Exercise 4

Cabin windows and cockpit windshields Cabin window

A cabin window consists of three panes:

1) an outer pane flush with the outside fuselage,

2) an inner pane which has a little hole in it,

3) a thinner, non-structural plastic pane called a scratch pane.

Figure 1: A typical commercial airplane passenger window.

Passengers can’t touch the inner pane (the one with the hole in it) or the outer pane, for safety reasons. Scratch pane isn’t actually part of the window assembly itself, but installed separately. The main thing to know is that aircraft cabin windows are not made of glass, but with something called stretched acrylic. It’s a lightweight material manufactured by a few global suppliers for the various aircraft flying today. Stretched acrylic is produced by stretching the base material of as-cast acrylic. It provides better resistance to crazing [hairline cracks], reduced crack propagation and improved impact resistance.

The inner and outer pane thickness is specific to each type of aircraft. Inner panes are generally thinner at approximately 5 mm thick and are only present as a fail-safe if the outer pane fails. The outer panes are thicker at approximately 10 mm thick and carry the pressure loads for the life of the window. The increased thickness is meant to allow for engagement with the airframe structure while maintaining the required strength. The air gap is approximately 6-7 mm and also varies for each aircraft.

Breather hole

Figure 2: The tiny breather hole in the inner pane with some frozen condensation.

As the plane gains altitude, the pressure acting on the outside the plane drops; the air is much less dense the higher your plane climbs. Because aircraft cabins are pressurized to about 1800 m for passenger comfort (and survival), there is more pressure inside the plane than acting on it from the outside. That pressure is bearing on the fuselage and the cabin windows. The little hole on the inner panel allows some of the cabin air to escape into the pocket between the inner and outer panes and equalize. This forces the outer pane to take all of the load, even if slowly. It’s a tiny hole so that as the plane ascends, the pressure slowly equalizes. The scratch plane covering the inner pane doesn’t block air from escaping into the hole. It’s not airtight. Air can flow into that space through multiple gaps in the interior wall assembly. Frost can form on the inner pane because moist air from the cabin seeps through the hole as the aircraft gains altitude. It freezes on contact with the very cold windows.

Flight deck windshields

Unlike cabin windows, the flight deck windshields are made with glass-faced acrylic — an outer layer of glass bonded to stretched acrylic. Then, there’s a layer between them, made of urethane. Each has anti-ice and anti-fog systems. In the case of the Boeing 787, there are then layers of stretched acrylic, just like the cabin windows, albeit much thicker — between 25 and 75mm thick depending on the aircraft.

Figure 3: Cockpit windshield on the Boeing 787.

Have you ever noticed that the cockpit windshields sometimes look like they have oil on them? That oily sheen is actually a coating of indium tin oxide, which is a conductive material between the layers and transmits heat. Accordingly, this thin coating is all that is needed to keep the windows nice and clear in frosty weather.

Figure 4: Rainbow effect induced from the thin coating of indium tin oxide in a windshield of a A321.

In the past this was accomplished with thin wires of a design similar to those in rear car windows, but the main manufacturers now use a coating of indium tin oxide. Just nanometers thick, this coating sits between glass plies and is completely transparent.

Thermal analysis of a cabin window: 1D steady state heat conduction with convective and radiative boundary conditions

Exercise a) – Single pane

Consider a window with an approximate size of 200x300 mm and a thickness of 15 mm, as if hypothetically realized with a single pane of acrylic Evonik PLEXIGLAS® GS 249 characterized by a thermal conductivity λ = 0,19 W / (m ° C).

The GS 249 is an example of material used for cabin windows in civil aircraft, glazing for pressurized aircraft cabins, cockpit glazing, canopies and windshields. It is certified to aviation standard. It is another cast acrylic specially developed to meet even higher demands by the aviation industry, but is additionally crosslinked. It offers higher resistance to media that cause stress cracking and a higher heat deflection temperature. It is also excellently suited for stretching, which makes it possible to improve its properties even further over the unstretched state. UV transmittance is less than 1 %.

