05. Transient Heat Conduction Lump Capacity Method
Content
5. TRANSIENT CONDUCTION Simplified Model: Lumped-Capacity Method – Kapacitní metoda Basic assumption: Neglect spatial temperature variation: T = T(t)
Criterion for Neglecting Spatial Temperature Variation Consider a wire cooled by convection: h kλ
α
ΔT
ro = δL 1
h kλ
α
ΔT
ro = δL
α = heat transfer coefficient λ = thermal conductivity L = r o = radius (characteristic dimension) ΔT = temperature drop across wire
Objective: Neglecting temperature drop ΔT across L • The Biot number combines the effects of λ, L and α: αL Bi = λ 2
• Criterion for neglecting ΔT in the lumped-capacity model:
αL Bi = < 0,1 λ
Physical significance of Bi:
(5.1)
L/λ internal resistance = Bi = 1/α external resistance
3
• Application to other geometries: Use Bi<0,1to justify applying the lumped-capacity model to: (1) Plate of thickness L, exchanging heat on both surfaces: L ≡ L/2 (2) Sphere of radius
L = ro
(3) Irregular shape of surface area S and volume V V L= S
(5.2)
4
Lumped-Capacity Analysis Consider a body which is exchanging heat with the ambient by convection S = surface area α = heat transfer coefficient T∞ = ambient temperature
Objective: Determine the transient temperature Assume: Bi < 0.1
T = T (t ) 5
Apply conservation of energy during time interval dt: E& in + E& g − E& out = ΔE
(5.3)
E& g = energy generated E& = energy added in E& = energy removed out
Δ E = energy change (storage, i.e. accumulation) Assume that the body is losing heat and assume no heat generation: E& g = E& in = 0. Equation (5.3) becomes − E& = ΔE (5.4) out
Neglect radiation and assume that heat is removed by convection 6
E& out = αS(T − T∞ )
(5.5)
For incompressible material dT ΔE = ρ c V dt
(5.6)
ρ = density c = specific heat V = volume (5.5) and (5.6) into (5.4) dT − α S (T − T∞ ) = ρ c V dt or
− α S (T −T∞ ) = ρ c V
d(T- T ) ∞
dt
7
d(T − T )
Separate variables
αS =− dt (T − T ) ρcV ∞
(5.7)
∞
• This is the lumped-capacity equation for all bodies exchanging heat by convection. Valid for Bi < 0.1
• Initial condition:
T − T∞ = To − T∞
(5.8)
Solution: Assume constant c, α, ρ and T∞ . Integrate of (5.7) . ⎛ T −T ⎞ αS ∞ ⎟ ⎜ ln =− t ⎜T −T ⎟ ρcV ⎝ o ∞⎠ T −T
⎛ αS ⎞ t⎟ = exp⎜ − T −T ⎝ ρcV ⎠ o ∞ ∞
(5.9) 8
Introducing new dimensionless temperature: Θ=
T −T
To − T∞
And time constant: Rewrite equation (5.9): Θ
1,0
τ
Θ ∈ 1,0
∞
ρcV τ= αS Θ=e
−
t τ
Higher accumulation capacity ⇒ higher time constant ⇒ longer response to changes
t 9
• This is the lumped capacity solution. Valid for: (1) Bi < 0.1 (2) constant c, α and T∞ and ρ (3) Q& zdr = 0 (4) No radiations
Q& = α S (T −T∞ )
Objective: Determine amount of heat exchanged between the body and ambiance in a certain time interval t
t
0
0
(
)
Q = ∫ Q& dt = αS ∫ T − T∞ dt t
(
)
Q = αS ∫ To − T∞ e 0
−
[J]
t τ dt
10
After integration and substituting for time constant τ t ⎛ − ⎞ Q = ρcV To − T∞ ⎜ 1 − e τ ⎟ ⎜ ⎟ ⎝ ⎠
(
)
Θ=
T −T
∞
To − T∞
Heat transferred for time t → ∞ Overview of formulas
Θ=e
Θ=e
−
t τ
Characteristic length as V/S
αSt αt αL λ t αL at = = = = = Bi.Fo 2 2 τ ρcV ρcL λ ρc L λ L t
-BiFo
11
Overview of formulas
Θ=e
−
t τ
Characteristic length as L=V/S
V α Bi = S λ
Θ = e -BiFo Characteristic length as a distance where maximum temperature difference occurs Plane wall – half thickness L=δ/2
