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2 INTRODUCTION
The purpose of this project is to analyze a design for and determine the feasibility of a portable 5
watt Stirling engine charger. The final design is driven by solar energy using an acrylic-based solar mirror
and water-based cooling. The heat is transferred to brass cylinders containing the hot pistons of the
Stirling engine. The Stirling cycle is achieved through the expansion of air in the hot cylinders, the
compression of air in the cold cylinders, and a cyclical volume change in each pair of cylinders. For
reference, an ideal Stirling cycle can be seen below in figure 1.
The cylinder volume change is induced through the rotation of a swash-plate, which separates the
hot and cold cylinders by a phase of 90o
. The main user interface is a dry bag on the side of the life ring,
where a phone can be left to charge during use. The main components of the design can be seen in figure
2 below.
Figure 2. Overview of the Solar Stirling Charger. The main components of the design are highlighted with arrows.
Figure 1. An ideal Stirling cycle, which consists of four phases: heating, expansion, cooling, and compression [3].](https://image.slidesharecdn.com/148977f3-5204-441f-b187-3e9faed48778-161221152128/75/Formal-Final-Report-3-2048.jpg)
![4
The engine in the center of the design consists of four pairs of cylinders that are alpha type Stirling
engines. An alpha type Stirling engine has two cylinders that contain the working gas and the hot and cold
pistons, which are separately located in the respective cylinder. It is the simplest configuration of the
Stirling engine, however it has the disadvantage of requiring seals around both of the pistons [1]. An
example of this configuration can be seen in figure 3. The configuration of the designed engine and its
main components can be seen in figure 4. A description of each component can be found in Section 3
Individual Components.
A more complete description of the design and how it works can be found in Section 5 How it Works.
The following sections of this paper will discuss the feasibility of this design, including the analysis of the
system dynamics and thermodynamics, critical failure points, design for manual assembly, costing analysis,
safety, and intellectual property.
Figure 3. Example of an alpha type Stirling engine [1].
Figure 4. Overview of the engine. The main components are highlighted with arrows.](https://image.slidesharecdn.com/148977f3-5204-441f-b187-3e9faed48778-161221152128/75/Formal-Final-Report-4-2048.jpg)
![13
Coupler
Figure 14. Coupler.
The purpose of the coupler is to provide a mechanical connection between the rotating swashplate
drive shaft and the generator shaft. The specific coupler selected is capable of attaching shafts of various
sizes, making it ideal for this system. The coupler is supplied by McMaster. The cost for two coupler hubs
and one spider is $10.03 per assembly.
Hot Cylinder Support
Figure 15. Hot cylinder support.
The hot cylinder support is required to supply stability to several components. It contains
indentations to mate with the generator support in sub-assembly C. The support also contains grooves
designed to accommodate the hot cylinder assembly and align it as needed with the hot pistons. The hot
cylinder support also requires alignment with the regenerator for proper fluid flow, which will be
accomplished using dowel pins. The geometry of the part facilitates the insertion of seals required to
contain the fluid. Upon considering all thermal and mechanical performance requirements for the part,
brass was selected as the optimum material. The part will be cast in order to reduce manufacturing costs.
The hot cylinder support is cast out of brass and weighs 1.27 kg. This component has average complexity
and is manufactured using water glass lost wax investment casting. Additional machining may be
necessary in order to ensure accuracy of the holes. This results in a cost of $13.12. All estimates for brass
casting take into account that brass is more expensive than steel and the price estimator is for steel. The
relative cost is found via the Ashby chart in Fig 6.20 and 6.28 [2].](https://image.slidesharecdn.com/148977f3-5204-441f-b187-3e9faed48778-161221152128/75/Formal-Final-Report-13-2048.jpg)
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𝐼 =
[
1
4
𝑀𝑠 𝑟 2
0 0
0
1
4
𝑀𝑠 𝑟 2
0
0 0
1
2
𝑀𝑠 𝑟 2
]
and 𝜔 𝑠
0
𝐺
is the angular velocity about the vertical axis defined as
𝜔 𝑠
0
𝐺
= [
0
0
𝜃̇
]
In addition to the dynamics of the swashplate, the dynamics of the pistons also needs to be
considered. The pistons move vertically with relation to a Newtonian reference frame and experience
forces due to the pressure inside the vessels. This motion can be described with the kinematics of the
swashplate previously analyzed, however, there is friction involved with this motion. The presence of air-
tight seals generates friction with the surface of the piston rods and reduces the force acting on the
swashplate, diminishing the final torque output.
Figure 32. Piston FBD.
The moments acting on the system directly depend on the pressures inside of the hot cylinders.
The force supplied by the hot cylinders is defined as
𝐹𝑑,𝑖 = (𝑃𝑖 ∗ 𝐴 𝑝 + 𝐹𝑓,𝑖) 𝑢̂1
And the force due to the pressure from the cool system is
𝐹𝑝,𝛿 = (𝑃𝑖,𝛿 ∗ 𝐴 𝑝 − 𝐹𝑓,𝛿) 𝑢̂1
Friction on the piston is to be modeled as viscous [11]
𝐹𝑓 = −(𝜇 ∗ 𝑟 ∗ sin 𝛽 ∗ tan 𝛽 ∗ 𝜃̇ ∗ sin 𝜃) 𝑢̂1
𝐹𝑓 𝐹𝑓
𝑃𝑖 ∗ 𝐴 𝑝 𝑃𝑖 ∗ 𝐴 𝑝
g](https://image.slidesharecdn.com/148977f3-5204-441f-b187-3e9faed48778-161221152128/75/Formal-Final-Report-25-2048.jpg)
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where 𝜇 is the coefficient of friction between the piston and the seals. This value was found to be 0.58
for 304 stainless steel [12].
The rotational friction due to the ball-and-socket joint at the base of the swashplate was modeled as
a linear torque opposing the rotation of the shaft, as shown below.
𝜏 𝑓 = 𝑎 ∗ 𝜃̇ 𝑢̂1
Combining all of the force relations and performing the cross product with their respective
displacements relative to the center of the swashplate, the moments in the three principle coordinates
become,
∑ 𝑀 𝑥
4
1
= 𝑟 ∗ sin 𝜃 ∗ cos 𝛽 ∗ (𝐹𝑝,𝛿 − 𝐹𝑝,𝑖 − 2 ∗ 𝑀 𝑝 ∗ 𝑔 − 2 ∗ 𝐹𝑓) − 𝑎 ∗ sin 𝛽 ∗ 𝜃̇,
∑ 𝑀 𝑦
4
1
= 𝑟 ∗ cos 𝜃 ∗ (𝐹𝑝,𝛿 − 𝐹𝑝,𝑖 − 2 ∗ 𝑀 𝑝 ∗ 𝑔) − 𝐿 ∗ 𝐹𝑦,
∑ 𝑀𝑧
4
1
= 𝑟 ∗ sin 𝜃 ∗ sin 𝛽 ∗ (𝐹𝑝,𝛿 − 𝐹𝑝,𝑖 − 2 ∗ 𝑀 𝑝 ∗ 𝑔 − 2 ∗ 𝐹𝑓) − 𝑎 ∗ cos 𝛽 ∗ 𝜃̇.
Applying these relations to (2), the following is obtained
[
−
1
4
𝑀𝑠 𝑟 2
sin 𝛽 𝜃̈
1
4
𝑀𝑠 𝑟 2
sin 𝛽 𝜃̇2
1
2
𝑀𝑠 𝑟 2
cos 𝛽 𝜃̈
]
= [
𝑟 ∗ cos 𝛽 ∗ (𝐹𝑝,𝛿 − 𝐹𝑝,𝑖 − 2 ∗ 𝑀 𝑝 ∗ 𝑔 − 2 ∗ 𝐹𝑓) ∗ sin 𝜃 − 𝑎 ∗ sin 𝛽 ∗ 𝜃̇
𝑟 ∗ cos 𝜃 ∗ (𝐹𝑝,𝛿 − 𝐹𝑝,𝑖 − 2 ∗ 𝑀 𝑝 ∗ 𝑔) − 𝐿 ∗ 𝐹𝑌
𝑟 ∗ sin 𝛽 ∗ (𝐹𝑝,𝛿 − 𝐹𝑝,𝑖 − 2 ∗ 𝑀 𝑝 ∗ 𝑔 − 2 ∗ 𝐹𝑓) ∗ sin 𝜃 − 𝑎 ∗ cos 𝛽 ∗ 𝜃̇
].
Based on these results, one can obtain a relationship for the angular acceleration of the system.
𝜃̈ =
2 ∗ tan 𝛽 ∗ (𝐹𝑝,𝛿 − 𝐹𝑝,𝑖 − 2 ∗ 𝑀 𝑝 ∗ 𝑔 − 2 ∗ 𝐹𝑓) sin 𝜃 − 𝑎 ∗ 𝜃̇
𝑀𝑠 ∗ 𝑟
(3)
Due to the complexity of the expression, it was evaluated through an iterative process to obtain
the steady-state value for the angular velocity as the process begins. Figure 33 illustrates the result of
these values.](https://image.slidesharecdn.com/148977f3-5204-441f-b187-3e9faed48778-161221152128/75/Formal-Final-Report-26-2048.jpg)
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5.2.2.4 Non-Ideal Stirling Cycle
Now, the pressure and volume can be plotted parametrically, producing the actual P-V diagram
of the Solar Stirling Charger, as seen in figure 39.
Figure 39. P-V diagram for theoretical Stirling cycle.
In the actual Stirling cycle, the heating and cooling processes are no longer constant volume
processes. However, the effective temperatures of the air in the hot, cold cylinders, and the regenerator
are assumed to be constant. To define these temperatures, a heat transfer analysis was conducted as
explained in the following section.
5.2.2.5 Heat Transfer
This section explains the details of the heat transfer analysis of the Solar Stirling Charger in the
following order: First, the heat input is described from the Sun to the hot cylinders followed by a discussion
of the heat losses. Second, the effect of the regenerator on transmitting heat to and from the air is
investigated. Third, the heat dissipation from the cold cylinders is described, followed by the work output
resulting from an energy balance of the Stirling engine.
Solar Intake
On a clear day, the sun shines with the solar flux Gatm of 893 W/m2
, [13]. The effective sky
temperature 𝑇̅𝑠𝑘𝑦 on a clear day is 354 K (177.53 ◦
F) as defined by (&), where 𝜎 is the Stefan-Boltzmann
constant (5.67 × 10−8
𝑊/(𝑚2
∙ 𝐾4
)), [14].
𝐺 𝑎𝑡𝑚 = 𝜎𝑇̅𝑠𝑘𝑦
4
(18)](https://image.slidesharecdn.com/148977f3-5204-441f-b187-3e9faed48778-161221152128/75/Formal-Final-Report-32-2048.jpg)
![33
On such a day, the dome with an outer surface area 𝐴 𝑑𝑜𝑚𝑒 of 0.2530 m2
(392.1 in2
) receives the
following in solar radiation:
𝑞 𝑠𝑜𝑙𝑎𝑟 = 225.9 W
As shown in figure 40, the Sun’s rays strike the dome of the Solar Stirling Charger, and are then
reflected by the Paraboloid which redirects the rays to the dome’s focus, where the hot cylinders stand.
Figure 40. Diagram of solar intake to the hot cylinders.
To calculate how much solar flux is transmitted into the engine is as follows:
For acrylic, the transmissivity 𝜏 𝑎𝑐𝑟𝑦𝑙𝑖𝑐 is on average 90% between 𝜆1 = 0.4 µm and 𝜆2 = 1.1 µm,
as approximated by the similar material in figure 41. Using the Sun as a blackbody emitter, the product of
𝜆𝑇𝑠𝑢𝑛 can be looked up in Table 12.2 of [14], wherein the effective temperature of the sun Tsun is 5800 K,
[14]. After linear interpolation, the fraction of radiation 𝐹(0→𝜆) for each wavelength was figured to be:
𝐹(0→𝜆1) = 0.124509
𝐹(0→𝜆2) = 0.769234
Figure 41. Transmission spectrum of acrylic, [Arkema].
𝑞 𝑠𝑜𝑙𝑎𝑟
𝑞𝑡𝑟𝑎𝑛𝑠
𝑞 𝑟𝑒𝑓
Paraboloid
Dome
Hot
Cylinders](https://image.slidesharecdn.com/148977f3-5204-441f-b187-3e9faed48778-161221152128/75/Formal-Final-Report-33-2048.jpg)
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The fraction of radiation for the specified range of wavelengths 𝐹(𝜆1→𝜆2) is the difference of these
two fractions. The product of this and the acrylic transmissivity gives the total transmissivity [14],
𝜏 𝑡𝑜𝑡𝑎𝑙 = 𝐹(𝜆1→𝜆2) ∗ 𝜏 𝑎𝑐𝑟𝑦𝑙𝑖𝑐 (19)
𝜏 𝑡𝑜𝑡𝑎𝑙 = (0.769234 − 0.124509) ∗ 0.90
𝜏 𝑡𝑜𝑡𝑎𝑙 = 0.5803
Thus, the solar flux transmitted through the dome is [14],
𝑞𝑡𝑟𝑎𝑛𝑠
″
= 𝜏 𝑡𝑜𝑡𝑎𝑙 ∗ 𝐺 𝑎𝑡𝑚 (20)
𝑞𝑡𝑟𝑎𝑛𝑠
″
= 518.2 𝑊/𝑚2
The solar Paraboloid receives 𝑞 𝑝𝑎𝑟𝑎
𝑞 𝑝𝑎𝑟𝑎 = 𝑞𝑡𝑟𝑎𝑛𝑠
″
∗ 𝐴 𝑝𝑎𝑟𝑎 (21)
𝑞 𝑝𝑎𝑟𝑎 = 177.2 𝑊
where 𝐴 𝑝𝑎𝑟𝑎 is the exposed surface area of the Paraboloid, 0.3419 m2
(530 in2
).
The solar Paraboloid reflects 68% of this sunlight, based on Table D from [16]. Though the
manufacturer gave a reflectivity of 90%, the 68% value is being used as a more conservative value.
𝑞 𝑟𝑒𝑓 = 0.68 ∗ 𝑞 𝑝𝑎𝑟𝑎 (22)
𝑞 𝑟𝑒𝑓 = 120.5 𝑊
This 𝑞 𝑟𝑒𝑓 is the heat being directed to the hot cylinders.The temperature of the outside of the hot
cylinders is given by their grey body absorption.
𝑇𝑜𝑢𝑡,ℎ = 𝑞 𝑟𝑒𝑓/(𝜎𝜀 𝑏 𝐴 𝑎𝑏𝑠)1 4⁄
(23)
Where 𝜎 is the Stefan-Boltzmann constant, 𝜀 𝑏 is the emissivity of brass (oxidized) from [18] which
is 0.61, and 𝐴 𝑎𝑏𝑠 is the surface area of the hot cylinders (see figure 42) that is absorbing sunlight.
Figure 42. The geometry of the hot cylinder assembly, wherein cylinder/sun blocker height Lh = 2 in, inner diameter Dih = 0.925
in, cylinder wall thickness th = 0.1 in, cylinder outer diameter Doh = 1.125 in, and sun blocker width Wsb = 0.68 in. The absorbing
area as the combined, exterior surface area of the hot cylinders and the sun blockers connecting them.
Dih
Doh
Lh
Wsb
th
qref
qref
Hot CylinderSun Blocker](https://image.slidesharecdn.com/148977f3-5204-441f-b187-3e9faed48778-161221152128/75/Formal-Final-Report-34-2048.jpg)
![35
Based on the geometry of the cylinders and the sun blockers, as seen in figure 42, the absorbing area is:
𝐴 𝑎𝑏𝑠 = 4[(𝐿ℎ 𝑊𝑠𝑏) + 0.75(𝜋𝐷 𝑜ℎ 𝐿ℎ)] (24)
Thus, the temperature of the outside of the hot cylinders is:
𝑇𝑜𝑢𝑡,ℎ = 709.2 𝐾 (816.89 °𝐹)
Solar Losses
Already, there are two losses accounted for during the radiation portion of the heat input: one
during the transmission of light through the dome, 𝑞𝑡𝑟𝑎𝑛𝑠,𝑙𝑜𝑠𝑠 and another during reflection of light off
the Paraboloid, 𝑞 𝑟𝑒𝑓,𝑙𝑜𝑠𝑠. These are as follows:
𝑞𝑡𝑟𝑎𝑛𝑠,𝑙𝑜𝑠𝑠 = (𝐺 𝑎𝑡𝑚 − 𝑞𝑡𝑟𝑎𝑛𝑠
″ ) ∗ 𝐴 𝑑𝑜𝑚𝑒 (25)
𝒒 𝒕𝒓𝒂𝒏𝒔,𝒍𝒐𝒔𝒔 = 𝟗𝟒. 𝟖 𝑾
Wherein 𝐴 𝑑𝑜𝑚𝑒 is outer surface area of the dome 392.1.
𝑞 𝑟𝑒𝑓,𝑙𝑜𝑠𝑠 = 𝑞 𝑝𝑎𝑟𝑎 − 𝑞 𝑟𝑒𝑓 (26)
𝒒 𝒓𝒆𝒇,𝒍𝒐𝒔𝒔 = 𝟓𝟔. 𝟕 𝑾
Conduction and Convection Losses
Heat leaving the hot cylinders to the ambient conditions directly outside of the Paraboloid (assumed
to be 8 degrees above Tsky, based on [19]) is controlled by the following thermal resistances:
Conduction resistance of air under the dome, 𝑅𝑡,𝑐𝑜𝑛𝑑,𝑝𝑎𝑟𝑎,𝑎𝑖𝑟.
