1. Engineering
2. Mechanical Engineering
3. to improve personal protection gear for the rural fire service...

# Question: to improve personal protection gear for the rural fire service...

###### Question details

To improve personal protection gear for the Rural Fire Service, you (an engineer) have been tasked with modelling the current structure of the firefighting coat which is shown below.

The representative thermal conductivities and dimensions are tabulated below.

The gaps between the exterior and the thermal liner are of 1 mm thickness. Heat is transferred across these gaps of still air via radiation and conduction. The linearised radiation coefficient for a gap is approximately hrad=σ(T1+T2)(T21+T22)4σT3avg${h}_{rad}=\sigma \left({T}_{1}+{T}_{2}\right)\left({T}_{1}^{2}+{T}_{2}^{2}\right)\approx 4\sigma {T}_{avg}^{3}$, where Tavg${T}_{avg}$ is the mean temperature of the surfaces comprising the gap. The radiation heat flux across the gaps of air can be expressed as q′′rad=hrad(T1T2)${q}_{rad}^{″}={h}_{rad}\left({T}_{1}-{T}_{2}\right)$ for both air gaps.

(a)  Model the coat structure using a thermal resistance diagram and calculate the thermal resistance per unit area (m2${}^{2}\cdot$K/W) for each layer as well as for the radiation and conduction processes in the gaps. Assume Tavg470K${T}_{avg}\approx 470\phantom{\rule{thinmathspace}{0ex}}\text{K}$.

Thermal resistance per unit area = .

(b) A flashover is the stage of a fire at which all surfaces/objects within a space have been heated to their ignition temperature, and a flame breaks out almost all at once over the surface of all objects in the space. The typical radiation heat flux on the fire-side of the coat in this situation is 0.25W/cm2$0.25\phantom{\rule{thinmathspace}{0ex}}\text{W}/{\text{cm}}^{2}$. What is the outer surface temperature of the coat corresponding to the inner surface temperature of 66o${66}^{o}$C which results in burn injury?

Outer surface Temperature = .