Heat Transfer Notes
Eb= blackbody emisibibility
= σT 4
T in K units W/m^2
E= εσT 4 qemit =EA Watts energy emitted recognize the direction qnet
Eᵬ= W/m^2 M
Radiation Amount of energy emitted from green color
10 nm 450 nm ~470 nm ∞
Eb ∫ Eᵬ, b ^dᵬ
0
∞
E= ∫ ƐE ᵬ, b d ᵬ
0
∞
∞
0
2
E= ∫ ƐE ᵬ, b d ᵬ ∫ ƐE ᵬ, b d ᵬ
G Gij
ᵬ
⍴ ᵭ+ᵬ=1 Ɛ= Erad/ ( σT 4 ) Blackbody surface
Jb=Eb
Ji= ƐiσT 4 + ∑
Geometrical alpha = epsilon
Radiosity
black body surface ji~~~ji Ji=Ebi Net work surface 1 node potential on )
Eb1
Eb2 eb3 connect the dots with resistors 1/A1F12
Eb1 to Eb2 net work surface 1 q1 on EB1 flows into q12 and q13 parallel q1=q12+q13 q1=energy flow from 1 to 2 and 1 to 3
three major parts outline Radiation arbitrary surface Gfi reflection pGji ƐalphaGji given part of the problem find the rest
alpha = Ɛ when exists qnet, rad =? energy conservation at steady state relationship between terms calculation problems
View factor Calculations F14=a
F12, F13, F21, F41 ect Fij sigma Fij=1
AiFij=AjFji
Fi l +m = Fil + Fim Ji Gij qij
i, Ji, Ai
qij=(JiJj)/(1/AiFij) blackbody
Ji=Ei(T)
function of temperature Radiation network to find flux or potential 40 to 50 % ison radiation Heat exchanger
U fouling factor will increase the resistance of heat transfer. how to include into the overall equation (mcp)h Th, in Tc, out
(mcp)c Tc,in You want to know the cooling effect energy conservation Q= {mcp(Th,iTh,o)}h
={mcp(Tc,oTc,i) }c Delta Tlm log mean difference Q=UA DTcm Q=HA Dt
A=Q/(U delta T) 2pirL Delta Tlm’= DeltaTlm*F
NTU
effectiveness
ƐNTUmethod
Ɛ=?(Q/Qmax) NTU= UA/mcpmin qmax/Qmax design the amount of heat exchange required for a problem eplison is known NTU =UA/ mcpmin Length of a pipe thermal analysis NTU > eplison
= σT 4
T in K units W/m^2
E= εσT 4 qemit =EA Watts energy emitted recognize the direction qnet
Eᵬ= W/m^2 M
Radiation Amount of energy emitted from green color
10 nm 450 nm ~470 nm ∞
Eb ∫ Eᵬ, b ^dᵬ
0
∞
E= ∫ ƐE ᵬ, b d ᵬ
0
∞
∞
0
2
E= ∫ ƐE ᵬ, b d ᵬ ∫ ƐE ᵬ, b d ᵬ
G Gij
ᵬ
⍴ ᵭ+ᵬ=1 Ɛ= Erad/ ( σT 4 ) Blackbody surface
Jb=Eb
Ji= ƐiσT 4 + ∑
Geometrical alpha = epsilon
Radiosity
black body surface ji~~~ji Ji=Ebi Net work surface 1 node potential on )
Eb1
Eb2 eb3 connect the dots with resistors 1/A1F12
Eb1 to Eb2 net work surface 1 q1 on EB1 flows into q12 and q13 parallel q1=q12+q13 q1=energy flow from 1 to 2 and 1 to 3
three major parts outline Radiation arbitrary surface Gfi reflection pGji ƐalphaGji given part of the problem find the rest
alpha = Ɛ when exists qnet, rad =? energy conservation at steady state relationship between terms calculation problems
View factor Calculations F14=a
F12, F13, F21, F41 ect Fij sigma Fij=1
AiFij=AjFji
Fi l +m = Fil + Fim Ji Gij qij
i, Ji, Ai
qij=(JiJj)/(1/AiFij) blackbody
Ji=Ei(T)
function of temperature Radiation network to find flux or potential 40 to 50 % ison radiation Heat exchanger
U fouling factor will increase the resistance of heat transfer. how to include into the overall equation (mcp)h Th, in Tc, out
(mcp)c Tc,in You want to know the cooling effect energy conservation Q= {mcp(Th,iTh,o)}h
={mcp(Tc,oTc,i) }c Delta Tlm log mean difference Q=UA DTcm Q=HA Dt
A=Q/(U delta T) 2pirL Delta Tlm’= DeltaTlm*F
NTU
effectiveness
ƐNTUmethod
Ɛ=?(Q/Qmax) NTU= UA/mcpmin qmax/Qmax design the amount of heat exchange required for a problem eplison is known NTU =UA/ mcpmin Length of a pipe thermal analysis NTU > eplison