Humans need to wear protective suits during space walks not only to provide an a
ID: 1886241 • Letter: H
Question
Humans need to wear protective suits during space walks not only to provide an atmosphere for breathing, but also to control thermal regulation of the body. The insulating materials of a spacesuit can be modelled as consisting of layers of a fibrous material with a total thickness 5.0 mm and a thermal conductivity of 0.0020 Js-mK-1 Consider an astronaut performing activities on the moon for which temperatures can range from -173°C in darkness to 130°C in daylight. We can assume that the temperature of the outer surface of the spacesuit is equal to the ambient temperature. A thermal regulation (heating/cooling) system connected to the spacesuit maintains the inside temperature of the suit at 22°C. Discuss the following (including quantitative calculations where relevant) (a) At what rate must energy be added/removed by the thermal regulation system to account for the heat loss/gain through the insulating layer of the suit? For this calculation, ignore any heat generated by the body. Consider the extreme temperatures only (b) Suppose the thermal regulation system failed. Estimate how much the temperature of the human body in the suit would change in 1 hour. You should assume that the heat is added/removed from the body uniformly throughout the body (c) Your calculations above have neglected the heat generated by the human body itself. Would this have a significant effect on your calculations of the temperature change of the body? Justify your answer. Note: In obtaining your answers you will need to do some research and/or make some estimations of relevant parameters such as body surface area, specific heat capacity, etc. Quote any sources that you useExplanation / Answer
Rate of heat transfer is given by = k*A*delT/t where
k - thermal conductivity; A - area of cross section, t - thickness and delT is temperature difference
https://www.medicinenet.com/g00/script/main/art.asp?articlekey=39851&i10c.encReferrer=aHR0cHM6Ly93d3cuZ29vZ2xlLmNvLmluLw%3D%3D&i10c.ua=1&i10c.dv=14 -- reference for data on human body surface area
Assuming surface area of 1.5 m^2
Rate of heat transfer when outside temperature is -173C
= - 0.0020* 1.5 * (22+173)/0.005 = 117 J/s should be added to the system
Rate of heat transfer when outside temperature is 130C
= 0.0020* 1.5 * (130 - 22)/0.005 = 64.8 J/s -- removed from the system
b)
for the first case above where dQ/dt = 117J/s, Heat lost over one hour = 117*60*60 = 421200 J
Q = C* M * DT where C is specif heat capacity, M is mass of the body and DT is change in temperature
https://chemistry.stackexchange.com/questions/44493/which-is-the-specific-heat-of-human-body -- source for specific heat capacity of human body
=> 421200 = 3500 * 75 * DT M - 75Kg average human weight
=> DT = 1.6 K
for the second case DQ/DT = 64.8
Heat generated in one hour = 64.8*60*60 = 233280 J
233280 = 3500 * 75 * DT M - 75Kg average human weight
=> DT = 0.88K increase in temperature
c. Thermal conductivity of human body is around 0.22 kJ/smK. This would affect the rate at which the heat transfer happens and hence the temperature change of the body
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