fv2001 fluid dynamics of fire - camtec

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FV2001 Fluid Dynamics of Fire

Chapter 1 : Heat and Mass Transfer

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(1) Heat Transfer

Heat transfer is an energy flow from one location to another because of temperature difference. There are 3 modes in which heat may be transferred by:conduction, convection and radiation.

(A) Conduction

Heat energy is transferred from one part of a substance to another part of the substance, or from one substance to another in physical contact with it and the energy is transferred as a result of molecular collision in a stationary fluid or by atomic vibrations in solids.

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In fire engineering, conduction is associated with ignition, flame spread over combustible solids, and fire resistance of materials.

Fourier’s law of conduction

Fourier’s law states that the rate of flow of heat through a single homogeneous solid is proportional to the area A of the section at right angles to the direction of heat flow, and to the change of temperature with respect to the length of the path of the heat flow, dt/dx. i.e. rate of heat flow, or [W or J/s]

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The constant λ(or k) is called the thermal conductivity of the material the thermal conductivity of a substance is defined as the heat flow per unit area per unit time when the temperature decreases by one degree in unit distance. The unit is W/m.K. The thermal conductivities of some materials encountered in engineering are shown in the following table:

Substance Thermal conductivity, λ or k(W/m.K)Pure copper 386Pure aluminum 229Cast iron 52Mild steel 48.5

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Substance Thermal conductivity, λ or k(W/m.K)Concrete 0.85 to 1.4Building brick 0.35 to 0.7

Materials with high thermal conductivities are good conductors of heat, whereas materials with low thermal conductivities are good thermal insulator.

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Conduction through a plane surface or wall

From Fourier’s law of conduction

Q Qk

x

t1t2

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Conduction through a composite wall

Consider 1-D steady conduction heat transfer,

t1

t4k1

k3

x1 x3x2

k2

t2t3

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For n layers, the heat flux due to conduction is

where R is called the thermal conductance in m2.K/W.

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Example 1

A brick wall 250 mm thick is faced with concrete 50mm thick. The brick has a coefficient of thermal conductivity of 0.69 W/m.K while that of the concrete is 0.93 W/m.K. If the temperature of the exposed brick face is 30 oC and that of the concrete is 5 oC,determine the heat lost/hr through a wall 10 m long and 5 m high. Determine the interface temperatures.

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The General Heat Conduction Equation

A general equation in which a three dimensional heat flow varying with time is considered in a material with heat generation. The general Fourier equation becomes

The term is called the thermal diffusivity.

The thermal diffusivity α [m2/s] is a property of the material. If there is no heat generation within the material, i.e. qg = 0, the thermal diffusivity is the property which represents how fast temperature equalization occurs.

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Typical values of α are shown in the table below.Note that metals and stagnant gases have high values, and insulators have low values of α.

Materials Thermal diffusivity α [m2/s] Aluminum 94 × 10-6

Copper 112 × 10-6

Carbon steel 15 × 10-6

Brick 0.3 ×10-6

Water 0.15 ×10-6

Air 22 ×10-6

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(3) Transient(unsteady) heat conduction of a wall

The conduction equation becomes

This is the 1-D unsteady heat conduction equation.

(a) Sudden temperature rise on a semi-infinite surface

If the solid is initially at a uniform temperature T0, the appropriate initial condition is t = 0, T = T0. The left face of the surface is suddenly raised to temperatureTs at time zero and held at that value.

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The two required boundary conditions arex = 0; T = Ts

The solution for the 1-D unsteady heat conduction (a) with the initialand boundary conditions are

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For a semi-infinite surface, the backface boundary condition has a negligible effect on the solution of the unsteady 1-D heat transfer problem.

A wall or slab of thickness L can be treated as a semi-infinite solid with little error provided that

δt is the thermal penetration depth defined by Babrauskas(2002) in the Handbook of Fire Protection Engineering and is useful for the analysis of ignition and flame spread of a solid surface.

