Fourier’s Law

Fourier looked at heat flowing through building  materials, and his experimental results produced ideas such as thermal conductivity, resistance and the thermal transmission co-efficient or u-value.

In a slab of material, if we say that the quantity of heat passing through it in t₁ seconds is q₁ joules, then we can measure q₂ later at the point t₂.  This is simply:

heat flow rate equals the difference in quantities of heat divided by the difference in time

This is expressed in mathematical shorthand like this:

heat flow rate = Δq / Δt

heat flow rate = [ (q₂-q₁) / (t₂-t₁) ]

and joules / seconds equals watts

Experimentally, Fourier found that

• heat flow rate was proportional to the surface area
• heat flow rate was proportional to the time difference
• heat flow rate was proportional to the thickness of the slab

he concluded that it therefore must be the case that

heat flow rate is proportional to the product of the area and temperature difference divided by the thickness.

It is always the case in mathematics that, for example, when

a is in proportion to b

we can say that

a = k . b

where k is the constant of proportion.

Thus, Fourier found the mathematical constant that is now known as thermal conductivity.

q = k . (A . ΔΦ / T)

q = heat flow rate (W)

k = thermal conductivity

A = area (m²)

Φ = temperature difference [C-C]

T= thickness (m)

We can find out the units of thermal conductivity easily enough –

W = k . (m² . [C-C]) / (m)

W = k . m² . C . m^-1

W = k . m. C

k = W . m^-1 C^-1

The next step for Fourier was to figure out Resistance to heat flow…

He concluded that resistance equalled thickness divided by thermal conductivity:

R = T / k

This lets us figure out composite structure resistance, which is handy as plaster is stuck to brick and so forth –

R_TOTAL= (T₁ / k₁) + (T₂ / k₂) etc.

The thermal transmission co-efficient or u-value is the reciprocal of resistance:

u = 1/R

for composite materials.

u_TOTAL = (1 / R₁) + (1 / R₂) … etc

HEAT TRANSFER EQUATIONS:

E = A . ΔΦ . u

E = A . ΔΦ / Total resistance

E = A . ΔΦ  . R_total ^-1

E = thickness . Area . temp diff / Resistance

E = T. A . ΔΦ  . R^-1

E = volume   .  temperature diff  / Resistance

E = V . ΔΦ  . R^-1

E = amount of heat in watts

HEAT TRANSFER DEFINITIONS

CONDUCTIVITY

k = W . m^-1 C^-1

The quantity of heat which is conducted through unit area of a slab of material of unit thickness with unit difference of temperature between the faces in unit time.

CONDUCTANCE

c = W . m^-2 C^-2

The quantity of heat conducted through unit area of a complete structure with unit difference in temperature in unit time.

c = k . thickness^-1

RESISTIVITY
r = m.C. W^-1
(i) reciprocal of conductivity
(ii) resistance to heat transfer through unit thickness of a material of unit surface area an a face to face temperature of 1C in unit time.
(iii) the temperature differential between faces of unit thickness of a slab of material of unit area for the transmission of unit quantity of heat in unit time.

RESISTANCE
R = m².C. W^-1
(i) reciprocal of conductance
(ii) equivalent to the temperature differential between faces of a thickness of slab of material of unit area for the transmission of unit quantity of heat in unit time.

R = thickness / k

TIP
unit thickness = …ivity
overall thickness = ….ance

Compound Interest

In design work, there is a business side of things that depends on loans and investments.  There is also a business side of things where people invest different amounts for different periods of time in helping you — and you might have to figure out their share of things in the future.

There is also the idea of inflation and cost of living — as you have to design to a cost, a budget, you have to sometimes predict the future costs involved.

This is a small introduction to the basics of compound interest, included here in the hope that it may help.

Flow Rate Calculations

Mass flow rate is (kg.s^-1)
It is the product of volume flow rate (m³.s^-1) and specific mass (kg . m^-3), where the cubic metres simply cancel each other out.