Conduction resistance of the Paraboloid 𝑅𝑡,𝑐𝑜𝑛𝑑,𝑝𝑎𝑟𝑎,𝑎𝑐𝑟𝑦𝑙𝑖𝑐 and the dome, 𝑅𝑡,𝑐𝑜𝑛𝑑,𝑑𝑜𝑚𝑒 .
Convection resistance of the wind, 𝑅𝑡,𝑐𝑜𝑛𝑣,𝑤𝑖𝑛𝑑.
The overall resistance of the losses 𝑅𝑡,𝑙𝑜𝑠𝑠 is given as
𝑅𝑡,𝑙𝑜𝑠𝑠 = 𝑅𝑡,𝑐𝑜𝑛𝑑,𝑝𝑎𝑟𝑎,𝑎𝑖𝑟 + (𝑅𝑡,𝑐𝑜𝑛𝑑,𝑝𝑎𝑟𝑎,𝑎𝑐𝑟𝑦𝑙𝑖𝑐
−1
+ 𝑅𝑡,𝑐𝑜𝑛𝑑,𝑑𝑜𝑚𝑒
−1
)
−1
+ 𝑅𝑡,𝑐𝑜𝑛𝑣,𝑤𝑖𝑛𝑑 (27)
Losses through Parabolic Resistances
The air space underneath the dome was approximated to be one-dimensional conduction through
a Paraboloid of the equation
𝑧 𝑝𝑎𝑟𝑎 =
𝑟𝑝𝑎𝑟𝑎
2
2𝑏
(28)
Where z is the height of the Paraboloid, r is the radial distance from the central axis, and b is the
constant 7.1 in.](https://image.slidesharecdn.com/148977f3-5204-441f-b187-3e9faed48778-161221152128/75/Formal-Final-Report-35-2048.jpg)
![36
Using the radial version of Fourier’s law of conduction from [14],
𝑞 𝑟 = −𝑘𝐴 (
𝑑𝑇
𝑑𝑟
) (29)
Where 𝑞 𝑟 is the radial heat transfer, k is the conductivity, A is the surface area normal to the
heat transfer, and
𝑑𝑇
𝑑𝑟
is the radial temperature gradient.
With the surface area of a Paraboloid 𝐴 𝑝𝑎𝑟𝑎 given by
𝐴 𝑝𝑎𝑟𝑎 = ((
𝑟
𝑏
)
2
+ 1)
3
2
(30)
Wherein
𝑎 =
2(7.12
𝜋)
3
𝑖𝑛2
After rearranging substituting (30) into (29) and solving for the heat transfer, the following
relationship is derived:
𝑞 𝑟,𝑝𝑎𝑟𝑎 =
𝑘𝑎(𝑇1 − 𝑇2)
𝐼
(31)
Where I is the integral
𝐼 𝑝𝑎𝑟𝑎 = ∫
𝑑𝑟
((
𝑟
𝑏
)
2
+ 1)
3
2
𝑟2
𝑟1
(32)
Which solves to
𝐼 𝑝𝑎𝑟𝑎 =
𝑟
√(
𝑟
𝑏
)
2
+ 1
]
𝑟2
𝑟1 (33)
Based on (31), conduction thermal resistance of the Paraboloid is given as
𝑅𝑡𝑐𝑜𝑛𝑑,𝑝𝑎𝑟𝑎 =
𝐼
𝑘𝑎
(34)
This formula can be applied to both the conduction resistance of the air and the acrylic Paraboloid
as such:
TABLE VII
THERMAL RESISTANCES OF PARABOLOIDS
MEDIUM
THERMAL
CONDUCTIVITY,
K, (W/(M∙K))
RPARA, 1 , (IN) RPARA,2 , (IN)
THERMAL
RESISTANCE, RT,
(K/W)
Air at 600K 46.9 x 10-3
10.916 11.084 33.29
Acrylic Paraboloid 0.2 1.8225 10.916 0.05](https://image.slidesharecdn.com/148977f3-5204-441f-b187-3e9faed48778-161221152128/75/Formal-Final-Report-36-2048.jpg)
![38
Losses due to Wind
Based on grey body emission, the temperature of the outside of the dome 𝑇𝑜𝑢𝑡,𝑑𝑜𝑚𝑒 is
𝑇𝑜𝑢𝑡,𝑑𝑜𝑚𝑒 = (
𝐺 𝑎𝑡𝑚
𝜎𝜀 𝑑𝑜𝑚𝑒
+ 𝑇𝑠𝑘𝑦
4
)
1
4
(38)
𝑇𝑜𝑢𝑡,𝑑𝑜𝑚𝑒 = 427 𝐾
Where 𝜀 𝑑𝑜𝑚𝑒 is the emissivity of the acrylic dome which is 0.9 as given by [14].
From [14], the convection heat transfer coefficient due to wind ℎ 𝑤𝑖𝑛𝑑 is
ℎ 𝑤𝑖𝑛𝑑 = 0.22(𝑇𝑜𝑢𝑡,𝑑𝑜𝑚𝑒 − 𝑇𝑎𝑚𝑏)
1 3⁄
(39)
ℎ 𝑤𝑖𝑛𝑑 = 0.8835
𝑊
𝑚2 ∙ 𝐾
And the convection resistance of the wind is, thus, given by
𝑅𝑡,𝑐𝑜𝑛𝑣,𝑤𝑖𝑛𝑑 = 1/(ℎ 𝑤𝑖𝑛𝑑 𝐴 𝑑𝑜𝑚𝑒) (40)
𝑅𝑡,𝑐𝑜𝑛𝑣,𝑤𝑖𝑛𝑑 = 4.4747 𝐾/𝑊
Altogether, the resistance of the heat losses is
𝑅𝑡,𝑙𝑜𝑠𝑠 = 37.8 𝑊/𝐾
The heat loss 𝑞𝑙𝑜𝑠𝑠 from the hot cylinders to the ambient can be given by
𝑞𝑙𝑜𝑠𝑠 = (𝑇𝑜𝑢𝑡,ℎ − 𝑇𝑎𝑚𝑏)/𝑅𝑡,𝑙𝑜𝑠𝑠 (41)
𝒒𝒍𝒐𝒔𝒔 = 𝟗. 𝟏𝟕𝟗𝟓
Now, all the heat losses can be summarized in the table below.
TABLE VIII
SUMMARY OF HEAT LOSSES
LOSS TYPE TRANSMISSION LOSS REFLECTION LOSS CONDUCTION AND CONVECTION
Thermal Loss, W 94.8 56.7 9.2
Thus, taking the conduction and convection losses into account, the net heat input 𝑞 𝑛𝑒𝑡,𝑖𝑛 to the
hot cylinder assembly’s surface is
𝑞 𝑛𝑒𝑡,𝑖𝑛 = 𝑞 𝑟𝑒𝑓 − 𝑞𝑙𝑜𝑠𝑠 (42)
𝑞 𝑛𝑒𝑡,𝑖𝑛 = 111.3 𝑊
Dividing this values equally among the four hot cylinders gives, 𝑞𝑖𝑛,ℎ, which is ultimately the heat
input into each separate pair of cylinders.
𝒒𝒊𝒏,𝒉 = 𝟐𝟕. 𝟖 𝑾](https://image.slidesharecdn.com/148977f3-5204-441f-b187-3e9faed48778-161221152128/75/Formal-Final-Report-38-2048.jpg)
![39
Figure 45. The geometry of the hot cylinder assembly, wherein Lh = 2 in, Dih = 0.925 in, th = 0.1 in and Doh = 1.125 in, Wsb = 0.68
in.
Heat Transfer through the Hot Cylinder
Assuming one-dimensional conduction through a cylinder, the thermal resistance of each hot cylinder is
𝑅𝑡,𝑐𝑜𝑛𝑑,ℎ =
ln (
𝐷 𝑜ℎ
𝐷𝑖ℎ
)
2𝜋𝑡ℎ 𝑘 𝑏
(43)
𝑅𝑡,𝑐𝑜𝑛𝑑,ℎ = 0.1115 𝐾/𝑊
Where 𝑘 𝑏 is conductivity of the brass hot cylinders (110 W/mK) from [14] and the hot cylinder
geometry is from figure 45.
Once the heat transverses the wall thickness of the hot cylinder, it encounters the forced
convection thermal resistance of the moving air inside. This thermal resistance is dependent on velocity
of the air, which has been approximated to the velocity of the hot side’s piston, 𝑢 𝑝𝑖𝑠𝑡𝑜𝑛(𝜃).
𝑢 𝑝𝑖𝑠𝑡𝑜𝑛(𝜃) = 𝑟(tan𝛽)(𝑠𝑖𝑛𝜃)𝜔 (44)
Wherein 𝑟 is the distance vector between the center of the drive shaft and the piston rod, 𝛽 is the
tilt of the swashplate, and 𝜃 and 𝜔 are the rotational angle and velocity of the drive shaft, respectively.
These variables are shown in figure 46.
Dih
Doh
Lh
Wsb
th
qnet,in
qnet,in
Hot CylinderSun Blocker](https://image.slidesharecdn.com/148977f3-5204-441f-b187-3e9faed48778-161221152128/75/Formal-Final-Report-39-2048.jpg)
![40
Figure 46. Swashplate and piston geometry.
Since the velocity of the piston is sinusoidal, the root-mean-square of the velocity is used to
represent the fluid velocity in the hot cylinder, 𝑢ℎ. At rotational speed achieved on a “clear day,” (2590
RPM), the air velocity is:
𝑢ℎ = 21.4 𝑚/𝑠
Furthermore, the Reynold’s number of the hot side air 𝑅𝑒 𝐷,ℎ is thus, using the thermal
properties of air at 600K, which are listed in Table IX [14]:
𝑅𝑒 𝐷,ℎ = 𝑢ℎ 𝐷𝑖ℎ/𝜈ℎ (45)
𝑅𝑒 𝐷,ℎ = 9525.7
Where Dih is the inner diameter of the hot cylinder and 𝜈ℎ is the kinematic viscosity of the hot air.
TABLE IX
THERMAL PROPERTIES OF AIR AT 600K
DENSITY,
𝛒 𝐡 (
𝐤𝐠
𝐦 𝟑)
DYNAMIC
VISCOSITY,
𝛍 𝐡 (𝐍 ∙ 𝐬/𝐦 𝟐
)
KINEMATIC
VISCOSITY,
𝛎 𝐡 (𝐦 𝟐
/𝐬)
PRANDTL
NUMBER,
Prh
THERMAL
CONDUCTIVITY,
𝐤 𝐡 (W/M K)
SPECIFIC
HEAT, 𝒄 𝒑,𝒉
(J/(KG K))
0.5804 305.8 x 10-7
52.69 x 10-6
0.685 46.9 x 10-3
1051
The Dittus-Boelter equation is used to determine the non-dimensional heat transfer coefficient,
the Nusselt number, of the hot cylinder flow [14]. The flow was assumed to be fully developed and
turbulent despite the Reynold’s number being slightly below the turbulence realm (𝑅𝑒 𝐷 ≥ 10,000).
𝑁𝑢 𝐷,ℎ = 0.023𝑅𝑒 𝐷
4 5⁄
Pr0.4
𝑟
𝛽
𝜃, 𝜔
Swashplate
Drive Shaft
Hot Piston](https://image.slidesharecdn.com/148977f3-5204-441f-b187-3e9faed48778-161221152128/75/Formal-Final-Report-40-2048.jpg)
![42
The surface temperature of the regenerator will play a large role in its performance, therefore
one must first consider its surface temperature due to the thermal radiation contained within the
Paraboloid-dome system.
𝑇𝑠𝑠 =
𝑞 𝑟𝑒𝑓
𝜎 ∗ 𝜀 𝑠𝑠 ∗ 𝑆𝐴 𝑟𝑒𝑔𝑒𝑛
(49)
The nature of the system dictates that convection is expected to be the dominant mode of heat
transfer in the system. To determine the convective thermal resistance, one must obtain the coefficient
of convective heat transfer, shown below.
ℎ 𝑟 =
𝑁𝑢 𝐷𝑟 ∗ 𝑘ℎ
𝐷𝑡𝑢𝑏𝑒
(50)
The geometry of the regenerator is already known, as is the conductance of the fluid, therefore
one only needs to determine the Nusselt number, the Dittus-Boelter relation for cooling was selected
based on the flow conditions [14].
𝑁𝑢 𝐷𝑟 = 0.023 ∗ 𝑅𝑒 𝐷𝑟
4
5
∗ 𝑃𝑟ℎ
0.3
(51)
The Reynolds for this flow can me calculated using the
𝑅𝑒 𝐷𝑟 =
4 ∗ 𝑚̇
𝜋 ∗ 𝐷𝑡𝑢𝑏𝑒 ∗ 𝜇ℎ
(52)
For calculating the Reynolds number the mass flow rate can be obtained based on fluid properties
and the fluid velocity based on the angular velocity of the system,
𝑚̇ = 𝜌ℎ ∗ 𝑢ℎ ∗ 𝐴 𝑡𝑢𝑏𝑒.
Using these relations, the coefficient of convective heat transfer can be obtained using 50 and the
convective thermal resistivity calculated as follows
𝑅𝑡,𝑐𝑜𝑛𝑣,𝑟 =
1
2 ∗ ℎ 𝑟 ∗ 𝜋 ∗ 𝐷𝑡𝑢𝑏𝑒 ∗ 𝐿 𝑡𝑢𝑏𝑒
(53)
The system specifics yield a value of 5.79 K/W.
The conductive thermal resistivity of the regenerator is calculated via the following relation,
approximating the pathway of the fluid to be cylindrical
𝑅𝑡,𝑐𝑜𝑛𝑑,𝑟 =
ln
𝑡 𝑟𝑒𝑔𝑒𝑛 + 𝐷𝑡𝑢𝑏𝑒
𝐷𝑡𝑢𝑏𝑒
2 ∗ 𝑘 𝑟 ∗ 𝜋 ∗ 𝐿 𝑡𝑢𝑏𝑒
(54)
This resistivity results in a value of 0.10 K/W.](https://image.slidesharecdn.com/148977f3-5204-441f-b187-3e9faed48778-161221152128/75/Formal-Final-Report-42-2048.jpg)
![43
The total heat flux across the regenerator shell can be expresses as
𝑞 𝑟 =
(𝑇ℎ − 𝑇𝑠𝑠)
(𝑅𝑡,𝑐𝑜𝑛𝑑 + 𝑅𝑡,𝑐𝑜𝑛𝑣)
(55)
yielding a total flux of 3.54 W/m2
across the regenerator.
Based on this analysis, the fluid temperature at the cool end of the regenerator can be calculated
with the following relation
𝑇𝑐 =
−𝑞 𝑟
𝑚̇ ∗ 𝐶 𝑝,ℎ
+ 𝑇ℎ
(56)
Therefore, the cool temperature across the regenerator is expected to be 528K.
Heat Rejection
The following section discusses the heat flow when the air enters the cool cylinder and is cooled
by the water that the Stirling Solar Charger is suspended in.
Once the effective temperature of the air in the cool cylinder is known, the heat rejected from the
system can be calculated following the reverse method that was applied to the hot cylinder. First, internal
convective heat transfer can be applied to get a convective thermal resistance. Using the geometry in
figure 48, the thermal properties shown in Table X [14] assuming the velocity of the air in the cool cylinder
𝑢 𝑐 is the same as that of the air in the hot cylinder, 𝑢ℎ, the Reynold’s 𝑅𝑒 𝐷,𝑐 number can calculated as
such:
𝑅𝑒 𝐷,𝑐 = 𝑢 𝑐 𝐷𝑖𝑐/𝜈𝑐 (57)
𝑅𝑒 𝐷,𝑐 = 12,599
Where 𝐷𝑖𝑐 is the inner diameter of the cool cylinder (0.9 in) and 𝜈𝑐 is the kinematic viscosity.
Once again, using the Dittus-Boelter equation (for cooling), the Nusselt number of the cool
cylinder air can be calculated [14]
𝑁𝑢 𝐷,𝑐 = 0.023𝑅𝑒 𝐷,𝑐
4 5⁄
Prc
0.3 (58)
𝑁𝑢 𝐷,ℎ = 39.1
Then, the convection heat transfer coefficient ℎ 𝑐 can be found:
ℎ 𝑐 = 𝑁𝑢 𝐷,𝑐 𝑘 𝑐/𝐷𝑖𝑐 (59)
ℎ 𝑐 = 45.0 𝑊/𝑚2
𝐾
and then the convection resistance 𝑅𝑡,𝑐𝑜𝑛𝑣,𝑐 can be found:
𝑅𝑡,𝑐𝑜𝑛𝑣,𝑐 = 1/(ℎ 𝑐 𝜋𝐷𝑖𝑐 𝐿 𝑐) (60)
𝑅𝑡,𝑐𝑜𝑛𝑣,𝑐 = 8.12 𝐾/𝑊
Where 𝜋𝐷𝑖𝑐 𝐿 𝑐 is the inner surface area of the cool cylinders.](https://image.slidesharecdn.com/148977f3-5204-441f-b187-3e9faed48778-161221152128/75/Formal-Final-Report-43-2048.jpg)
![44
Figure 48. Geometry of the cool cylinder assembly: Lc = 1.5 in, Dic = 0.9 in, Doc = 1.125 in, and tc = (Doc-Dic)/2.