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(b) Constant surface heat flux to the surface

If at time t = 0, the surface issuddenly exposed to a constant heat flux qs (by radiation from a high temperature source or fire). TheSolution to the equation (a) with the boundary conditions becomes

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Example 2

A 150 mm-thick concrete firewall has a black silicone paint surface. The wall is suddenly exposed to radiant heat source that can be approximately as a black body at 1000 K. How long will it take for the surface to reach 500 K if the initial temperature of the wall is 300 K?Data: For the concrete, absorptivity The Stefan-Boltzman constant is 5.67 × 10-8 W/m2.K4

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(b) Convection

Convection is the energy transferred between a solid surface and the adjacent fluid that is in motion, and it involves the combined effects of conduction and fluid motion.

The movement of the fluid may be caused by differences in density resulting from the temperature differences as in natural convection (or free convection) or the motion may be produced by mechanical means, as in forced convection.

In fire engineering, convection is related to movement of smoke and combustion gases in fire plumes.

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Newton’s Law of Cooling

Newton’s Law of cooling states that the convection heat transfer from a solid surface of area A, at a temperature tw, to a fluid of temperature tf, is given by

where h is convective heat transfer coefficient in W/m2.K and has different value for different situation.

The heat transfer from one side of a wall to the other side often involves the effect of combined conduction and convection:

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Convection

with hA

Convection

with hB

wall

k

Conduction

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Example 3

A mild steel tank of wall thickness 10 mm contains water at 90 oC. Calculate the rate of heat loss per m of tank surface area when the atmospheric temperature is 15 oC. The thermal conductivity of mild steel is 50 W/m.K, and the convective heat transfer coefficients for the inside and outside of the tank are 2800 and 11 W/m2.K, respectively. Calculate also the temperature of the outside surface of the tank.

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Composite wall and the electrical analogy

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By ohm’s law, we have

Similarly, we can define thermal resistance R by

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The overall heat transfer

For n number of layers of walls, the total thermal resistance becomes

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Thermal resistances R for conduction and convection for plane walls,

The total thermal resistance with conduction and convection RT = Rcond + Rconv

The total heat transfer is given by

where △Toverall is the outer and inner fluid

temperature difference.

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Example 4

A furnace wall consists of 125 mm wide refractory brick and 125 mm wide insulating firebrick separated by an air gap. The outside wall is covered with a 12 mm thickness of plaster. The inner surface of the wall is at 1100 oC and the room temperature is 25 oC. Calculate the rate at which heat is lost per m2 of wall surface. The heat transfer coefficient from the outside wall surface to the air in the room is 17 W/m2K, and the resistance to heat flow of the air gap is 0.16 K/W, the thermal conductivity of refractory brick, insulating firebrick, and plaster are 1.6, 0.3, and 0.14 W/m.K respectively. Calculate also each interface temperature, and the temperature of the outside surface of the wall.

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1100 oC

Furnace Air

PlasterFirebrick

25 oC

Refractory brick

Air gap

t2t3

t4t5

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Convection heat transfer correlations

The primary difficulty in convection heat transfer analysis is to determine the convective heat transfer coefficient h. Dimensional analysis is used to obtain a functional relationship(correlation) between the convective heat transfer coefficient h and the relevant physical properties and kinematic/ dynamicsparameters of the flow situation. The relevant physical properties and kinematic/dynamic parameters of the flow situation can be grouped as appropriate dimensionless numbers such Reynolds number(Re), Nusselt number(Nu), Prandtl number(Pr), and Grashof number(Gr), etc.

