2 heat transfer analysis
DESCRIPTION
heatTRANSCRIPT
1
2.2 Heat Transfer in steelwork
Protected and
unprotected steelwork
Unprotected Steelwork
Tfi
Ts
2
Reference
Read Page I-43 to I-49Implementation of Eurocodes Handbook 5: Design of buildings for the heat situationPdf files download from IVLE
Heat stored per unit volume
Cross-section area (A) x length
Fire exposure perimeter length (Hp) around cross-section x length
( ) ssfis
s ATThdt
dTCV −=ρ
L
TsTfi
Heat input from fire per unit area (Heat Flux )
.Q
3D Transient Heat Transfer
3
( ) tTTVA
ChT sfi
s
ss Δ−=Δ
ρΔT≤5 seconds
Tfi = fire temperature (oC) at particular time t (sec)Ts = steel temperature, assumed to be uniform, at time tAs/V = section factor (m-1) of the exposed steel member per unit length
h = heat transfer coefficient per unit area per degree Celsius= hr + hc
hc = 25 W/(m2K) for ISO fire= 35 for natural fire and advanced
fire model= 50 for hydrocarbon fire
εr = resultant emissivity= εf εm
= 1.0x0.7 = 0.7 σ = Stephan Boltzmann constant
= 5.67 x 10-8 W/m2K4
fireMaterial surface
4 4r fi s
rfi s
(273 T ) (273 T )h
(T T )
⎡ ⎤ε σ + − +⎣ ⎦=−
Section Factor
p p ps H L H perimeter length (H ) around cross-sectionA V AL A Cross Section Area
= = =
L
TsTfi
4
Hp/A Concept
Heating Rate in a fire depends on-
• The perimeter of steel exposed Hp
• The cross-sectional area of the section A
Hp/A Concept
Low AHigh HP
Fast Heating
Low HPHigh A
Slow Heating
5
Section factor Hp/A -unprotected steel members
perimeterc/s area
exposed perimeterc/s area
h
b
2(b+h)c/s area
I section with side plates
reinforcement
exposed perimeter
Total c/s area
exposed plate
Total c/s area
exposed flange
Total c/s area
Section factor Am/V -inherently protected systems
6
Am/V for unprotected steel
( ) tTTVA
ChT sfi
s
ss Δ−=Δ
ρFunction of Ts
H = hr + hc, Function of Ts
Ts
t
ΔTs
Δt
Steel temperature time curve
Tfi
tFire temperature time curve
Temperature-Time Relationship
tt+0.5Δt
7
Specific heat of steel EC3:Part 1-2
Step by Step Method
Calculate ΔTs with values of Tfiand Ts from this
row
Tfi – TsFire temp half way through
time step (at t1 + Δt/2)
Ts from previous time step + ΔTs
from previous row
t2 = t1 + Δt::::
Calculate ΔTs with values of Tfiand Ts from this
row
Tfi – TsoFire temp half way through time step (at
Δt/2)
Initial steel temp Tso
t1 = Δt
Change in Steel Temp ΔTs from
Equation
Difference in Tem (Tfi – Ts)
Fire Temp (Tfi)Steel Temp (Ts)Time (t)
( ) tTTVA
ChT sfi
s
ss Δ−=Δ
ρ
h, Cs are calculated based on this fire temp at half time step
8
Use the step-by-step procedure to calculate the steel temperature of an unprotected beam exposed to ISO fire. The beam section factor is 200 m-1. Use a convective heat transfer coefficient hc = 25 W/m2K and emissivity 0.6. The density of steel is 7850 kg/m3 and the specific heat is 600 J/kgK. Use a time step of 0.5 minutes.
Example 1:
Example 1: use the step by step method to calculate the steel temperature of an unprotected beam exposed to ISO fire
( ) tTTVA
ChT sfi
s
ss Δ−=Δ
ρ
( )2 2r r fi s fi sh (T 273) (T 273) (T T 546)= ε σ + + + + +
Hp/A=As/V= 200m-1 ; hc = 25 W/m2K; εr = 0.6; σ = 5.67 x 10-8 W/m2K4
ρ = 7850kg/m3; Cs=600 J/kgK
ISO Fire: Fire Temperature Tfi=345log(8t+1)+20
9
:::::3.0
:::::2.5
24.0380.9461.280.32.252.0
21.5366.9425.858.81.751.5
18.2338.7379.340.61.251.0
13.8284.7311.626.80.750.5
6.8164.6184.620.00.250.0
Changein steel
temperature
DifferenceIn
temperature
ISO firetemperatureat half step
Tfi
Steeltemperature
Ts
Timeat halfstep
Time(minutes)
HW: Repeat the problem using time step of 1 minute and observe the difference.
