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

To calculate the steady-state surface temperature, use the formula Q/t = (k × A × ΔT) / d. The time required for the surface temperature to come within 10°C of its steady-state value can be calculated using the formula τ = (ρ × cp × V) / (h × A).

Step-by-step explanation:

In order to calculate the steady-state surface temperature of the heating element, we need to consider the rate of heat transfer through conduction. The formula for heat conduction is Q/t = (k × A × ΔT) / d where Q/t is the rate of heat transfer, k is the thermal conductivity, A is the surface area, ΔT is the temperature difference, and d is the thickness of the material.

Using the given information, we can rearrange the formula to solve for ΔT. Plugging in all the values, we have ΔT = (Q/t × d) / (k × A).

To calculate the specific values, we have Q/t = 2000 W/m, d = 10 mm = 0.01 m, k = 240 W/m·K, and A = π * (D/2)^2, where D is the diameter of the heating element.

For part (a), we can plug in the values and calculate the steady-state surface temperature. For part (b), we can calculate the time required for the surface temperature to come within 10°C of its steady-state value using the formula for time constant, τ = (ρ × cp × V) / (h v A), where ρ is the density, cp is the specific heat, V is the volume, h is the heat transfer coefficient, and A is the surface area.

Answer 2

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.

Step-by-step explanation:

Given data

D = 10 mm

k = 240
`(W)/(m K)`

`\rho` = 2700
`(Kg)/(m^(3) )`

`C_(p)` = 900
`(J)/(kg K)`

Ambient temperature
`T_(o)` = 27 °c = 300 K

Heat transfer per unit length Q = 2000
`(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
`(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

`(T - T_(o) )/(T_(s) - T_(o) ) =`
`e^(-at)` ------- (2)

Here

`a = (h A)/(\rho V C)`

Put all the values in above equation we get

`a = (6h)/(\rho D C)`

`a = ((6)(105.2))/((2700) (0.01) (900))`

a = 0.026

From equation (2)

`(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.