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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
QuickField.com >
Applications > Industrial > Thermal models
Applications > Sample problems> Examples gallery
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