. Air at 200 C blows over a 50 cm x 75 cm plain carbon steel (AISI 1010) hot plate with a constant surface temperature of 2500 C. The convection heat transfer coefficient is 25 W/m2K. Assume that 300 W is lost from the surface by radiation. Calculate the inside plate temperature. Assume that the plate is 2 cm thick. State all assumptions

Respuesta :

Answer:

The inside temperature, [tex]T_{in}[/tex] is approximately 248 °C.

Explanation:

The parameters given are;

Temperature of the air = 20°C

Carbon steel surface temperature 250°C

Area of surface = 50 cm × 75 cm = 0.5 × 0.75 = 0.375 m²

Convection heat transfer coefficient = 25 W/(m²·K)

Heat lost by radiation = 300 W

Assumption,

Air temperature = 20 °C

Hot plate temperature = 250 °C

Thermal conductivity K = 65.2 W/(m·K)

Steady state heat transfer process

One dimensional heat conduction

We have;

Newton's law of cooling;

q = h×A×([tex]T_s[/tex] - [tex]T_{\infty[/tex]) + Heat loss by radiation

= 25×0.325×(250 - 20) + 300

= 2456.25 W

The rate of energy transfer per second is given by the following relation;

[tex]P = \dfrac{K \times A \times \Delta T}{L}[/tex]

Thermal conductivity K = 65.2 W/(m·K)

Therefore;

[tex]2456.25 = \dfrac{65.2 \times 0.375 \times (250 - T_{in})}{0.02}[/tex]

[tex]T_{in} = 250 - \dfrac{2456.25 \times 0.02}{65.2 \times 0.375} = 247.99 ^{\circ}C[/tex]

The inside temperature, [tex]T_{in}[/tex] = 247.99 °C  ≈ 248 °C.

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