To determine: the internal and external surface temperatures, the heat flux and the stationary thermal power transmitted through the window in conditions in which the temperature of the

cabin is 𝑇"#(%)= 295.15 K ( 22°C ). Suppose an airplane cruises at a Mach number of 0.8 whose

length is 40 m, the external convective heat transfer coefficient is determined to be 332.47 W/(m2 K) 1. The external radiation heat transfer to the outside can be meaningful at the cruise altitude. Consider the extreme case when an airplane flies in the night-time, so the solar radiation is eliminated. The radiative heat loss to the sky by a grey body is:

𝑞( = 𝜎𝜀 𝑇,- − 𝑇,#(()-

where 𝑞( is radiative heat loss flux, 𝜀 is the emissivity around 0.92, 𝑇, is external surface

temperature, and 𝑇,#(() is the mean radiant temperature of the sky. At a cruising altitude of nine to

eleven km above sea level, the sky temperature 𝑇,#(() is about 56 K lower than the free static air

temperature 𝑇,#(%) = 208.15 K ( -65 °C ). The thermal boundary condition on the outer shell skin

is set as the combination of heat convection with free air and radiation to the sky.

Assume ℎ" = 10 W/(m2°C) as coefficient of convective heat exchange on the inner surface of the window, including in it the weak effect of the thermal irradiation.

                                                                                                                         1 T. Zhang, L. Tian, C. Lin , S. Wang. Insulation of commercial aircraft with an air stream barrier along fuselage. Building and Environment, vol. 57 (2012) pp. 97-109

𝜆1 = 0,19 W/(m2 K)

𝑠1 = 15 mm

ℎ" = 10 W/(m2 K)

ℎ, = 332.47 W/(m2 W)

𝜎 = 5.67 10-8 W/(m2 K4)

𝜀 = 0.92

Window area 𝐴1=0.06 m2

𝑇"#(%)= 295.15 K

𝑇,#(%) = 208.15 K

𝑇,#(() = 152.15 K

 

 

Hypothesis

The two surfaces of the window remain at constant temperatures. These are isothermal surfaces. To calculate the heat fluxes we assume the hypothesis of:

§   Stationary heat transmission because the internal and external temperatures are assumed to be constant

§   One-dimensional heat transmission because the temperature gradient is significant only in the direction from inside to outside

§   Constant thermal conductivity

Solution

𝑞′ = ℎ" 𝑇𝑖∞(𝑐) − 𝑇"                                                  

𝑞 = − 𝜆𝑤𝑠𝑤(𝑇, − 𝑇")                                                        

𝑞99 = ℎ, 𝑇, − 𝑇𝑒∞𝑐 + 𝜎𝜀 𝑇𝑒4 − 𝑇𝑒∞

(𝑟)4                                                  

(i) 𝑞9 = 𝑞       ⟹         ℎ" 𝑇"#% − 𝑇" = − @A

BA(𝑇, − 𝑇")      ⟹

𝑇" =CD

EFGAHIJA

+CI#K

EF JAGAHI

= 0.56  𝑇, + 130.21

(e) 𝑞 = 𝑞99    ⟹        − @ABA

𝑇, − 𝑇" = ℎ, 𝑇, − 𝑇,#% + 𝜎𝜀 𝑇,- − 𝑇,#

(()-

(i) in (e)        ⟹        5.22  10ST  𝑇,- + 338.04  𝑇, − 70853.39 = 0

𝑇, = 209.30  𝐾        (  −63.85  °𝐶  )

𝑇" = 247.42  𝐾        (  −25.73  °𝐶  )

To note is the negative temperature value on the internal surface although the indoor temperature is 22 °C. This is to be avoided because it can cause condensation or frost on the inner surface when the humidity in the cabin is high.

External surface temperature 𝑇, is near the free static air temperature 𝑇,#% , the radiative

contribution is small.

