Mechanical Engineering: Heat Transfer Analysis of a Cylindrical System

This document provides a comprehensive solution to a heat transfer assignment, focusing on a cylindrical system with multiple layers of materials. The solution begins by calculating the overall heat transfer rate and determining temperatures at various points within the system. It then explores the impact of emissivity on the calculations. Furthermore, the assignment delves into the concept of critical radius of insulation, explaining its significance and how to determine it. The document also compares polyethylene and glass wool as insulation materials, considering their thermal conductivities and costs to determine which is more efficient for the given conditions. The solution includes relevant formulas, calculations, and a discussion of factors affecting thermal conductivity, making it a valuable resource for students studying heat transfer.

Solution
Given
Length (l) = 5 m
Radius r2 = 11 mm
Thickness (t) =0.9 mm
Radius r1 = 10.1 mm
Radius r3 = 15 mm
Radius r4 = 16mm
Temperature inside Ti = 80c
K cu (1-2) =13.1 W/mk
h1 = 20 W/m^2 k
h4 = 63 W/m^2 k
K gw (2-3) =0.033 W/mk
K al (2-3) =10 W/mk
Temperature outside To = 17c
Task 1
a) Determine overall heat transfer rate?
Q=2πl( ti - to )
{
( 1
h1∗r1 ) +
ln r2
r1
kcu
+
ln r3
r2
kgw
+
ln r4
r3
kal
+ ( 1
h4∗r4 ) }
Put all the values in formula, we get
r1
r2
r3
r4
copper
glasswool
Aluminium
skin

Q= 1978.2
4.95+ 0.0065+9.398+0.0064 +0.992
Q = 128.84 W
b) Calculate temp. At r2, r3 , r4
For t2
t1−t2=
Q∗ln r2
r1
2 πl kcu
80-t2 =0.026
t2 = 79.97C
For t3
t2−t3=
Q∗ln r3
r2
2 πlk gw
79.97- t3 = 38.56
t3 = 41.4C
For t4
t3−t4=
Q∗ln r 4
r3
2 πl k al
41.4- t3 = 0.026
t3 = 41.37C

Task 2
a) Now due to emissivity
∈ = 0.6 for copper surface
hr =∈ σ ( T1 +T 2 ) ( T1
2 +T 2
2 )
hr =0.6∗5.67∗10−8∗( 80+79.97 )∗(802+ 79.972)
hr =0.0696 W
m2 k
Rradiation= 1
hr∗A
Qnet = 2πl( ti - to )
{
( 1
h1∗r1 ) +
ln r2
r1
kcu
+
ln r3
r2
k gw
+
ln r 4
r3
kal
+
( 1
h4∗r4 ) + 1
hr∗r1
}
Qnet = 1978.2
4.95+0.0065+ 9.398+0.0064+0.992+ 0.0007
Qrad=128.84 W
b) Calculate temp. At r2, r3 , r4
For t2
t1−t2=
Qnet∗ln r2
r1
2 πl kcu
80-t2 =0.026
t2 = 79.974C
For t3

t2−t3=
Qnet∗ln r3
r2
2 πlk gw
79.974- t3 = 38.54
t3 = 41.42C
For t4
t3−t4=
Qnet∗ln r 4
r3
2 πl k al
41.42- t3 = 0.026
t3 = 41.394C
c) Emissivity is defined as the ratio of energy radiated from surface of material
to that radiated from a blackbody.
Emissivity is a dimensionless number between 0¿ ϵ <1 ,ϵ=¿ 0 (for a perfect
reflector) and ϵ =1 (for a perfect emitter).
In black bodyϵ ¿ 1 , So no loss in heat while radiation. In non black body 0¿ ϵ <1 ,
So there is some loss in heat while radiation.
Task 3
a)
 Critical radius of insulation is radius at which the heat transfer rate is
maximum and the resistance is minimum.
 For cylinder, adding the insulation to the cylindrical pipe increase the
conduction resistance but decreases the convection resistance.
 For cylinder
rc= k
h
Critical radius = rc

Thermal conductivity of material =K
Heat transfer coefficient = h
b) Insulation Material- A material which retards the flow of heat with
reasonable effectiveness is known as ‘Insulation’. Insulation serves the
following two purposes : (i) It prevents the heat flow from the system to the
surroundings ; (ii) It prevents the heat flow from the surroundings to the
system.
 polyethylene insulation
Polyethylene is a thermoplastic compound. Its dielectric constant is
value is very low therefore good in insulation purpose. It is moisture
resistance , low density and excellent weather resistant .
Dielectric constant value =2.3
As according to formula
rc= k
h

So critical radius depend on the thermal conductivity(k) of material
Factors affecting thermal conductivity Some of the important factors which
affect thermal conductivity (k) of the insulators (the value of k should be
always low to reduce the rate of heat flow) are as follows :
1. Temperature. For most of the insulating materials, the value of k increases
with increase in temperature.
2. Density. There is no mathematical relationship between k and ρ (density).
The common understanding that high density insulating materials will have
higher values of k in not always true.
3. Direction of heat flow. For most of the insulating materials (except few like
wood) the effect of direction of heat flow on the values of k is negligible.
4. Moisture. It is always considered necessary to prevent ingress of moisture in
the insulating materials during service, it is however difficult to find the effect
of moisture on the values of k of different insulating materials.
5. Air pressure. It has been found that the value of k decreases with decrease
in pressure.
6. Convection in insulators. The value of k increases due to the phenomenon
of convection in insulators.
For polyethylene
K =0.5 W/mk
For Glasswool
K=0.04 W/mk
As for same value of h = 10 W/(m^2 k) for air , Thermal conductivity of
polyethylene is greater than that of Glass wool.
For polyethylene rc=50 mm
For Glass wool rc=4 mm
So according to the calculation , for same environmental condition for
maximum heat transfer

The critical radius for glass wool is less than that of polyethylene.
Cost of glass wool = 9$ / Kg
Cost of polyethylene =1.5 $ /Kg
The cost of glass wool is greater than that of polyethylene.
Finally it is concluded that the glass wool is more efficient to use as compared
to polyethylene for same condition.
Reference
B. Yoda, K. Muraki, Development of EHV cross-linked polyethylene insulated
power cables, May 1972.
F. H. Kreuger, "Endurance tests with polyethylene insulated cables", CIGRE,
1968.