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)

11

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

12

( )( )

( ) 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!

13

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

16

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)

17

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

19

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.