The project analyzes the radiator of a 1999 Suzuki Baleno, focusing on calculating the overall heat transfer coefficient and heat transfer area, comparing theoretical results with experimental data. Instruments used include an anemometer and laser temperature sensor, with calculations revealing a significant percentage error of 51.58% between theoretical and experimental values. The findings have practical applications in cooling internal combustion engines and electronic devices.

1. Heat & Mass Transfer
Semester Project
Fall-2017 (5th
Semester, Batch-2015)
Working on the radiator of Suzuki Baleno 1999
Submitted To:
Dr. S Kamran Afaq
Submission Date:
January 15, 2018.
Sr. No Registration No. Student Name Contribution
1 15-ME-071 Muhammad Usama Experimental data
2 15-ME-075 Muhammad Abdullah Khalid Complete Project report
3 15-ME-072 Muhammad Haseeb Adil Calculations
4 15-ME-074 Khubaib Rizwan Print outs and calculation
Department of Mechanical Engineering
HITEC University Taxila Cantt.

2. Problem Statement / Project Statement:
 To calculate the overall heat transfer coefficient (U).
 To calculate the heat transfer area and compare it with the experimental data being collected.
Project Introduction:
For the save use of IC engine we must safely transfer the extra heat from engine through some source.
Radiator is use for transferring heat from engine. In this project we find out that how heat of engine is
transfer from radiator in atmosphere. We found that different areas of radiator are included in heat
transfer.
Objectives:
Our Group has analyzed a Radiator of a 1300 CC Suzuki Baleno.
Our Objective was to calculate the Heat Transfer Area and compare it with Experimental Data.
The Method used to calculate the Heat Transfer is “LMTD”.
Instruments Used:
We use following instruments.
 Anemometer:
To Measure the Speed of the Air coming out of Radiator Fan
 Digital Pyrometer Laser Temperature Sensor:
To accurately determine the Temperature at the inlet and outlet of Radiator.
 Vernier Caliper and Measuring Tape:
To correctly measure the dimensions of the Radiator understudy.

3. Diagram:
Assumptions:
 Radiator Type is “Cross Flow with both Fluids Unmixed”
 Mass Flow Rate had to be assumed due to the unavailability of the Specifications of Water Pump.
 Heat Flow is considered to be Uniform.
 Frictional Effects inside the Radiator Tubes are neglected.
 The Radiator is taken as Steady Flow Device.
 Fluid Properties are assumed to be Constant.
 No Axial Heat Transfer.
 No Change in Operating Temperatures.
 Fins present on the radiator is Assumed to be of Rectangular Cross Section.
Technical Data / Available Data:
Radiator Length = 36 cm = 0.36 m
Radiator Width = 62cm = 0.62 m
Radiator Depth = 2.8cm = 2.8e-2 m
𝑚̇ Of water = 0.34 kg/s
Velocity of air = 8 m/s
Tube Thickness = 0.028m
Tube Depth = 0.0135m
Fin Thickness = 1 mm = 0.001 m
Fin depth = 0.0135 m
Fin width = 0.008 m
Spacing b/w Fins = 0.002m
Total tubes= 62

4. Inlet Temperature of Water= T (hot in) = 85o
C
Outlet Temperature of Water= T (hot out) = 61o
C
Inlet Temperature of Air = T (cold in) =18o
C (atmospheric)
Outlet Temperature of Air = T (cold out) =?
Water Properties: (from table A9)
T(avg) =(85+61)/2=73o
C
ρ = 975.82 Kg/m3
Cp = 4191.8 J/kg.K
k = 0.6654 W/m.K
Pr = 2.45
ѵ = 4.43e-7 m^2/s
Air Properties
T(avg)= (33.68+18)/2=25.84 o
C
ρ = 1.2085 Kg/m3
Cp = 1005 J/Kg.K
ρ = 1.2085 Kg/m3
K = 0.0261 W/m.K
μ = 1.872 × 10-5
Kg/m.s
Pr = 0.7124
Calculations /Results:
Q̇ = ṁ Cp ΔT
As, Q̇ c = Q̇ h
Q̇ h = ṁw Cp-h (ΔT) h
Q̇ h = (0.34) (4191.8) (85-61)
Q̇ h = 34.205 kW
Now;
Q̇ h = ṁaCp-c (ΔT) c
34.205e3 = ṁaCpc (ΔT) c
ṁa = ρAV

