Answer:
Q = 378.247 Bt/hr
Explanation:
given data:
diameter of container = 3 m
so r = 1.5 m
T1 = 50°C
T2 = 100°C
depth y = 3 ft
Heat transfer is given as Q
[tex]Q = SK\Delta T[/tex]
Where
S = Shape factor for the object
[tex]S = \frac{4\pi r}{1-\frac{r}{2y}}[/tex]
[tex]S = \frac{4\pi *1.5}{1-\frac{1.5}{2*3}}[/tex]
S = 25.132 ft
[tex]Q = SK\Delta T[/tex]
Q = 25.132*0.301 *(100-50)
Q = 378.247 Bt/hr