RAD1.ppt Radiation Heat Transfer

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RAD1.ppt Radiation Heat Transfer
(Chapter in Outline Notes Radiation Heat
Transfer )
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Components of Course What Stage are We Up To?

• Types of exchangers, revision of OHTCs, fouling
  factors.
• Heat exchanger selection.
• Thermal performance analysis (NTUs) for co-
  counter-current exchangers.
• Multi-pass exchangers (ST).
• Condensation boiling.
• Radiation.

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Outline

• Key concepts (also in notes)
• Simple, two-surface, example (not in notes)
• Key points from three-surface example in lecture
  notes

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What Types of Problems Can Be Answered at this
Stage?

• Furnace power requirements
• Making a solar absorber power requirements
• Temperatures of surfaces in ovens
• Distillation reboiler heating requirements by
  radiation
• Analysis of solar heating panels

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Examples in Outline Notes

• View factors examples 1,2,3
• Simple examples examples 4,5,6
• Two surfaces examples 7, 8
• Three surfaces example 9

(Chapter in Outline Notes Radiation Heat
Transfer )
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Key Concepts in Heat Transfer by Thermal Radiation
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Basic Considerations

• Thermal radiation
• 10-1 to 102 microns
• Surfaces emit, absorb, may transmit radiation

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Blackbody

• Theoretical concept, but useful in practice
• Gives estimate of maximum absorption and emission
  for surface
• Blackbody emissive power (W/m2) depends on
  temperature (T) of surface

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From Whence Does Eb?T4 Come?

• Can plot Eb? monochromatic emissive power
  spectral blackbody emissive power power at each
  wavelength against wavelength
• This relationship changes with temperature
• Curve gets higher, more power
• Peaks at lower wavelength, higher frequency (more
  lower wavelengths at higher temperatures)

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• Integrating this over all wavelengths gives
  Eb?T4

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Relevance?

• Driving forces Heat transfer by radiation is
  driven by differences in emissive power
  (proportional to T4), not just temperature
  differences (convection conduction)

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Real Surfaces
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Real Surfaces

• At thermal equilibrium
• emissivity of surface absorptivity

• transmissivity of solid surfaces 0
• emissivity is the only significant parameter
• emissivities vary from 0.1 (polished surfaces) to
  0.95 (blackboard)

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View Factors

• if I was walking around surface A1 and I looked
  everywhere around me that I could, how much of my
  view would surface A2 take up?
• hence only part of radiation emitted by surface
  A1 reaches surface A2
• assumes uniform intensity of radiation in all
  directions (non-uniform intensity is beyond scope
  of course)

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Complication

• In practice, we cannot just consider the
  emissivity or absorptivity of surfaces in
  isolation
• Radiation bounces backwards and forwards between
  surfaces
• Use concept of radiosity (J) emissive power
  for real surface, allowing for emissivity,
  reflected radiation, etc

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All Real Surfaces are Grey
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Surface Resistance
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View Factor Resistance
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Relevance?

• Heat-transfer coefficients
• view factors (can surfaces see each other?
  Radiation is line of sight )
• emissivities (can surface radiate easily? Shiny
  surfaces cannot)

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Basic Concepts of Analysis

• Analogies with electrical circuit analysis
• Blackbody emissive power voltage
• Resistance resistance
• Heat-transfer rate current

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Simple Two-Surface Example
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Key Points for Two-Surface Example

• How to do view factor arithmetic
• How to use the concepts of view factors, surface
  resistances and view factor resistances to solve
  radiation problems
• How to develop radiation networks
• Application storage of very cold (cryogenic)
  fluids (e.g. N2), Q3, Tut 4

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Example What is the Nett Heat-Transfer Rate?

• Situation-
• hemisphere (surface 1), emissivity 0.1,
  temperature 700 K, above
• 1 m diameter disk (surface 2) , emissivity 0.5,
  temperature 500 K

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Step 1 Sketch the Situation
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Step 2 Sketch Radiation Network

• Surface and view factor resistances important
• One surface resistance for each surface
• One view factor resistance if one surface can see
  another

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Step 3 View Factor Concept Ant on Surface

• Surface 1 (hemisphere)
• When looking towards surface 2 (disk), can see
  both surfaces 1 and 2 (concave surface)
• 0 lt F11 lt 1, 0 lt F12 lt 1
• Surface 2 (disk)
• When looking towards surface 1, hemisphere,
  cannot see itself (flat or convex)
• F22 0