Hp: 𝑇,#( = 𝑇,#

%

Under this hypothesis:

𝑇, = 209.59  𝐾        (  −63.56  °𝐶  )

𝑇" = 247.58  𝐾        (  −25.57  °𝐶  )

The heat flux from the cabin to the external environment is:

𝑞 = 𝑞9 = 𝑞99 = − @ABA

𝑇, − 𝑇" = 481.21  W/m2

The thermal power lost by each windows is:

𝑄 = 𝑞  𝐴1 = 28.87  𝑊

The radiative flux introduces non-linearity into the calculus. It is like to have a convective boundary condition with a coefficient temperature dependent:

𝑞( = 𝜎𝜀 𝑇,- − 𝑇,#(()-

𝑞( = 𝜎𝜀 𝑇,] + 𝑇,#(()] 𝑇,] − 𝑇,#

(()]

𝑞( = 𝜎𝜀 𝑇,] + 𝑇,#(()] 𝑇, + 𝑇,#

(() 𝑇, − 𝑇,#(()

𝑞( = 𝜎𝜀 𝑇,^ + 𝑇,]  𝑇,#(() + 𝑇,  𝑇,#

(()] + 𝑇,#(()^ 𝑇, − 𝑇,#

(()

𝑞( = ℎ 𝜎  , 𝜀  , 𝑇,  , 𝑇,#(() 𝑇, − 𝑇,#

(()

in our case, the coefficient h is 1.9, negligible if compared to ℎ,.

It is possible to introduce the concept of thermal resistance R

𝑞 =𝑇"#(%) − 𝑇,#

(%)

𝑅  

 where R is the global thermal resistance calculated as:

 

𝑅 =1ℎ"+𝑠1𝜆1

+1

ℎ,(%) + ℎ,

(() =110 +

0,0150,19 +

1332.47 + 1.9 = 0,182  

𝑚]𝐾𝑊  

 

𝑅%bcd,"          𝑅%bce              𝑅%bcd,, + 𝑅(fe,,    

One more time, the thermal power lost through the windows is:

𝑄 = 𝐴1𝑇𝑖∞(𝑐) − 𝑇𝑒∞

(𝑐)

𝑅= 0.06

295.15 − 208.150,182 = 28.7  𝑊  

Exercise b) – Double pane

Redo the previous exercise, assuming that the window is constituted of two layers of acrylic, the external 10 mm and the internal one 5 mm thick, separated by a thick air gap of 7 mm.

𝜆1, = 0,19 W/(m2 K)

𝑠1, = 10 mm

𝜆1" = 0,19 W/(m2 K)

𝑠1" = 5 mm

𝜆1f"( = 0,026 W/(m2 K)

𝑠1f"( = 7 mm

Solution

In this case the thermal resistance of the window is:

𝑅1 =𝑠g𝜆g

^

ghE

=𝑠1"

𝜆1"+𝑠1f"(

𝜆1f"(+𝑠1,

𝜆1,=0.0050.19 +

0.0070.026 +

0.010.19 = 0,348  

𝑚]𝐾𝑊  

The heat flux through the windows is:

𝑞 = − CDSCIiA

                                                       

Following the same procedure as in exercise a), the internal temperature 𝑇" is linked to the external temperature 𝑇, with the relationship:

𝑇" =𝑇,

1 + 𝑅1  ℎ"+

𝑇"#%

1 + 1𝑅1  ℎ"

= 0.22  𝑇, + 229.27

𝑇, is solution of the equation:

5.22  10ST  𝑇,- + 342.35  𝑇, − 72206.40 = 0

𝑇, = 210.61  𝐾        (  −62.54  °𝐶  )

𝑇" = 275.60  𝐾        (  +2.45  °𝐶  )

The temperature of the internal surface of the window is higher than previous case and in particular greater than zero. Such temperature prevents condensation phenomena.

The heat flux from the cabin to the external environment is:

𝑞 = 𝑞9 = 𝑞99 = − CDSCIiA

= 186.75  W/m2

The thermal power lost by each windows is:

𝑄 = 𝑞  𝐴1 = 11.21  𝑊

The window is more efficient in terms of thermal insulation if compared to the previous case, with the same weight of material.

The inner surface of the external pane reaches a temperature 𝑇] calculated as:

𝑞 = −𝑇, − 𝑇]𝑠1,𝜆1,

     ⟹         𝑇] = 𝑇, +𝑠1,

𝜆1,𝑞 = 220.44  𝐾        (  −52.71  °𝐶  )

For this reason, the humidity of the air that flows through the tiny hole freezes immediately when in contact with the cold surface as in figure 2.