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 —

RTOTAL= (T1 / k1) + (T2 / k2) etc.

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

u = 1/R

for composite materials.

uTOTAL = (1 / R1) + (1 / R2) … etc

HEAT TRANSFER EQUATIONS:

E = A . ΔΦ . u

E = A . ΔΦ / Total resistance

E = A . ΔΦ  . Rtotal-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

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