Electrical resistivity & conductivity
This is a seldom used chapter and it looks a bit tricky when one wants to dig deeper.
In a nutshell, resistivity (or specific resistance) is defined as the resistance of a substance per unit of volume. This last words are the trick.
The formula to use : resistance = resistivity * Length / Area
If one uses the same units for L and A , the resistivity is expressed as "Ohms.L" (L = length unit)
For instance Ohm.mm or Ohm.cm or Ohm.m or even Ohm.in
and the product : Ohm.L * L / L^2 gives Ohm as a result.
From time to time, because of the definition ("resistance per unit of volume"),
one will find "Ohms/cm3" instead of Ohm.cm
This is obviously a mistake.
To settle, there's nothing better than an example. Let's take the copper resistivity at 20°C :
17.2 10^ -6 Ohm.mm or 1.72 10^ -6 Ohm.cm or 17.2 10^ -9 Ohm.m or 677 10^ -9 Ohm.inch
Another trap : a table may choose different units. For copper, I've seen : 0.0172 Ohms.mm2/m
This is the same value as seen before but it has to be divided by 1000 to become homogeneous.
Resistivity is sometimes defined as the resistance of a mil-foot (a wire 1 ft long with a uniform section of 0.001 inch in diam.)
If you apply the previous figures, you should find 10.346 Ohms for copper.
Do it as an exercise !
As for conductivity, it is of course the reciprocal of resistivity.
Let's take a resistivity of 17.2 10^ -6 Ohm.mm : the conductivity will be 58 139.5 / (Ohm).(mm)
The unit "1 / Ohm" was originally called the Mho. Therefore : 58 139.5 Mho/mm.
It is still used sometimes.
About 30 years ago, the name Siemens replaced "Mho". If we convert the whole line above, we get for copper :
58 139.5 S/mm or 581 395 S/cm or 58 139 535 S/m or 1 477 105 S/inch
Conductivity is mainly used for compounds with a higher resistance, like electrolytic solutions.
Example : very pure water at 20°C will have a conductivity of 4.2 microSiemens/m or 42 nanoS/cm or 4.2 nanoS/mm
( And 4.2 microS/m is equivalent to 238 095 Ohm.m or 23.8 MegaOhm.cm)
Note : add 1 ppm of NaCl and the conductivity goes to 2 microSiemens/cm (equiv. to 0.5 MOhm.cm) !
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Viscosity
Dynamic (or absolute) viscosity
Let's remember that this is not a highly scientific text written by an academic zombi to impress distinguished collegues, but rather a memento for engineers.
We won't bother either with official SI notations : exponants will be kept to a minimum (too tricky in HTML.)
Imagine a liquid pressed in a thin layer between two plates : if one tries to move one plate versus the other, the liquid may exert a resistance. To achieve a constant speed, a constant force will be needed.
To simplify, consider 2 layers of liquid slipping on each other, each one "attached" to its plate. They are separated by a surface dS and the total thickness of the fluid is dN. The difference of speed between them is dV. If the dynamic viscosity is VIS, the force needed is dF, equal by definition to :
dF = VIS * dV * dS / dN
hence, the dynamic viscosity will be expressed as : VIS = dF * dN / (dV * dS)
in units, we have : Force * Length / (Length/Time * Surface)
Let's put that in good order. Remains : Force * Time / Surface
or, in SI units : newtons * sec / m^2 - also called poiseuille (Pl - not yet BIPM approved)
As 1 N / m^2 = 1 pascal, we can also express it in Pa.s
For the sake of history, there are other ways to express the dynamic viscosity.
Remember the first law of dynamics : F = ma
hence a force may be replaced with : Mass * Length / Time^2
and VIS may be expressed in : Mass * Length * Time / (Time^2 * Surface)
Some order again, and we find : Mass / (Time * Length)
or, for instance : kg / (sec * m) (which, in SI, has the same value as the N*sec/m^2)
This is obviously the kg mass and not the kg force !
Knowing this, we may now consider the various equivalencies.