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Dimensionless number(parameter) is the ratio of different forces in a flow field. Some dimensionless numbers related to convection heat transfer are:

Parameter Definition Physical interpretation

Re (Reynoldsnumber)

Pr(Prandtlnumber)

Ratio of inertia to viscous force

Ratio of momentum and thermal diffusivity

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Parameter Definition Physical interpretation

Nu (Nusseltnumber)

Gr(Grashopnumber)

Ratio of convection heat transfer to conduction in a fluid slab with thickness l

Ratio of buoyancy to viscous force

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For forced convection, the correlation formulae is

The dimensionless group, is the Nusselt number is the Prandtl number Pr

is the Reynolds number

i.e. or

For example, the heat transfer from a flat plate for forced convection,

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When evaluating Nu ,Pr and Re, it is necessary to take the fluid properties at a suitable mean temperature. For cases in which the temperature of the bulk of fluid is not very different from the temperature of the solid surface or wall, then fluid properties are evaluated at the mean bulk temperatureof the fluid tb. When the temperature difference is large, a mean film temperature is used, defined by tf

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Example 5

Calculate the heat transfer coefficient for water flowing through a 25 mm diameter tube at the rate of 1.5 kg/s, when the mean bulk temperature is 40 oC. For turbulent flow of liquid, take where the characteristic dimension of length is the diameter and all properties are evaluated at mean bulk temperature. For water at

(data obtained from the table in page 10 of the table “Thermodynamic and Transport properties of Fluids, Rogers & Mayhew”)

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For Natural(Free) convection, the correlation formulae is

For natural convection with low velocity,

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Most of natural fire and flows associated with fire are in the domain of natural convection. Although wind can be an important factor for fires such as forest fires, forced flow convection is not generally relevant. In natural convection we can represent a characteristicvelocity

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Example 6

A vertical flat plate of length 1 m and temperature Tw = 800 oC is exposed to air at Tf = 20 oC. Calculate the convection coefficient h. Assume Pr = 0.7. The film temperature is 410 oC and the properties of air at that temperature are k = 57.8 x 10-3 W / m.K and ν = 82.3 x 10-6 m2 / s.

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(c) Radiation

Radiation is the energy transferred through electromagnetic waves emitted due to the agitation of the molecules of a substance and without the need for any physical contact between bodies. It requires no medium for its propagation and will pass through a vacuum. If two bodies at different temperatures are placed near each other, the body at the lower temperature will receive more radiant energy then it is radiating.

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Thermal radiation0.1-100 μm

Solar radiation0.1-3.0 μm

Wavelength(μm)

Ultraviolet 0.01-0.4μm

Visible range 0.4-0.7μm

Infrared 0.7-1000 μm

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Black body

A black body is an ideal body, which absorbs all the radiation, which fall upon it. That is, for a black body, absorptivity equal to unity, i.e., and . For a non-black body, .

In practice, when a body is placed in large surroundings, the surroundings are approximately black to thermal radiation.

The blackbody is the perfect absorber and emitter of radiation energy and it is a standard for comparing the actual radiative properties of real surfaces.

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The Stefan-Boltzmann law

The emissive power of a black body is directly proportional to the fourth power of its absolute temperature and this is known as the Stefan-Boltzmann law, i.e. where emissive power of the black

body in W/m2

absolute temperature of the body in KStefan-Boltzmann constant in W/m2.K4

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Example 7

A smoke layer below the ceiling of a compartment has a temperature of 600 oC. If the smoke is assumed as a black body, determine the radiation emitted of the smoke.

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A black body, being the best absorber of radiation, is also the best possible emitter of radiation energy. That is

for two bodies at the same temperature T

where Eb is the emissive power of a black body and E is the emissive power of a non-black body.

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Kirchoff’s Law

The emissivity of a body radiating energy at a temperature T, is equal to absorptivity of the body when receiving energy from a source at a temperature, T.

i.e.,

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(b) Gray body

Because the variation of emissive power with wavelength is very irregular for a real body, an ideal surface (i.e. the gray body) is therefore defined in order to simplify calculations. A gray body has a constant emissivity over the whole waveband at all temperatures, where the ratio of the emissive power of a body to the emissive power of a black body is called the emissivity,

For a black body and for non-black body .