TABLE X
THERMAL PROPERTIES OF AIR AT 500K
DENSITY,
𝛒 𝐜 (𝐤𝐠/𝐦 𝟑
)
DYNAMIC
VISCOSITY,
𝛍 𝐜 (𝐍 ∙ 𝐬/𝐦 𝟐
)
KINEMATIC
VISCOSITY,
𝛎 𝐜 (𝐦 𝟐
/𝐬)
PRANDTL
NUMBER,
PrC
THERMAL
CONDUCTIVITY,
𝐤 𝐜 (W/M K)
SPECIFIC
HEAT, 𝒄 𝒑,𝒄
(J/(KG K))
0.6964 270.1 x 10-7
38.7 x 10-6
0.684 26.3 x 10-3
1040
The conduction resistance of the cold cylinders, which are made of stainless steel (thermal
conductivity, kss = 15.1 W/m K, [14].
𝑅𝑡,𝑐𝑜𝑛𝑑,𝑐 =
ln (
𝐷 𝑜𝑐
𝐷𝑖𝑐
)
2𝜋𝑡 𝑐 𝑘 𝑠𝑠
(61)
𝑅𝑡,𝑐𝑜𝑛𝑑,𝑐 = 0.8231 𝐾/𝑊
Where 𝐷 𝑜𝑐, 𝐷𝑖𝑐, and 𝑡 𝑐 are the geometric properties of the cold cylinder.
On the outside of the cold cylinder heat is being dissipated via natural convection. The following
relation can be used to obtain the Nusselt number for natural convection of a short, vertical cylinder, [20].
𝑁𝑢 𝐷,𝑤 =
4
3
[
7𝐺𝑟 𝐷,𝑤 𝑃𝑟𝑤
2
5(20 + 21Prw)
]
1 4⁄
+
4(272 + 315Prw)
35(64 + 63Pr 𝑤)𝐷 𝑜𝑐
(62)
Where 𝐺𝑟 𝐷,𝑤 is the Grashof number of the cooling water about the cool cylinder. Technically, the
Grashof number should be calculated with the temperature of the outer surface of the cylinder but since
this value is unknown, the effective temperature of the cool cylinders 𝑇𝑐 is used instead. The thermal
properties of water listed in Table XI are based on sea water at 17°C (290K). For simplicity, these properties
were used despite the calculated water temperature, which is calculated as such with respect to the solar
flux.
𝐷𝑖𝑐
𝐷 𝑜𝑐
𝐿 𝑐
𝑡 𝑐
Cool
Cylinder
Support
Cool
Cylinder
qout
qout](https://image.slidesharecdn.com/148977f3-5204-441f-b187-3e9faed48778-161221152128/75/Formal-Final-Report-44-2048.jpg)
![45
𝑇 𝑤𝑎𝑡𝑒𝑟 = (𝐺 𝑎𝑡𝑚/𝜎𝜀 𝑤)1/4
(63)
For clear day conditions and with water having an emissivity of 𝜀 𝑤 = 0.97, [14].
𝑇 𝑤𝑎𝑡𝑒𝑟 = 356.97 𝐾
TABLE XI
THERMAL PROPERTIES OF WATER AT 290K
SPECIFIC VOLUME,
𝐯 𝐰 (𝐦 𝟑
/𝐤𝐠)
DYNAMIC
VISCOSITY,
𝛍 𝐰 (𝐍 ∙ 𝐬/𝐦 𝟐
)
KINEMATIC
VISCOSITY,
𝛎 𝐰 (𝐦 𝟐
/𝐬)
COEFFICIENT
OF
THERMAL
EXPANSION,
𝜷 𝒘
PRANDTL
NUMBER,
PRW
THERMAL
CONDUCTIVITY,
𝐤 𝐰 (W/M K)
1.001 x 10-3
1.080 x 10-3
1.0811 x 10-6
1.740 x 10-4
7.56 0.598
Now, with this temperature, thermal properties of water, and acceleration due to gravity g as
9.81 m2
/s, the Grashof number 𝐺𝑟 𝐷,𝑤 can be calculated as follows:
𝐺𝑟 𝐷,𝑤 = 𝑔𝛽 𝑤(𝑇𝑐 − 𝑇 𝑤)/𝜈 𝑤 (64)
𝐺𝑟 𝐷,𝑤 = 5.868 × 106
Now, the Nusselt number of the water can be calculated using (62):
𝑁𝑢 𝐷,𝑤 = 53.8
Then, the heat transfer coefficient of the natural convection can be formulated:
ℎ 𝐷,𝑤 = 𝑁𝑢 𝑤 𝑘 𝑤/𝐷 𝑜𝑐 (65)
ℎ 𝐷,𝑤 = 1.1265 × 103
𝑊/𝑚2
𝐾
From here, the convection thermal resistance of naturally cooling water 𝑅𝑡,𝑐𝑜𝑛𝑣,𝑤 can be
calculated with:
𝑅𝑡,𝑐𝑜𝑛𝑣,𝑤 = 1/(ℎ 𝑤 𝜋𝐷 𝑜𝑐 𝐿 𝑐) (66)
𝑅𝑡,𝑐𝑜𝑛𝑣,𝑤 = 0.2595 𝐾/𝑊
Now, that all three thermal resistances of the cool side have been found (convection inside the
cool cylinder 𝑅𝑡,𝑐𝑜𝑛𝑣,𝑐, conduction through the cool cylinder 𝑅𝑡,𝑐𝑜𝑛𝑑,𝑐, and natural convection around the
cool cylinder 𝑅𝑡,𝑐𝑜𝑛𝑣,𝑤), the heat rejection of a single cool cylinder 𝑞 𝑜𝑢𝑡,𝑐 can be calculated:
𝑞 𝑜𝑢𝑡,𝑐 = (𝑇𝑐 − 𝑇 𝑤𝑎𝑡𝑒𝑟)/(𝑅𝑡,𝑐𝑜𝑛𝑣,𝑐 + 𝑅𝑡,𝑐𝑜𝑛𝑑,𝑐 + 𝑅𝑡,𝑐𝑜𝑛𝑣,𝑤) (67)
𝑞 𝑜𝑢𝑡,𝑐 = 18.72 𝑊
The total heat rejection 𝑞 𝑜𝑢𝑡 is four times this amount to account for all the cool cylinders:
𝒒 𝒐𝒖𝒕 = 𝟕𝟒. 𝟖𝟕 𝑾](https://image.slidesharecdn.com/148977f3-5204-441f-b187-3e9faed48778-161221152128/75/Formal-Final-Report-45-2048.jpg)
![48
𝑆 𝑒 = 𝑘 𝑎 𝑘 𝑏 𝑆 𝑒
′
= (0.883)(0.9)(0.5)(68) = 27𝑘𝑝𝑠𝑖
DE Goodman equation to find smaller shaft diameter:
𝑑 = {
16𝑛
𝜋
(
2(𝐾𝑓 𝑀 𝑎)
𝑆 𝑒
+
[3(𝐾𝑓𝑠 𝑇 𝑚)2
]
1/2
𝑆 𝑢𝑡
)}
1/3
(69)
𝑛 = 1.5, 𝑘 𝑓 = 1.7, 𝑀 𝑎 = 𝑇 𝑚 = 0.4425 𝑙𝑏𝑓 ∗ 𝑖𝑛, 𝑆 𝑒 = 27𝑘𝑝𝑠𝑖, 𝑆 𝑢𝑡 = 68 𝑘𝑝𝑠𝑖
𝑑 = {
16(1.5)
𝜋
(
2(1.7)(0.4425)
27000
+
[3(1.5)(0.4425)2]1/2
68000
)}
1/3
= .08217𝑖𝑛
Next standard size up is chosen (.09375 in.)
For a
𝐷
𝑑
≈ 1.2, 𝐷 = .125 𝑖𝑛 (standard size)
. 125
. 09375
= 1.33
Fillet radius
𝑟 =
𝑑
10
= .01 𝑖𝑛,
𝑟
𝑑
= 0.11 𝑖𝑛
𝑘 𝑓 = 1 + 𝑞(𝑘 𝑡 − 1)
𝑘 𝑓𝑠 = 1 + 𝑞 𝑠(𝑘 𝑡𝑠 − 1)
Using Figure A-15-9, A-15-8
𝑘 𝑡 = 1.6, 𝑘 𝑡𝑠 = 1.4
Using Figure 6-20
𝑞 = 0.5, 𝑘 𝑓 = 1 + 0.5(1.6 − 1) = 1.3
Using Figure 6-21
𝑞 𝑠 = 0.55, 𝑘 𝑓𝑠 = 1 + 0.55(1.4 − 1) = 1.22
Now must adjust 𝑘 𝑎 and 𝑘 𝑏 values:
𝑘 𝑎 = 0.883
𝑘 𝑏 = 1.24𝑑−0.107
= 1.24(. 09375)−0.107
= 1.597](https://image.slidesharecdn.com/148977f3-5204-441f-b187-3e9faed48778-161221152128/75/Formal-Final-Report-48-2048.jpg)
![77
8 COST ANALYSIS
The landed cost is totaled to be $832.22 and can be used along with the direct labor, indirect labor,
burden labor, and facility expenses to approximate the total cost per engine if 8,000 engines are
manufactured per month. In order to ensure an even number of workstation production, the actual
number of engines produced per month is 8,104. For this analysis, one shift that consists of 160 hours per
month per workstation will be used.
Based on the production layout discussed previously, the facility expenses are approximated. The
calculated assembly time of fifteen minutes per engine is used to approximate the number of work
stations necessary. The electricity cost is estimated based on the average commercial rate in Florida [7].
The cost per square foot is found to be $7.50 per sq. ft. per year [8] and the millage rate is found to be
approximately $24.55/1k$/year [9]. The price for umbrella insurance is estimated to be $3,000. The
product liability rate is determined based on the average for a new business [10]. All other values are
decided based on experience.
TABLE XIII
FACILITY EXPENSES
ITEM VALUE UNITS
Sq. ft. per workstation 200 (ft2
/WS)
Number of workstations 13 (WS)
Sq. ft. for workstation 2600 (ft2
)
Indirect sq. ft. factor 100 (%)
Total sq. ft. (raw) 5200 (ft2
)
Total sq. ft. 5700 (ft2
)
Sq. ft. cost 7.50 ($/ft2
/year)
Electricity cost 0.08 ($/kwh)
Power per workstation 300 (W/WS)
Brazing power 10 (kW)
AC power 35 (kW)
Cost for lights 100.21 ($/month)
Cost for AC/heat 449.68 ($/month)
Brazing power 256.96 ($/month)
Value of property 200000.00 ($)
Millage rate 24.55 ($/1k$/year)
Umbrella insurance rate 3000.00 ($/year)
Product liability rate 17 (%/year)
Estimated annual sales 0.19 ($M/year)
Rent 3562.50 ($/month)
Electricity 806.85 ($/month)
Phone/internet 500.00 ($/month)
Water 300.00 ($/month)
Waste 500.00 ($/month)
License 100.00 ($/month)
Property text 409.17 ($/month)
Security 200.00 ($/month)
Insurance 2970.00 ($/month)
Supplies 1000.00 ($/month)
Total 10348.52 ($/month)](https://image.slidesharecdn.com/148977f3-5204-441f-b187-3e9faed48778-161221152128/75/Formal-Final-Report-77-2048.jpg)
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10 EVALUATION
Based on the analysis, the product meets the specifications determined in the preliminary phase of
the design process, i.e., a portable 5 W Stirling engine powered USB charger. These specifications can be
seen in Section 3 Performance Specifications.
From the solar input (about 225 W based on the calculation above in Section 5), the solar to thermal
efficiency is 16.2%. The thermal to mechanical efficiency is 44.6%, which results in 16.27 W of power from
the engine. The mechanical to electrical efficiency is 50%, creating a final electrical output of 8.11 W.
Based on these efficiencies, the design has a 3.6% overall efficiency. This exceeds the 5 W specification
for the device. It is thought that the excess power can be used to charge additional devices, remain as a
safety buffer (for charging in low light/cloudy conditions), or allow the design to be reduced/optimized in
further prototypes.
The final weight of the assembly is approximately 45 lbs., and it measures 25 in. in diameter. This
puts the device within the parameters of man portable according to MIL-STD 1472D [4] and can be carried
with two hands. This weight is considered to be fatiguing for a single person to carry, but the user would
only have to physically carry it a short while, as the device is buoyant. Nevertheless, weight reduction is a
point for future optimization.
The stress analysis of the system shows that the factor of safety is higher than 4 on the critical
components, which will provide a long fatigue life and leaves room for optimization. Fatigue life due to
heating, solar radiation damage, salt water corrosion, and impact strength have not been considered and
should be added in further evaluations.
When producing 8 000 units, final cost of the device comes to $845.14 with an overhead of 110%.
The sale of this product is primarily for high volume orders (e.g. FEMA or cruise lines) so this is not an
unreasonable price range. The ideas used in this design are open to the possibility of patents which secure
sales of this device from copycat devices on the market.
Overall the device is a viable product and is ready for optimization and testing in further prototyping
stages.](https://image.slidesharecdn.com/148977f3-5204-441f-b187-3e9faed48778-161221152128/75/Formal-Final-Report-81-2048.jpg)
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12 REFERENCES
[1] Israel Urieli, “Chapter 2a – Alpha Stirling Engines”,
https://www.ohio.edu/mechanical/stirling/engines/engines.html
[2] M. Ashby, “Materials Selection in Mechanical Design”, Butterworth-Heinemann
[3] Wikipedia, “Stirling Cycle Color”, https://commons.wikimedia.org/wiki/File:Stirling_Cycle_color.png
[4] U.S. Military, “MIL-STD 1472D”
[5] Mike Griffis, “handling-times1.pdf”
[6] Mike Griffis, “insertion-times1.pdf”
[7] US Energy Information Administration, “Electric Power Monthly”,
https://www.eia.gov/electricity/monthly/epm_table_grapher.cfm?t=epmt_5_6_a
[8] CoStar Group, Inc., “Loopnet”,
http://www.loopnet.com/Listing/19531124/13301-US-HWY-441-Alachua-FL/
[9] Florida Tax Watch, “Report and Recomendations of the Florida Tangible Personal Property Tax Task
Force”, http://www.floridataxwatch.org/resources/pdf/20111128tpptaskforce.pdf
[10] AIG, “Product Liability Insurance”,
http://www.aig.com/producer-compensation/casualty-and-liability/product-liability-insurance
[11] Anil Rao, “Dynamics of Particles and Rigid Bodies”, published 2011
[12] British Stainless Steel Association, “Frictional Properties of Stainless Steels”
www.bssa.org.uk/topics.php?article=99
[13] C. Johnson. “Solar Energy - How Much Energy Comes From the Sun.” Internet:
http://mb-soft.com/public2/energyso.html, Jan. 2, 2016. [Mar. 25, 2016].
[14] T. L. Bergmann, A. S. Lavine, F. P. Incropera, and D. P. Dewitt. Fundamentals of Heat and Mass
Transfer, 8th
ed. Hoboken, NJ: John Wiley & Sons, 2011.
[15] Arkema. “Spectral Transmission Plexiglas V044” Internet:
http://www.plexiglas.com/en/acrylic-resins/optical-and-weathering-properties/, Nov. 20, 2015. [Mar.
30, 2016].
[16] Cyro Industries. “Acrylite Light Transmission and Reflectance.” Internet:
https://www.tapplastics.com/uploads/pdf/acrylite%20light%20transmission.pdf, May 25, 2014. [Mar.
30, 2016].](https://image.slidesharecdn.com/148977f3-5204-441f-b187-3e9faed48778-161221152128/75/Formal-Final-Report-97-2048.jpg)
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[17] Patriot Solar Group. “Solar Reflective Material.” Internet:
http://www.patriotsolargroup.com/dataSheets/SolarReflectiveMaterial.pdf, Jun. 8, 2010. [Mar. 30,
2016].
[18] Auburn. “Heat Transfer: Radiation.” Internet:
http://www.auburn.edu/academic/classes/matl0501/coursepack/radiation/text.htm, Jul. 21, 2010.
[Apr. 12, 2016].
[19] R. Forristall. (2003, Oct.). “Heat Transfer Analysis and Modeling of a Parabolic Trough Solar Receiver
Implemented in Engineering Equation Solver,” National Renewable Energy Laboratory.
