Length: mm
Mass: tonne (T)
Density: T/mm^3
Time: second
Energy / Work: millijoule (mJ)
Power: milliwatt (= mW = mJ/s)
Force: Newton
Modulus / Stress: N/mm^2 (=MPa)
Temperature: Kelvin (for temperature differentials, 1 K = 1
degree C)
Conductivity: mW/mm/K (= mJ/s/mm/K)
Specific Heat: mJ/T/K
Flux: mW/mm^2 (= mJ/s/mm^2)
Convection: mW/mm^2/K (= mJ/s/mm^2/K)
In a heat transfer analysis it is the product of density and
specific heat that is important. So, as
long as consistent mass units are used for the two (and the length,
power/energy, and temperature units are consistent with the thermal
conductivity and the rest of the model), it should be OK. So, the following sets of units will both
work:
k [W/mm C], rho [kg/mm^3], C [J/kg C]
k [W/mm C], rho [g/mm^3], C [J/g C]
When working with coupled thermal-electric problems it's
almost always easiest (and safest) to work with standard SI units (m, kg, sec,
N, Volts, Amps, Joules, etc.), because you need to make sure both the thermal
and electrical unit systems are consistent.
But, if you really
want to use millimetres for your length scale, the following
units should be consistent:
Length -> mm
Density -> g/mm^3
Thermal Conductivity -> W/mm K
Specific Heat -> J/g K
Heat Flux -> W/mm^2
Heat Source -> W/mm^3
Film Coefficient -> W/mm^2 K
Note that energy and power units are Joules and Watts,
respectively. You can get away with using grams as the mass unit since only the
product of density and specific heat appears in the heat transfer equations,
and hence the mass units cancel out. You
couldn't get away with this in a coupled thermal-structural problem - you'd
have to use mega-grams. Assuming
potentials are defined in Volts, then the units for current can be derived
using the relationship:
P = iV (i.e., [Watts] = [current units]*[Volts])
Since 1 Watt = 1 Joule/sec, and 1 Volt = 1 Joule/Coulomb,
the units of current will be 1 Coulomb/sec, which is an Ampere.
Finally, since V = iR = i*(L/(sigma*A)) (where sigma is the
electrical conductivity, L is length, and A is area), the units of electrical
conductivity must be Amps/Volt/mm = (Ohm mm)^-1.
Note that the ABAQUS documentation tells you what your units
need to be once you've selected the base units (i.e., time, length, etc.). For example, the documentation for
*ELECTRICAL CONDUCTIVITY states that the units must be C T^-1 L^-1 phi^-1:
i.e. Coulomb/second/mm/Volt = Amperes/Volt/mm = (Ohm mm)^-1.
http://abaqus-users.1086179.n5.nabble.com/clarification-needed-on-quot-units-quot-td9492.html