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View Factor Arithmetic

• F21 1 - F22 1
• F12 A2 F21 / A1
• A1 1.571 m2 A2 0.785 m2
• F12 0.5

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Step 4 Evaluate Resistances
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Step 5 Evaluate Total Resistance

• Total resistance 5.7301.2731.273
• 8.28 m-2

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Step 6 Evaluate Emissive Powers ( Voltages)

• Hemisphere (surface 1) and disk (surface 2)

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Step 7 Evaluate Heat-Transfer Rate ( Current)
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Three-Surface Example
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Key Points for Three-Surface Example

• How to use standard view factor charts
• How to treat adiabatic ( well-insulated) walls
• Application performance analysis of solar energy
  collectors

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Example

• Heating panels are located uniformly on the roof
  of a furnace, which is being used to dry out a
  bed of sand, which is situated on the floor. The
  furnace is cubical in shape, with the sides being
  1 m long.

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• The sand on the floor stays at a constant
  temperature of 290K due to evaporation of
  moisture.
• The temperature of the panels is 800K.
• The emissivity of the heating panels is 0.9, and
  that of the floor (sand) is 0.6.
• How much energy needs to be supplied to the
  panels to keep them at this temperature?

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Situation
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In What Context Might This Calculation Be Carried
Out?

• Suppose that you knew that the panels may burn
  out due to overheating if the panel temperature
  rises above 800K.
• Such a burn out would mean replacing the panels
  (expensive) and might also be a safety hazard
  (possibility of fire).

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• You would want to limit the energy input,
  because the panel temperature will rise as the
  energy input increases if the floor temperature
  stays the same.
• This calculation would then tell you the critical
  heat flux.

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• The insulation on the adiabatic walls will also
  degrade, possibly with hazardous consequences, if
  the wall temperature gets too high (say at 750
  K).
• We can also estimate the wall temperature.

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Interpretation

• All of the adiabatic walls see the same view of
  the other walls, so they can all be treated as
  one surface

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Treatment of Adiabatic Walls

• There is no heat flow through these walls (i.e.
  no equivalent of current), so
• The emissivity of these walls does not matter.
• The blackbody emissivity and the radiosity are
  the same, so
• The temperature can be estimated from the
  radiosity.
• These walls are just blank nodes in a radiation
  network.

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General Procedure

• Draw up radiation network (always first)
• In any order
• calculate view factors, then view factor and
  surface resistances, then total resistances
• calculate blackbody emissive powers (voltages)
• Calculate heat flows

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Radiation Network
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Explanation of Radiation Network

• If a surface can see another surface, then there
  must be a view factor resistance between the two
  surfaces.
• If the surface is not a blackbody, then it must
  have a surface resistance.
• For an adiabatic surface, the surface resistance
  does not matter.

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Evaluate View Factors

• Need to use symmetry as much as possible. Here
  roof (surface 1) sees the same view of the walls
  (surface 3) as the floor (surface 2), so F13
  F23
• Need to use standard chart for F12 (roof-floor) -
  eg chart in Hewitt et al Fig. 2.88, p. 138
• L, X, Y 1 m F12 0.23

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• F11 0 (flat surface cannot see itself)
• F11 F12 F13 1 (enclosed space)
• F13 1- F11 - F12 1 - 0 - 0.23 0.77

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Evaluate Resistances
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Evaluate Total Resistance
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Evaluate Emissive Powers

• Roof (surface 1) and floor (surface 2)

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Evaluate Heat Flow
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What Temperature are the Adiabatic Walls at the
Moment?

• Eb1, Eb2, J1, J2 and J3 all have units of W m-2
• There is no practical difference between the
  radiosity (J3) and the blackbody emissive power
  of the adiabatic walls.
• If we know J3, then we can calculate T3 from-

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Find J1, J2

• Resistance from Eb1 to J1 0.111 m-2
• Resistance from J2 to Eb2 0.667 m-2
• Total resistance 2.404 m-2
• Total voltage drop 23224-401 22823 Wm-2

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J3

• The resistances 1/(A1F13) and 1/(A2F23) are
  equal, so J3 is half way between J1 and J2
• J3 (222006730)/2 14450 Wm-2

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Conclusions for Example

• Maximum energy flux for panel temperature of 800
  K 9.5 kW
• Temperature of walls at this energy flux 711 K
  (safe if maximum temperature is 750 K)

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Conclusions

• Basic mechanisms
• Radiation networks electrical analogies
• View factors
• View factor surface resistances
• Adiabatic walls
• Two and three-surface problems (examples)