5. ṁa= 1.225*(0.36*0.616)*8
ṁa=2.17 kg/s
34205=2.17*1005* ΔT
ΔT=15.68o
C
T(cold out)-T(cold in)= 15.68
T(cold out)= 15.68+18
T(cold out)=33.68o
C
Number of fins = length/center to center distance
=0.36/0.003 = 120
Total fins = 61*120 = 7320
Experimental area:
ATotal = AFin + AUnfin + A1 + A2 + A3
AFin = (Pl + ATip) NFins
AFin = [(0.028+0.001)*2*0.008 + (0.001*0.028)] *7320
AFin = 3.60 m2
AUnfin = (ANofin) NTubes - NFins (ATip)
AUnfin = (0.36*0.028) (62) – (7320) (0.001*0.028)
AUnfin = 0.624-0.2049 = 0.4190*2

6. AUnfin = 0.838 m2
A1 = (0.002*0.36)
A1 = 7.2e-4 m2
A1 = A2
A2 = 7.2e-4 m2
A3 = 0.0135*0.002
A3 = 2.7e-5 m2
A3 = A4
A4= 2.7e-5 m2
Now,
ATotal = AFin + AUnfin + A1 + A2 + A3 + A4
ATotal = 3.60 + 0.838 + 2*7.2e-4 + 2*2.7e-5
ATotal =4.439 m2
Theoretical area:
A = (Q̇ ) / (U × (ΔTlm))
ΔTlm cross flow both fluid un mixed
(ΔTlm) counter flow = (ΔT1 - ΔT2) / ln (ΔT1/ ΔT2)
ΔT1 = Th-in – Tc-out = 85-33.68 = 51.32 o
C
ΔT2 = Th-out – Tc-in = 61-18 = 43o
C
ΔTlm = (51.32-43)/ ln (51.32/43)
So, (ΔTlm) Counter Flow = 47.03o
C
For (F),
T1 = 85 o
C t1 = 18o
C
T2 = 61 o
C t2 =33.68 o
C
P = (t2 - t1) / (T1 - t1)
P = (33.68-18) / (85-18) = 0.234
R = (T1 - T2) / (t2 - t1)
R = (85-61) / (33.68-18) = 1.53

7. So,
F = 0.97 (From Graph Using P & R)
Now,
(ΔTlm) Cross Flow = (F) (ΔTlm) Counter Flow
(ΔTlm) Cross Flow = (0.97) (47.03)
(ΔTlm) Cross Flow= 45.619o
C
For hi:
Geometry = Rectangular
Hydraulic Diameter (Dh)
Dh = 4(Ac) / P
Dh = 4(0.002*0.0135) / (0.002+0.0135)×2
Dh = 3.48e-3m
As, ṁ = ρAv
v = ṁ / ρA
v = 0.34/(975.82*(0.002*0.013)) = 12.90m/s
Reynolds Number (Re)
Re = V Dh / v
Re = (12.90*3.48e-3)/4.43e-7
Re = 101449.06
As, Re > 10,000 So, Turbulent Flow
Nu =0.023*Re^0.8*Pr^0.3 0.3 for cooling
Nu=0.023*(101449.06^0.8)*(2.45^0.3)
Nu=304.42
Now, Nu = hi Dh / K
So, hi = Nu k / Dh
= (304.42*0.665)/3.48e-3
hi = 58172.60 W/m2
.o
C
For ho:
Hydraulic Diameter (Dh)
Dh = 4(Ac) / P
Dh = 4(0.002*0.36) / (0.002+0.36)×2

8. Dh = 3.97e-3m
Reynolds Number (Re)
Re = V Dh / v
Re = (8*3.97e-3)/15.63e-6
Re = 2036.03
As, Re <2300 So, Laminar Flow + external forced convection
From table
Nu =0.228*Re^0.731*Pr^(1/3)
Nu=0.228*(2036.03^0.731)*(0.712^0.333)
Nu=53.39
Now, Nu = h0 Dh / K
So, h0 = Nu k / Dh
= (53.39*0.0261)/3.97e-3
hi = 351 W/m2
.o
C
U =
𝟏
𝟏
𝒉 𝒊
+
𝟏
𝒉 𝟎
U =
𝟏
𝟏
𝟑𝟓𝟏
+
𝟏
𝟓𝟖𝟏𝟕𝟐
U = 348.89 W/m2
.o
C
A = 34205/(348.89*45.61)
A = 2.149 m2
Percentage Error of Theoretical & Experimental Values (if any)
Error = (4.439-2.149)/4.439
Error = 51.58%

9. Practical / Industrial Applications of the Project: (if any)
Radiators are used for cooling internal combustion engines, mainly in automobiles but also in piston-engine
aircraft, railway locomotives, motorcycles, stationary generating plants and other places where such engines
are used. Tiny radiators known as heat sinks are used to convey heat from the electronic components into a
cooling air stream. Radiators are also found as components of some spacecraft.
Project Pictures: