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## A long, cylindrical, electrical heating element of diameter D= 10 mm, thermal conductivity k= 240 W/m·K, density rho= 2700 kg/m3, and specif

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.

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Physics
5 months
2021-08-02T05:58:17+00:00
2021-08-02T05:58:17+00:00 1 Answers
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## Answers ( )

Answer:The steady-state surface temperature = 905.5 KThe 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 KTherefore 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 secTherefore the time required to come within 10°C of its steady-state value is 137.4 sec.