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Unformatted text preview: CHEM ENG 2A04 RADIATION ENERGY CONSERVATION CONDUCTION
Thursday January 10, 2008 Thursday Radiation
Heat transfer at a gas/surface interface involves radiation emission from the surface and may also involve the absorption of radiation incident from the surroundings (irradiation,) as well as convection ( if Ts ≠ T∞ ) .
G Energy outflow due to emission: E = ε Eb = εσ Ts4 σ : Stefan-Boltzmann constant ( 5.67×10-8 W/m 2 ⋅ K 4 ) E : Emissive power ( W/m 2 ) ε : Surface emissivity ( 0 ≤ ε ≤ 1) Eb : Emissive power of a blackbody (the perfect emitter Energy absorption due to irradiation: Gabs = α G
Gabs :Absorbed incident radiation (W/m 2 ) α : Surface absorptivity ( 0 ≤ α ≤ 1)
G : Irradiation ( W/m 2 ) Irradiation – Special Case
Irradiation: Special case of surface exposed to large surroundings of uniform temperature, Tsur
4 G = Gsur = σ Tsur If α = ε , the net radiation heat flux from the surface due to exchange with the surroundings is:
4 ′′ qrad = ε Eb ( Ts ) − α G = εσ ( Ts4 − Tsur ) CONSERVATION OF ENERGY (FIRST LAW OF THERMODYNAMICS)
• An important tool in heat transfer analysis, often providing the basis for determining the temperature basis temperature of a system. of • Alternative Formulations Alternative Time Basis:
At an instant or Over a time interval Type of System:
Control volume or Control surface APPLICATION TO A CONTROL VOLUME Surface Phenomena E in E out :
, g g rate of thermal and/or mechanical energy transfer across the control surface due to heat transfer, fluid flow and/or work interactions. Volumetric Phenomena Eg : g rate of thermal energy generation due to conversion from another enegy form (e.g., electrical, nuclear, or chemical); energy conversion process occurs within the system. E st : g rate of change of energy storage in the system. APPLICATION TO A CONTROL VOLUME • At an Instant of Time:
Conservation of Energy E in − E out + E g g g g g dEst = ≡ E st dt Each term has units of J/s or W. • Over a Time Interval Ein − Eout + Eg = ∆Est Each term has units of J. THE SURFACE ENERGY BALANCE
A special case for which no volume or mass is encompassed by the control surface. Conservation Energy (Instant in Time):
g g Ein − E out = 0
• Applies for steady-state and transient conditions. (1.12) • With no mass and volume, energy storage and generation are not pertinent to the energy balance, even if they occur in the medium bounded by the surface. THE SURFACE ENERGY BALANCE - EXAMPLE
Consider surface of wall with heat transfer by conduction, convection and radiation.
′′ ′′ ′′ qcond − qconv − qrad = 0 T1 − T2 4 k − h ( T2 − T∞ ) − ε 2σ T24 − Tsur = 0 L ( ) METHODOLOGY OF FIRST LAW ANALYSIS
• On a schematic of the system, represent the control surface by dashed line(s). • Choose the appropriate time basis. • Identify relevant energy transport, generation and/or storage terms by labeled arrows on the schematic. • Write the governing form of the Conservation of Energy requirement. • Substitute appropriate expressions for terms of the energy equation. • Solve for the unknown quantity. Heat flow is dependent on three factors: three T(x) – ‘A’ – surface area surface perpendicular to the flow of T1 heat, heat, T2 – ‘k’ – capacity of material to capacity transfer heat across its body q (thermal conductivity) (thermal – ‘ΔT/L’ – temperature gradient temperature L x 0 across the thickness of the wall, L Flow Rate of Heat,
qx, is a vector quantity [units of Watts] These three factors are related together through FOURIER’S LAW: through
dT q x = −kA dx The driving force of conduction is the The temperature gradient
Here temp. gradient is an gradient average average T T Here temp. gradient is a local value x x Thermal Conductivity (k): Thermal
Units: W/m-K Often given as a constant, though it does vary with Often temperature
– so it is generally important you get a value valid for the temperature range you are interested temperature
In general, high k values are good conductors while low k values are better insulators. low k – high for metals high k – low of plastics, glass, wood Table – compiled from Appendix A of text
Material Brick Cork Glass Temp (K) 300 300 300 k (W/m-K) Material Temp (K) 300 300 300 k (W/m-K) 0.026 401 237 0.72 Air 0.039 Copper 1.4 Aluminium More accurate relationship for ‘k’ is: More K = K o (1 + aT )
– where ‘Ko’ and ‘a’ are constants. ‘a’ can be and negative or positive depending on the material negative Temp. Profile: Linear or Not? Temp. Consider the following:
– Is there an internal source of heat? – Does the surface area perpendicular to the heat Does flow change with dx? (think of radial flow across a sphere or cylinder) sphere Ex. For a constant heat flow Ex.
