Outcome
Outcome
Outcome
OUTCOME 1 - TUTORIAL 1
Radiation: nature of radiation; Stefan-Boltzman law; black and grey body radiation; emissivity;
absorptivity; correction for overall heat transfer coefficient
You should judge your progress by completing the self assessment exercises. On completion of this
tutorial the student should be able to do the following.
Calculate the thermal resistance for flat, cylindrical and spherical layers.
©D.J.Dunn www.freestudy.co.uk 1 1
CONDUCTION
INTRODUCTION
Heat exchangers are used in so many fields of engineering from central heating systems to large
industrial boilers, from sugar refining to oil distillation, from refrigerators to air conditioners and so
on. The list is endless. The design of suitable units is based on theory and empirical data and this
covers a vast range of work. The purpose of these tutorials is to provide the basic level of
understanding and some advanced theory for anyone studying the topic from beginner’s level. The
tutorial only covers steady state heat transfer.
More advanced studies of conduction will enable you to study the following topics.
This occurs when the temperature in a material is changing with time. Calculating the temperature
at any point is beyond the scope of this tutorial.
When a block of material is maintained at constant but different temperatures on all its faces, a
steady flow of heat may be produced at each face but the temperature at any given point on the
block is difficult to find and is also beyond the scope of this tutorial.
CONDUCTION THEORY
Heat transfer occurs from one body to another by three methods, conduction, convection and
radiation. Most heat exchangers will use elements of all three. A net amount of heat is always
transferred from the hotter body to the colder body.
This tutorial covers conduction, the process by which heat is passed on through solids, liquids and
gasses from one molecule to another.
©D.J.Dunn www.freestudy.co.uk 2
FOURIER’S CONDUCTION LAW
Calculate the heat transfer through a flat copper plate 200 mm tall by 300 mm wide and 25 mm
thick when the surface temperatures are 150oC and 55oC.
SOLUTION
©D.J.Dunn www.freestudy.co.uk 3
ELECTRICAL ANALOGY AND SIGN CONVENTION
If we make the analogy that Φ is equivalent to electric current I and Δθ is equivalent to voltage ΔV,
we may make an analogy with Ohm’s law
ΔV Δθ t
Electrical Resistance R Thermal Resistance R
I Φ kA
We should note that Δθ and ΔV are both strictly negative quantities because Δ means the last figure
minus the first figure so Δθ = (θc – θh). If we stick to this, we must put a minus sign in front of the
previous equations. If we simply let Δθ represent the temperature difference Δθ = (θ h – θc) and
remember that heat flows from hot to cold, we will have no difficulty solving basic problems.
CALCULUS FORM
In reality, the temperature drops in the direction of heat flow so if we measure the distance from the
hot side as x, the heat flow through a very thin layer thickness dx will be:-
dθ
Φ k A
dx
This is important when analysing non-steady flow conduction and in the following work.
dr
2π kL
dθ Ln r RR
o
2π kL
io
Ri
r Φ θi
i
Φ
R
Ln o
2π kL
θ o θ i Φ 2 π k L θ o θ i Φ
2πkL
θ i θ o
Ri Φ R Ro
Ln o Ln
Ri i
R
Calculate the heat transfer through a copper tube 5 m long with inner diameter 80 mm and
outer diameter 100 mm. The inside temperature is 200 oC and the outside temperature is 70oC.
SOLUTION
2πkL
Φ θ i θ o 2 π x 385 x 5 200 70 7 046 438 W
R 50
Ln o Ln
Ri 40
©D.J.Dunn www.freestudy.co.uk 4
CONDUCTION THROUGH A HOLLOW SPHERE
The derivation is similar to that for a tube. The surface area of a sphere is 4πr2
k A dθ 4 π r 2 k dθ
Φ
dx dr
dr 4 π k dθ
Rearrange to get 2
r Φ
Now integrate between limits.
r
θo
4 k
Ro
4πk
dr
dθ
1 R o
Ri θθθio
Ri r 2
Φ θi
Φ
R -o1 R i-1
4πk
Φ
θ o θi Φ
4πk
R i-1 R -o1
θ i θ o
A spherical steel reaction vessel has an outer radius of 1.5 m and is covered in lagging 200 mm
thick. The thermal conductivity of the lagging is 0.1 W/m K. The temperature at the surface of
the steel is 340oC and the surface temperature of the lagging is 45oC. Calculate the heat loss.
