Question

cylindrical electric resistance heater has a diameter of 1cm and length of 0.25m. When air at 25oC flows across the heater a heat-transfer coefficient of 25W/(m2 . oC) exists at the surface. (JUSTIFY ANY ASSUMPTIONS YOU IMPOSE!) a) If the electrical input to the heater is 5W, what is the steady state surface temperature of the heater if radiation is assumed negligible

Answer 1

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 $(T_{\infty})$ = 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:

$Q = hA(T - T_{\infty})$

$5= 25 \times (\pi \times (1\times 10^{-2}) \times 0.25) (T -0.25)$

By solving the above calculation:

T ( surface temperature of the heater) = 50.46° C  122.83° F