Chemical Engineer
17 de Feb de 2016
1 de 15

• 1. Presented By : Asim Farooq farooq9284682@gmail.com Department Bsc Engg Tech Chemical (SUIT) Peshawar Radiation Heat Transfer
• 2. Contents  Heat Transfer  Heat transfer by Radiation  Stefan Boltzmann law  Examples on Stefan Boltzmann Law
• 3. Heat Transfer  Transferred from higher temperature objects to objects at a lower temperature. Figure. 1 Figure. 2 Ho t Cold Heat Transfer At thermal equilibrium
• 4. How Heat Can Be Transferred  Conduction  Convection  a The above are the modes of heat transfer but we have to focus on only. Radiation Radiation
• 5. Radiation Heat Transfer  Energy of radiation transferred in the form of electromagnetic waves.  Radiation mode of heat transfer is independent of medium.  It occurs more effectively in vacuum.
• 6. Radiation continues….  We can feel the heat of the Sun, although it is 150 million km away from the Earth.  Because no particles are involved while reaching the heat from sun to earth.
• 7. Thermal Radiations  All objects at non zero temperature emits radiations.  The hotter the object more radiations will be emitted. More more hotterLess Hotter More Hotter
• 8. Stefan-Boltzmann Law Heat rate through radiation mode can be calculated by Stefan Boltzmann Law which is q = εσ T4 A Where q = heat transfer per unit time (W) ε = emissivity power σ = 5.6703 10-8 (W/m2K4) - The Stefan-Boltzmann Constant T = Surface temperature in Kelvin (K) A = area of the emitting body (m2)
• 9. Example: 01 A Metal ball 3 cm in radius is heated in a furnace to 5000C. If its emissivity is 0.5, at what rate does it radiate energy? Given: Radius (r) = 3 cm Temperature (T) = 5000C Emissivity (ε) = 0.5 To Find: Radiation heat rate (q)
• 10. Example continuous…. Solution: Step1: First, calculate the Surface area of the ball (A), A = 4 π r2 = (4 x 3.142)(0.03 m)2 = (12.568) (0.03 m) 2 = 0.0113112 m2 Step2: Now, calculate the Temperature (T), T = 500 0C + 273 = 773 K.
• 11. Example continuous…. Step3: Finally, calculate the Radiation heat rate (q), q = ε σ A T4 q = 0.5 x 5.67 x 10-8 x 0.0113112 x (773)4 q = 114.37 W.
• 12. Example:02 Statement: I have a system in space that is generating a lot of waste heat which I need to get rid of by radiation into space. The radiant heat rate is 1.0×10³ W. My external radiator has a surface area of 1.0 m by 2.0 m, and has an emissivity of 0.99. What is the equilibrium temperature of my radiator in °C?
• 13. Example continuous….. Solution: Since I am given the amount of radiation that I need to emit, I can determine Qemitted without having to worry about Qincoming. From the equation Qemitted = q/A = 1.0×10³ W/(1.0 m × 2.0 m) = 500W/m²
• 14. Example continuous….. Qemitted = q/A = ε σ T4 So that, T = [Qemitted/ ε σ ] = {500 W/m²/[0.99 × 5.67×10- 8 W/(m²·K4)]} = 307.2 K We were asked for °C, so convert from K to °C: T = 307.2 - 273.15 = 34°C