The old unit was the poise (Po), equal to 1 g / (sec * cm) or 1 dyne * sec / cm^2
Its subdivision, the centipoise (cPo) was used more often
(because the dynamic viscosity of water at 20°C or 68°F is about 1 centipoise)
Adapting the units, we find : 10 poises = 1000 centipoises = 1 N*sec / m^2
Therefore 1 centipoise = 1 mN.s/m^2 = 1mPa.s
Other units :
- Using the old kg force, we have : 1 kgf*sec / m^2 = 9.80665 N*sec / m^2
- 1 lbmass /(ft.sec) = 1 poundal.sec/sq.ft = 1.48816 N*sec / m^2
- dividing by 3600, 1 lbmass /ft.hr = 1/2419 N*sec / m^2 (or 413.4 micropoiseuilles)
- 1 lbforce.sec/sq.ft = 47.88 N*sec / m^2
Note : the viscosity definition is equally valid for gases (although the figures are much smaller.)
For instance, air at 20 C has a viscosity of 18*10^-6 N*sec / m^2 (or 0.018 centipoises or 18 micropoiseuilles)
Funny detail : the viscosity of the gases increases with temperature and the viscosity of the liquids decreases.
Maybe you'll hear the name fluidity : it is the inverse of viscosity. An old unit was the rhe, equal to 1/poise
Relative viscosity
Mainly used for liquids : is the ratio of its viscosity to that of water.
You may also encounter a specific viscosity, which
is the ratio of a fluid viscosity to that of a standard fluid (maybe water, as in the relative viscosity) but in a more general way : it will be, for instance, the viscosity of a solute versus the solvant in a solution.
Kinematic viscosity
This is another brand. The kinematic viscosity is the absolute viscosity divided by the density.
The unit is therefore : (kg/(sec*m)) / (kg/m^3) - reduced to m^2 / sec
- 1 sq.ft/sec = 0.0929 m^2/sec
- the old cgs unit was called the stoke = 1 cm^2/sec = 1/10000 m^2/sec
The interest lies in the easier measure of the kinematic viscosity : by the time (in sec) taken for a given volume of liquid to flow through a standard capillary tube.
There are several different instruments. Sure, you have heard of Saybolt, Redwood or Engler viscometers.
Funny detail : do you know that the kinematic viscosity of air is much higher than water ?
For air : 18 * 10^-6 / 1.2 = 15 * 10^-6 m^2/sec (at 20 °C)
For water : 10^-3 / 1000 = 10^-6 m^2/sec (or 15 times less than air)
Other pages ? a kinematic viscosity conversion program, a simple lesson on viscosity, some viscosity conversion tables (for oil) ...
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Thermal conductivity
Heat transfer is actually very similar to the study of electricity, at least when pure conduction is considered. The same laws apply, if we convert voltage into difference of temperature and amperage into heat flow.
The notions of resistivity - resistance - conductivity - conductance are exactly the same.
One can even calculate the flow of heat from a changing source (e.g. with a sinusoidal temperature.)
There are however some differences : heat is much easier to lose than electricity and one cannot discount surface exchange (convection) and radiation, but in electricity we also have to cope with magnetic fields and thermo-ionic valves !
For now, let's look at conduction alone
The standard formula reads : Q = k * S * T / d
- with Q = quantity of heat (can be expressed in : W or J/sec - J/h - kcal/h - kcal/sec - BTU/hr - etc.)
- S = surface through which the heat circulates (in m2, sq.ft., sq.in., square centimeters, etc.)
- d = thickness of the material through which the heat has to travel ( meters, inches, etc.)
- T = difference of temperature between the two ends, across "d" in other words (in °F or °C)
The specific conductivity "k" represents the amount of heat that can be transmitted during a given time across a section of material with a unit surface and a unit thickness, under a difference of temperature of 1 deg. (It is experimental, of course).
In a nutshell, the heat transmission increases with "k", "S" and "T", and decreases with "d".
If we apply the formula to a piece with given sizes, we get its conductance = k * S / d
and the formula becomes : Q = conductance * T
We can now define the resistance "R", equal to 1 / conductance
and we can write : T = R * Q ( compare to Ohm's Law : U = R * I)
As you guessed, there are lots of different values for "k" according to the units used.