[20] B. Bebhart, Y. Jaluria, R. L. Mahajan, and B. Sammakia. “Vertical Axisymmetric Flows” in Buoyancy
Induced Flows and Transport, vol. 2. S. Tamburrino and M. Prescott, New York: Hemisphere Publishing
Company, 1988, pp. 133 – 159.](https://image.slidesharecdn.com/148977f3-5204-441f-b187-3e9faed48778-161221152128/75/Formal-Final-Report-98-2048.jpg)
Formal Final Report
- 1. Solar Stirling Charger SparkeIndustries, Group 7 “All it takes is an idea." Contributors: Ben Gajus, Juan Pablo Lopez, Allison Johnson, Nicholas Natale, Kendall Wade, Samantha Webster
- 2. 2 1 TABLE OFCONTENTS 2 Introduction..................................................................................................................................................3 3 Performance Specifications..........................................................................................................................5 4 Individual Components.................................................................................................................................6 4.1 Sub-Assembly A....................................................................................................................................7 4.2 Sub-Assembly B..................................................................................................................................11 4.3 Sub-Assembly C..................................................................................................................................18 5 How it Works..............................................................................................................................................21 5.1 Nominal Use.......................................................................................................................................21 5.2 Analysis...............................................................................................................................................22 5.2.1 Kinematics and Dynamics ..........................................................................................................23 5.2.2 Thermodynamics and Heat Transfer .........................................................................................27 5.3 Failure.................................................................................................................................................47 5.3.1 Theoretical Analysis...................................................................................................................47 5.3.2 SolidWorks Finite Element Analysis...........................................................................................52 6 Design for Manual Assembly......................................................................................................................54 6.1 Steps to Assemble..............................................................................................................................54 6.2 Work Station Layout...........................................................................................................................70 7 Tolerancing and Closure.............................................................................................................................71 7.1 Shaft and Coupler...............................................................................................................................71 7.2 Starter and Slot ..................................................................................................................................72 7.3 Key and Gear......................................................................................................................................73 7.4 Hot Cylinder and Hot Support............................................................................................................74 7.5 Paraboloid and Engine .......................................................................................................................75 8 Cost Analysis...............................................................................................................................................77 9 Intellectual Property...................................................................................................................................80 10 Evaluation...............................................................................................................................................81 11 Appendix ................................................................................................................................................82 11.1 Tables for Shaft Failure Analysis ........................................................................................................82 11.2 Manual Assembly Time Table ............................................................................................................85 11.3 Machining Tolerances ........................................................................................................................96 12 References..............................................................................................................................................97
- 3. 3 2 INTRODUCTION The purposeof this project is to analyze a design for and determine the feasibility of a portable 5 watt Stirling engine charger. The final design is driven by solar energy using an acrylic-based solar mirror and water-based cooling. The heat is transferred to brass cylinders containing the hot pistons of the Stirling engine. The Stirling cycle is achieved through the expansion of air in the hot cylinders, the compression of air in the cold cylinders, and a cyclical volume change in each pair of cylinders. For reference, an ideal Stirling cycle can be seen below in figure 1. The cylinder volume change is induced through the rotation of a swash-plate, which separates the hot and cold cylinders by a phase of 90o . The main user interface is a dry bag on the side of the life ring, where a phone can be left to charge during use. The main components of the design can be seen in figure 2 below. Figure 2. Overview of the Solar Stirling Charger. The main components of the design are highlighted with arrows. Figure 1. An ideal Stirling cycle, which consists of four phases: heating, expansion, cooling, and compression [3].
- 4. 4 The engine inthe center of the design consists of four pairs of cylinders that are alpha type Stirling engines. An alpha type Stirling engine has two cylinders that contain the working gas and the hot and cold pistons, which are separately located in the respective cylinder. It is the simplest configuration of the Stirling engine, however it has the disadvantage of requiring seals around both of the pistons [1]. An example of this configuration can be seen in figure 3. The configuration of the designed engine and its main components can be seen in figure 4. A description of each component can be found in Section 3 Individual Components. A more complete description of the design and how it works can be found in Section 5 How it Works. The following sections of this paper will discuss the feasibility of this design, including the analysis of the system dynamics and thermodynamics, critical failure points, design for manual assembly, costing analysis, safety, and intellectual property. Figure 3. Example of an alpha type Stirling engine [1]. Figure 4. Overview of the engine. The main components are highlighted with arrows.
- 5. 5 3 PERFORMANCE SPECIFICATIONS TableI contains the specifications determined for this project. These specifications will be compared with the analytical model outputs of the design in order to support its feasibility. TABLE I PERFORMANCE SPECIFICATIONS DESCRIPTION SPECIFICATION Able to be carried with two hands Less than 2.5 ft. Access to USB port Must have convenient user interface with USB port Power output to USB port 5 W Thermal Efficiency 20% Must be portable Under 50 lb.
- 6. 6 4 INDIVIDUAL COMPONENTS Thissection will introduce and discuss the details of each component, which can be found in the bill of materials and exploded views in the following sections. The descriptions will include what the component is made of, why the particular material was used, how the component was made, and how much each component costs. All custom parts are manufactured via casting. The prices of all casted parts are estimated using a cost estimator created by Dandong Funding Engineering Machinery Co. This cost estimator can be used for various types of steel, complexities, and casting processes. Figure 5. Full assembly exploded view.
- 7. 7 4.1 SUB-ASSEMBLY A Sub-assemblyA contains all components that encase sub-assembly B and C for transportation and storage. This assembly also contains the reflective Paraboloid and the dry bag. The Paraboloid is attached to the floatation tube using the clip straps. The dome is then attached to the Paraboloid using the dome seal, clip-on nuts, and countersink screws. Figure 6. Line drawing of sub-assembly A. TABLE II BILL OF MATERIALS FOR SUB-ASSEMBLY A ITEM NO. DESCRIPTION NUMBER QTY. 1 Paraboloid PH-A-001 1 2 Dome Cover PH-A-002 1 3 Floatation Tube PH-A-003 1 4 Rope PH-A-004 1 5 Dry Bag PH-A-005 1 6 Clip Strap PH-A-006 3 7 Dome Seal PH-A-007 1 8 Clip-On Nut PH-A-008 4 9 1/4"-20 X 1/2" Countersunk Screw PH-A-009 4
- 8. 8 Paraboloid Figure 7. Paraboloid. TheParaboloid reflects sunlight on to the hot cylinders of the engine; this heats up the air in the cylinders and begins the expansion phase of the Stirling cycle. It is an acrylic-based solar mirror that is manufactured via injection molding. Acrylic was the material of choice because it is lighter and cheaper than glass, as well as easier to manufacture. This part is parabolic in shape, where it is 22 in. in diameter and 8.5 in. tall. It has four extrusions on the lip that will serve as fastening points for the dome cover and three extruded slots on the lower portion for fastening the flotation device with clip straps. Dome Cover Figure 8. Dome cover. The dome cover protects the engine from the environment and provides a barrier between the engine and the user. This part is made of acrylic in order to minimize weight and provide a clear cover in which radiation can still enter the dome at a high transmission rate. It is 22 in. in diameter in order to match the shape of the Paraboloid and 1.5 in. tall to provide clearance for sub-assembly C. The dome cover has four extrusions on its lip for attachment to the Paraboloid and a small cut-out for the USB cord to exit the dome. This part is injection molded in order to support its complex geometry.
- 9. 9 Flotation Tube Figure 9.Flotation device. The flotation device, as the name implies, allows the Paraboloid assembly to float in a body of water. This allows only the cold cylinders to be submerged in water and helps to keep water away from the rest of the device. The floatation device also helps to stabilize the rest of the assembly in cases of uneasy water surface conditions. The ropes on the flotation device allow easy retrieval of the assembly, and also allows towing of the assembly. This part is made out of Elvaloy/Hypalon which is an extremely buoyant material and is more than enough to keep the forty five pound device afloat. The flotation device has a 24 in. outer diameter and is 4 in. in width. This 24 in. flotation device is bought from Wholesale Marine for $64.99. Rope There is 10 ft. of extra rope that allows the assembly to be tied to various devices in order to tow or maintain the assembly. The rope is made out of nylon which ensures that the rope is sturdy and can sustain large pulling forces. The rope is bought from Wholesale Marine for $0.14 per foot. One assembly requires ten feet which results in $1.40 per assembly. Dry Bag Figure 10. Dry bag. The dry bag is a waterproof storage vessel that is used to house the charging electronic device. The dry bag also houses the safety equipment needed to handle the device and is used to store other loose articles or devices. The 2 L dry bag is made out of a proprietary composite material that is water proof. The dry bag is purchased as a three pack from Walmart for $9.97. Due to only one dry bag being used per assembly, the total cost for one assembly is $3.32.
- 10. 10 Clip Strap Figure 11.Clip Strap. The clip strap allows the floatation device to be attached to the Paraboloid via the Paraboloid extrusions. The clip strap also helps to stabilize the assembly via tight attachment to the floatation device. The clip strap is made out of nylon which ensures that the attachment is sturdy and can sustain large pulling forces. A clip strap is 1 ft. long, 1 in. wide, and made out of Nylon webbing. An additional clip for fastening is purchased from Strap Works. The nylon webbing cost $0.18 per foot and each clip is $0.29. Three of these sets are necessary for each assembly, this results in a total of $1.41 per assembly. Dome Seal The dome seal is used to ensure that no water penetrates the inside of the dome when the dome cover is placed on top of the Paraboloid. The seal is made out of fabric-reinforced high-temperature silicone foam and has an adhesive backing so it is easy to place on top of the Paraboloid. The dome seal foam is priced at $65.28 per 30 ft. For one assembly, 6 ft. of the 1/8 in. thick foam strip is needed and the cost for the seal will be $12.46. Clip-on Nut The clip-on nut is a self-locking nut that slides on to the Paraboloid in a spring like fashion. This is used to fasten the back of the screws given that the whole the screws go through are not threaded and thus secure the dome cover. These 1/4'’-20 steel clip on nuts are bought in packages of 25 for $18.08 from The Cliphouse. One assembly will require four nuts and these will cost $2.89 total. 1/4”-20 x ½” Countersunk Screw The countersink screws are standard ½ in. long screws that are used to fasten the dome cover to the Paraboloid. Countersunk screws were specifically chosen because a countersunk is normally used with removable paneling and this made the most sense for the removable dome cover. These stainless steel dome screws can be purchased in packages of 1000 for $100.00 from Bolt-Depot. One assembly requires four screws which results in a cost of $0.40 per assembly.
- 11. 11 4.2 SUB-ASSEMBLY B Thefollowing section will explore the components of sub-assembly B, along with their role in the system. Sub-assembly B contains the most essential components of the Stirling engine. The assembly contains the bottom half of the coupler, which attaches to the coupler connection for the generator in order to transmit rotation between the two shafts. The drive shaft sits within the regenerator and depends on a ball-and-socket joint for attachment with the cool cylinder support. The drive shaft contains the swashplate, which aids in the transfer of thermal energy to mechanical energy. The rest of the components directly interface with the fluid within the engine. The hot pistons sit within the hot cylinders and are able to be manipulated by the fluid pressure. As the system cycles through, the fluid is shuttled through the regenerator and is cooled down while heading towards the cool cylinders. The cool cylinders contain the other half of the pistons. The motion of the pistons as the fluid travels between the cylinders transfers the necessary forces to the swashplate to keep the cycle running. Figure 12. Upper half exploded view of sub-assembly B.
- 12. 12 Figure 13. Lowerhalf exploded view of sub-assembly B. TABLE III BILL OF MATERIALS FOR SUB-ASSEMBLY B ITEM NUMBER DESCRIPTION PART NUMBER QTY. 1 Drive Shaft Flexible Shaft Coupler PH-B-001 1 2 Hot Cylinder Support PH-B-002 1 3 Swashplate Drive Shaft PH-B-003 1 4 Regenerator PH-B-004 1 5 Cool Cylinder Support PH-B-005 1 6 Hot Piston PH-B-006 4 7 Cool Piston PH-B-007 4 8 1/4"-20 X .75" Countersunk Machine Screw PH-B-008 4 9 O-Ring PH-B-009 16 10 Hot Piston Rod Seal PH-B-010 8 11 Cool Piston Seal PH-B-011 4 12 Hot Cylinders PH-B-012 4 13 External Retaining Ring PH-B-013 1 14 Shaft Key PH-B-014 1 15 Starter Gear PH-B-015 1
- 13. 13 Coupler Figure 14. Coupler. Thepurpose of the coupler is to provide a mechanical connection between the rotating swashplate drive shaft and the generator shaft. The specific coupler selected is capable of attaching shafts of various sizes, making it ideal for this system. The coupler is supplied by McMaster. The cost for two coupler hubs and one spider is $10.03 per assembly. Hot Cylinder Support Figure 15. Hot cylinder support. The hot cylinder support is required to supply stability to several components. It contains indentations to mate with the generator support in sub-assembly C. The support also contains grooves designed to accommodate the hot cylinder assembly and align it as needed with the hot pistons. The hot cylinder support also requires alignment with the regenerator for proper fluid flow, which will be accomplished using dowel pins. The geometry of the part facilitates the insertion of seals required to contain the fluid. Upon considering all thermal and mechanical performance requirements for the part, brass was selected as the optimum material. The part will be cast in order to reduce manufacturing costs. The hot cylinder support is cast out of brass and weighs 1.27 kg. This component has average complexity and is manufactured using water glass lost wax investment casting. Additional machining may be necessary in order to ensure accuracy of the holes. This results in a cost of $13.12. All estimates for brass casting take into account that brass is more expensive than steel and the price estimator is for steel. The relative cost is found via the Ashby chart in Fig 6.20 and 6.28 [2].
- 14. 14 Drive Shaft Figure 16.Drive shaft. The swashplate drive shaft is the selected form of power transmission between the pistons and the generator. This part contains a swashplate at an angle of 20⁰ relative to the horizontal. The tilt of the swashplate generates enough instability for the part to rotate about the central axis when a force is applied towards the radial extremes. The forces experienced by the swashplate are generated by both the hot and cool pistons. The plate contains troughs on either side for alignment with the pistons. For proper support, the base of the part contains the cup of a ball-and-socket joint, which is meant to attach with the cool cylinder support. The swashplate acts as a flywheel for the system, using its inertia to keep the cycle in motion and ease the flow of the pistons. The shaft diameter is reduced at the tip for attachment with the coupler via a keyway. The tip of the shaft also attaches to the starter gear to provide the initial momentum for motivating the system. Material analysis indicates that 304 stainless steel will provide the necessary momentum properties, as well as strength, for the part to perform as needed. The most effective method for manufacturing the part is casting due to the low tolerances and geometry of the part. This component is cast using silica sol lost wax investment casting and possesses normal complexity. Taking into account that simple machining may be needed to ensure smooth troughs, the final cost of the drive shaft is $6.51. Regenerator Figure 17. Regenerator. The regenerator serves multiple purposes in the assembly. The main function is to provide a shuttling path for the fluid within the engine, while providing a method for storing and releasing thermal energy to and from the fluid to improve engine efficiency. Another function of the regenerator for this design is to
- 15. 15 provide a protectiveenclosure for the rotating drive shaft. An effective regenerator is required to absorb heat from the fluid as it flows from the hot cylinders to the cold cylinders, retain as much of the heat as possible, and release the heat back to the fluid as it returns towards the hot cylinders. To accomplish all of these tasks, the regenerator will be made from 304 stainless steel. The general shape of the regenerator is a hollow cylinder. This shell will have the pathways for the fluid as well as threaded holes for attachment with the hot and cold cylinder supports. In order to fit the rotating drive shaft inside, the inner diameter is 3.50 in. with a thickness of 0.50 in. to accommodate the fluid pathways as well as the threads. To achieve this complex geometry the regenerator will be cast. The regenerator is cast using silica sol lost wax investment casting. This cast is considered complex due to the internal holes. There is also additional machining needed to generate the threaded holes. This part weighs 2.22 kg and the total cost is $20.50. Cool Cylinder Support Figure 18. Cool cylinder support. The cold cylinder support is the foundation of the engine assembly. The part is fastened to the regenerator for proper alignments and allows the air to be shuttled into the chambers. The chambers guide the cooling pistons and hold the shuttled air as it is cooled by the water. The center support holds the drive shaft and allows it to rotate as it is driven by the piston. The piece is made of 304 stainless steel to be corrosion resistant and strong. It measures about 5 in. in diameter and 3.5 in. tall and, due to the complex geometry, the part will be cast which results in a cost of $14.78. Hot Piston Figure 19. Hot piston. The hot pistons displace the air in the hot cylinders and drive the swashplate. The pistons are cast out of 304 stainless steel for strength. Each piston is about 3.5 in. long, 1 in. in diameter, weighs 0.06 kg, and costs $5.53.
- 16. 16 Cool Piston Figure 20.Cool piston. The cool pistons compress the air in the cool cylinders and drive the swashplate. The pistons are cast out of 304 stainless steel for strength and corrosion resistance. Each piston is about 3.5 in. long, 1 in. in diameter, weighs 0.06 kg, and costs $5.53. Piston Seals Figure 21. Piston seal. The piston seals prevent the air moved by the pistons from escaping from the chamber except by the desired channels. The part is used on both the hot and cool pistons. The high temperature piston rod seals are purchased from McMaster-Carr for $17.88 per seal and approximately ½ in. wide and 1/ 8 in. thick. Hot Cylinders Figure 22. Hot cylinders. The brass cylinders transfer the heat from the sunlight focused on its outer walls into the air contained inside to fuel the Stirling cycle. They are welded to the hot cylinder support and hold the hot pistons in line. The hot cylinders are cast out of brass and cost $9.73.
