Question

(a) While shopping at a thrift store, you purchase a cheap water-heater. The heater has a total surface area of 3 m3 which you decide to insulate using a special kind of insulation material with conductivity of 0.08 W/m K. For normal operation, you want to maintain the inside surface of the insulation at 75C while the outside is at 20C. Additionally, you want to limit heat transfer loss to 200 W. Given your requirements, how thick does the insulation need to be?

(b) The heaters on a Space X spacecraft suddenly fail. Given your expertise insulating thrift store waterheaters, you are called in to help. You discover that without the heaters, the spacecraft is losing heat via radiation at a rate of 100 kJ/h. The onboard systems generate heat at 75 kJ/h. The air within the spacecraft is at a temperature of 25C and a pressure of 100 kPa, and occupies a 10 m3 volume. Your job is to determine how long it will take before the air temperature within the spacecraft reaches - 20C.

Answer 1

a) Power loss= (coefficient of thermal conductivity * Area * temp. gradient)/thickness of insulating material.

Since our power loss has to be limited to 200W & with given constraints as area= 3m²? , Temp. gradient= 75? - 20? = 55?, coefficient of thermal conductivity = 0.08 W/mK.

? 200 = (08*3*55)/d ,where d= thickness of insulation required to limit the power loss to 200W.Solving the equation we can find d= 0.066m i.e 66mm.

b) In this part, the heaters have failed & the air inside the rocket is radiating heat at 100KJ/H. The onboard systems are generating the heat at 75KJ/H. As a result, every hour,the air is losing 25 KJ/H to outside.

The energy to be lost by the given mass of air from 25 to -20 can be calculated from mc_pT.

Mass of air inside the rocket =Volume occupied by the air * density of air at 100kPa, 25.

Mass of air inside the rocket = 10m³ * 1.225 kg/m³ ? ?Mass of air inside the rocket = 12.25 kg

Hence, heat to be lost from 12.25 kg of air to loss temp from 25 to -20?= 12.25 * c_p ?* 55,

Considering c_?p?= 0.718 kJ/kg .K,heat to be lost from 12.25 kg of air to loss temp from 25 to -20?=?483.75 kJ.

As we know heat loss rate is 25 kJ/H, hence time taken for the air temp to reduce from 25? to -20?=(483.75kJ/ 25kJ/H)= 19.35 hours.