The official SI units would be the watt, the meter, the square mater and the kelvin (or degree Celsius, don't worry : we only use differences of temperature at this stage !)
Applying these units to the formula, we see that "k" should be expressed in W / m * K
The table here under gives some equivalencies for most of the units human mischief could invent.
Just select the line with the unit you have and read the factor in the column you want.
Example : 1 BTU/hr.ft.°F is equal to 1.73 W/m.K
Note : - in "BTU/hr.ft.°F", the surface is in sq.ft. and the thickness in ft.
- in "BTU.in/hr.ft2.°F", the surface is still in sq.ft but the thickness is expressed in inches !
Heat conductivity equivalency
| |
W/m.K | kcal/h.m.C | kcal.cm/h.m2.C | BTU/hr.ft.F | BTU.in/hr.ft2.F | W/cm.K |
|
W/m.K | 1 | 0.86 | 86 | 0.578 | 6.94 | 0.01 |
|
kcal/h.m.C | 1.16 | 1 | 100 | 0.672 | 8.06 | 0.0116 |
|
kcal.cm/h.m2.C | 0.0116 | 0.01 | 1 | 0.00672 | 0.0806 | 0.000116 |
|
BTU/hr.ft.F | 1.73 | 1.49 | 149 | 1 | 12 | 0.0173 |
|
BTU.in/hr.ft2.F | 0.144 | 0.124 | 12.4 | 0.0833 | 1 | 0.00144 |
|
W/cm.K | 100 | 86 | 8600 | 57.8 | 694 | 1 |
As an example, the thermal conductivity of copper should read :
- 380 W/m.K
- 327 kcal/h.m.C
- 32700 kcal.cm/h.m2.C
- 220 BTU/hr.ft.F
- 2640 BTU.in/hr.ft2.F
- 3.8 W/cm.K
(These values are never very accurate, they may vary widely from reference to reference and there is no need to carry too many decimals in the table.)
Resistance is generally expressed as R and its converse - conductance - is expressed as U (especially in the building industry where it's used to indicate the heat losses of a house.)
When different materials are stuck together, one has to calculate each thermal resistance and add them together to find the total resistance. R = R1 + R2 + ...
If one uses the thermal conductances, it has to become : 1/U = 1/U1 + 1/U2 + ...
Never add conductances !
Precision : you may add conductances if you have several possible routes for the heat, like for resistors in parallel in electricity. But the general problem is more often about sticking together different materials in series - like resistors in series in electricity.
Note : Conductance will obviously be expressed in "W/K" and resistance in "K/W".
Special notes on building insulation
In this case, "resistance" is often used per unit of surface and is therefore slightly different from what we just defined. You've heard of "R20" etc.
A wall with a "R" of 1 will have a conductance of 1/1 = 1 BTU/hr.(sq.ft).°F
meaning that 1 square foot of a wall with such a resistance to heat will lose 1 BTU per hour per deg. F across the wall.
Example : we have 14°F outside and 68°F inside : we will lose 54 BTU per hour and per sq.ft. of wall.
If we had a "R20", the conductance would be 1/20 = 0.05 BTU/hr.(sq.ft).°F
and, in the same conditions of temperature, we would lose 54/20 = 2.7 BTU per hour and sq.ft.
Note : a builder will also talk about a "U factor" of 1/20 = 0.05
You may also encounter "RSI" (hopefully more and more) - standing for "Resistance Système International" and obviously its converse - conductance - is expressed in W/m2.K
A simple arithmetics shows that 1 R = 0.1761 RSI
In other words, a wall with R20 is also a wall with 3.522 RSI.
Similarly, a wall with 1 RSI will have a conductance of 1 and a loss of 1 watt per square meter per degree (celsius or kelvin)
For the same temperatures (here -10°C and +20°C), we lose 30 W per square meter
If RSI = 3.522 then the loss is : 30/3.522 = 8.518 W per square meter.
Check up : 8.518 * 0.3048^2 * 3600 / 1054 = 2.7 BTU/hr.(sq.ft) - good !