Θ = e -BiFo
Cylinder – radius R, L=R
Θ = e -2BiFo
Sphere – radius R, L=R
Θ = e -3BiFo
αL Bi = λ 12
Criterion for Neglecting Spatial Temperature Variation Consider a wire cooled by convection: h kλ
α
ΔT
ro = δL 1
h kλ
α
ΔT
ro = δL
α = heat transfer coefficient λ = thermal conductivity L = r o = radius (characteristic dimension) ΔT = temperature drop across wire
Objective: Neglecting temperature drop ΔT across L • The Biot number combines the effects of λ, L and α: αL Bi = λ 2
• Criterion for neglecting ΔT in the lumped-capacity model:
αL Bi = < 0,1 λ
Physical significance of Bi:
(5.1)
L/λ internal resistance = Bi = 1/α external resistance
3
• Application to other geometries: Use Bi<0,1to justify applying the lumped-capacity model to: (1) Plate of thickness L, exchanging heat on both surfaces: L ≡ L/2 (2) Sphere of radius
L = ro
(3) Irregular shape of surface area S and volume V V L= S
(5.2)
4
Lumped-Capacity Analysis Consider a body which is exchanging heat with the ambient by convection S = surface area α = heat transfer coefficient T∞ = ambient temperature
Objective: Determine the transient temperature Assume: Bi < 0.1
T = T (t ) 5
Apply conservation of energy during time interval dt: E& in + E& g − E& out = ΔE
(5.3)
E& g = energy generated E& = energy added in E& = energy removed out
Δ E = energy change (storage, i.e. accumulation) Assume that the body is losing heat and assume no heat generation: E& g = E& in = 0. Equation (5.3) becomes − E& = ΔE (5.4) out
Neglect radiation and assume that heat is removed by convection 6
E& out = αS(T − T∞ )
(5.5)
For incompressible material dT ΔE = ρ c V dt
(5.6)
ρ = density c = specific heat V = volume (5.5) and (5.6) into (5.4) dT − α S (T − T∞ ) = ρ c V dt or
− α S (T −T∞ ) = ρ c V
d(T- T ) ∞
dt
7
d(T − T )
Separate variables
αS =− dt (T − T ) ρcV ∞
(5.7)
∞
• This is the lumped-capacity equation for all bodies exchanging heat by convection. Valid for Bi < 0.1
• Initial condition:
T − T∞ = To − T∞
(5.8)
Solution: Assume constant c, α, ρ and T∞ . Integrate of (5.7) . ⎛ T −T ⎞ αS ∞ ⎟ ⎜ ln =− t ⎜T −T ⎟ ρcV ⎝ o ∞⎠ T −T
⎛ αS ⎞ t⎟ = exp⎜ − T −T ⎝ ρcV ⎠ o ∞ ∞
(5.9) 8
Introducing new dimensionless temperature: Θ=
T −T
To − T∞
And time constant: Rewrite equation (5.9): Θ
1,0
τ
Θ ∈ 1,0
∞
ρcV τ= αS Θ=e
−
t τ
Higher accumulation capacity ⇒ higher time constant ⇒ longer response to changes
t 9
• This is the lumped capacity solution. Valid for: (1) Bi < 0.1 (2) constant c, α and T∞ and ρ (3) Q& zdr = 0 (4) No radiations
Q& = α S (T −T∞ )
Objective: Determine amount of heat exchanged between the body and ambiance in a certain time interval t
t
0
0
(
)
Q = ∫ Q& dt = αS ∫ T − T∞ dt t
(
)
Q = αS ∫ To − T∞ e 0
−
[J]
t τ dt
10
After integration and substituting for time constant τ t ⎛ − ⎞ Q = ρcV To − T∞ ⎜ 1 − e τ ⎟ ⎜ ⎟ ⎝ ⎠
(
)
Θ=
T −T
∞
To − T∞
Heat transferred for time t → ∞ Overview of formulas
Θ=e
Θ=e
−
t τ
Characteristic length as V/S
αSt αt αL λ t αL at = = = = = Bi.Fo 2 2 τ ρcV ρcL λ ρc L λ L t
-BiFo
11
Overview of formulas
Θ=e
−
t τ
Characteristic length as L=V/S
V α Bi = S λ
Θ = e -BiFo Characteristic length as a distance where maximum temperature difference occurs Plane wall – half thickness L=δ/2
Θ = e -BiFo
Cylinder – radius R, L=R
Θ = e -2BiFo
Sphere – radius R, L=R
Θ = e -3BiFo
αL Bi = λ 12