Determine the time required to heat the steel to 550oC
( ) tTTVA
ChT sfi
s
ss Δ−=Δ
ρ
Protected Steelwork
Tfi
Ts
10
Convective/Radiant Boundary Condition
dp
T1 T2
Tf
Ts( ).
.
f i 1 f 1i
QQ h T T T Th
= − ⇒ − =
( ).
s 2 s
.
2 ss
Q h T T
QT Th
= − ⇒
− =
. .p2 1
p 1 2p p
dT TQ k T T Qd k−
= − ⇒ − =
.p
f sf p s
d1 1Q T Th k h
⎛ ⎞+ + = −⎜ ⎟⎜ ⎟
⎝ ⎠
Total thermal resistanceTf
Ts
Fire Protection
Assume Tp = 0.5(Ts+Tfi) Protection temperature = average of steel and fire
( )( )
fi
sspp
psfis Tt
Ckth
VATTT Δ
⎟⎟⎠
⎞⎜⎜⎝
⎛+
−Δ⎟⎠⎞
⎜⎝⎛ ++
−=Δ
121
211//1
/
φφρ
VA
tCC p
pss
pp
ρρ
φ =
tp
Ts
Tf
Protection
.p
f sf p s
d1 1Q T Th k h
⎛ ⎞+ + = −⎜ ⎟⎜ ⎟
⎝ ⎠ Heat transfer from the fire through the protection to the steel
( )ATTkdh
Q ssfpp
con −+
=//1
1.
ppppssreq AtCVCQ +=.
ρρ sTt
ΔΔ
fi s1 T T2 t
Δ + Δ⎛ ⎞⎜ ⎟Δ⎝ ⎠
Heat required to increase temperature of the steel and protection
(1)
(2)
Equating (1) and (2)
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Eurocode 3 Part 1.2
( )( )
( ) fi
sspp
psfis Tet
Ckt
VATTT Δ−−Δ
⎟⎠⎞
⎜⎝⎛ +
−=Δ 1
311/
/ 10/φ
φρ
( )( )
fi
sspp
psfis Tt
Ckth
VATTT Δ
⎟⎟⎠
⎞⎜⎜⎝
⎛+
−Δ⎟⎠⎞
⎜⎝⎛ ++
−=Δ
121
211//1
/
φφρ
Δt ≤ 30 seconds
Ap=As
Thermal resistance of fire is small compared to the protection material
Note: ΔTs ≥0 if : ΔTfi >0 VA
tCC p
pss
pp
ρρ
φ =
One dimensional heat flow through a thick protection material
Typical heating curve of a protected steel section
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( )( )
( ) fi
sspp
psfis Tet
Ckt
VATTT Δ−−Δ
⎟⎠⎞
⎜⎝⎛ +
−=Δ 1
311/
/ 10/φ
φρ
φ defines the relative amount of heat stored in the protective materialIt ignores the surface radiation and convection effects, which are important in unprotected sections, but small in in comparison with the insulation capacity of the protection materials.Due to the second term, the steel temperature may decrease at the early stage of heating. In this case, the increase in steel temperature should be taken as zero.Δt should not exceed 30 sec.
VA
tCC p
pss
pp
ρρ
φ =
>0
Note that Structural steel temperature cannot be higher than the fire
temperature!