Note : this is only part of the problem. The full thermal loss of a wall has also to take into account the input of heat into the wall and the loss to the outside air, through convection and radiation. See later.
An insulating material will be defined as (for instance) R 3.9 per inch, meaning its specific conductivity is 0.2564 BTU.in/hr.ft2.°F (and you need about 5 inches to get R 20)
Using the above table, we see that it corresponds to 0.03695 W/m.K
and the specific resistivity is : 1/0.03695 = 27 RSI per meter.
To complicate matters, people use quite often "RSI per millimeter" ! So we get : 0.027 RSI/mm
"R" maybe, but not quite "SI" !
Surface exchange
One of the trickiest chapters. Let's try to keep it VERY simple or we'll get carried away.
If we consider only exchanges between a solid and a gas (e.g. air), we have two different forms of heat transmission, and it is always difficult to calculate the exact part played by each one : convection (or exchange with the gas passing by) and radiation towards the surroundings.
Convection is a function of the gas speed vs. the solid.
You'll find lots of more or less accurate formulas. Let's give some of them :
- k = 2 + 10 * sq.rt.(speed) (for metallic surfaces)
- k = 1 + 5 * sq.rt.(speed) (for bricks, etc.)
- k = 10 + 5 * sq.rt.(speed) (for bricks and high temp. gas - metallurgy !)
"k" is the specific conductivity, to be handled like any other. It is expressed in kcal/h.m2.°C and has to be multiplied by 1.163 to get W/m2.K (or 0.2048 to get BTU/hr.ft2.°F)
The gas speed is expressed in meter/sec.
Example : outside air at 5 m/s on a brickwall : k = 14 W/m2.K
Equivalent RSI = 1/14 = 0.071
inside air at +/- 0 m/s : k = 1.163 W/m2.K
Equivalent RSI : 1/1.163 = 0.86
If the wall had an RSI of 3.522, the total RSI becomes 3.522 + 0.071 + 0.860 = 4.453
One can see that in this case the real obstacle to heat transmission is the wall itself, not really the transmission between air and wall.
The heat transmission will be 1/4.453 = 0.224 W/m2.K (or 0.224 W/m2 for each degree C between inside and outside temperature. If we have -10°C and +20°C, the loss will be 6.72 W/m2.)
Knowing the overall resistance and the extreme temperatures, you can use the individual resistances to calculate the temperatures at each layer, exactly like in electricity.
There are other formulas : (in W/m2.K)
- k = 11 + 4 * speed (for bricks and concrete)
- k = 11 + 3 * speed (for wood)
These are experimental and give a much higher result : they obviously include some form of radiation effect.
Radiation is also quite tricky. A scientist will say that a black body will emit heat according to the fourth power of its absolute temperature :
k = 5.67 * 10^-8 * T^4
k in W/m2.K4 and T in kelvin.
The problem lies with the "black body" : there are very few of them ! Generally, a solid will emit between 10 and 90 % of what a black body will do !
Anyway, to use that formula, you also have to consider the temperature of the surroundings, in this form :
k = 5.67 * 10^-8 * (T^4 - t^4)
with "T" for the absolute temp. of the emitter and "t" for the surroundings.
Example : a wall at 10°C in winter (-5°C outside) will emit a maximum of 71 W/m2 (work it by yourself)
In this case, we have two different forms of heat losses and the conductances for convection and radiation have to be added - not the resistances.
This is quite an interesting problem of fluxes - you'll have to use a node equation - that's fun !
While we are at it, I'll give a personal experimental formula for heat losses vs a surface temperature, in calm air at room temp. :
k = 0.00021 * t^3 - 0.05973 * t^2 + 23.1238 * t - 861.1
in W/m2 ("t" is expressed in °C here !)
Valid from ambient to 400°C and quite useful to estimate the loss of a furnace or a duct of hot gas inside a building.
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Dynamics variations
The basic law reads : Force = mass * acceleration or (for the friends only) : F = m*a
As we may have several different units for all three factors, imagine the variations !
Generally, people love having a "1" for the acceleration.
We have already seen most used systems in the main page. Let's summarize :
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Last updated : Oct. 2, 1999
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