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.

Respuesta :

Answer:

The maximum power is 23.89 k watt.

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

[tex]\Delta Q=mc_{p}\Delta T[/tex]

[tex]\Delta Q=\rho\times V\times c_{p}\times\Delta T[/tex]

[tex]\Delta Q=8960\times\dfrac{\pi}{4}\times(0.015)^2\times0.3\times385\times(423-293)[/tex]

[tex]\Delta Q=23897.6\ J[/tex]

[tex]\Delta Q=23.89\ kJ[/tex]

We need to calculate the power

Using formula of power

[tex]P=\dfrac{\Delta Q}{dt}[/tex]

[tex]P=23.89\ k watt[/tex]

Hence, The maximum power is 23.89 k watt.