Fundamentals of Heat and Mass Transfer
Fundamentals of Heat and Mass Transfer
7th Edition
ISBN: 9780470501979
Author: Frank P. Incropera, David P. DeWitt, Theodore L. Bergman, Adrienne S. Lavine
Publisher: Wiley, John & Sons, Incorporated
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Chapter 12, Problem 12.1P

Consider an opaque horizontal plate that is well insulated on its back side. The irradiation on the plate is 2500   W / m 2 , of which 500   W / m 2 is reflected. The plate is at 227 C and has an emissive power of 1200   W / m 2 . Air at 127 C flows over the plate with a heat transfer convection coefficient of 15   W / m 2 K . Determine the emissivity, absorptivity, and radiosity of the plate. What is the net heat transfer rate per unit area?

Expert Solution & Answer
Check Mark
To determine

The value of emissivity, absorptivity and the radiosity of the plate.

The value of the net heat transfer rate per unit area.

Answer to Problem 12.1P

The value of emissivity is, ε=0.33

The value of absorptivity is, α=0.8

The value of radiosity is, J=1700Wm2

The value of the net heat transfer rate per unit area is, qnet=700Wm2.

Explanation of Solution

Given:

The irradiation on the plate is, G=2500W/m2

The reflected irradiation is, Gref=ρ.G=500W/m2

The emissive power of any surface, E=1200W/m2

The temperature of the plate,

  TPlate=227CTPlate=227+273TPlate=500K

The temperature of the air over the plate

  Tair=127CTair=127+273Tair=400K

Calculation:

The energy balance equation as per the incident radiation on a surface is,

  α+ρ+τ=1    (1)

Where,

  α= absorptivity

  ρ= reflectivity

  τ= transmissivity

For an opaque surface transmissivity, τ=0

Now the equation (1) become as,

  α+ρ=1 (2)

The emissive power of a black surface, Eb=σTplate4.

Where,

The Stefan Boltzmann constant, σ=5.67×108W/m2K4

Emissivity(ε)-It is the ratio of emissive power of any surface (E) to the emissive power of the black surface.

Mathematically,

  ε=EEbε=EσT plate4ε=12005.67× 10 8× ( 500 )4ε=12003543.75ε=0.33

Reflectivity(ρ)-It is the ratio of the reflected radiation (Gref) to the irradiation (G) on the surface.

Mathematically,

  ρ=G refGρ=5002500ρ=0.2

Absorptivity(α)-The amount of irradiation that is absorbed by the surface is known as the absorptivity of that surface.

From equation (2), absorptivity will be calculated as,

  α+ρ=1α=1ρα=10.2α=0.8

Radiosity (J) -It is a radiant flux, at which radiation leaves from the surface.

  J=E+ρGJ=1200+0.2×2500J=1700Wm2

The net heat transfer rate per unit area is calculated as:

  qnet=qinqoutqnet=G(G ref+E+q convection)qnet=G[Gref+E+(hΔt)]qnet=G[Gref+E+h( T plate T air)]qnet=2500[500+1200+15(500400)]qnet=700Wm2

Therefore, the value of the net heat transfer rate per unit area is, 700Wm2.

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Fundamentals of Heat and Mass Transfer

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