Answer:
a.) 1.453MW/m2, b.) 2,477,933.33 BTU/hr c.) 22,733.33 BTU/hr d.) 1,238,966.67 BTU/hr
Explanation:
Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2
Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr
1W = 3.41 BTU/hr
Given parameters:
thickness, t = 7.5mm = 7.5/1000 = 0.0075m
Temperatures 150 C = 150 + 273 = 423 K
50 C = 50 + 273 = 323 K
Temperature difference, T = 423 - 323 = 100 K
We are assuming steady heat flow;
a.) Heat flux, Q" = kT/t
K= thermal conductivity of the material
The thermal conductivity of brass, k = 109.0 W/m.K
Heat flux, ![Q" = \frac{109 * 100}{0.0075} = 1,453,333.33 W/m^{2} \\ Heat flux, Q" = 1.453MW/m^{2} \\](https://tex.z-dn.net/?f=Q%22%20%3D%20%5Cfrac%7B109%20%2A%20100%7D%7B0.0075%7D%20%3D%201%2C453%2C333.33%20W%2Fm%5E%7B2%7D%20%5C%5C%20Heat%20flux%2C%20Q%22%20%3D%201.453MW%2Fm%5E%7B2%7D%20%5C%5C)
b.) Area of sheet, A = 0.5m2
Heat loss, Q = kAT/t
Heat loss, ![Q = \frac{109*0.5*100}{0.0075} = 726,666.667W](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7B109%2A0.5%2A100%7D%7B0.0075%7D%20%3D%20726%2C666.667W)
Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr
c.) Material is now given as soda lime glass.
Thermal conductivity of soda lime glass, k is approximately 1W/m.K
Heat loss, ![Q=\frac{1*0.5*100}{0.0075} = 6,666.67W](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B1%2A0.5%2A100%7D%7B0.0075%7D%20%3D%206%2C666.67W)
Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr
d.) Thickness, t is given as 15mm = 15/1000 = 0.015m
Heat loss, ![Q=\frac{109*0.5*100}{0.015} =363,333.33W](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B109%2A0.5%2A100%7D%7B0.015%7D%20%3D363%2C333.33W)
Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr