Question

A cartridge electrical heater is shaped as a cylinder of length L = 300 mm and outer diameter D = 30 mm. Under normal operating conditions, the heater dissipates 2 kW while submerged in a water flow that is at 20 C and provides a convection heat transfer coefficient of ° W/m2 h = 5000 K. Neglecting heat transfer from the ends of the heater, determine its surface temperature Ts and the thermal resistance due to convection. If the water flow is inadvertently terminated while the heater continues to operate, the heater surface is exposed to air that is also at C but for which W/m2 20° h = 50 K. What are the corresponding thermal resistance due to convection and surface temperature? What are the consequences of such an event?

Answer 1

when water is surrounding
`T_s = 34.147 degree C`

when air is surrounding
`T_s = 1434.7 degree C`

From above calculation we can conclude that air is less effective than water as heat transfer agent

Step-by-step explanation:

Given data:

length = 300 mm

outer diameter = 30 mm

dissipitated energy = 2 kw = 2000 w

heat transfer coefficient IN WATER = 5000 W/m^2 K

heat transfer coefficient in air = 50 W/m^2 K

we know that q_{convection} = P

From newton law of coding we have

`q_(convection) = hA(T_s - T_(\infty))`

`T_s` is surface temp.

`T_(\inft)` - temperature at surrounding

`P = = hA(T_s - T_(\infty))`

`(P)/(\pi hDL) = (T_s - T_(\infty))`

solvinfg for
`T_S` w have

`T_S = T_(\infty) + (P)/(\pi hDL)`

`T_S = 20 + (2000)/(\pi * 5000 * 0.03* 0.3)`

`T_s = 34.147 degree C`

When air is surrounding we have

`T_S = T_(\infty) + (P)/(\pi hDL)`

`T_S = 20 + (2000)/(\pi * 50 * 0.03* 0.3)`

`T_s = 1434.7 degree C`

from above calculation we can conclude that air is less effective than water as heat transfer agent