Technical data > Thermic calculations >
Mould
heating example > Calculation of the thermal balance
3. Necessary power to balance heat losses at the surface of the plates, of the mould and the frame of the press.
where :
l = coefficient of conduction
of the rigid insulator
= 0,65 W/m.K
e = insulator
thickness
= 0,015 m
S = total insulated surface ( upper and lower plates)
= (0,25 x 0,4) x 2
= 0,2 m²
DT =
180°C
Losses by convection of the lateral surfaces
of the mould and the plates:
where :
h.DT = power dispersed per
m² on an upright metallic surface multiplied by
the DT (180°C).
= 900W/m²
S = upright surface of the plates and the mould
= (0,15 + 0,25) x 2 x 0,15 + (0,25 + 0,4) x 2 x 0,05 x2
= 0,25 m²
Losses by convection below the upper plate:
Where :
h.DT =
850W/m²
S = (0,25 x 0,4) – (0,15 x 0,25)
= 0,0625 m²
Losses by convection above the lower plate :
Where :
h.DT =
1650W/m²
S = 0,0625 m²
Losses by radiation of every surface of the mould and the plates:
Where:
a = coefficient
of emission for hardening steel slightly
oxidized
= 0,7
s = constant of StefanBoltzmann
= 5,675 . 10^{8} W/m².K^{4}
^{
}K = Kelvin = degree Celsius + 273
T = Temperature of the mould and the plates in K
= 200 + 273
= 473 K
S = total surface of the mould and of the radiant plates
= 0,375 m²
Total loss per hour:
Conduction 
Fa 
= 1560 W 
Convection lateral side 
Fb 
= 225 W 
Convection lower side 
Fc 
= 53 W 
Convection upper side 
Fd 
= 103 W 
Radiation 
Fe 
= 745 W 

Total loss 
F3 
= 2686 W 