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Step by Step Method may be used to calculate the steel temperature
Tf, t2+ Δt -Tf, t2
Tf, 0+ Δt -Tf, 0
ΔTf
Calculate ΔTs with values of TfiΔTf and Ts from
this row
Tfi – TsFire temp half way through
time step (at t1 + Δt/2)
Ts from previous
time step + ΔTs from previous
row
t2 = t1+ Δt
Calculate ΔTs with values of TfiΔTf and Ts from
this row
Tfi – TsoFire temp half way through time step (at
Δt/2)
Initial steel temp Tso
t1 = Δt
Change in Steel Temp ΔTs from
Equation
Difference in Tem
(Tfi – Ts)
Fire Temp (Tfi)
Steel Temp (Ts)
Time (t)
14
Definition of Section Factor for various
types of fire protection
Thermal properties of fire protection materials
VA
tCC p
pss
pp
ρρ
φ =
ρp Cpkp
15
Section Factor Hp/A
15
10
400
150
Case 1
Concrete slab
Hp=2*400+150*3-2*10 = 1230 mm, A=2*15*150+(400-15*2)*10=8200mm2 Hp/A=0.15mm-1 = 150 m-1
r=150
r=160
Case 2
Hp=2πRo=320π, A=π(Ro2-Ri
2)=2900πHp/A = 0.1103mm-1 = 110.3 m-1
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Protected Steel Temperature Rise
• Not less than 0• Not higher than in unprotected steel
Example 2: use the step by step method to calculate the steel temperature of an protected beam exposed to ISO fire. The beam is protected with 50mm of lightweight insulating materials which has thermal conductivity of
kp = 0.2 W/mK, specific heat Cp = 1100 J/kgK and density ρp 300 kg/m3.
( )( )
( ) fi
sspp
psfis Tet
Ckt
VATTT Δ−−Δ
⎟⎠⎞
⎜⎝⎛ +
−=Δ 1
311/
/ 10/φ
φρ
VA
tCC p
pss
pp
ρρ
φ =
Properties of the steel section:Hp/A=As/V= 200m-1
ρs = 7850kg/m3; Cs=600 J/kgKISO Fire: Fire Temperature Tfi=345log(8t+1)+20
Calculate ΔTs with values of Tfi and Ts
from this row
Tfi – TsFire temp half way through time step
(at t1 + Δt/2)
Ts from previous time step + ΔTs from
previous row
t2 = t1 + Δt
Calculate ΔTs with values of Tfi and Ts
from this row
Tfi – TsoFire temp half way through time step
(at Δt/2)
Initial steel temp Tsot1 = Δt
Change in Steel Temp ΔTs from Equation
Difference in Tem (Tfi – Ts)
Fire Temp (Tfi)Steel Temp (Ts)Time (t)
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Some Approximate Solutions
Temperatures in Concrete Slab, Tslab
heffθc
Heated lower sideofof slab
Depthx
mm
Temperature θ [°C] after an ISOfire duration in min. of
30' 60' 90' 120' 180' 240'5101520253035404550556080100
c
5354704153503002502101801601401251108060
705642581525469421374327289250200175
100140
738681627571519473428387345294271220160
754697642591542493454415369342270210
738689635590
330260
469430
549508
395305
520495
645550
740700670
x
Values are for normal weight concrete based on ISO Standard fire.For lightweight concrete multiply value by 0.9
18
( ) 0
2
18log345 TtT
TCT
fi
fis
++=
=
Tslab refers to the Table in the previous two slidesC1 is given in the next slideFR = Fire rating in minutes
t (mm)
Temperature in Concrete Filled Square/Circular Columns
1120
120)(02.012
1
≤−
××−=
==
FRmmtC
TTTCT
crebar
sc
Multiplication Factor C1 values
*0.5
* For square section, an equivalent depth of half the cover depth should be used tocalculate the corner rebar temperature because of heating from both sides
Diameter or size of square section in mm
Distance of centre of layer from outer surface in mm
10 30 50 70 >70
1.08 1.22 1.41 1.60 1.801.05 1.14 1.22 1.36 1.501.03 1.09 1.18 1.25 1.35
200300400500 1.02 1.07 1.12 1.18 1.25
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Slim Floor Beam
Plate TipPlate
Lower flange
Upper flange
T=C1-C2tf-C3tp
Tuf=1.5t +20
tp
tf
Slim Floor Beam
t =standard fire exposuretime in minutes
20
Without air gap With an air gap of 4mm Fire
rating (min)
Location of steel C1
(oC) C2 (oC/mm)
C3 (oC/mm)
C1 (oC)
C2 (oC/mm)
C3 (oC/mm)
30
plate
570
5.0
5.0
730
1.5
8.0
plate tip 675 2.0 7.5 700 1.5 7.0 lw. flange 550 3.0 4.4 450 2.0 5.8 60
plate
850
3.0
3.0
850
1.5
2.0
plate tip 920 2.0 3.5 900 2.0 2.0 lw. flange 850 3.8 3.5 750 3.5 3.8 90
plate
930
1.0
1.0
930
1.0
1.0
plate tip 980 0.5 2.0 970 1.0 1.0 lw. flange 925 1.8 1.8 840 2.0 1.8 120
plate
980
0
0
980
0
0
plate tip 1010 0 0 1010 0 0 lw. flange 980 1.4 1.4 920 1.5 1.7
Asymmetrical Beams
Use slim floor results, plate thickness =0, no air gap
21
Homework 2Q1 Calculate the section factor of a steel H-section column of dimension 300 x 300mm.
