Answer:
Explanation:
From the information given:
The diameter of the cylindrical heater (d) = 1 cm
The length of the cylindrical heater (l) = 0.25 m
The ambient air temperature [tex](T_{\infty})[/tex] = 25° C= (273+25)K = 298 K
The convective heat transfer coefficient (h) = 25 W/m² °C
The electric input Q = 5W
As stated in the question that if radiation is being neglected:-
Let also assume that;
the heat transfer takes place at a steady-state
1-D flow takes place
No external heat generation; &
No force convection takes place;
Then; the heat transfer through the convection can be calculated as:
[tex]Q = hA(T - T_{\infty})[/tex]
[tex]5= 25 \times (\pi \times (1\times 10^{-2}) \times 0.25) (T -0.25)[/tex]
By solving the above calculation:
T ( surface temperature of the heater) = 50.46° C 122.83° F
