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.

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

= 2700

= 900

Ambient temperature = 27 °c = 300 K

Heat transfer per unit length Q = 2000

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

Q = h ( d l ) ( - )  ----- (1)

Since the convection heat transfer coefficient at 450 K is

h =  105.2

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

2000 = 105.2 (3.14 × 0.01 × 1) ( - 300)

= 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

------- (2)

Here

Put all the values in above equation we get

a = 0.026

From equation (2)

0.028 =

-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.

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.