The column is exposed to fire on all four sides. Make calculations for a) box type protection b) spray-on protection.
Q2 Use the step-by-step procedure to calculate the steel temperature of an unprotected beam exposed to ISO fire and hydrocarbon fire. The beam section factor is 250 m-1. The density of steel is 7850 kg/m3 and the specific heat is in accordance with Eurocode3. Use a time step of 0.5 minutes. Plot the temperature-time curve of ISO fire and hydrocarbon and the corresponding steel temperature.
Q3 Use the step-by-step procedure to calculate the steel temperature of a protected beam exposed to ISO fire. The beam is same as Q 2. The beam is protected with 50mm lightweight insulating material which has thermal conductivity of 0.2 W/mK, and specific heat 1100 J/kgK and density 300 kg/m3. Plot the temperature-time curve of ISO fire and steel temperature.
Q4 Use the step-by-step procedure to calculate the steel temperature of an unprotected beam exposed to a parametric fire. The beam is same as Q 2. The fire compartment is made from lightweight concrete with density 2000 kg/m3, thermal conductivity of 0.8 W/mK, and specific heat 840 J/kgK . The room is 5 m square and 3 m high with one window 2.4m wide and 1.5m high. The fuel load is 900MJ/m2 floor area. Plot the
temperature-time curve of the design fire and steel temperature.
Q5 Repeat Q4 with the beam protected with 25mm and 50mm lightweight insulating material which has thermal conductivity of 0.2 W/mK, and specific heat 1100 J/kgKand density 300 kg/m3. Plot the temperature-time curve of the design fire and steel temperature with different thickness of fire protection.
Q6 Determine the temperature in the reinforcement bars of a rectangular infilled column under 60 minutes ISO fire as shown in the following figure.
Note: Q4 and Q5 require calculation of fire temperature in a compartment, which will be taught in chapter 3.
75mm
RHS 200 x 100 x 10mm
30mm
22
Additional HW1a Use the step-by-step procedure to calculate the steel temperature
of an unprotected beam 406×140×46UB exposed to ISO fire and hydrocarbon fire. The beam section factor is 250 m-1. The density of steel is 7850 kg/m3 and the specific heat is in accordance with Eurocode 3. Use a time step of 0.5 minutes. Plot the temperature-time curve of ISO fire and hydrocarbon and the corresponding steel temperature.
1b Use the step-by-step procedure to calculate the steel temperature of a protected beam exposed to ISO fire. The beam is same as 1b. The beam is protected with 35mm lightweight insulating material which has thermal conductivity of 0.15 W/mK, and specific heat 1050 J/kgK and density 300 kg/m3. Plot the temperature-time curve of ISO fire and steel temperature.
For Q1a and b determine the time required to heat the steel to 550oCPlot the temperature time curves for the fire, unprotected steel and protected steel
Reading Assignment: Chapter 6, Y C Wang Book
23
CE6705 Fire EngineeringAssignment 1: Heat Transfer
(Due Date: 17 March 2007, Monday, Time: 1800)
Q1 Use the step-by-step procedure to calculate the steel temperature of an unprotected column section 305×305×97UC exposed to ISO fire and hydrocarbon fire. The density of steel is 7850 kg/m3 and the specific heat of steel is in accordance with Eurocode 3. Use a reasonable time step and plot the temperature-time curve of ISO fire, hydrocarbon fire and the corresponding steel temperature. Determine the time required to heat the column to 550oC.
Q2 Use the step-by-step procedure to calculate the steel temperature of a protected column section exposed to ISO fire and hydrocarbon fire. The section is same as Q1. The column is protected with 35mm thick normal weight concrete. Assume a reasonable time step and plot the temperature-time curve of ISO fire, hydrocarbon fire and steel temperature. Determine the time required to heat the column to 550oC.
Q3 Discuss the effectiveness of the fire protection materials in Q2. What is the fire resistant time, if the column temperature is to be limited to 550oC.
State clearly all your assumptions. Professional report submission is expected. Submit detailed calculations including spreadsheet calculations in the appendix.