculator/dvjsbbpk2x

• Left graph: Temp on \( x \)-axis; rate of change of heat on \( y \)-axis
• https://www.desmos.com/calculator/dvjsbbpk2x