Thermoelectric cooler powered by Bicycle's Dynamo

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ME 332 Thermodynamics II
CEP#7
MEMBERS
 HASSAN SHOAB SIDDIQUI 2014114
 M.KAMRAN ABBASI 2014233
 M.MUDASSAR ABBAS 2014236
 M.UMER FAROOQ 2014260
 SYED MANSOOR JAN 2014361

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Table of Contents
PROBLEM STATEMENT...........................................................................................................................1
SOLUTION APPROACH AND ASPECTS CONSIDERED/ CODE CONCLUSION ............................................1
PART 1...................................................................................................................................................1
ASSUMPTIONS ....................................................................................................................................2
ANALYSIS.............................................................................................................................................2
PART 2...................................................................................................................................................5
RESULTS AND DISCUSSION.....................................................................................................................5
MATHEMATICAL MODELLING.............................................................................................................5
DATA ...............................................................................................................................................5
ASSUMPTIONS ................................................................................................................................6
EXPERIMENTATION.............................................................................................................................8
PROBLEM STATEMENT
You are working on a thermoelectric cooler that is plugged into USB connection powered by
your bicycle’s dynamo, and are designing one such cooler which would claim to cool any
suitably weighed (User defined) drink from one temperature to other or to heat a well-sealed
cup of coffee from one temperature to other, in about an optimum time in a well-insulated
cup holder. You would determine if such claim is feasible and what is the distance needed to
be travelled in order for cooling and heating to occur. Also determine the rate of heat removal
from the drink, and the rate of heat supply to the hot drink. In addition to that, if heating a cup
of coffee and cooling a cup of water are demanded at the same time, plot the temperatures of
both of the cups with respect to time/distance covered.
SOLUTION APPROACH AND ASPECTS CONSIDERED/ CODE
CONCLUSION
PART 1
Cool a suitably weighed drink from one temperature to other
Heat a well-sealed cup of coffee from one temperature to other in a well-
insulated cup holder
Q) Distance needed to be travelled?

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Q) Qc =? ; Qh=?
ASSUMPTIONS
1) A continuous power of 60 W is supplied by the bicyclist travelling at 15km/h
2) The cup holder is well insulated meaning no hear loss by convection
3) When cup drink is cooled, the heat from other hot junction of peltier device is
removed through forced convection
ANALYSIS
Assuming that a beverage is to be cooled from 40° C to 5° C (350ml) and the cup of
coffee is to be heated from 40 C to 60 C(350ml)
Qc = mcΔT …………………….(1)
Qh = mcΔT …………………….(2)
M = ρV = (1kg/L)(0.35L) =0.35 kg
Eq(1) reveals that
Qc = (0.35kg)(4.18kJ/kg C)(35° C)
Qc = 51.205 kJ
Eq(2) reveals that
Qh = (0.35kg)(4.18kJ/kg C) (20° C)
Qh = 29.26 kJ
According to TEC-I 12706 Datasheet, assuming a reasonable COP of 0.5 at 6W
and DT= 40 C
(DT is temperature difference of the two junctions of Peltier device)

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COP = Q/Win = Q/P
Q = P(COP) = 6 (0.5) W
Q = 3 W
Qc
.
= Qc/Δt
Δt = 51.205kJ/3W = 17068s
Δt = 284.5min =4.74 h = t1
Qh
.
= Qh/Δt’
Δt’ = 29.26kJ/3W = 9753 s
Δt’ = 162.56 min = 2.71 h = t2
This shows the cyclist would have to travel long to sufficiently heat or cool a cup of coffee or
beverage respectively.
An average cyclist pedals at about 15.5 km/h (Source: Wikipedia)
 Distance
D1 = 15.5 km/h X 4.74 h = 73.47 km
D2 = 15.5 km/h X 2.71 h = 42.005 km

