Answer:
[tex]\dot Q = 524.957 W[/tex], [tex]T_{out} = 317.048 K[/tex].
Explanation:
a) The rate of heat loss is determined by following expression:
[tex]\dot Q = \frac{T_{ins, in} - T_{air}}{R_{th}}[/tex]
Where [tex]R_{th}[/tex] is the thermal resistance throughout the system:
[tex]R_{th} = R_{cond} + R_{conv || rad}[/tex]
Since convection and radiation phenomena are occuring simultaneously, the equivalent thermal resistance should determined:
[tex]\frac{1}{R_{conv||rad}} = \frac{1}{R_{conv}} + \frac{1}{R_{rad}}[/tex]
[tex]R_{conv||rad} = \frac{R_{conv}\cdot R_{rad}}{R_{conv} + R_{rad}}[/tex]
Where:
[tex]R_{conv} = \frac{1}{h_{conv} \cdot A} \\R_{rad} = \frac{1}{h_{rad} \cdot A}[/tex]
Outer surface area is given by:
[tex]A = 2 \cdot \pi \cdot r_{out} \cdot L[/tex]
Heat resistance associated to conduction through a hollow cylinder is:
[tex]R_{cond} = \frac{\ln \frac{r_{out}}{r_{in}} }{2 \cdot \pi L \cdot k}[/tex]
According to an engineering database, calcium silicate has a thermal conductivity k = [tex]0.085 \frac{W}{m \cdot K}[/tex]. Then, needed variables are calculated (L = 1 m):
[tex]r_{out} = 0.08 m, r_{in} = 0.06 m[/tex]
[tex]A \approx 0.503 m^2[/tex]
[tex]h_{conv} = 25 \frac{W}{m^{2}\cdot K}\\h_{rad} = 30 \frac{W}{m^2 \cdot K}[/tex]
[tex]R_{conv} = 0.080 \frac{K}{W}\\R_{rad} = 0.066 \frac{K}{W}[/tex]
[tex]R_{conv||rad} = 0.036 \frac{K}{W}[/tex]
[tex]R_{cond} = 0.539 \frac{K}{W}[/tex]
[tex]R_{th} = 0.575 \frac{K}{W}[/tex]
[tex]\dot Q = 524.957 W[/tex]
The temperature on the outside surface of the calcium silicate can be determined from the following expression:
[tex]\dot Q = \frac{T_{in}-T_{out}}{R_{cond}}[/tex]
Then,
[tex]T_{out}=T_{in}-\dot Q \cdot R_{cond}\\T_{out} = 317.048 K[/tex]
