Heat Transfer Analysis of Intercooler: Design and Performance

This report provides a comprehensive heat transfer analysis of an intercooler, treating it as a compact heat exchanger. The analysis considers both air and water cooling mechanisms to determine intercooler effectiveness and pressure drops, assuming clean heat exchanger surfaces without fouling. Geometric characteristics are derived from given parameters and tables, including length, width, and fin details. Fluid properties for both air and water are established based on temperature conditions. Calculations include Reynold's number, heat transfer coefficients, fin effectiveness, and surface effectiveness. The report concludes with the determination of the heat exchanger's effectiveness, outlet temperatures, and total pressure drop, accounting for entrance, core, and exit losses. References to relevant academic publications are included to support the analysis.

SOLUTION TO THE HEAT TRANSFER PROBLEM
The intercooler under consideration is assumed to be a compact heat exchanger type(Pawar,
Pg8-16 & McQuiston and Parker, Pg 20). Hence analysis will be based on the geometric
characteristics of the HX. The performance is dual based implying that analysis will have to
consider both air and water cooling mechanisms. However, separation is only for analysis;
design is expected to integrate both units in a seamless fashion. Notably, at the end of this
analysis, the following shall be realized:
1) Intercooler effectiveness
2) Pressure drops by considering both water and air flows
However, it must be assumed that the HX surfaces are clean and that there is no allowance for
fouling (Shah and Sekulic, Pg 99-220).
ANALYSIS
(1) Surface characteristics
This can be determined from the given parameters and also by considering table 9-4c
Now, from fig b-1, the geometric characteristics of the intercooler can be deduced
hence l= 1.625m, w=0.61m and B= 0.483m (note that the arrangement is such that air
coolers are packed between the water coolers.
The temperature conditions for the given surface characteristics (11.32-737SR): Tia=
400K, Toa= 297K (for air side) and For water side : Tiw= 300K and Tow= 288K

Therefore from table 9-4c, we move to the last row and select the surface
characteristics as illustrated in table 1 below:
Surface
design
Tube
arrangement
Fin
type
Tube
length
Tube
widt
h
Fin/in Hydra
ulic
diamt
er
Fin
Thi
ck
ne
ss
Fin
are
a/t
ota
l
are
a
Fre
efl
ow
/
fro
nta
l
are
a
HT/
TVa
11.32-
737SR
Staggered Ruff
ed
18.7x10-3 2.5x
10-3
11.32 5.51x1
0-5m
0.1
02
x1
0-
5m
0.7
8
886
m2/
m3
(2) Fluid properties
For this refer to Fig A-2, A-9 and A-18
For air entering temperature of 2600F, we can use figure A-2 to deduce the fluid
properties hence:
K= 0.02Btu/hrft2oF/ft
Cp= 0.25Btu/lboF
μr= 0.06lbs/hrft
Pr= 0.68
For humidity correction factor (HCF) : this is obtained for density and specific heat
capacity SHC
Hence HCF for density= 0.99 and HCF for SHC = 1.015
For an inlet temperature of water of 288K (60oF),
μw= 2.5
Pr= 8.0
ρt= 62
kl= 0.320
Cp= 1.0Btu/lboF

(3) Reynold’s Number
Re= 10-3x 4rhG/μ
= 0.001x5.51x9.81/2.5= 21.582x10-3
(4) From 10.97, Stpr2/3 is determined for a Re= 21.582 x(0.001) this lies between 0.009 and
0.030 hence we obtain the average value= (0.009+0.030)/2= 0.0195
The frictional factor f is obtained from fig 6.6 such that it is f= 0.008
(5) Heat transfer coefficient
For the air side: Pr= 0.68, Re= 104 hence Nu= 30
For the water side: Pr= 8.0, Re= 104and Nu= 90 (that is: (80+100)/2)
(6) Fin effectiveness
From figure 2.13 (straight fins) with an inlet temp of 260oF
The scale below the figure can be used to estimate “m(ro-ri)” parameter hence it is 0.4,
Besides, 4rh= 5.51x10-5
Hence rh= (5.51x10-5/4= 1.3775x10-5 and rp= 1.00x10-5
Therefore rh/rp= 1.3775
Hence tracing using the two parameters above, we obtain ηf= 0.92
(7) Surface effectiveness is given by:
Atbundles/Atotal x100=0.1843 /0.29463= 62.55%
And Ntu= AU/Cmin
Cmin/Cmax= 25/50= 0.5
Cmin= 25kg/s
A= bxw= 0.61x0.483= 0.29463
1/U= 1/90+1/30= 4/90; U= 90/4= 22.5 (0verall HX coefficient )
(8) HX effectiveness
Using Cmin/Cmax=0.5, Ntu= 0.265 approx. 0.5 hence έ= 40%
(9) For outlet temperature , refer to section (1) hence To= 288K (for water side) and 297K
(for air side)
(10) The total pressure drop is determined by considering entrance, core and exit
losses hence:
For entrance loss: P’i= (1+Г2i+Kc)x1/2G2/ρi
Where G= m’/A= 50/0.29463= 169.7 approx. 170
Г=0.78 (from the surface characteristics)
Kc= 0.32, ρi=62 henc P’i= (1-0.782+0.32)x1/2x1702/62= 0.7116x233.065= 165.85

Core loss: P’c= 4fL/Dhx1/2G2/ρm
= 4x0.008x1.625/0.61x0.5x1702/62= 19.87
Exit losses:
Pe’= -(1-Г2-ke)x0.5G2/ρ= -(1-0.782-0.32)x0.5x1702/62= -0.0716x233.065= -16.69
Hence Total pressure drops, P’t= 165.85+19.87-16.69= 169.033Pa
REFERENCE
Pawar ,Nilesh. “Thermal Design and Development of Intercooler”. Dhole Patil College of
Engineering: Wagholi. Online Journal, 2016, Pg 8-16. http://ijettjournal.org/Special
%20issue/ICGTETM-2016/ICGTETM_2016_paper_7.pdf
[Accessed: 15th November 2018]
McQuiston, F.C. and Parker, J.D. “Heating, Ventilation, and Air Conditioning: Analysis
and Design, 6th Edition, Wiley: New York NY, 2004, Pg 20. [Accessed: 15th November]
Shah, Ramesh K. and Sekulic, Dusan P. Fundamentals of Heat Exchanger Design. Wiley:
New York NY, 2003, Pg 99-220. http://teguhhady.lecturer.pens.ac.id/FUNDAMENTAL
%20OF%20HEAT%20EXCHANGER%20DESIGN.pdf . [Accessed: 15th November 2018]