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Since the time required to cool or heat a drink is not optimum, the alternate method is to
power the peltier device using a battery of sufficient power. The energy by the cyclist can
stored in the battery when he/ she goes for long sessions. This power can later be used to cool
or heat a drink in optimum time.
Power required to cool/heat a drink in optimum time (30min)?
Since
Qc
.
= Qc/Δt
Qc
.
= 51.205kJ/30x60s = 28.45 W
Qh
.
= Qh/Δt’
Qh
.
= 29.26kJ/30x60s = 16.25W
P = Q / COP
Pc = 28.45 x 2 W = 56.9 W
Ph = 16.25 x 2 W = 32.5 W
Therefore if powered by a battery, the battery specifications should atleast be
Energy (Wh) = 60 W x 0.5h = 30 Wh
For a battery of 12V
Battery rating = (30Wh / 12V) x 1000mA = 2500 mAh

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PART 2
If heating a cup of coffee and cooling a cup of water are demanded at the
same time, plot the temperatures of both of the cups with respect to
time/distance covered.
RESULTS AND DISCUSSION
MATHEMATICAL MODELLING
DATA
Cup temperature = 40o
C
Ambient temperature = 40o
C
Required temperature = 5o
C
TH = Hot side temperature = 40 + 10 = 50o
C (for effective heat transfer)
TL = Cold side temperature = 5 - 2 = 3o
C (for effective heat transfer)
Volume of water = 300 mL
Mass of water = ρ.V = 1000 x 0.30 x 10 -3
= 0.3 kg
Specific heat of water (Constant) = 4.186 KJ/Kg.o
C
ΔT = 40 – 5 = 35o
C
Energy to be taken out = Qc = mcΔT = 0.3 x 4.186 x 35 =44 KJ

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ASSUMPTIONS
1. Cup is well insulated.
2. The convection rate is very fast.
3. The heat is transferred to hot side as long and quickly as the heat is absorbed from
cold side.
4. The resistance of the wires are too small.
Newton’s Law of heating / cooling
𝑑𝑇
𝑑𝑡
∝ ∆𝑇
where T = T- Tplate
𝑑𝑇
𝑑𝑡
= 𝑘∆𝑇
We assume that Tplate is constant = 5o
C
𝑑𝑇
∆𝑇
= 𝑘𝑑𝑡
∫
𝑑𝑇
∆𝑇
= 𝑘
∆𝑇
35
∫ 𝑑𝑡
𝑡
0
ln (
∆𝑇
35
) = 𝑘𝑡
∆𝑇 = ∆𝑇 𝑚𝑎𝑥 × 𝑒 𝑘𝑡
When ∆𝑇 𝑚𝑎𝑥 = ∆𝑇𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 35o
C
From experiment from cooling
k = −0.008 /s
∆𝑇 = 35𝑒−0.008𝑡
According to conduction law,
𝑄̇= (k/d) (∆T)
dQ/dT = k(∆T/d)
dQ = k(∆T/d) dt

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∫ 𝑑𝑄
−44000
0
= k/d ∫ ∆T dt
𝑡
0
-44000 = (k/d) ∫ 35 e^ct dt
𝑡
0
-44000 = (k/d)(35/c)(ect
)
Converting 35 o
C into 308 k,
ect
= (-cd (44000))/308
t = (1/c) ln ((-cd(44000))/308)
For cup,
Thickness = d = 2mm = 0.002 m
K for aluminum = 205 w/k
Value of ‘c’ found = -0.008
t = (1/(-0.008)) ln ((-(-0.008)(0.002)(44000))/(308 x 205))
t = (-1/0.008)(-11.404)
t = 1425.52 s
t = 23.76 = 24 min (approx.)
Power required is,
𝑄 𝑐
̇ = Qc /t = (44000/(24 x 60)) = 30.60 w
P = 𝑄 𝑐
̇ /(COP)
Let, we selected COP = 0.5
P = (30/0.5) = 60 w
P = 60 w
Choose model: CP 1.4-127-06
Imax = 6 A Vmax = 16 V
Qc max
. = 50 w
Then at COP = 0.5
I/Imax = 0.8

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I = Imax x 0.8 = (6) (0.8) = 4.8 A
P = VI ⇒ V = (P/I) = (60/4.8) = 12.5
V = 12.5 V
Now, for cooling
T = (∆Tmax)/(e0.008 t
) – ϕ (I2
R)
And for heating,
T = (∆Tmax)/(e0.008 t
) + ϕ (I2
R)
Assumption: At 15.5 km/h dynamo gives 6w,
So, we calculate energy as follow,
Energy required = (60 w) (24) (60) = 86400 J
So, now
6t = 86400 sec ⇒ t = 14400 sec
Now,
V = 15.5 km/h
S = v t = (15.5 x 14400)/(3.6)
⇒ s = 62 km
Hence, impossible.
EXPERIMENTATION
To simulate the output of the thermoelectric peltier device powered by a bicycle dynamo, a
device TEC I 12706 was connected to a GW Instek variable supply which was set at 6W
which is the maximum power output of commercially available models of a bicycle
dynamometer.
Initially, the voltage was set to 12 V and current to 0.5 A. But a problem arose, the resistance
of peltier device at 25 degrees is approx. 2A meaning the device would reach saturation state
at the current supply of the variable power supply dropped to zero. Therefore after certain
trial and errors, the power was adjusted to provide 6W. In actual practice we would require a
Buck converter to step down the voltage of the dynamo and step up the current.

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The experiment shows measurement of surface temperatures of the junctions with time.
Time 0
1
0
2
0
3
0
4
0
5
0
6
0
7
0
8
0
9
0
10
0
11
0
12
0
Temperatur
e
Cold
Junction
1
8
1
0
1
4
1
7
2
0
2
4
2
5
2
9
3
2
3
5 37 40 41
Hot Junction
1
8
4
2
4
7
5
4
6
1
6
5
6
9
7
3
7
7
8
0 84 86 96
As noted in the above plots, the temperature of the cold junction also begins to rise when heat
is not removed from the hot junction. For this some sort of forced convection is not provided
for example a CPU fan.
0
10
20
30
40
50
0 10 20 30 40 50 60 70 80 90 100110120
Temperature Cold Junction
0
20
40
60
80
100
120
0
10
20
30
40
50
60
70
80
90
100
110
120
Temperature Hot Junction

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Time (sec) 0 10 20 30 40 50 60
Temperature 18 8 6 4 8 11 10
A can of water (350ml) was then heated to determine the time it takes to heat water to a
sufficiently temperature. As evident from the experimental results shown below, the power
from a cyclist is not sufficient to boil 350ml of water or would take a very large amount of
time making the process impractical.
Time (min) 0 5 10 15 20 25 30 50 60 65 70
Temperature 21.1 24 26 27.6 28.5 29.5 30.3 32.8 33.6 34 34.3
0
2
4
6
8
10
12
14
16
18
20
0 10 20 30 40 50 60
Temperature of cold junction using heat sink

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In another experiment, a cup of water at 55 degrees was desired to be heated to at least 60
degrees but the heat loss to the air by convention was way higher than that provided by the
peltier device meaning that insulation would be necessary especially considering that fact in
actual application heat would be lost by forced convention.
Time (min) 0 6 9 12 15 18 21 24 27
Temperature 53 50.4 49.7 49.2 48.6 48.1 47.5 47 46.7
0
5
10
15
20
25
30
35
40
0 5 10 15 20 25 30 50 60 65 70
Temperaature
Time
Temperature Heating
43
44
45
46
47
48
49
50
51
52
53
54
1 2 3 4 5 6 7 8 9
Temperaature
Time
Temperature Heating

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The following two experiments show the attempts to cool a beverage 350ml and 250ml
respectively.
Time (min) 1 2 3 4 5 6 7 8 9
Temperature 19.4 18.7 18.3 18 17.9 17.8 17.7 17.6 17.8
Time (min) 0 3 6 10 15 20 25 30
Temperature 28.8 27.8 27.1 26.5 25.9 25.4 25.1 24.8
16.5
17
17.5
18
18.5
19
19.5
20
1 2 3 4 5 6 7 8 9
Temperaature
Time
Temperature Cooling
22
23
24
25
26
27
28
29
30
0 3 6 10 15 20 25 30
Temperaature
Time
Temperature Cooling