SOLUTIONS TO CHAPTER 15: FIRE AND EXPLOSION HAZARDS OF FINE POWDERS EXERCISE 15.1: It is proposed to protect a section of duct used for pneumatically transporting a food product in powder form in air by adding a stream of carbon dioxide. The air flowrate in the present system is 3 m3/s and the air carries 2% powder by volume. If the minimum oxygen for combustion (by replacement of oxygen with carbon dioxide) of the powder is 13% by volume, what is the minimum flowrate of carbon dioxide which must be added to ensure safe operation? SOLUTION TO EXERCISE 15.1: The current total air flow of 3 m3/s includes 2% by volume of plastic powder and 98% air (made up of 21% oxygen and 79% nitrogen by volume). In this stream the flow rates are therefore: powder: 0.060 m3/s 2% by volume oxygen: 0.6174 m3/s 20.58% by volume nitrogen: 2.323 m3/s 77.42% by volume At the limit, the exit oxygen concentration of the flowing mixture should be 13% by volume. Hence, using a simple mass balance assuming constant densities, volume flow of O2total volume flow

= 0.61743.0 + n = 0.13

from which, the minimum required flow rate of added carbon dioxide, n = 1.75 m3/s. EXERCISE 15.2: A combustible dust has a lower flammability limit in air at 20C of 1.2% by volume. A dust extraction system operating at 3 m3/s is found to have a dust concentration of 1.5% by volume. What minimum flow rate of additional air must be introduced to ensure safe operation? SOLUTION TO EXERCISE 15.2: Assume that the dust explosion hazard will be reduced by bringing the dust concentration in the extract to below the lower flammability limit. In 3 m3/s of extract, the flow rates or air and dust are:

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air: 2.955 m3/s 98.5% by volume dust: 0.045 m3/s 1.5% by volume At the limit, the dust concentration after addition of dilution air will be 1.2%, hence: volume of dusttotal volume

= 0.0453.0 + n = 0.012

from which, the minimum required flow rate of added dilution air, n = 0.75 m3/s. EXERCISE 15.3: A flammable pharmaceutical powder suspended in air at a concentration within the flammable limits and with an oxygen concentration above the minimum oxygen for combustion flows at 40 m/s through a tube whose wall temperature is greater than the measured ignition temperature of the dust. Give reasons why ignition does not necessarily occur. SOLUTION TO EXERCISE 15.3: Although the tube wall temperature is greater than the measured minimum ignition temperature and the energy available is likely to be greater than the minimum ignition energy, the residence time for the suspension in the tube must also be sufficient for suspension to reach the minimum ignition temperature. The point to note is that the heat transfer conditions here are quite different from those found in the equipment used to determine explosion characteristics of the dust. EXERCISE 15.4: A fine flammable plastic powder is leaking from a pressurised container at a rate of 0.5 litres/min into another vessel of volume 2 m3 and forming a suspension in the air in the vessel. The minimum explosible concentration of the dust in air at room temperature is 1.8% by volume. Stating all assumptions, estimate: a) the delay from the start of the leak before explosion occurs if there is no

ventilation b) the delay from the start of the leak before explosion occurs if the air ventilation

rate in the second vessel is 0.5 m3/h c) the minimum safe ventilation rate under these circumstances.

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SOLUTION TO EXERCISE 15.4: Mass balance on the dust in the second vessel:

=

air withvessel of out

dust offlow of ratevessel the into

dust offlow of ratevessel second in dust

of onaccumulati of rate

assuming constant gas density,

VdCdt

= Fin QoutC (Solution Manual-Equation 15.4.1) (where Fin is the leak rate of dust in m3/h, Qout is the air ventilation rate in m3/h, V is the volume of the second vessel in m3 and C is the dust concentration in m3/m3 in the second vessel at time t). (Leak rate of dust into vessel,

) F in = 0.5 litres / min = 0.5 10-3 60 m3 / h = 0.03 m3 / h

(a) If there is no ventilation, Qout = 0 and so VdCdt

= Fin = 0.03 Integrating, C = 0.015t + constan t With the boundary condition that C = 0 at t = 0, C = 0.015t and so, when C = 0.018 m3/m3, t = 1.2 hours. (b) Rearranging the general equation (Solution Manual-Equation 15.4.1) and integrating with the initial condition, C = 0 at t = 0,

hours 0.015

2CQ

-0.015ln

Q2t

out

out

=

Assuming the explosion occurs when the dust concentration reaches the lower flammability limit, 1.8%. Therefore, substituting Qout = 0.5 m3/h and C = 0.018 m3/m3 in the above equation:

time required = hours 1.43= 0.015

2018.05.0-0.015

ln 5.0

2t

= .

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(c) To ensure safety, the limiting ventilation rate is that which gives a dust concentration of 1.8% in the second vessel at steady state (ie. when dC/dt = 0). Under this condition, from Solution Manual-Equation 15.4.1: 0 = Fin Qout,safeC With Fin = 0.03 m3/h and C = 0.018 m3/m3: Qout, safe = 1.667 m3/h Hence, the minimum ventilation rate for the second vessel = 1.667 m3/h.

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