Question

A long, cylindrical, electrical heating element of diameter D= 10 mm, thermal conductivity k= 240 W/m·K, density rho= 2700 kg/m3, and specific heat cp= 900 J/kg·K is installed in a duct for which air moves in cross flow over the heater at a temperature and velocity of 27°C and 20 m/s, respectively. (a) Neglecting radiation, estimate the steady-state surface temperature when, per unit length of the heater, electrical energy is being dissipated at a rate of 2000 W/m. (b) If the heater is activated from an initial temperature of 27°C, estimate the time required for the surface temperature to come within 10°C of its steady-state value. Evaluate the properties of air at 450 K.

Answer 1

Answer:

The steady-state surface temperature = 905.5 K

The time required to come within 10°C of its steady-state value is 137.4 sec.

Explanation:

Given data

D = 10 mm

k = 240$\frac{W}{m K}$

$\rho$ = 2700 $\frac{Kg}{m^{3} }$

$C_{p}$ = 900 $\frac{J}{kg K}$

Ambient temperature $T_{o}$ = 27 °c = 300 K

Heat transfer per unit length Q = 2000 $\frac{W}{m}$

(a). We know that at steady state the heat transfer per unit length is given by

Q = h ($\pi$ d l ) ($T_{s}$ - $T_{o}$)  ----- (1)

Since the convection heat transfer coefficient at 450 K is

h =  105.2 $\frac{W}{m^{2} K }$

Put all the values in equation (1), we get

2000 = 105.2 (3.14 × 0.01 × 1) ($T_{s}$ - 300)

$T_{s}$ = 905.5 K

Therefore the steady-state surface temperature = 905.5 K

(b).The time required for the surface temperature to come within 10°C of its steady-state value is given by

$\frac{T - T_{o} }{T_{s} - T_{o} } =$ $e^{-at}$ ------- (2)

Here

$a = \frac{h A}{\rho V C}$

Put all the values in above equation we get

$a = \frac{6h}{\rho D C}$

$a = \frac{(6)(105.2)}{(2700) (0.01) (900)}$

a = 0.026

From equation (2)

$\frac{300-283}{905-300} = e^{-0.026t}$

0.028 = $e^{-0.026t}$

-3.57 = -0.026 t

t = 137.4 sec

Therefore the time required to come within 10°C of its steady-state value is 137.4 sec.