Heat Exchanger Design, Insulation, and Heat Transfer Analysis

This assignment solution addresses the design and analysis of a heat exchanger, focusing on heat transfer principles and insulation strategies. The solution begins by calculating the heat flux through a tank wall and determining the optimal insulation thickness to minimize heat loss. It then proceeds to calculate the required flow rate and tube length within the heat exchanger, considering parameters such as cold water temperature, flow rate, and tube dimensions. Finally, the solution calculates the necessary insulation thickness for a heated pipe using both polystyrene and glass fiber, aiming to minimize heat loss. The solution incorporates relevant formulas and calculations to provide a comprehensive analysis of heat transfer phenomena within the given industrial application scenario. The assignment also requires the student to provide references and adhere to specific academic integrity guidelines.

ASSIGNMENT SOLUTION
TASK 1
The heat exchanger is designed to reclaim the
heat from saturated water at 300 K. The low
temperature pipes are to be housed in a tank
made from 5mm thick aluminium plate with a
thermal conductivity of 237 Wm-1K-1. The tank
is insulated with pine wood having a thermal
conductivity of 0.12 Wm-1K-1. Assuming one
dimensional heat transfer, determine:
• The heat flux of the tank wall.
• The required thickness of the insulation to minimise heat loss.
Solution
Given data
T1 = 300 K
Plate thickness (dx) = 5 mm
Kal = 237 Wm-1K-1
Kpw = 0.12 Wm-1K-1
Assume low temperature pipe at ambient temperature
T2 = 273 K
Heat flux of the tank wall (W/m^2)
q=k al
dt
x
q=237 ( 300−273 )∗1000
5
q=1279.8 kW
m2
The required thickness (x1) of the insulation to minimise heat loss
When insulation is added then heat flux formula, differential of heat flux is equal to
zero .
q= dt
x1
k pw
+ x
kal
q= 27
x1
0.12 + 5
1000∗23

dq
d x1
=0
For minimum heat flux , its differential is zero
x1=1.25 mm
TASK 2
Cold water, at an average temperature of 273 K and flow rate of 2.5 Kgs-1, enters
the heat exchanger and passes through ten 25mm diameter thin walled copper
tubes. Calculate:
• The hot saturated water flow rate required to fully recover all available
heat.
• The length of each tube
Solution
Assume hot water temperature to be 400 to 300 deg C
Hot water flow rate ( ˙mh ¿
From the overall energy balance, the heat transfer required of the exchanger is
q= ˙mc∗cpc∗(tco−tc i)
q=2.5∗4181∗273
q=2.853 MW
q= ˙mh∗c ph∗(th o−th i)
2853532.5= ˙mh∗4181∗(400−300)
˙mh=6.825 kg /sec
The length of each tube
Reynold number
ℜ= ˙4 mc / N
πDμ
ℜ= 4∗2.5 /10
π 25
1000∗855∗10−6
ℜ=14899.2
As flow is turbulent therefore use turbulent equation.

Nu=0.023∗ℜ
4
5∗Pr0.4
Nu=0.023∗14889.2
4
5∗5.830.4
Nu=108.8 7
Overall heat transfer coefficient U
hi= Nu
D ∗K
hi= 108.87
0.025 ∗0.613
hi =2700W m−1 k−1
U = 1
1
2700 + 1
8 0 00
U =2019W m−1 k−1
Length of tube
L= q
UNπDF∗∆ T m
∆ T m= ( 400−300 )−273
ln ( ( 400−300 )−273)
∆ T m=¿33.5 deg C
L= 2853532.5
2019∗10∗π∗0.025∗1∗33.5
L=53.74 meters

TASK 3
The heated water is to pass through 10m of insulated 100mm diameter thin walled
copper pipe with a thermal conductivity of 401 Wm-1K-1 to an insulated storage tank.
Calculate;
• The thickness of polystyrene insulation k = 0.027 Wm-1K-1 that would be
needed to minimise heat loss from the pipe.

• The thickness of glass fibre insulation k = 0.046 Wm-1K-1 that would be
needed to minimise heat loss from the pipe.
Solution
Consider heat transfer coefficient for hot water is h =0.5 Wm-2K-1
The thickness of polystyrene insulation k = 0.027 Wm-1K-1 that would be needed to
minimise heat loss from the pipe.
Critical radius of insulation = k /h
= 0.027/0.5
= 0.054*1000
Critical radius of insulation = 54 mm
Critical thickness of insulation = 54-50 = 4 mm
Critical thickness of insulation of polystyrene insulation is 4 mm
The thickness of glass fibre insulation k = 0.046 Wm-1K-1 that would be needed to
minimise heat loss from the pipe.
Critical radius of insulation = k /h
= 0.046/0.5
= 0.092*1000
Critical radius of insulation = 92 mm
Critical thickness of insulation = 92-50 = 38 mm
Critical thickness of insulation of glass fibre insulation is 34 mm
Assessment
You are required to attempt all aspects of the assignment. The grading criteria
associated with assessment are outlined below. You must produce a fully
annotated list of the printed material consulted, together with any electronic
databases you have used.
Remember to include a reference list / bibliography and to apply the correct use of
the IEEE(2006) referencing system.

To avoid penalties associated with Academic Misconduct and Plagiarism you should
refer to the Exeter College Regulations, as detailed in your course handbook, or
contact your course tutor for guidance.
Word length
There is no word limit for this assignment. However, you must attempt the
Assessment Criteria as fully as you can and ensure that you answer every question
to its fullest extent appropriate to your level of study.
Support Texts
See Moodle for appropriate texts and websites
ASSESSMENT CRITERIA
LO3 Examine the principles of heat transfer to industrial applications
Pass:
P6 - Determine the heat transfer through composite walls.
P7 - Apply heat transfer formulae to heat exchangers
Merit:
M3 – Explore heat losses through lagged and unlagged pipes.
Distinction:
D3 – Distinguish the differences between parallel and counter flow recuperator heat
exchangers.