- 17. 17 Pull Start Gear Figure23. Shaft gear. This 304 stainless steel gear is one component of the rack-pinion method of starting the device. The gear is attached to the main shaft and helps begin the Stirling cycle in the engine. The starter gear is purchased from Stock Drive Products Sterling Instrument for $9.83 each. Piston O-rings Figure 24. Piston O-ring. The piston O-rings help provide a tight seal to the chamber walls and help decrease friction. The piston O-rings are made out of stainless steel to withstand the high temperatures experienced in the cylinders. The O-rings are purchased from Darcoid Nor-Cal Seal and the price is approximated at $10.00 per seal. Screws The regenerator screws are bought in packages of 1000 from Bolt-Depot for $236.00. Only eight screws are needed per assembly, therefore, the cost for these screws is $1.89 per assembly. External Retaining Ring The external retaining ring is bought from McMaster as a pack of 10 for $8.46. Each assembly requires one snap ring which totals $0.85 per assembly. Shaft Key The shaft key is manufactured by casting UNS G10180 CD Low carbon mild steel and weighs approximately 0.002 kg. This geometry is very simple and costs $5.17.
- 18. 18 4.3 SUB-ASSEMBLY C Thefollowing section will explore the components of sub-assembly C, along with their role in the system. Sub-assembly C contains all of the electronic components necessary for the device. The coupler present in this assembly is used to attach the generator to the mechanical drive shaft described in sub- assembly B. The generator is fastened to the generator support which is epoxied to the starter cover. The power output from the generator is delivered to the buck converter and finally output using a DC to USB port. Figure 25. Sub-assembly C. TABLE IV BILL OF MATERIALS FOR SUB-ASSEMBLY C ITEM NUMBER DESCRIPTION PART NUMBER QTY. 1 Generator Support PH-C-001 1 2 Generator Flexible Shaft Coupling PH-C-002 1 3 Starter Cover PH-C-003 2 4 Generator PH-C-005 1 5 Buck Converter PH-C-006 1 6 DC to USB Port PH-C-007 1 7 Generator and Electronics Cover PH-C-008 1 8 3M x 10mm Hex Head Screw PH-C-009 2
- 19. 19 Generator Support Figure 26.Generator support. The generator support is used to stabilize the generator and provide a mounting location for the generator as well as all of the electrical components. The geometry must be such that the generator and the screws can be inserted properly. The cover must also be able to fit onto the top of the support in order to protect the electronic components. The overall size of the support is 3.50 in. by 3.50 in. by 4.61 in. The generator support is cast out of brass using water glass lost wax investment casting with no additional machining. The material for this component is brass to allow for it to be welded to the hot cylinder support. This support weight 0.26 kg and costs $3.08. Generator Flexible Shaft Coupler Figure 27. Generator flexible shaft coupler. This half of the coupler mates with the coupler previously described in sub-assembly B. Starter Cover Figure 28. Starter cover.
- 20. 20 The purpose ofthe starter cover is to guide the pull key starter into the starter gear. This cover is cast out of stainless steel using water glass lost wax investment casting. The overall dimensions of the cover are 3.50 in. by 3.50 in. by 0.10 in. Each cover weights 0.06 kg and does not possess complex geometry nor does it need additional machining. This results in each cover costing $1.07 each. Each assembly requires two starter covers which totals $2.14. Generator The generator is used to convert the mechanical energy of the shaft into electrical energy that can ultimately be used to charge a modern day cellphone. The generator is purchased from RobotShop for $5.88. Buck Converter The Buck converts the AC current from the generator to a controlled DC current, which is supplied to the DC to USB port. The buck converter is supplied by Amazon for $10.66. DC to USB Port The DC to USB converter is the direct interface between the engine and the device being charged. This converter is DROK brand and purchased from Amazon for $5.80. Generator and Electronics Cover Figure 29. Generator and electronics cover. The generator and electronics cover is used to insulate the sensitive electronic components from the extreme temperatures experienced by the engine. It is made using a ceramic sheet from CeraMaterials at a rate of $12.78 for 864 sq-in. Each assembly requires 37 sq-in and totals to $0.55 per assembly. 3M x 10mm Hex Head Screw Two 3M x 10mm Hex Head Screw are used in order to secure the generator to the generator support. These screws are purchased in packages of 100 from Amazon for $6.57 per package. Two screws are needed per assembly which totals to $0.13.
- 21. 21 5 HOW ITWORKS 5.1 NOMINAL USE This device has several requirements that must be met in order to function as intended. These requirements are as follows: - Ensure reflective surface and clear dome is free of dirt and debris. - Place assembly in an area with direct sunlight. - Placed in a body of water at least 2 feet deep. - Have dome securely attached to Paraboloid. - Attach entire assembly to a fixed structure using the rope provided. - Place electronic device in dry bag and close tightly. In addition to these requirements, it is critical to visually inspect the device in order to ensure seals are functioning as intended and that the flotation device and clip straps are in working condition. The formal instruction manual for this product will contain safety warnings in the following format: WARNING ELECTRIC SHOCK, FIRE OR BURN INJURIES CAN OCCUR IF THIS EQUIPMENT IS NOT USED PROPERLY. TO REDUCE RISK OF INJURY: Do not operate unattended. Assume all surfaces are hot while in use Only handle hot components while in a stable position and wearing safety gloves Keep face and eyes away from hot surfaces Do NOT attempt to start while swimming in the water Do NOT allow foreign materials into the dome and engine components Keep loose articles away from rotating components Use Open dry bag and plug in electronic, retrieve safety gloves, and starter pull key Reseal bag and place device in the water, within reach, and in direct sunlight Use provided safety gloves to open dome cover Insert starter pull key and pull vigorously Repeat Step 4 until motor begins turning on its own Replace cover and seal tightly Ensure device is properly tethered to prevent drifting Check the charge on devices regularly Follow maintenance procedures after each use
- 22. 22 Maintenance 1. Remove devicefrom sunlight and place in safe area 2. Let it cool completely (checking with hand near but not touching the surfaces) 3. Empty dry bag and wipe out with a cloth 4. Rinse external surfaces with clean water; wipe dry 5. Dry and store in a cool, dry place away from sunlight 5.2 ANALYSIS This section will detail the approach used to analyze the kinematics, dynamics, and thermodynamics of the design. This is not an exhaustive analyzation, and there is a potential for future calculations. The table below contains the definitions and variables used throughout the following calculations. TABLE V VARIABLE SYMBOLS AND DEFINITIONS SYMBOL DEFINITION Ap Incident area of a single piston g Gravity r Instantaneous distance from the nth piston to the swashplate pivot Mp Mass of a single piston Ms Mass of swashplate Pi Pressure realized by piston Pi,δ Pressure of 90o offset cool cylinder Ts Swashplate swivel torque Tilt angle of swashplate Rotation angle of swashplate d Diameter of hot and cool piston h Height of hot piston L Distance from hot cylinder support to the center of the swashplate tswash Thickness of swash plate s Distance between the origin and center of the track VT Total volume of one pair of cylinders ka Surface condition modification factor kb Size modification factor
- 23. 23 kc Load modificationfactor kd Temperature modification factor ke Reliability factor kf Miscellaneous-effects modification factor Se ’ Rotary-beam test specimen endurance limit Se Endurance limit at the critical location of a machine part in the geometry and condition of use 5.2.1 Kinematics and Dynamics The overall geometry of the system is the driving factor for the design of many of the components. The selected method of power transmission is a swashplate mechanism, which will rotate along the central axis of the Paraboloid. This motion is due to the 20⁰ tilt of the plate portion of the drive shaft and the pistons exerting forces on the swashplate. Figure 30. Shaft and piston diagram to illustrate volume changes. The nature of the Stirling cycle is one of dynamic volume. The volume for each of the four alpha pairs depends on the angle of rotation of the shaft. The kinematics of the system will play a large role in the behavior of the engine. Figure 30 depicts the relationship between one piston and the swashplate drive shaft. 𝑟𝑝 = 𝑟 ∗ tan 𝛽 ∗ cos 𝜃 𝑢̂1 + 𝑟 𝑢̂2 Using the transport theorem, the velocity of the point as viewed by the plate is 𝑒 𝑧 𝑒 𝑥 𝑢̂1 𝑢̂2 r h 𝛽
- 24. 24 𝑣 𝑝 =−𝑟 ∗ sin 𝛽 ∗ tan 𝛽 ∗ 𝜃̇ ∗ sin 𝜃 𝑒 𝑥 + 𝑟 ∗ sin 𝛽 ∗ 𝜃̇ ∗ sin 𝜃 𝑒 𝑧 And the acceleration becomes 𝑎 𝑝 = −𝑟 ∗ sin 𝛽 ∗ tan 𝛽 ∗ (𝜃̈ sin 𝜃 + 𝜃̇2 cos 𝜃)𝑒 𝑥 + 𝑟 ∗ sin 𝛽 ∗ (𝜃̈ sin 𝜃 + 𝜃̇2 cos 𝜃)𝑒 𝑧 (1) Figure 31. Shaft FBD. Euler’s 2nd Law was used to analyze the angular momentum of the system 𝑑 𝑑𝑡 ( H 𝑆 0 𝐺 ) = ∑ 𝑀 (2), where the angular momentum, H 𝑆 0 𝐺 , is defined to be H 𝑆 0 𝐺 = 𝐼 ∙ 𝜔 𝑠 0 𝐺 I is the moment of inertia about the body-fixed coordinate system, 𝛽 𝑒 𝑧 𝑒 𝑥 𝑢̂1 𝑢̂2 g 𝜏𝑓 𝐹 𝑑, 𝑖 𝐹𝑝,𝛿 r 𝐹𝑦
- 25. 25 𝐼 = [ 1 4 𝑀𝑠 𝑟2 0 0 0 1 4 𝑀𝑠 𝑟 2 0 0 0 1 2 𝑀𝑠 𝑟 2 ] and 𝜔 𝑠 0 𝐺 is the angular velocity about the vertical axis defined as 𝜔 𝑠 0 𝐺 = [ 0 0 𝜃̇ ] In addition to the dynamics of the swashplate, the dynamics of the pistons also needs to be considered. The pistons move vertically with relation to a Newtonian reference frame and experience forces due to the pressure inside the vessels. This motion can be described with the kinematics of the swashplate previously analyzed, however, there is friction involved with this motion. The presence of air- tight seals generates friction with the surface of the piston rods and reduces the force acting on the swashplate, diminishing the final torque output. Figure 32. Piston FBD. The moments acting on the system directly depend on the pressures inside of the hot cylinders. The force supplied by the hot cylinders is defined as 𝐹𝑑,𝑖 = (𝑃𝑖 ∗ 𝐴 𝑝 + 𝐹𝑓,𝑖) 𝑢̂1 And the force due to the pressure from the cool system is 𝐹𝑝,𝛿 = (𝑃𝑖,𝛿 ∗ 𝐴 𝑝 − 𝐹𝑓,𝛿) 𝑢̂1 Friction on the piston is to be modeled as viscous [11] 𝐹𝑓 = −(𝜇 ∗ 𝑟 ∗ sin 𝛽 ∗ tan 𝛽 ∗ 𝜃̇ ∗ sin 𝜃) 𝑢̂1 𝐹𝑓 𝐹𝑓 𝑃𝑖 ∗ 𝐴 𝑝 𝑃𝑖 ∗ 𝐴 𝑝 g
- 26. 26 where 𝜇 isthe coefficient of friction between the piston and the seals. This value was found to be 0.58 for 304 stainless steel [12]. The rotational friction due to the ball-and-socket joint at the base of the swashplate was modeled as a linear torque opposing the rotation of the shaft, as shown below. 𝜏 𝑓 = 𝑎 ∗ 𝜃̇ 𝑢̂1 Combining all of the force relations and performing the cross product with their respective displacements relative to the center of the swashplate, the moments in the three principle coordinates become, ∑ 𝑀 𝑥 4 1 = 𝑟 ∗ sin 𝜃 ∗ cos 𝛽 ∗ (𝐹𝑝,𝛿 − 𝐹𝑝,𝑖 − 2 ∗ 𝑀 𝑝 ∗ 𝑔 − 2 ∗ 𝐹𝑓) − 𝑎 ∗ sin 𝛽 ∗ 𝜃̇, ∑ 𝑀 𝑦 4 1 = 𝑟 ∗ cos 𝜃 ∗ (𝐹𝑝,𝛿 − 𝐹𝑝,𝑖 − 2 ∗ 𝑀 𝑝 ∗ 𝑔) − 𝐿 ∗ 𝐹𝑦, ∑ 𝑀𝑧 4 1 = 𝑟 ∗ sin 𝜃 ∗ sin 𝛽 ∗ (𝐹𝑝,𝛿 − 𝐹𝑝,𝑖 − 2 ∗ 𝑀 𝑝 ∗ 𝑔 − 2 ∗ 𝐹𝑓) − 𝑎 ∗ cos 𝛽 ∗ 𝜃̇. Applying these relations to (2), the following is obtained [ − 1 4 𝑀𝑠 𝑟 2 sin 𝛽 𝜃̈ 1 4 𝑀𝑠 𝑟 2 sin 𝛽 𝜃̇2 1 2 𝑀𝑠 𝑟 2 cos 𝛽 𝜃̈ ] = [ 𝑟 ∗ cos 𝛽 ∗ (𝐹𝑝,𝛿 − 𝐹𝑝,𝑖 − 2 ∗ 𝑀 𝑝 ∗ 𝑔 − 2 ∗ 𝐹𝑓) ∗ sin 𝜃 − 𝑎 ∗ sin 𝛽 ∗ 𝜃̇ 𝑟 ∗ cos 𝜃 ∗ (𝐹𝑝,𝛿 − 𝐹𝑝,𝑖 − 2 ∗ 𝑀 𝑝 ∗ 𝑔) − 𝐿 ∗ 𝐹𝑌 𝑟 ∗ sin 𝛽 ∗ (𝐹𝑝,𝛿 − 𝐹𝑝,𝑖 − 2 ∗ 𝑀 𝑝 ∗ 𝑔 − 2 ∗ 𝐹𝑓) ∗ sin 𝜃 − 𝑎 ∗ cos 𝛽 ∗ 𝜃̇ ]. Based on these results, one can obtain a relationship for the angular acceleration of the system. 𝜃̈ = 2 ∗ tan 𝛽 ∗ (𝐹𝑝,𝛿 − 𝐹𝑝,𝑖 − 2 ∗ 𝑀 𝑝 ∗ 𝑔 − 2 ∗ 𝐹𝑓) sin 𝜃 − 𝑎 ∗ 𝜃̇ 𝑀𝑠 ∗ 𝑟 (3) Due to the complexity of the expression, it was evaluated through an iterative process to obtain the steady-state value for the angular velocity as the process begins. Figure 33 illustrates the result of these values.
- 27. 27 The steady-state torqueproduced by the system with a load of 0.06 N-m is calculated as 𝑇 = 0.06 𝑁𝑚 ∗ 𝜃̇, which results in a mechanical power of 16.3 watts. 5.2.2 Thermodynamics and Heat Transfer The following section goes into more detail on the thermal analysis of the Solar Stirling Charger to show how the power output spec of 5 W is feasible thermodynamically. First is an overview of the Stirling cycle and how it applies to the Solar Stirling Charger, showing how the air volume and pressure varies with the rotational angle of the motor shaft. Second, the overall heat transfer is discussed, explaining how sunlight is converted into a usable power output along with any heat losses incurred in the process and calculated efficiencies. For these analyses, the following weather conditions were considered: TABLE VI CLEAR SKY CONDITIONS CONDITION VALUE Solar Radiation, Gatm (W/m2) 893 Effective Sky Temperature, Tsky (K) 354 (177.53 °F) Finally, the thermodynamic inputs, outputs, and efficiencies of the system were iterated at various weather conditions to show how the Solar Stirling Charger’s performance changes and whether it still meets spec. Figure 33. Steady state angular velocity of the drive shaft.
- 28. 28 5.2.2.1 The StirlingCycle The ideal Stirling cycle (figure 34b) consists of four phases, which follow the air’s movement throughout a pair of cylinders: a. Isothermal expansion (1 -> 2) The air in the hot cylinders receives heat input Qin from sunlight, causing the air to expand and exert force on the hot pistons. b. Isochoric cooling (2 -> 3) The air flows from the hot cylinder to the cool cylinder through the regenerator with some of its heat, Qr, being transferred to the regenerator. c. Isothermal compression (3 -> 4) The air in the in the cool cylinder rejects Qout into the water, causing the air to become denser, enabling the piston to compress the air. d. Isochoric heating (4 -> 1) The air flows from the cool cylinder to the hot cylinder through the regenerator with the heat being stored in the regenerator, Qr, being returned to the air. Figure 34. a) Heat movement throughout system. b) Ideal P-V diagram. 5.2.2.2 Volume The total volume of the Solar Stirling Charger is dependent on three air volumes: the air in the hot and cold cylinders, VH and VC, and the regenerator, VR. The following volume equations were derived from kinematics and are dependent on 𝜃, the rotation angle of the drive shaft. In these equations, 𝛿 is the phase difference between the hot and cold pistons (𝜋/2 rad), d is the diameter of a piston, s is the distance vector from the centroid of the swashplate to the connection point of the piston rod, 𝛽 is the tilt of the swashplate, 𝛾 is the local tilt of s, 𝜙 is the global tilt of s, and tswash is the thickness of the swashplate. 𝑉𝐻(𝜃) = 0.25𝜋𝑑2(𝑠 ∗ sin 𝜙)(1 − cos 𝜃) (4) 𝑉𝐶(𝜃) = 0.25𝜋𝑑2(𝑠 ∗ sin 𝜙)(1 − cos(𝜃 − 𝛿)) (5) Qin Qout Qr a) b)
- 29. 29 Wherein the geometryis defined as follows: 𝛽 = 𝜋 9 𝑟𝑎𝑑 (6) 𝑠 ∗ sin(𝛾) = 𝑡 𝑠𝑤𝑎𝑠ℎ/2 (7) 𝜙 = 𝛽 + 𝛾 (8) Some of these geometric parameters are labeled in figures 35 and 36. Whereas, the volume of the air in the regenerator is a constant: 𝑉𝑅 = 0.0348 𝑖𝑛3 Figure 35. Geometry for volume model: a) Swashplate, b) Close-up. Figure 36. Geometry for volume model: hot cylinders (aka bulbs). 𝜃 a) b) Swashplate Drive Shaft s s 𝛽 𝛾 tswash Stroke Hot Cylinders d
- 30. 30 Using the precedingrelations, the total volume of a pair of cylinders 𝑉𝑇 can be expressed as thus: 𝑉𝑇 = 𝑉𝐻(𝜃) + 𝑉𝐶(𝜃 − 𝛿) + 𝑉𝑅 (9) The cyclical change of these volumes for a single pair of cylinders are plotted in figure 37. Figure 37. Volume of one cylinder pair vs. crank angle of swashplate. 5.2.2.3 Pressure Assuming ideal gas and constant pressure throughout the engine, the relationship between pressure P and volume V can be expressed as follows: 𝑃𝑉 = 𝑚𝑅𝑇 (10) Wherein m is mass of the air, R is the gas constant for air, and T is the air temperature. Applying (10) to each section of the engine (hot (subscript H) and cold (subscript C) cylinders and regenerator (subscript R)), the air mass can be calculated, individually. 𝑚 𝐻 = 𝑃𝑉𝐻 𝑅𝑇 𝐻 (11) 𝑚 𝐶 = 𝑃𝑉𝐶 𝑅𝑇𝐶 (12) 𝑚 𝑅 = 𝑃𝑉𝑅 𝑅𝑇𝑅 (13) The same relationship can be applied to the total mass (mT) of the engine at ambient conditions (P0 = 14.7 psi, T0 = 77°F), which is a constant: 𝑚 𝑇 = 𝑃0 𝑉𝑇 𝑅𝑇0 (14)
- 31. 31 The total masscan, of course, be written as the sum of the individual air masses. 𝑚 𝑇 = 𝑚 𝐻 + 𝑚 𝐶 + 𝑚 𝑅 (15) Therefore, (12) through (15) can be plugged into (16), giving 𝑚 𝑇 = 𝑃𝑉𝐻 𝑅𝑇 𝐻 + 𝑃𝑉𝐶 𝑅𝑇𝐶 + 𝑃𝑉𝑅 𝑅𝑇𝑅 (16) After rearranging, pressure is a function of volume which is a function of shaft rotation angle 𝜃. 𝑃 = 𝑚 𝑇 ∗ 𝑅 𝑉𝐻 𝑇 𝐻 + 𝑉𝐶 𝑇𝐶 + 𝑉𝑅 𝑇𝑅 (17) Using the geometry of the engine, pressure can be plotted as thus: Figure 38. Pressure of cylinder pair vs. crank angle of swashplate.
- 32. 32 5.2.2.4 Non-Ideal StirlingCycle Now, the pressure and volume can be plotted parametrically, producing the actual P-V diagram of the Solar Stirling Charger, as seen in figure 39. Figure 39. P-V diagram for theoretical Stirling cycle. In the actual Stirling cycle, the heating and cooling processes are no longer constant volume processes. However, the effective temperatures of the air in the hot, cold cylinders, and the regenerator are assumed to be constant. To define these temperatures, a heat transfer analysis was conducted as explained in the following section. 5.2.2.5 Heat Transfer This section explains the details of the heat transfer analysis of the Solar Stirling Charger in the following order: First, the heat input is described from the Sun to the hot cylinders followed by a discussion of the heat losses. Second, the effect of the regenerator on transmitting heat to and from the air is investigated. Third, the heat dissipation from the cold cylinders is described, followed by the work output resulting from an energy balance of the Stirling engine. Solar Intake On a clear day, the sun shines with the solar flux Gatm of 893 W/m2 , [13]. The effective sky temperature 𝑇̅𝑠𝑘𝑦 on a clear day is 354 K (177.53 ◦ F) as defined by (&), where 𝜎 is the Stefan-Boltzmann constant (5.67 × 10−8 𝑊/(𝑚2 ∙ 𝐾4 )), [14]. 𝐺 𝑎𝑡𝑚 = 𝜎𝑇̅𝑠𝑘𝑦 4 (18)
- 33. 33 On such aday, the dome with an outer surface area 𝐴 𝑑𝑜𝑚𝑒 of 0.2530 m2 (392.1 in2 ) receives the following in solar radiation: 𝑞 𝑠𝑜𝑙𝑎𝑟 = 225.9 W As shown in figure 40, the Sun’s rays strike the dome of the Solar Stirling Charger, and are then reflected by the Paraboloid which redirects the rays to the dome’s focus, where the hot cylinders stand. Figure 40. Diagram of solar intake to the hot cylinders. To calculate how much solar flux is transmitted into the engine is as follows: For acrylic, the transmissivity 𝜏 𝑎𝑐𝑟𝑦𝑙𝑖𝑐 is on average 90% between 𝜆1 = 0.4 µm and 𝜆2 = 1.1 µm, as approximated by the similar material in figure 41. Using the Sun as a blackbody emitter, the product of 𝜆𝑇𝑠𝑢𝑛 can be looked up in Table 12.2 of [14], wherein the effective temperature of the sun Tsun is 5800 K, [14]. After linear interpolation, the fraction of radiation 𝐹(0→𝜆) for each wavelength was figured to be: 𝐹(0→𝜆1) = 0.124509 𝐹(0→𝜆2) = 0.769234 Figure 41. Transmission spectrum of acrylic, [Arkema]. 𝑞 𝑠𝑜𝑙𝑎𝑟 𝑞𝑡𝑟𝑎𝑛𝑠 𝑞 𝑟𝑒𝑓 Paraboloid Dome Hot Cylinders
- 34. 34 The fraction ofradiation for the specified range of wavelengths 𝐹(𝜆1→𝜆2) is the difference of these two fractions. The product of this and the acrylic transmissivity gives the total transmissivity [14], 𝜏 𝑡𝑜𝑡𝑎𝑙 = 𝐹(𝜆1→𝜆2) ∗ 𝜏 𝑎𝑐𝑟𝑦𝑙𝑖𝑐 (19) 𝜏 𝑡𝑜𝑡𝑎𝑙 = (0.769234 − 0.124509) ∗ 0.90 𝜏 𝑡𝑜𝑡𝑎𝑙 = 0.5803 Thus, the solar flux transmitted through the dome is [14], 𝑞𝑡𝑟𝑎𝑛𝑠 ″ = 𝜏 𝑡𝑜𝑡𝑎𝑙 ∗ 𝐺 𝑎𝑡𝑚 (20) 𝑞𝑡𝑟𝑎𝑛𝑠 ″ = 518.2 𝑊/𝑚2 The solar Paraboloid receives 𝑞 𝑝𝑎𝑟𝑎 𝑞 𝑝𝑎𝑟𝑎 = 𝑞𝑡𝑟𝑎𝑛𝑠 ″ ∗ 𝐴 𝑝𝑎𝑟𝑎 (21) 𝑞 𝑝𝑎𝑟𝑎 = 177.2 𝑊 where 𝐴 𝑝𝑎𝑟𝑎 is the exposed surface area of the Paraboloid, 0.3419 m2 (530 in2 ). The solar Paraboloid reflects 68% of this sunlight, based on Table D from [16]. Though the manufacturer gave a reflectivity of 90%, the 68% value is being used as a more conservative value. 𝑞 𝑟𝑒𝑓 = 0.68 ∗ 𝑞 𝑝𝑎𝑟𝑎 (22) 𝑞 𝑟𝑒𝑓 = 120.5 𝑊 This 𝑞 𝑟𝑒𝑓 is the heat being directed to the hot cylinders.The temperature of the outside of the hot cylinders is given by their grey body absorption. 𝑇𝑜𝑢𝑡,ℎ = 𝑞 𝑟𝑒𝑓/(𝜎𝜀 𝑏 𝐴 𝑎𝑏𝑠)1 4⁄ (23) Where 𝜎 is the Stefan-Boltzmann constant, 𝜀 𝑏 is the emissivity of brass (oxidized) from [18] which is 0.61, and 𝐴 𝑎𝑏𝑠 is the surface area of the hot cylinders (see figure 42) that is absorbing sunlight. Figure 42. The geometry of the hot cylinder assembly, wherein cylinder/sun blocker height Lh = 2 in, inner diameter Dih = 0.925 in, cylinder wall thickness th = 0.1 in, cylinder outer diameter Doh = 1.125 in, and sun blocker width Wsb = 0.68 in. The absorbing area as the combined, exterior surface area of the hot cylinders and the sun blockers connecting them. Dih Doh Lh Wsb th qref qref Hot CylinderSun Blocker
- 35. 35 Based on thegeometry of the cylinders and the sun blockers, as seen in figure 42, the absorbing area is: 𝐴 𝑎𝑏𝑠 = 4[(𝐿ℎ 𝑊𝑠𝑏) + 0.75(𝜋𝐷 𝑜ℎ 𝐿ℎ)] (24) Thus, the temperature of the outside of the hot cylinders is: 𝑇𝑜𝑢𝑡,ℎ = 709.2 𝐾 (816.89 °𝐹) Solar Losses Already, there are two losses accounted for during the radiation portion of the heat input: one during the transmission of light through the dome, 𝑞𝑡𝑟𝑎𝑛𝑠,𝑙𝑜𝑠𝑠 and another during reflection of light off the Paraboloid, 𝑞 𝑟𝑒𝑓,𝑙𝑜𝑠𝑠. These are as follows: 𝑞𝑡𝑟𝑎𝑛𝑠,𝑙𝑜𝑠𝑠 = (𝐺 𝑎𝑡𝑚 − 𝑞𝑡𝑟𝑎𝑛𝑠 ″ ) ∗ 𝐴 𝑑𝑜𝑚𝑒 (25) 𝒒 𝒕𝒓𝒂𝒏𝒔,𝒍𝒐𝒔𝒔 = 𝟗𝟒. 𝟖 𝑾 Wherein 𝐴 𝑑𝑜𝑚𝑒 is outer surface area of the dome 392.1. 𝑞 𝑟𝑒𝑓,𝑙𝑜𝑠𝑠 = 𝑞 𝑝𝑎𝑟𝑎 − 𝑞 𝑟𝑒𝑓 (26) 𝒒 𝒓𝒆𝒇,𝒍𝒐𝒔𝒔 = 𝟓𝟔. 𝟕 𝑾 Conduction and Convection Losses Heat leaving the hot cylinders to the ambient conditions directly outside of the Paraboloid (assumed to be 8 degrees above Tsky, based on [19]) is controlled by the following thermal resistances: Conduction resistance of air under the dome, 𝑅𝑡,𝑐𝑜𝑛𝑑,𝑝𝑎𝑟𝑎,𝑎𝑖𝑟. Conduction resistance of the Paraboloid 𝑅𝑡,𝑐𝑜𝑛𝑑,𝑝𝑎𝑟𝑎,𝑎𝑐𝑟𝑦𝑙𝑖𝑐 and the dome, 𝑅𝑡,𝑐𝑜𝑛𝑑,𝑑𝑜𝑚𝑒 . Convection resistance of the wind, 𝑅𝑡,𝑐𝑜𝑛𝑣,𝑤𝑖𝑛𝑑. The overall resistance of the losses 𝑅𝑡,𝑙𝑜𝑠𝑠 is given as 𝑅𝑡,𝑙𝑜𝑠𝑠 = 𝑅𝑡,𝑐𝑜𝑛𝑑,𝑝𝑎𝑟𝑎,𝑎𝑖𝑟 + (𝑅𝑡,𝑐𝑜𝑛𝑑,𝑝𝑎𝑟𝑎,𝑎𝑐𝑟𝑦𝑙𝑖𝑐 −1 + 𝑅𝑡,𝑐𝑜𝑛𝑑,𝑑𝑜𝑚𝑒 −1 ) −1 + 𝑅𝑡,𝑐𝑜𝑛𝑣,𝑤𝑖𝑛𝑑 (27) Losses through Parabolic Resistances The air space underneath the dome was approximated to be one-dimensional conduction through a Paraboloid of the equation 𝑧 𝑝𝑎𝑟𝑎 = 𝑟𝑝𝑎𝑟𝑎 2 2𝑏 (28) Where z is the height of the Paraboloid, r is the radial distance from the central axis, and b is the constant 7.1 in.
