Question 1
Heat transfer is an engineering discipline that involves the study of generation, use, conversion and exchange of thermal energy
between various mechanisms. It is concerned more with the quantity and time (rate) of energy transfer.
Mode
Description
Law
governing
Terms used
Conduction
This is heat transfer between
objects/substances in direct contact. It
occurs at a molecular level because of
a temperature gradient. An example
is an empty sufuria resting on a hot
cooking coil getting hot to the touch.
Fourier’s law
Convection
This is heat transfer in fluids through
a medium. It can be through random
molecular motion or bulk motion of a
fluid. It can be natural or forced. An
example is how water boils
Radiation
Heat transfer along Electromagnetic
waves. It does not require a
medium and can take place through
vacuum. An example is the earth
being warmed by the sun.
Newton’s law of cooling
Stafan- Boltzmann law
q – Heat flux
A – Area in m2
K – thermal conductivity in W/mk
T – Temperature in kelvin
X – thickness in meters
q – Heat flux
A – Area in m2
h – heat transfer coefficient in W/m2k
T – Temperature in kelvin
Eb – Emissivity for a black body
𝜎 – Stefan-Boltzmann constant
Equation
𝑞 = 𝑘𝐴
∆𝑇
∆𝑥
𝑞 = ℎ𝐴(𝑇1 − 𝑇∞ )
Ts – Absolute Temperature in
Kelvin
𝐸𝑏 = 𝜎 ∙ 𝑇𝑠4
Question 2 Solution
Step 1: What we know
Heat transfer occurs along a temperature gradient from high temperature region to low temperature region. The heat transferred from
the inner surface of the brick wall to the outer surface through conduction is equal to the heat transferred to the surroundings through
convection and radiation.
Step 2: What is to be determined
Step 3: Schematic diagram
The temperature at the inner surface of the brick wall.
qradiation
T1
Step 4: Physical properties used
Ta
qconduction
Ta = 25oC
T 2= 100oC
∆x = 0.15m
Atmosphere
Furnace
T2
h
k
ε
k= 1.2W/mk
qconvection
𝜀 = 0.8
Wall
h = 20W/m2k
x
KEY
Step 5: Assumptions used
 Steady- state condition
 One- dimensional conduction
 No uniform heat generation
Red- conduction
Orange-convection
Yellow- radiation
 Isotropic wall material
Step 6:ANALYSIS
Energy balance equation
Energy in= Energy out
Energy in – Energy out = 0
qconduction – qconvection – qradiation = 0
∆T
kA ∆x - hA(T − T∞ ) - εσ ∙ ∆Ts4 = 0
kA
T 1 −T 2
L
- hA(T2 − Ta ) - εσ ∙ (T24 − Ta4 ) = 0
1.2W/mkA
T 1 −373
0.15m
- 20W/m2kA(373 − 298) – 0.8x5.6704 10−8 (3734 − 2984 ) = 0
Assuming area to be unity and Simplifying gives:
1.2
T 1 −373
0.15m
= 𝑇1 =
- 1500 – 514 = 0
(1500 +514 .5309 )×0.15
1.2
+ 373 = 624.8 𝑘