SOLUTION
4 π x 0 .1
Φ 340 45 4.7 kW
1.51 1.7 1
1. Calculate the heat loss through flat sheet of glass 2 m x 1 m and 5 mm thick when the surface
temperatures are 20oC and 5 oC.
(Answer 6 kW)
2. A steam pipe has an external diameter of 75 mm and it is covered with lagging 25 mm thick
with a thermal conductivity of 0.09 W/m K. The surface temperature of the lagging is 300 oC on
the inside and 80oC on the outside. Calculate the heat loss per metre length.
(Answer 243.5 W/m)
3. A steel pipe has an inner diameter of 50 mm and outer diameter of 100 mm. The outside
temperature is 400 oC and the inside temperature is 120 oC. The thermal conductivity 60 W/m K.
Calculate the heat flow from the outside to the inside for 1 m length.
(Answer 152.3 kW/m)
4. A spherical vessel 2 m diameter is lagged to a depth of 300 mm. The thermal conductivity of
the lagging is 0.1 W/m K. The temperature of the inside and outside of the lagging is 180 oC and
40oC respectively. Calculate the heat loss.
(Answer 762.4 W)
©D.J.Dunn www.freestudy.co.uk 5
COMPOUND LAYERS
RB b
t
θ 2 θ3 ΦRB θ3 θ 2 Φ RB θ h Φ RA Φ RB
kb A
RC c
t
θ3 θ c ΦRA
kc A
θ c θ3 ΦRC θ h ΦRA ΦRB ΦRC θ c θ h ΦRA RB RC
θh θc θ θc
Φ h
RA RB RC R
All we need to do is calculate the resistance of each part and add them up.
R
Ln o
Cylindrical layers R
Δθ
i Spherical layers R
R o
Δθ R i-1 R -1
Φ 2πk L Φ 4 πk
When we have compound layers (this may also include convection), it is convenient to use the
equation Φ = - U A Δθ where U is the overall heat transfer coefficient and Δθ is the temperature
change across the entire layer. This will be covered and used in the following tutorials.
A wall with an area of 25 m2 is made up of four layers. On the inside is plaster 15 mm thick,
then there is brick 100 mm thick, then insulation 60 mm thick and finally brick 100 mm thick.
The thermal conductivity of plaster is 0.1 W/m K.
The thermal conductivity brick is 0.6 W/m K
The thermal conductivity the insulation is is 0.08 W/m K
The inner surface temperature of the wall is 18oC and the outer is -2oC.
Calculate the heat loss and the temperature at the interface between the plaster and the brick.
SOLUTION
t 0.015
Plaster R1 R4 1 6 x 10 3 K/W
k1 A 0.1 x 25
t 0.1
Brick R2 2 6.67 x 10 3 K/W
k 2 A 0.6 x 25
t 0.06
Insulation R3 3 30 x 10 3 K/W
k 3 A 0.08 x 25
Total Resistance R = R1 + R2 + R3 + R4 = 49 x 10-3 K/W
θ θ c 18 - (-2) 20
Φ h -3
405 W (Answer calculated with all memory retained)
R 49 x 10 49 x 10 -3
Temperature drop over plaster = ΦR1 = 405 x 6 x 10-3 = 2.43 K Hence the temperature at the
interface is 18 – 2.43 = 15.57 oC
©D.J.Dunn www.freestudy.co.uk 6
WORKED EXAMPLE No. 5
A steel pipe 120 mm inside diameter has a wall 10 mm thick. It is covered with insulation 20
mm thick. The thermal conductivity of steel is 60 W/m K and for the insulation is 0.09 W/m K.
The pipe carries steam at 150oC and the outer surface temperature of the insulation is 0oC.
Calculate the heat loss per metre length. Calculate the temperature at the pipe’s outer surface.
SOLUTION
R 70
Ln o Ln
R i 60 4.089 x 10 4 K/W
R
Pipe
2πkL 2 π x 60 x 1
R 90
Ln o Ln
Insulation R i 70 0.444 K/W
R
2πkL 2 π x 0.09 x 1
Total resistance =0.445 K/W Φ = Δθ/R = (150 – 0)/0.445 = 337.2 W
2. A nylon pipe has a bore of 50 mm and a wall thickness of 5 mm. It is covered with insulation 10
mm thick. The thermal conductivity of the nylon is 0.25 W/m K and for the insulation it is
0.1 W/m K. The inside of the pipe carries hot water at 50 oC and the outer surface of the pipe is
at 10oC. Calculate the heat loss per metre length and the temperature between the two materials.
(Answer 70 W and 41.9oC)
©D.J.Dunn www.freestudy.co.uk 7