Answer:
1. 
2. 
Explanation:
1.
Given:
- height of the window pane,

- width of the window pane,

- thickness of the pane,

- thermal conductivity of the glass pane,

- temperature of the inner surface,

- temperature of the outer surface,

<u>According to the Fourier's law the rate of heat transfer is given as:</u>

here:
A = area through which the heat transfer occurs = 
dT = temperature difference across the thickness of the surface = 
dx = t = thickness normal to the surface = 


2.
- air spacing between two glass panes,

- area of each glass pane,

- thermal conductivity of air,

- temperature difference between the surfaces,

<u>Assuming layered transfer of heat through the air and the air between the glasses is always still:</u>



Answer:

Explanation:
Given data:
Diffusion constant for nitrogen is 
Diffusion flux 
concentration of nitrogen at high presuure = 2 kg/m^3
location on which nitrogen concentration is 0.5 kg/m^3 ......?
from fick's first law

Take C_A as point on which nitrogen concentration is 2 kg/m^3

Assume X_A is zero at the surface


Answer:
The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²
Explanation:
Here we have the heat Q given as follows;
Q = 15 × 1075 × (1100 -
) = 10 × 1075 × (850 - 300) = 5912500 J
∴ 1100 -
= 1100/3
= 733.33 K

Where
= Arithmetic mean temperature difference
= Inlet temperature of the gas = 1100 K
= Outlet temperature of the gas = 733.33 K
= Inlet temperature of the air = 300 K
= Outlet temperature of the air = 850 K
Hence, plugging in the values, we have;

Hence, from;
, we have
5912500 = 90 × A × 341.67

Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².