(kg.s^-1) = (m³.s^-1) . (kg . m^-3)

Using the units this way enables you to determine the formula (rather than remember it).

For example, if you want Volume flow rate (m³.s^-1)
and you know the specific mass to be 1.205.kg.m^-3, then you only need look at the units to see what it is you need to measure.

we have (m³.s^-1) and (kg.m^-3), and it’s clear that if you divide (m³.s^-1) by (kg.m^-3) , the m^-3 and m³ will cancel leaving kg and s^-1 which is (kg.s^-1) — which is mass flow rate.

If you want to determine the Volume Flow Rate for ventilating a room, and you do not have specific mass or mass flow rates to hand, you can improvise, using the reasoning of units.

You would be likely to know the space’s volume (floor area multiplied by height), and the air changes per hour, and this is all you need:

Space volume is m³, and air changes are “per hour” or hr^-1. To get per hour to per second, divide air change rate by 3600. s.hr^-1

Room volume x air change rate/3600 = Volume Flow rate

m³ . ( a . hr^-1/3600. s.hr^-1) =>

m³ . ( a /3600. s.) =>

m³ . ( a /3600)  . s^-1 =>

m³. s^-1 . = Volume Flow rate.

By changing the subject, you can determine the air change rate for the room if you know the volume flow rate, or you can figure out the room volume if you know the air changes and volume flow rate.

Obviously by mixing the equations you can determine the mass flow rate if you know a room volume an air change rate and specific mass.

Ventilation Grille Sizing

The aspect ratio is

Length
______   of the grille.
Breadth

(Total volume flow rate [m³ . s^-1]).(100)
______________________________________________= Area (m²)
(No. inlets).( inlet velocity [m . s^-1]).( per cent free area)

Fluids Constants

SPECIFIC HEAT CAPACITY OF DRY AIR
1.01 . kJ . kg^-1 . C^-1 da

SPECIFIC HEAT CAPACITY OF WATER VAPOUR
1.89 . kJ. kg^-1

SPECIFIC HEAT CAPACITY OF ATMOSPHERIC AIR (stp)
1.012 . kJ . kg^-1 . C^-1

MEAN SPECIFIC HEAT CAPACITY OF WATER
4.185 . kJ. kg^-1

LATENT HEAT OF WATER VAPOUR AT 273K (0C)
2500 . kW

SPECIFIC MASS OF WATER (stp)
1000 . kg . m^-3

SPECIFIC MASS OF ATMOSPHERIC AIR AT 293K (20C)
1.205 . kg . m^-3

VELOCITY OF SOUND IN AIR
330 . m . s^-1

ATMOSPHERIC PRESSURE

The standard atmosphere (1.atm) =

101 325 Pa  — or 101.325 kPa.

The following units are equivalent, but only to the number of decimal places shown:

• 760 mm Hg (Torr),
• 29.92 in Hg,
• 14.696 PSI,
• 1013.25 millibars.

One standard atmosphere is standard pressure used for pneumatic fluid power (ISO R554), and in the aerospace industry(ISO 2533) and petroleum industry(ISO 5024) .

Temperature Correcting

Because the specific mass of atmospheric air at 293 . K

= [1.205 . (293. K/293 . K) ] . kg . m^-3

= 1.205 . kg . m^-3

So the specific mass of atmospheric air at a temperature of 29.C

= [1.205 . (293 /273+29) ] . kg . m^-3

= [1.205 . (293 /302) ] . kg . m^-3

= 1.169 . kg . m^-3

which is the specific mass corrected for the higher temperature.

Composition of Air

COMPOSITION		% by volume
N2	Nitrogen	78.03
O2	Oxygen	20.99
A	Argon	0.94
CO2	Carbon dioxide	0.03
H2	Hydrogen	0.01
others		parts per million
Ne	Neon	0.001230
He	Helium	0.000400
Kr	Krypton	0.000050
Xe	Xenon	0.000006