Respuesta :
Answer:
The rate of heat loss through the wall is 1700 watts.
Explanation:
The complete statement of the problem is:
The wall of an industrial furnace is constructed from 0.15-m-thick, fireclay brick having a thermal conductivity of 1.7 W/m·K. Measurements made during steady state operation reveal temperatures of 1400 and 1150 K at the inner and outer sur- faces, respectively. What is the rate of heat loss through a wall that is 0.5 m by 1.2 m on a side?
Given that wall of the industrial furnace is under steady conditions of heat transfer and whose configuration is a flat element, we use the equation of conductive heat transfer rate ([tex]\dot Q[/tex]), measured in watts:
[tex]\dot Q = \frac{k\cdot w\cdot h}{l}\cdot (T_{i}-T_{o})[/tex]
Where:
[tex]k[/tex] - Thermal conductivity, measured in watts per meter-Kelvin.
[tex]w[/tex] - Width of the wall, measured in meters.
[tex]h[/tex] - Height of the wall, measured in meters.
[tex]l[/tex] - Thickness of the wall, measured in meters.
[tex]T_{i}[/tex] - Inner surface temperature, measured in Kelvin.
[tex]T_{o}[/tex] - Outer surface temperature, measured in Kelvin.
If we know that [tex]k = 1.7 \,\frac{W}{m\cdot K}[/tex], [tex]w = 1.2\,m[/tex], [tex]h = 0.5\,m[/tex], [tex]l = 0.15\,m[/tex], [tex]T_{i} = 1400\,K[/tex] and [tex]T_{o} = 1150\,K[/tex], the steady state heat transfer is:
[tex]\dot Q = \left[\frac{\left(1.7\,\frac{W}{m\cdot K} \right)\cdot (1.2\,m)\cdot (0.5\,m)}{0.15\,m} \right]\cdot (1400\,K-1150\,K)[/tex]
[tex]\dot Q = 1700\,W[/tex]
The rate of heat loss through the wall is 1700 watts.
The rate of heat transfer is 1700 W
Heat Transfer:
Given that the wall of the industrial furnace is under steady conditions of heat transfer.
The conductive heat transfer rate is given by:
[tex]\frac{Q}{t}=\frac{kA\Delta T}{d}[/tex]
here Q/t is the rate of heat transfer
A is the area,
d is the thickness of the wall here,
ΔT is the temperature difference, and
k is the thermal conductivity
Now,
area of the wall A = 0.5×1.2 = 0.6 m²
d = 0.15 m
k = 1.7 W/mk
and ΔT = 1400 - 1150 = 250K
So,
[tex]\frac{Q}{t}=\frac{1.7\times0.6\times250}{0.15}\\\\\frac{Q}{t}=1700\;W[/tex]
Learn more about heat transfer:
https://brainly.com/question/15043671?referrer=searchResults
