Question

A heater 0.015 m in diameter and 0.3 m long is submerged horizontally in oil at 20 oC. To avoid oil fumes, the heater surface should not exceed Ts= 150 oC. Calculate the maximum power that should be supplied to the heater.

Answer 1

The maximum power is 23.89 k watt.

Step-by-step explanation:

Given that,

Diameter = 0.015 m

Long = 0.3 m

Initial temperature = 20°C

Final temperature = 150°C

Suppose the material is copper and here no maintain at time so we assuming heat supplied per unit time

We need to calculate the energy

Using formula of energy

`\Delta Q=mc_(p)\Delta T`

`\Delta Q=\rho* V* c_(p)*\Delta T`

`\Delta Q=8960*(\pi)/(4)*(0.015)^2*0.3*385*(423-293)`

`\Delta Q=23897.6\ J`

`\Delta Q=23.89\ kJ`

We need to calculate the power

Using formula of power

`P=(\Delta Q)/(dt)`

`P=23.89\ k watt`

Hence, The maximum power is 23.89 k watt.