The emissive power of a gray body is

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Fig.1 : Radiation Spectrum

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Consider a body 1 of emissivity ε at a temperature , completely surrounded by large(black) surroundings at a lower temperature

Energy emitted by body 1 and the energy emitted by the blacksurroundingsEnergy absorbed by body 1

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If the body 1 is gray, at all temperature. Then the energy absorbed by body 1

Then the heat transferred from the body 1 to its surrounding per unit area of the body

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If body 1 is not gray, at same temperature T

then the energy absorbed by body 1

Then the heat transferred from the body 1 to its surrounding per unit area of the body

where is the emissivity at temperature and is the emissivity at

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Example 8

A body of 1100 oC in black surroundings at 550 oChas an emissivity of 0.4 at 1100 oC and an emissivity of 0.7 at 550 oC. Calculate the rate of heat loss by radiation per m2,(a) when the body is assumed to be gray

with emissivity of 0.4;(b) when the body is not gray.

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Wien’s displacement Law

As the temperature is increased the maximum black body radiation intensity shifts towards the shorter wavelengths as shown in next slide:

The wavelength corresponding to the maximum radiation intensity for a given temperature T can be found from the Wien’s Displacement Law:

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Fig.2 : Wien’s displacement Law

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Example 9

Radiant energy fire detector in one of three wave-length bands:

Ultra-violet (UV) - 0.1 to 0.35 μmVisible - 0.37 to 0.75 μmInfra-red (IR) - 0.75 to 200 μm

Select the most type of detector for fire in a smouldering carpet in which the temperature is about 600 oC.

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(2) Mass transfer

In mass transfer, we deal with the movement of a species in a multi-component system. The driving potentials for the transfer of mass can be obtained in various ways. Here we shall confine our discussion to molecular diffusion and convective mass transfer.

(a) Molecular Diffusion - Fick’s Law

Mass transfer by molecular diffusion of a species through another stationary medium is analogous to heat transfer by conduction. Fick’s Law relates the diffusion rate or mass flow mA of a species A to its driving potential: the concentration gradient ∂CA/∂x.

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Fick’s Law gives (1)

where A is the normal area and D is called the diffusion coefficient or diffusivity. The unit of diffusion coefficient D is found to be m2/s. It has the dimensions of kinematic viscosity or thermal diffusivity α.

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(b) Convective Mass Transfer

In molecular diffusion, bulk velocities are insignificant. Most physical applications of mass transfer involve the bulk motion of fluids. This gives rise to convective mass transfer which is similar to the transfer of heat by convection. Accordingly, we define the mass transfer coefficient hM by the relation:

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(c) Analogy between Momentum, Heat and Mass Transfer

To illustrate the analogy, equations for the transfer of shear stress, heat flux and mass flux can be rewritten as follows:

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Then if v = α or Pr = 1, the velocity and temperature distributions in the flow will be the same. In that case, the heat transfer coefficient h can be determined from the knowledge of the knowledge of the friction factor f by the similarity relation called Reynolds analogy :

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Similarly, the velocity and concentration profiles will have the same shape if v = D or Sc = 1 ( Eq. (4) gives Sc = 1). Thus, the Schmidt number plays the same roles in mass transfer as does the Prandtl number in heat transfer. An equivalent of the Nusselt number in heat transfer is the Sherwood number defined by

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Combining Eq. (11) and (15) to eliminate , we getjH = jM (16)

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We obtain a similarity relation between heat transfer h and mass transfer coefficients hM:

If Le = 1, the temperature and concentration profiles are the same. Incidentally for air and water vapour mixtures at atmospheric pressure, the value of the Lewis number Le is approximately equal to unity. This simplifies the design of air-conditioning and combustion equipment.

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Example 10

Air at 35 oC and 1 atm flows at a velocity of 30 m/s over a flat plate 0.5 m long. Calculate the mass transfer coefficient of water vapour from the plate into air. Assume the concentration of vapour in air as very small. The diffusion coefficient of water vapour into air is 0.256 × 10-4 m2/s. The heat transfer coefficient is given by jH = 0.0296Re-0.2.

