Question

Consider a 400 mm × 400 mm window in an aircraft. For a temperature difference of 90°C from the inner to the outer surface of the window, calculate the heat loss rate through L = 12-mm-thick polycarbonate, soda lime glass, and aerogel windows, respectively. The thermal conductivities of the aerogel and polycarbonate are kag = 0.014 W/m ⋅ K and kpc = 0.21 W/m ⋅ K, respectively.
Evaluate the thermal conductivity of the soda lime glass at 300 K. If the aircraft has 130 windows and the cost to heat the cabin air is $1/kW ⋅ h, compare the costs associated with the heat loss through the windows for an 8-hour intercontinental fight.

Answer 1

Answer:

HEAT LOST

polycarbonate = 252 W

soda lime glass = 1680 W

aerogel = 16.8 W

COST associated with heat loss

polycarbonate = $ 262.08

soda lime glass =  $ 1,747.2

aerogel =  $ 17.472

The cost associated with heat loss is maximum in Soda Lime and minimum in Aerogel

Explanation:

Given that;

surface area for each window = 0.4m * 0.4m = 0.16m^2

DeltaT = 90°C, L = 12mm = 0.012m

thermal conductivity of soda line can be gotten from tables in FUNDAMENTALS OF HEAT AND MASS TRANSFER

so at 300K

KsL = 1.4 W/mK

Kag = 0.014 W/mK

Kpc = 0.21 W/mK

Now HEAT LOSS

for polycarbonate;

Qpc  = -KA dt/dx

NOTE (  heat flows from high temperature region to low temperature regions. so the second temperature would be smaller compared to the initial causing a negative in the change in temperature)

so Qag  = (0.21 * 0.16 * 90) / 0.012

= 252 W

for soda lime glass;

Qsl  = (1.4 * 0.16 * 90) / 0.012

= 1680 W

for aerogel

Qaq  = (0.014 * 0.16 * 90) / 0.012

= 16.8 W

Now for COST associated with heat lost

for polycarbonate;

cost = Qpc * 130 * 8 * 1/1000

= 252 * 130 * 8 * 1/1000

= $ 262.08

for soda lime glass;

cost = 1680 * 130 * 8 * 1/1000

= $ 1,747.2

for aerogel

cost = 16.8 * 130 * 8 * 1/1000

= $ 17.472

Therefore the cost associated with heat loss is maximum in Soda Lime and minimum in Aerogel