- 36. 36 Using the radialversion of Fourier’s law of conduction from [14], 𝑞 𝑟 = −𝑘𝐴 ( 𝑑𝑇 𝑑𝑟 ) (29) Where 𝑞 𝑟 is the radial heat transfer, k is the conductivity, A is the surface area normal to the heat transfer, and 𝑑𝑇 𝑑𝑟 is the radial temperature gradient. With the surface area of a Paraboloid 𝐴 𝑝𝑎𝑟𝑎 given by 𝐴 𝑝𝑎𝑟𝑎 = (( 𝑟 𝑏 ) 2 + 1) 3 2 (30) Wherein 𝑎 = 2(7.12 𝜋) 3 𝑖𝑛2 After rearranging substituting (30) into (29) and solving for the heat transfer, the following relationship is derived: 𝑞 𝑟,𝑝𝑎𝑟𝑎 = 𝑘𝑎(𝑇1 − 𝑇2) 𝐼 (31) Where I is the integral 𝐼 𝑝𝑎𝑟𝑎 = ∫ 𝑑𝑟 (( 𝑟 𝑏 ) 2 + 1) 3 2 𝑟2 𝑟1 (32) Which solves to 𝐼 𝑝𝑎𝑟𝑎 = 𝑟 √( 𝑟 𝑏 ) 2 + 1 ] 𝑟2 𝑟1 (33) Based on (31), conduction thermal resistance of the Paraboloid is given as 𝑅𝑡𝑐𝑜𝑛𝑑,𝑝𝑎𝑟𝑎 = 𝐼 𝑘𝑎 (34) This formula can be applied to both the conduction resistance of the air and the acrylic Paraboloid as such: TABLE VII THERMAL RESISTANCES OF PARABOLOIDS MEDIUM THERMAL CONDUCTIVITY, K, (W/(M∙K)) RPARA, 1 , (IN) RPARA,2 , (IN) THERMAL RESISTANCE, RT, (K/W) Air at 600K 46.9 x 10-3 10.916 11.084 33.29 Acrylic Paraboloid 0.2 1.8225 10.916 0.05
- 37. 37 Figure 43. Geometryof the Paraboloid. Losses through the Dome Similarly the conductive resistance of the dome (spherical cap) can be calculated using radial Fourier’s law (31), where surface area 𝐴 𝑑𝑜𝑚𝑒 is given by 𝐴 𝑑𝑜𝑚𝑒 = 𝜋(𝑎 𝑑𝑜𝑚𝑒 2 + ℎ 𝑑𝑜𝑚𝑒 2 ) (35) Wherein the geometry is given by figure 44. Figure 44. Geometry of the dome, 𝑎 𝑑𝑜𝑚𝑒 = 11 𝑖𝑛, ℎ 𝑑𝑜𝑚𝑒,1 = 1.75 𝑖𝑛, 𝑡 𝑑𝑜𝑚𝑒, = 0.2 𝑖𝑛, and. ℎ 𝑑𝑜𝑚𝑒,2 = ℎ 𝑑𝑜𝑚𝑒,1 + 𝑡 𝑑𝑜𝑚𝑒, For Fourier’s law, 𝑎 𝑑𝑜𝑚𝑒 is kept constant and ℎ 𝑑𝑜𝑚𝑒 is assumed to change (from ℎ1,𝑑𝑜𝑚𝑒 to ℎ2,𝑑𝑜𝑚𝑒), wherein the radial heat transfer is given as 𝑞 𝑟,𝑑𝑜𝑚𝑒 = 𝑘𝑎 𝑑𝑜𝑚𝑒 𝜋Δ𝑇 arctan( ℎ2,𝑑𝑜𝑚𝑒 𝑎 𝑑𝑜𝑚𝑒 ) − arctan ( ℎ1,𝑑𝑜𝑚𝑒 𝑎 𝑑𝑜𝑚𝑒 ) (36) Thus, the conduction thermal resistance of the dome is given by: 𝑅𝑡,𝑐𝑜𝑛𝑑,𝑑𝑜𝑚𝑒 = arctan( ℎ2,𝑑𝑜𝑚𝑒 𝑎 𝑑𝑜𝑚𝑒 ) − arctan ( ℎ1,𝑑𝑜𝑚𝑒 𝑎 𝑑𝑜𝑚𝑒 ) 𝑘𝑎 𝑑𝑜𝑚𝑒 𝜋 (37) 𝑅𝑡,𝑐𝑜𝑛𝑑,𝑑𝑜𝑚𝑒 = 0.1007 𝐾/𝑊 Wherein the geometric variables are in figure 44 and the conductivity k is of acrylic (0.2 W/mK). 𝑧 𝑝𝑎𝑟𝑎 𝑟𝑝𝑎𝑟𝑎 _ Air Space Acrylic ℎ 𝑑𝑜𝑚𝑒,1 𝑎 𝑑𝑜𝑚𝑒 𝑡 𝑑𝑜𝑚𝑒
- 38. 38 Losses due toWind Based on grey body emission, the temperature of the outside of the dome 𝑇𝑜𝑢𝑡,𝑑𝑜𝑚𝑒 is 𝑇𝑜𝑢𝑡,𝑑𝑜𝑚𝑒 = ( 𝐺 𝑎𝑡𝑚 𝜎𝜀 𝑑𝑜𝑚𝑒 + 𝑇𝑠𝑘𝑦 4 ) 1 4 (38) 𝑇𝑜𝑢𝑡,𝑑𝑜𝑚𝑒 = 427 𝐾 Where 𝜀 𝑑𝑜𝑚𝑒 is the emissivity of the acrylic dome which is 0.9 as given by [14]. From [14], the convection heat transfer coefficient due to wind ℎ 𝑤𝑖𝑛𝑑 is ℎ 𝑤𝑖𝑛𝑑 = 0.22(𝑇𝑜𝑢𝑡,𝑑𝑜𝑚𝑒 − 𝑇𝑎𝑚𝑏) 1 3⁄ (39) ℎ 𝑤𝑖𝑛𝑑 = 0.8835 𝑊 𝑚2 ∙ 𝐾 And the convection resistance of the wind is, thus, given by 𝑅𝑡,𝑐𝑜𝑛𝑣,𝑤𝑖𝑛𝑑 = 1/(ℎ 𝑤𝑖𝑛𝑑 𝐴 𝑑𝑜𝑚𝑒) (40) 𝑅𝑡,𝑐𝑜𝑛𝑣,𝑤𝑖𝑛𝑑 = 4.4747 𝐾/𝑊 Altogether, the resistance of the heat losses is 𝑅𝑡,𝑙𝑜𝑠𝑠 = 37.8 𝑊/𝐾 The heat loss 𝑞𝑙𝑜𝑠𝑠 from the hot cylinders to the ambient can be given by 𝑞𝑙𝑜𝑠𝑠 = (𝑇𝑜𝑢𝑡,ℎ − 𝑇𝑎𝑚𝑏)/𝑅𝑡,𝑙𝑜𝑠𝑠 (41) 𝒒𝒍𝒐𝒔𝒔 = 𝟗. 𝟏𝟕𝟗𝟓 Now, all the heat losses can be summarized in the table below. TABLE VIII SUMMARY OF HEAT LOSSES LOSS TYPE TRANSMISSION LOSS REFLECTION LOSS CONDUCTION AND CONVECTION Thermal Loss, W 94.8 56.7 9.2 Thus, taking the conduction and convection losses into account, the net heat input 𝑞 𝑛𝑒𝑡,𝑖𝑛 to the hot cylinder assembly’s surface is 𝑞 𝑛𝑒𝑡,𝑖𝑛 = 𝑞 𝑟𝑒𝑓 − 𝑞𝑙𝑜𝑠𝑠 (42) 𝑞 𝑛𝑒𝑡,𝑖𝑛 = 111.3 𝑊 Dividing this values equally among the four hot cylinders gives, 𝑞𝑖𝑛,ℎ, which is ultimately the heat input into each separate pair of cylinders. 𝒒𝒊𝒏,𝒉 = 𝟐𝟕. 𝟖 𝑾
- 39. 39 Figure 45. Thegeometry of the hot cylinder assembly, wherein Lh = 2 in, Dih = 0.925 in, th = 0.1 in and Doh = 1.125 in, Wsb = 0.68 in. Heat Transfer through the Hot Cylinder Assuming one-dimensional conduction through a cylinder, the thermal resistance of each hot cylinder is 𝑅𝑡,𝑐𝑜𝑛𝑑,ℎ = ln ( 𝐷 𝑜ℎ 𝐷𝑖ℎ ) 2𝜋𝑡ℎ 𝑘 𝑏 (43) 𝑅𝑡,𝑐𝑜𝑛𝑑,ℎ = 0.1115 𝐾/𝑊 Where 𝑘 𝑏 is conductivity of the brass hot cylinders (110 W/mK) from [14] and the hot cylinder geometry is from figure 45. Once the heat transverses the wall thickness of the hot cylinder, it encounters the forced convection thermal resistance of the moving air inside. This thermal resistance is dependent on velocity of the air, which has been approximated to the velocity of the hot side’s piston, 𝑢 𝑝𝑖𝑠𝑡𝑜𝑛(𝜃). 𝑢 𝑝𝑖𝑠𝑡𝑜𝑛(𝜃) = 𝑟(tan𝛽)(𝑠𝑖𝑛𝜃)𝜔 (44) Wherein 𝑟 is the distance vector between the center of the drive shaft and the piston rod, 𝛽 is the tilt of the swashplate, and 𝜃 and 𝜔 are the rotational angle and velocity of the drive shaft, respectively. These variables are shown in figure 46. Dih Doh Lh Wsb th qnet,in qnet,in Hot CylinderSun Blocker
- 40. 40 Figure 46. Swashplateand piston geometry. Since the velocity of the piston is sinusoidal, the root-mean-square of the velocity is used to represent the fluid velocity in the hot cylinder, 𝑢ℎ. At rotational speed achieved on a “clear day,” (2590 RPM), the air velocity is: 𝑢ℎ = 21.4 𝑚/𝑠 Furthermore, the Reynold’s number of the hot side air 𝑅𝑒 𝐷,ℎ is thus, using the thermal properties of air at 600K, which are listed in Table IX [14]: 𝑅𝑒 𝐷,ℎ = 𝑢ℎ 𝐷𝑖ℎ/𝜈ℎ (45) 𝑅𝑒 𝐷,ℎ = 9525.7 Where Dih is the inner diameter of the hot cylinder and 𝜈ℎ is the kinematic viscosity of the hot air. TABLE IX THERMAL PROPERTIES OF AIR AT 600K DENSITY, 𝛒 𝐡 ( 𝐤𝐠 𝐦 𝟑) DYNAMIC VISCOSITY, 𝛍 𝐡 (𝐍 ∙ 𝐬/𝐦 𝟐 ) KINEMATIC VISCOSITY, 𝛎 𝐡 (𝐦 𝟐 /𝐬) PRANDTL NUMBER, Prh THERMAL CONDUCTIVITY, 𝐤 𝐡 (W/M K) SPECIFIC HEAT, 𝒄 𝒑,𝒉 (J/(KG K)) 0.5804 305.8 x 10-7 52.69 x 10-6 0.685 46.9 x 10-3 1051 The Dittus-Boelter equation is used to determine the non-dimensional heat transfer coefficient, the Nusselt number, of the hot cylinder flow [14]. The flow was assumed to be fully developed and turbulent despite the Reynold’s number being slightly below the turbulence realm (𝑅𝑒 𝐷 ≥ 10,000). 𝑁𝑢 𝐷,ℎ = 0.023𝑅𝑒 𝐷 4 5⁄ Pr0.4 𝑟 𝛽 𝜃, 𝜔 Swashplate Drive Shaft Hot Piston
- 41. 41 𝑁𝑢 𝐷,ℎ =30.1 Which is then used to find the heat transfer coefficient for the hot side, ℎℎ: ℎℎ = 𝑁𝑢 𝐷,ℎ 𝑘ℎ/𝐷𝑖ℎ (46) ℎℎ = 60.2 𝑊/𝑚2 𝐾 Where 𝑘ℎ is the thermal conductivity of the air (see Table IX). From which the convection thermal resistance of the air in the hot cylinder is: 𝑅𝑡,𝑐𝑜𝑛𝑣,ℎ = 1/(ℎℎ 𝜋𝐷𝑖ℎ 𝐿ℎ) (47) 𝑅𝑡,𝑐𝑜𝑛𝑣,ℎ = 4.43 𝐾/𝑊 Where the term 𝜋𝐷𝑖ℎ 𝐿ℎ is the inner surface area of the hot cylinder. Putting the thermal resistances of the hot cylinder together with the heat input 𝑞𝑖𝑛,ℎ gives the temperature difference from outside the cylinder 𝑇𝑜𝑢𝑡,ℎ to the effective temperature of the air in the hot cylinder, 𝑇ℎ. 𝑇ℎ = 𝑇𝑜𝑢𝑡,ℎ − 𝑞𝑖𝑛,ℎ ∗ (𝑅𝑡,𝑐𝑜𝑛𝑑,ℎ + 𝑅𝑡,𝑐𝑜𝑛𝑣,ℎ) (48) 𝑻 𝒉 = 𝟓𝟖𝟐. 𝟕𝟓 𝑲 (𝟓𝟖𝟗. 𝟐𝟕 °𝑭) Regenerator The regenerator plays an important role in improving the efficiency of the system as a whole. It functions as a heat bank. As the hot fluid flows to the cool cylinders, the regenerator absorbs some of the thermal energy in attempt to cool the fluid as much as possible before it reaches the compressive chambers. When the fluid returns towards the hot cylinders, the regenerator releases the thermal energy back to the fluid to raise its temperature to reduce the amount of energy required to expand the fluid and keep the cycle in motion. The following section will explore the various modes of heat transfer involved with the regenerator as the fluid travels between the hot and cool cylinders. Figure 47. Transparent view of the regenerator. Fluid flow 𝑇ℎ 𝑇𝑐 qref qref
- 42. 42 The surface temperatureof the regenerator will play a large role in its performance, therefore one must first consider its surface temperature due to the thermal radiation contained within the Paraboloid-dome system. 𝑇𝑠𝑠 = 𝑞 𝑟𝑒𝑓 𝜎 ∗ 𝜀 𝑠𝑠 ∗ 𝑆𝐴 𝑟𝑒𝑔𝑒𝑛 (49) The nature of the system dictates that convection is expected to be the dominant mode of heat transfer in the system. To determine the convective thermal resistance, one must obtain the coefficient of convective heat transfer, shown below. ℎ 𝑟 = 𝑁𝑢 𝐷𝑟 ∗ 𝑘ℎ 𝐷𝑡𝑢𝑏𝑒 (50) The geometry of the regenerator is already known, as is the conductance of the fluid, therefore one only needs to determine the Nusselt number, the Dittus-Boelter relation for cooling was selected based on the flow conditions [14]. 𝑁𝑢 𝐷𝑟 = 0.023 ∗ 𝑅𝑒 𝐷𝑟 4 5 ∗ 𝑃𝑟ℎ 0.3 (51) The Reynolds for this flow can me calculated using the 𝑅𝑒 𝐷𝑟 = 4 ∗ 𝑚̇ 𝜋 ∗ 𝐷𝑡𝑢𝑏𝑒 ∗ 𝜇ℎ (52) For calculating the Reynolds number the mass flow rate can be obtained based on fluid properties and the fluid velocity based on the angular velocity of the system, 𝑚̇ = 𝜌ℎ ∗ 𝑢ℎ ∗ 𝐴 𝑡𝑢𝑏𝑒. Using these relations, the coefficient of convective heat transfer can be obtained using 50 and the convective thermal resistivity calculated as follows 𝑅𝑡,𝑐𝑜𝑛𝑣,𝑟 = 1 2 ∗ ℎ 𝑟 ∗ 𝜋 ∗ 𝐷𝑡𝑢𝑏𝑒 ∗ 𝐿 𝑡𝑢𝑏𝑒 (53) The system specifics yield a value of 5.79 K/W. The conductive thermal resistivity of the regenerator is calculated via the following relation, approximating the pathway of the fluid to be cylindrical 𝑅𝑡,𝑐𝑜𝑛𝑑,𝑟 = ln 𝑡 𝑟𝑒𝑔𝑒𝑛 + 𝐷𝑡𝑢𝑏𝑒 𝐷𝑡𝑢𝑏𝑒 2 ∗ 𝑘 𝑟 ∗ 𝜋 ∗ 𝐿 𝑡𝑢𝑏𝑒 (54) This resistivity results in a value of 0.10 K/W.
- 43. 43 The total heatflux across the regenerator shell can be expresses as 𝑞 𝑟 = (𝑇ℎ − 𝑇𝑠𝑠) (𝑅𝑡,𝑐𝑜𝑛𝑑 + 𝑅𝑡,𝑐𝑜𝑛𝑣) (55) yielding a total flux of 3.54 W/m2 across the regenerator. Based on this analysis, the fluid temperature at the cool end of the regenerator can be calculated with the following relation 𝑇𝑐 = −𝑞 𝑟 𝑚̇ ∗ 𝐶 𝑝,ℎ + 𝑇ℎ (56) Therefore, the cool temperature across the regenerator is expected to be 528K. Heat Rejection The following section discusses the heat flow when the air enters the cool cylinder and is cooled by the water that the Stirling Solar Charger is suspended in. Once the effective temperature of the air in the cool cylinder is known, the heat rejected from the system can be calculated following the reverse method that was applied to the hot cylinder. First, internal convective heat transfer can be applied to get a convective thermal resistance. Using the geometry in figure 48, the thermal properties shown in Table X [14] assuming the velocity of the air in the cool cylinder 𝑢 𝑐 is the same as that of the air in the hot cylinder, 𝑢ℎ, the Reynold’s 𝑅𝑒 𝐷,𝑐 number can calculated as such: 𝑅𝑒 𝐷,𝑐 = 𝑢 𝑐 𝐷𝑖𝑐/𝜈𝑐 (57) 𝑅𝑒 𝐷,𝑐 = 12,599 Where 𝐷𝑖𝑐 is the inner diameter of the cool cylinder (0.9 in) and 𝜈𝑐 is the kinematic viscosity. Once again, using the Dittus-Boelter equation (for cooling), the Nusselt number of the cool cylinder air can be calculated [14] 𝑁𝑢 𝐷,𝑐 = 0.023𝑅𝑒 𝐷,𝑐 4 5⁄ Prc 0.3 (58) 𝑁𝑢 𝐷,ℎ = 39.1 Then, the convection heat transfer coefficient ℎ 𝑐 can be found: ℎ 𝑐 = 𝑁𝑢 𝐷,𝑐 𝑘 𝑐/𝐷𝑖𝑐 (59) ℎ 𝑐 = 45.0 𝑊/𝑚2 𝐾 and then the convection resistance 𝑅𝑡,𝑐𝑜𝑛𝑣,𝑐 can be found: 𝑅𝑡,𝑐𝑜𝑛𝑣,𝑐 = 1/(ℎ 𝑐 𝜋𝐷𝑖𝑐 𝐿 𝑐) (60) 𝑅𝑡,𝑐𝑜𝑛𝑣,𝑐 = 8.12 𝐾/𝑊 Where 𝜋𝐷𝑖𝑐 𝐿 𝑐 is the inner surface area of the cool cylinders.