FV2001 - Fluid Dynamics of Fire Ch1 - Heat and Mass Transfer

©E&S Professional Education Services/used solely by Dr. Albert Yau for teaching FV2001 in 2006 1

Example 1 With brick and concrete layer, n = 2

t1 = 30 oC, t3 = 5 oC, x1 = 0.25 m, k1 = 0.69 W/m.K, x2 = 0.05 m, k2 = 0.93 W/m.K,

( ) ( )=

= = + = + =

− −= = = × =

∑2

1 2t

n 2 1 2

1 3

t

Thetotal thermal resistance for conduction heat transfer,

x xx 0.25 0.05 m .KR 0.416

k k k 0.69 0.93 W

t t 30 5Q qA A 10 5 3000 W or 3 kW

R 0.416

Heat lost/hr = 3 × 3600 = 10800 kJ

11

1

o

1 2

1

1 12 1 1

1

xFor the brick wall, and R

A k

3000 0.25The interface temperature, 30 8.3 C

0.69 (10 5)

t tQR

QR Qxt t t

A k A

=

×= − =

× ×

−=

= − = −

Example 2 The heat flux from the fire

4 4 8 4 3 2sq T T 0.9 5.67 10 1000 51.0 10 W / m−= εσ = ασ = × × × = ×

For constant heat flux qs at the surface, at x = 0, T = Ts

( ) ( ) 2

0 s 0s ss 0

s

T T kq q4 t 4 tT T e 0 erfc(0) t

k k 2q−

−α α π− = − × = ⇒ = π π α

( )( )

2

6 3

500 300 1.4t 31.6 s

0.75 10 2 51.0 10− −

− ×π = = × ×

To check the validity of semi-infinite solid,

−α = × × × = < =64 t 4 0.75 10 31.6 0.02 m L 0.15 m

Example 3

Given: tA = 90 oC, tB = 15 oC, k = 50 W/m.K, hA = 2800 W/m2.K, hB = 11 W/m2.K

= = + + = + + = + +×

=

3A B

2

1 1 x 1 1 10 1R 0.000357 0.0002 0.0909

U h k h 2800 10 50 11

0.0915 [m .K / W]



Heat transfer per unit area −

= = − = A BA B

Q (t t )q U(t t )

A R

− = = 290 15

q 820 W/m0.0915

, i.e., Rate of heat loss per m2 of surface area = 0.82 kW

From equation (iii), = = −B 2 B

Qq h (t t )

A,

∴ = × −2820 11 (t 15) where t2 is the temperature of the outside surface of the tank.