T T r x Normally, we can make certain assumptions allowing us to simplify a problem: allowing
– Steady State Assumption
We say that the process is at steady-state if none of the temperatures are a function of time temperatures In some cases we are looking at “pseudo-steady state” – Dimensions
Heat flow is actually a vector existing in a 3-D space Heat
ˆ ˆ ∂T + ˆ ∂T + k ∂T ∇T = grad T = i j ∂x ∂y ∂z – Dimensions
But are all heat components significant? But
dT q x = −kA dx dT q y − kA dy dT q z = −kA dz In many cases, at least one or more of the components are considerably smaller than the others and may be ignored with little error others Consider Consider
A heated room, with air at 20oC but the floor is 18oC and the ceiling is 16oC. What is the inside wall and C. temperature? Most of the wall will be ~20oC, but in temperature? C, the local area around where the ceiling and floor meet the wall the temperature will be different. The heat flow term, q, will be a constant over 99% of the wall’s face – Does the 1% really necessitate that we calculate heat flow to the floor and ceiling as well as through the wall? Usually NOT. through Problem 1.43: Thermal processing of silicon wafers in a two-zone furnace. Determine (a) the initial rate of change of the wafer temperature and (b) the steady-state temperature. KNOWN: Silicon wafer positioned in furnace with top and bottom surfaces xposed to hot and cool zones, respectively.
FIND: (a) Initial rate of change of the wafer temperature from a value of Tw ,i = 300 K, and (b) steady-state temperature. Is convection significant? Sketch the variation of wafer temperature with vertical distance. SCHEMATIC: • Problem: Silicon Wafer (cont.) ASSUMPTIONS: (1) Wafer temperature is uniform, (2) Hot and cool zones have uniform temperatures, (3) Radiation exchange is between small surface (wafer) and large enclosure (chamber, hot or cold zone), and (4) Negligible heat losses from wafer to pin holder.
ANALYSIS: The energy balance on the wafer includes convection from the upper (u) and lower (l) surfaces with the ambient gas, radiation exchange with the hot- and cool-zone and an energy storage term for the transient condition. Hence, from Eq. (1.11c), g g g E in − E out = E st
or, per unit surface area ′′ ′′ ′′ ′′ qrad , h + qrad , c − qcv, u − qcv, l = ρ cd 4 4 4 4 εσ Tsur , h − Tw + εσ Tsur , c − Tw − hu ( Tw − T∞ ) − hl ( Tw − T∞ ) = ρ cd ( ) ( ) d Tw dt d Tw dt (a) For the initial condition, the time rate of change of the wafer temperature is determined using the foregoing energy balance with Tw = Tw,i = 300 K,
0.65 × 5.67 × 10−8 W / m 2 ⋅ K 4 15004 − 300 4 K 4 + 0.65 × 5.67 × 10 −8 W / m 2 ⋅ K 4 3304 − 3004 K 4 −8 W / m 2 ⋅ K ( 300 − 700 ) K − 4 W / m 2 ⋅ K ( 300 − 700 ) K = 2700 kg / m3 × 875 J / kg ⋅ K ×0.00078 m ( d Tw / dt ) i ( ) ( ) ( d Tw / dt ) i = 104 K / s Problem: Silicon Wafer (cont.) (b) For the steady-state condition, the energy storage term is zero, and the energy balance can be solved for the steady-state wafer temperature, Tw = Tw,ss .
4 4 0.65 σ 15004 − Tw,ss K 4 + 0.65 σ 3304 − Tw,ss K 4 ( ) ( ) −8 W / m 2 ⋅ K Tw,ss 700 K − 4 W / m 2 ⋅ K Tw,ss − 700 K = 0
Tw,ss = 1251 K ( ) ( ) To assess the relative importance of convection, solve the energy balances assuming no convection. With ( d Tw / dt ) i = 101 K / s and Tw,ss = 1262 K. , we conclude that the radiation exchange processes control the initial rate of change and the steady-state temperature. If the wafer were elevated above the present operating position, its temperature would increase, since the lower surface would begin to experience radiant exchange with progressively more of the hot zone. Conversely, by lowering the wafer, the upper surface would experience less radiant exchange with the hot zone, and its temperature would decrease. The temperature-distance relation might appear as shown in the sketch. Announcements Announcements
Assignment 1: Due January 16 at the end of class. at In-class Quiz: Monday January 21 In-class Office hours: Office
– Filip Bidzinski: Tuesday’s 1:30 to 2:30 in JHE/342 – Mid-Term 1: – Mid-Term 2: Mid-Terms (tentative dates): Mid-Terms
Tutorial 1: Thursday February 14 2008 Tutorial 2: Monday February 18 2008 Tutorial Monday Tutorial 1: Thursday March 20 2008 Tutorial 2: Monday March 24 2008 Tutorial: See you there Tutorial: ...
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