Vladimir Podnos Director of marketing and support, Tera Analysis Ltd.

Thermal problems in QuickField

Thermal engineering with QuickField

Sergey Ionin Support engineer, Tera Analysis Ltd.

Basics of the thermal problem setup

Thermal problems in QuickField

Vladimir Podnos, Director of Marketing and Support,

Tera Analysis Ltd.

QuickField

Magnetic analysis suite

Magnetic Problems

Magnetostatics

AC Magnetics

Transient Magnetics

Electric analysis suite

Electric Problems

Electrostatics and DC Conduction

AC Conduction

Transient Electric field

Thermostructural analysis suite

Thermal and

mechanical

problems

Steady-State Heat transfer

Transient Heat transfer

Stress analysis

Thermal analysis

Magnetic analysis suite

Magnetic Problems

Magnetostatics

AC Magnetics

Transient Magnetics

Electric analysis suite

Electric Problems

Electrostatics and DC Conduction

AC Conduction

Transient Electric field

Thermostructural analysis suite

Thermal and

mechanical

problems

Steady-State Heat transfer

Transient Heat transfer

Stress analysis

Geometry Physical parameters Result

Stages of solution

Thermal parameters Blocks: Thermal conductivity (may be temperature-

dependent or anisotropic);

Volume power of the heat source;

for time-domain

Specific heat (may be temperature-dependent);

Volume density.

Edges: Temperature, flux, convection, radiation;

Equal temperature, even and odd periodic

temperature.

Vertices: Temperature;

Heat sources.

Coupling: Heat Sources distribution may be imported

from Electromagnetic problems

Thermal parameters Blocks: Thermal conductivity (may be temperature-

dependent or anisotropic);

Volume power of the heat source;

for time-domain

Specific heat (may be temperature-dependent);

Volume density.

Edges: Temperature, flux, convection, radiation;

Equal temperature, even and odd periodic

temperature.

Vertices: Temperature;

Heat sources.

Coupling: Heat Sources distribution may be imported

from Electromagnetic problems

Basics of the thermal problem setup

Sergey Ionin Support engineer, Tera Analysis Ltd.

1. Heat transfer through the wall. Heat Losses.

2. Natural Convection. Calculation of the

convection coefficient.

3. Radiation. Heat Transfer in case of Radiation.

4. Complex problem with automation.

Basics of the thermal problem setup

1. Heat transfer through the wall.

Heat Losses.

t=15 0C

t=95 0C

Find:

1. Temperature of

the wall surface

2. Heat flux

t, 0С

Flux, W/m2

Theory* 94,52 795,2

QuickField 94,53 795,3

*Engineering Thermodynamics, Third Edition,

R.K.Rajput, example 15.6

Water Wall Air

λ = 42 W/(K · m)

α = 2850 W/(K· m2)

α = 10 W/(K· m2)

1. Heat transfer through the wall. Heat Losses.

2. Natural Convection. Calculation of the

convection coefficient.

3. Radiation. Heat Transfer in case of Radiation.

4. Complex problem with automation.

Basics of the thermal problem setup

Convection

Brick wall

Insulation

Plaster

Find:

1. Convection

coefficients

2. Temperature of the

wall

3. In what layer the

condensation may

occur (dew point

100C)

outside t = -10 0C

α = ?

inside t = +20 0C

α = ?

•Surface shape and orientation

•Specific dimensions, L

•Viscosity, μ

•Thermal conductivity, λ

Convection

Similarity theory formulas for the natural convection

from the vertical plate

•Density, ρ

•Thermal coefficient of volumetric expansion, β

•Heat capacity, Cp

•Temperature, T; Temperature difference ΔT

2

23

TgLGr

pCPr

PrGrRa

2

27/816/9

6/1

]Pr)/492.0(1[

387.0825.0

RaNu

Grasshof number

Prandtl number

Rayleigh number

Nusselt number

Convection coefficient

L

Nu

99

9

25

23

5

1022.21080.2793.0

1080.21087.1

003695.01581.9342.11

793.00238.0

10091087.1Pr

Ra

Gr

99

9

25

23

5

1061.11014.2752.0

1014.21087.1

003501.01581.9205.11

752.00250.0

10051087.1Pr

Ra

Gr

Convection

inside

59.31

0250.04.141

4.141

Nu

outside

80.31

0238.07.159

7.159

Nu

W/(K· m2)

W/(K· m2)

λ = 0,7 W/(K · m)

λ = 0,04 W/(K · m)

λ = 0,21 W/(K · m)

Convection

inside t = +20 0C

α = 3.59 W/(K· m2)

outside t = -10 0C

α = 3,80 W/(K· m2)

1. Heat transfer through the wall. Heat Losses.

2. Natural Convection. Calculation of the

convection coefficient.

3. Radiation. Heat Transfer in case of Radiation.

4. Complex problem with automation.

Basics of the thermal problem setup

Radiation

Straightening of the

tungsten filament

Radiating surface

Zero-flux surface

(radiated energy =absorbed energy)

Find:

1. Temperature of the filament

Q = P / V P – bulb power

V – filament volume

1. Heat transfer through the wall. Heat Losses.

2. Natural Convection. Calculation of the

convection coefficient.

3. Radiation. Heat Transfer in case of Radiation.

4. Complex problem with automation.

Basics of the thermal problem setup

Steam Pipe

Find:

1. Insulation thickness to limit

heat losses by 2.1 kW

r, mm

105 Theory*

QuickField 104,65

*Engineering Thermodynamics, Third Edition,

R.K.Rajput, example 15.11

Steam,

α = 100 W/(K· m2) Steel Pipe,

λ = 42 W/(K · m)

Insulation

λ = 0,8 W/(K · m)

Air

α = 30 W/(K· m2)

Similarity theory formulas for the natural convection

from the cylindrical surface

2

27/816/9

6/1

Pr/559.01

387.060.0

RaNu

Steam Pipe

2

23

TgLGr

pCPr

PrGrRa

Grasshof number

Prandtl number

Rayleigh number

Nusselt number

L

Nu Convection coefficient

More examples

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