o2

820t 15 89.6 C

11= + =

Example 4

Given: x1 = 0.125 m, x3 = 0.125 m, x4 = 0.012 m, t1 = 1100 oC, tB = 25 oC,

h = 17 W/m2.K, R2 =0.16 K/W, k1 = 1.6 W/m.K, k3 = 0.3 W/m.K, k4 = 0.14 W/m.K

For plane walls, A remains unchanged and Rcond = x/k

Thermal resistance of refractory brick = = =11

1

x 0.125R 0.0781 K/W

k 1.6

Thermal resistance of insulating firebrick = = =33

3

x 0.125R 0.417 K/W

k 0.3

Thermal resistance of plaster = = =44

4

x 0.012R 0.0857 K/W

k 0.14

Also using thermal resistance for a fluid film, R = 1/h for plane wall

Thermal resistance of air film on outside surface BB

1 1R K/W

h 17= =

Hence, total thermal resistance, T 1 2 3 4 BR R R R R R= + + + +

T

1 KR 0.0781 0.16 0.417 0.0857 0.8

17 W= + + + + =

Heat flux, A B2

T

Q t t 1100 25 Wq 1344 assuming unit area

A R 0.8 m− −

= = = =

1100 oC

t2

t3

t4

t5

tB x1 x3 x4

Furnace A

tA Outside B

R2



i.e. Rate of heat loss per m2 of surface area = 1344 W or 1.344 kW

− −= ⇒ = =1 2 2

1

t t 1100 tQ Q 1344

R 0.0781,i.e., o

2t 1100 1344 0.0781 995 C= − × =

− −= ⇒ = =2 3 2 3

2

t t t tQ Q 1344

R 0.16 i.e. o

3t 995 0.16 1344 780 C= − × =

o3 4 3 44

3

t t t tQ 1344 , i.e., t 780 1344 0.417 220 C

R 0.417− −

= ⇒ = = − × =

o4 5 4 55

4

t t t tQ 1344 , i.e., t 220 1344 0.0857 104 C

R 0.0857− −

= ⇒ = = − × =

Example 5

Volume flow rate,

••

= = = × =ρ

3f

mQ mv 1.5 0.001 0.0015 m / s

Velocity of water in tube, 2 2

Q 4Q 4 0.0015U 3.06 m / s

A d 0.025×

= = = =π π ×

Reynolds number of the flowing fluids,

−

ρ ×= = = =µ µ × ×d 6

f

Ud 1 Ud 3.06 0.025Re 117500, the flow is turbulent as Re > 4000.

v 0.001 651 10

The Nusselt number, ( ) ( )= = =0.8 0.40.8 0.4

dNu 0.0243Re Pr 0.0243 117500 4.3 496

62dNu .k 496 632 10

h 12.55 kW / m Kd 0.025

−× ×= = =

The convection heat transfer from water to the pipe is calculated by the Newton’s law

of cooling, ( )= − = × π× × × − = 2f wQ hA(T T ) 12.55 0.025 1 (40 20) 251 kW / m

Example 6

( ) ( ) ( ) ( )( ) ( )

( )−

−

β = =β −= = × = ××

×= = = × =

f

3239w f

22 6 2

31/31/3 9 2

1 1For air,

T 293 K

9.81 m / s 780 K 1 mg T T lRa Pr 0.7 3.94 10 .

v 293 K 82.3 10 m / s

Nu.k k 57.8 10 W / m.Kh 0.1 Ra 0.1 3.94 10 9.12 W / m K

l l 1 m

This is a typical convective coeffici 2ent values, and a range of 5 to 15 W/m K is to be

expected for these gas velocities and associated natural convection conditions.



Example 7

−= σ = × × + =4 8 4 2B

The emissive power of the smoke,

E T 5.67 10 (600 273) 26 kW / m.

This radiation flux from the smoke shall ignite almost all combustible materials in the

compartment and this may lead to flashover.

Example 8

o o1 2 T1 T2Given : T 1100 C 273 1373 K, T 550 C 273 823 K, 0.4, 0.7= + = = + = ε = ε =

( )( ) ( )

T1 T2

4 4 8 4 4T1 1 2

a If the body is gray, at all temperature

Q q T T 0.4 5.67 10 1373 823 70.2 kW

A−

ε = α ⇒ ε = ε

= = ε σ − = × × − =

( )4 4

T1 1 T2 2

8 4 8 4

b If the body is not gray,

Qq T T

A

0.4 5.67 10 1373 0.7 5.67 10 823 62400 W or 62.4 kW− −

= = ε σ − ε σ

= × × × − × × × =

The radiation heat transfer of the gray body is greater than that of the real body.

Example 9

= =λ µ

λ = µ

oGiven : T 600 C 873 K

From Wien's displacement law, .T = 2897.6 m.K for maximum radiation

2897.6 2897.6 = = 3.32 m

T 873Therefore IR would be the most suitable type.

Example 10

5

4

5

2 10Sc 0.682

D 1.146 0.256 10

uL 1.146 30 0.5Re 859500

2 10

−

−

−

µ ×= = =ρ × ×ρ × ×= = =µ ×

( ) 0.20.2 3M H

3M

M 2/3 2/3

j j 0.0296Re 0.0296 859500 1.925 10

j u 1.925 10 30h 0.075 m / s

Sc 0.682

−− −

−

= = = = ×

× ×= = =

fv2001 fluid dynamics of fire - camtec

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