- 44. 44 Figure 48. Geometryof the cool cylinder assembly: Lc = 1.5 in, Dic = 0.9 in, Doc = 1.125 in, and tc = (Doc-Dic)/2. TABLE X THERMAL PROPERTIES OF AIR AT 500K DENSITY, 𝛒 𝐜 (𝐤𝐠/𝐦 𝟑 ) DYNAMIC VISCOSITY, 𝛍 𝐜 (𝐍 ∙ 𝐬/𝐦 𝟐 ) KINEMATIC VISCOSITY, 𝛎 𝐜 (𝐦 𝟐 /𝐬) PRANDTL NUMBER, PrC THERMAL CONDUCTIVITY, 𝐤 𝐜 (W/M K) SPECIFIC HEAT, 𝒄 𝒑,𝒄 (J/(KG K)) 0.6964 270.1 x 10-7 38.7 x 10-6 0.684 26.3 x 10-3 1040 The conduction resistance of the cold cylinders, which are made of stainless steel (thermal conductivity, kss = 15.1 W/m K, [14]. 𝑅𝑡,𝑐𝑜𝑛𝑑,𝑐 = ln ( 𝐷 𝑜𝑐 𝐷𝑖𝑐 ) 2𝜋𝑡 𝑐 𝑘 𝑠𝑠 (61) 𝑅𝑡,𝑐𝑜𝑛𝑑,𝑐 = 0.8231 𝐾/𝑊 Where 𝐷 𝑜𝑐, 𝐷𝑖𝑐, and 𝑡 𝑐 are the geometric properties of the cold cylinder. On the outside of the cold cylinder heat is being dissipated via natural convection. The following relation can be used to obtain the Nusselt number for natural convection of a short, vertical cylinder, [20]. 𝑁𝑢 𝐷,𝑤 = 4 3 [ 7𝐺𝑟 𝐷,𝑤 𝑃𝑟𝑤 2 5(20 + 21Prw) ] 1 4⁄ + 4(272 + 315Prw) 35(64 + 63Pr 𝑤)𝐷 𝑜𝑐 (62) Where 𝐺𝑟 𝐷,𝑤 is the Grashof number of the cooling water about the cool cylinder. Technically, the Grashof number should be calculated with the temperature of the outer surface of the cylinder but since this value is unknown, the effective temperature of the cool cylinders 𝑇𝑐 is used instead. The thermal properties of water listed in Table XI are based on sea water at 17°C (290K). For simplicity, these properties were used despite the calculated water temperature, which is calculated as such with respect to the solar flux. 𝐷𝑖𝑐 𝐷 𝑜𝑐 𝐿 𝑐 𝑡 𝑐 Cool Cylinder Support Cool Cylinder qout qout
- 45. 45 𝑇 𝑤𝑎𝑡𝑒𝑟 =(𝐺 𝑎𝑡𝑚/𝜎𝜀 𝑤)1/4 (63) For clear day conditions and with water having an emissivity of 𝜀 𝑤 = 0.97, [14]. 𝑇 𝑤𝑎𝑡𝑒𝑟 = 356.97 𝐾 TABLE XI THERMAL PROPERTIES OF WATER AT 290K SPECIFIC VOLUME, 𝐯 𝐰 (𝐦 𝟑 /𝐤𝐠) DYNAMIC VISCOSITY, 𝛍 𝐰 (𝐍 ∙ 𝐬/𝐦 𝟐 ) KINEMATIC VISCOSITY, 𝛎 𝐰 (𝐦 𝟐 /𝐬) COEFFICIENT OF THERMAL EXPANSION, 𝜷 𝒘 PRANDTL NUMBER, PRW THERMAL CONDUCTIVITY, 𝐤 𝐰 (W/M K) 1.001 x 10-3 1.080 x 10-3 1.0811 x 10-6 1.740 x 10-4 7.56 0.598 Now, with this temperature, thermal properties of water, and acceleration due to gravity g as 9.81 m2 /s, the Grashof number 𝐺𝑟 𝐷,𝑤 can be calculated as follows: 𝐺𝑟 𝐷,𝑤 = 𝑔𝛽 𝑤(𝑇𝑐 − 𝑇 𝑤)/𝜈 𝑤 (64) 𝐺𝑟 𝐷,𝑤 = 5.868 × 106 Now, the Nusselt number of the water can be calculated using (62): 𝑁𝑢 𝐷,𝑤 = 53.8 Then, the heat transfer coefficient of the natural convection can be formulated: ℎ 𝐷,𝑤 = 𝑁𝑢 𝑤 𝑘 𝑤/𝐷 𝑜𝑐 (65) ℎ 𝐷,𝑤 = 1.1265 × 103 𝑊/𝑚2 𝐾 From here, the convection thermal resistance of naturally cooling water 𝑅𝑡,𝑐𝑜𝑛𝑣,𝑤 can be calculated with: 𝑅𝑡,𝑐𝑜𝑛𝑣,𝑤 = 1/(ℎ 𝑤 𝜋𝐷 𝑜𝑐 𝐿 𝑐) (66) 𝑅𝑡,𝑐𝑜𝑛𝑣,𝑤 = 0.2595 𝐾/𝑊 Now, that all three thermal resistances of the cool side have been found (convection inside the cool cylinder 𝑅𝑡,𝑐𝑜𝑛𝑣,𝑐, conduction through the cool cylinder 𝑅𝑡,𝑐𝑜𝑛𝑑,𝑐, and natural convection around the cool cylinder 𝑅𝑡,𝑐𝑜𝑛𝑣,𝑤), the heat rejection of a single cool cylinder 𝑞 𝑜𝑢𝑡,𝑐 can be calculated: 𝑞 𝑜𝑢𝑡,𝑐 = (𝑇𝑐 − 𝑇 𝑤𝑎𝑡𝑒𝑟)/(𝑅𝑡,𝑐𝑜𝑛𝑣,𝑐 + 𝑅𝑡,𝑐𝑜𝑛𝑑,𝑐 + 𝑅𝑡,𝑐𝑜𝑛𝑣,𝑤) (67) 𝑞 𝑜𝑢𝑡,𝑐 = 18.72 𝑊 The total heat rejection 𝑞 𝑜𝑢𝑡 is four times this amount to account for all the cool cylinders: 𝒒 𝒐𝒖𝒕 = 𝟕𝟒. 𝟖𝟕 𝑾
- 46. 46 5.2.2.6 Work Out Bythe first law of thermodynamics with no internal heat generation, the thermodynamic work output is the difference between net heat intake of the hot side 𝑞 𝑛𝑒𝑡𝑖𝑛 and the heat rejection of the cool side, 𝒒 𝒐𝒖𝒕: 𝑊𝑜𝑢𝑡 = 𝑞 𝑛𝑒𝑡𝑖𝑛 − 𝑞 𝑜𝑢𝑡 (68) 𝑊𝑜𝑢𝑡 = 111.3 𝑊 − 74.87 𝑾 𝒐𝒖𝒕 = 𝟑𝟔. 𝟒𝟒 𝑾
- 47. 47 5.3 FAILURE 5.3.1 TheoreticalAnalysis The drive shaft is considered a critical component of this design due to inherent loading conditions of the system. For this reason, the following failure analysis is performed on the drive shaft. It is determined through visual observation that the drive shaft will have critical points in the trough and in connection to the swashplate. In order to ensure high factors of safety and a long operating life, the drive shaft is analyzed using DE Goodman methodology. It is important to note that the swashplate-shaft system is modeled as a gear and shaft structure in order to easily evaluate the system. The shaft is analyzed in SolidWorks FEA and compared with the predicted theoretical model. All tables referenced in this section can be found in Section 11 Appendix. 5.3.1.1 Shaft Design Figure 49. Modeled drive shaft. From Table 7-1 Initial and initial guesses 𝐾𝑡 = 1.7, 𝐾𝑡𝑠 = 1.5, 𝐾𝑓 = 𝐾𝑡, 𝐾𝑓𝑠 = 𝐾𝑡𝑠 Using 1020 cold drawn steel as initial material 𝑆 𝑢𝑡 = 68 𝑘𝑝𝑠𝑖 Using Table 6-2 𝐾𝑎 = 𝑎𝑆 𝑢𝑡 𝑏 = 2.70(68)−0.265 = 0.883 Guess 𝐾𝑏 = 0.9, 𝑘 𝑐 = 𝑘 𝑑 = 𝑘 𝑒 𝑆 𝑒 ′ = 0.5𝑆 𝑢𝑡
- 48. 48 𝑆 𝑒 =𝑘 𝑎 𝑘 𝑏 𝑆 𝑒 ′ = (0.883)(0.9)(0.5)(68) = 27𝑘𝑝𝑠𝑖 DE Goodman equation to find smaller shaft diameter: 𝑑 = { 16𝑛 𝜋 ( 2(𝐾𝑓 𝑀 𝑎) 𝑆 𝑒 + [3(𝐾𝑓𝑠 𝑇 𝑚)2 ] 1/2 𝑆 𝑢𝑡 )} 1/3 (69) 𝑛 = 1.5, 𝑘 𝑓 = 1.7, 𝑀 𝑎 = 𝑇 𝑚 = 0.4425 𝑙𝑏𝑓 ∗ 𝑖𝑛, 𝑆 𝑒 = 27𝑘𝑝𝑠𝑖, 𝑆 𝑢𝑡 = 68 𝑘𝑝𝑠𝑖 𝑑 = { 16(1.5) 𝜋 ( 2(1.7)(0.4425) 27000 + [3(1.5)(0.4425)2]1/2 68000 )} 1/3 = .08217𝑖𝑛 Next standard size up is chosen (.09375 in.) For a 𝐷 𝑑 ≈ 1.2, 𝐷 = .125 𝑖𝑛 (standard size) . 125 . 09375 = 1.33 Fillet radius 𝑟 = 𝑑 10 = .01 𝑖𝑛, 𝑟 𝑑 = 0.11 𝑖𝑛 𝑘 𝑓 = 1 + 𝑞(𝑘 𝑡 − 1) 𝑘 𝑓𝑠 = 1 + 𝑞 𝑠(𝑘 𝑡𝑠 − 1) Using Figure A-15-9, A-15-8 𝑘 𝑡 = 1.6, 𝑘 𝑡𝑠 = 1.4 Using Figure 6-20 𝑞 = 0.5, 𝑘 𝑓 = 1 + 0.5(1.6 − 1) = 1.3 Using Figure 6-21 𝑞 𝑠 = 0.55, 𝑘 𝑓𝑠 = 1 + 0.55(1.4 − 1) = 1.22 Now must adjust 𝑘 𝑎 and 𝑘 𝑏 values: 𝑘 𝑎 = 0.883 𝑘 𝑏 = 1.24𝑑−0.107 = 1.24(. 09375)−0.107 = 1.597
- 49. 49 𝑆 𝑒 =(0.883)(1.597)(0.5)(68) = 47.945 𝑘𝑝𝑠𝑖 Finding the alternating Von Mises stress 𝜎 𝑎 ′ : 𝜎 𝑎 ′ = 32𝑘 𝑓 𝑀 𝑎 𝜋𝑑3 = 32(1.3)(0.4425) 𝜋(0.09375)3 = 7111.2 𝑙𝑏𝑓/𝑖𝑛3 Finding the midrange Von Mises stress 𝜎 𝑚 ′ : 𝜎 𝑚 ′ = √3 16𝑘 𝑓𝑠 𝑇 𝑚 𝜋𝑑3 = √3 16(1.22)(0.4425) 𝜋(0.09375)3 = 5779.5 𝑙𝑏𝑓/𝑖𝑛3 Checking fatigue factor of safety yields: 1 𝑛 𝑓 = 𝜎 𝑎 ′ 𝑆 𝑒 + 𝜎 𝑚 ′ 𝑆 𝑢𝑡 = 7111.2 47945 + 5779.5 68000 Which yields the following factor of safety: 𝑛 𝑓 = 4.286 5.3.1.2 Keyway Designed for 𝑟 𝑑 = 0.02 on 0.4375 in shaft 𝑟 = 0.02(0.4735) = .00875 𝑖𝑛 𝑘 𝑓 = 1 + 𝑞(𝑘 𝑡 − 1) 𝑘 𝑓𝑠 = 1 + 𝑞 𝑠(𝑘 𝑡𝑠 − 1) Using Table 7-1 𝑘 𝑡 = 2.14, 𝑘 𝑡𝑠 = 3.0 Figure 6-20 𝑞 = 0.45 𝑘 𝑓 = 1 + 0.45(2.14 − 1) = 1.513 Figure 6-21 𝑞 𝑠 = 0.5
- 50. 50 𝑘 𝑓𝑠 =1 + 0.5(3.0 − 1) = 2 Finding the alternating Von Mises stress 𝜎 𝑎 ′ : 𝜎 𝑎 ′ = 32𝑘 𝑓 𝑀 𝑎 𝜋𝑑3 = 32(1.513)(0.4425) 𝜋(0.4375)3 = 81.43 𝑙𝑏𝑓/𝑖𝑛3 Finding the midrange Von Mises stress 𝜎 𝑚 ′ : 𝜎 𝑚 ′ = √3 16𝑘 𝑓𝑠 𝑇 𝑚 𝜋𝑑3 = √3 16(2)(0.4425) 𝜋(0.4375)3 = 5779.5 𝑙𝑏𝑓/𝑖𝑛3 1 𝑛 𝑓 = 𝜎 𝑎 ′ 𝑆 𝑒 + 𝜎 𝑚 ′ 𝑆 𝑢𝑡 = 81.43 27000 + 93.27 68000 𝑛 𝑓 = 227.91 5.3.1.3 Key For 0.4375 in shaft 𝐹 = 𝑇 𝑟 = 0.4425 0.4375/2 = 2.02286𝑙𝑏𝑓 Distortion energy 𝑆𝑠𝑦 = 0.577(54) = 31.2 𝑘𝑝𝑠𝑖 Using UNS G10180 cold-drawn low-carbon mild steel as initial material 𝑆𝑠𝑦 𝑛 = 𝐹 𝑡𝑙 = 31200 1.5 = 2.022 0.125𝑙 𝑙 = .00184 (this is the minimum length of the key) 5.3.1.4 Snap Ring For 0.4375 in snap ring, Groove width = 0.029 in Depth = 0.025 in Diameter = 0.412 𝑟 𝑡 = 0.20
- 51. 51 𝑎 𝑡 = 0.029 0.025 = 1.16, 𝑟= (0.2)(0.025) = 0.005𝑖𝑛 From Figure A-15-16 𝑘 𝑡 = 4.4 From Figure 6-20 𝑞 = 0.45 𝑘 𝑓 = 1 + 𝑞(𝑘 𝑡 − 1) 𝑘𝑓 = 1 + 0.45(4.4 − 1) = 2.53 The fatigue factor of safety is then: 𝑛 𝑓 = 𝑆 𝑒 𝜎 𝑎 ′ = 27000 32𝑘 𝑓 𝑀 𝑎 𝜋𝑑3 = 27000 32(2.53)(0.4425) 𝜋(0.4375)3 = 198.27 From the analysis above, it can be seen that the modeled shaft diameters are sufficient. The shaft diameters of the design are much larger, and are therfore acceptable with respect to failure. Moreover, the keyway, key, and snap ring have large factors of safety, which shows that the chosen hardware is more than sufficient. Commercially availble couplers limit the shaft design parameters, which provides additonal opportunities for optimization
- 52. 52 5.3.2 SolidWorks FiniteElement Analysis Figure 50. Drive shaft modeled in SolidWorks with an applied torque. The FEA model assumes that the bottom of the shaft is fixed in all three planes and that the torque acts along the plane of the swashplate as shown above. These conditions lead to the stress results shown below. Unfortunately this critical analysis is not exact. Figure 51. The drive shaft is fixed on the bottom, which does not completely reflect the physical model of the shaft; however, it is able to reflect two critical points on the drive shaft.
- 53. 53 While there willbe critical points on the bottom of the shaft due to forces and friction with the ball and socket joint, the forces on the small area of the lower shaft shown above are due to the assumptions of the model and will not be as critical during normal operation. Another critical stress point is in the groove for the pistons to move through as the shaft rotates. Figure 52. Critical point in the trough of the swashplate. The stress point in the grove makes sense as that is the thinnest section of the part and has an inherent stress concentration due to the shape of the groves. In addition to this concentration, the real world shaft would also have stress points at the connection between the main shaft and the swashplate as this also would create a stress concentration at the sharp corner. The stress levels seen in this analysis are significantly lower than yield stress (by orders of magnitude) therefore the shaft will be considered to have an infinite fatigue life.
- 54. 54 6 DESIGN FORMANUAL ASSEMBLY 6.1 STEPS TO ASSEMBLE The steps required to assemble this product are outlined below in Table XX. The Boothroyd and Dewhurst method of assembly was used to accomplish this analyzation, and the overall assembly time is 924 seconds. TABLE XII MANUAL ASSEMBLY PROCESS Step Number Step Name Repetitions Total Time (s) Picture 1 Handle hot cylinder support 1 1.8 2 Handle and insert hot cylinders 1 2.63 3 Weld hot cylinder support 1 132
- 55. 55 4 Handle and insert hot cylinderseal 4 25.72 5 Handle hot cylinder assembly 1 1.8 6 Handle generator support 1 1.5
- 56. 56 7 Application of epoxy 4 50.9 8 Handleand insert starter cover 1 2.63 9 Handle generator support 1 1.8 10 Handle dry bag 1 1.8
- 57. 57 11 Application of epoxy 1 14.9 12 Handle floatation deviceand attach dry bag 1 7 13 Handle floatation device assembly 1 1.95 14 Handle cold piston 4 6
- 58. 58 15 Handle and insert ORings 8 53.52 16 Handle cold piston assembly 4 6 17 Handle cold cylinder support 1 1.8 18 Handle and insert cold seals 4 25.72
- 59. 59 19 Insert cold piston assembly 4 26 20 Handlecold cylinder assembly 1 1.8 21 Handle and insert swashplate 1 6.5
- 60. 60 22 Handle and insert regenerator 1 3.3 23 Handleand insert screws 4 32.9 24 Handle and insert hot cylinder support 1 3.3
- 61. 61 25 Handle and insert screws 432.9 26 Handle hot piston 4 6 27 Handle and insert O rings 8 53.52
- 62. 62 28 Handle hot piston assembly 4 6 29 Inserthot piston assembly 4 26 30 Handle and insert snap ring 1 6.69 31 Handle and insert key 1 2.63
- 63. 63 32 Handle and insert gear 14.3 33 Handle and insert bottom coupler 1 10.4 34 Handle sub assembly B 1 1.5
- 64. 64 35 Handle DC motor 1 1.8 36 Handleand solder buck converter 1 42.85 37 Handle and solder USB converter 1 12.85 38 Handle generator support assembly 1 1.5
- 65. 65 39 Handle and insert electronics assembly on generator support assembly 13.45 40 Handle and insert screws 4 34.1 41 Handle and insert USB cover 1 4.45
- 66. 66 42 Handle and insert top coupler 110.4 43 Handle sub assembly C 1 1.95 44 Handle Paraboloid 1 1.5 45 Handle and insert clip strap 3 12
- 67. 67 46 Handle and insert floatation device assembly 1 3.45 47 Placeand tighten straps around floatation device 3 16.5 48 Handle sub assembly A 1 3 49 Handle sub assembly B 1 1.5
- 68. 68 50 Handle and insert sub assemblyC 1 3 51 Weld legs of sub assembly C 4 168 52 Handle sub assembly A 1 3
- 69. 69 53 Application of epoxy 1 14.9 54 Handleand insert sub assembly B/C 1 4 55 Handle and insert water seal 1 16.59 Total: 924 Further details on the manual assembly can be found in Section 11 Appendix.
- 70. 70 6.2 WORK STATIONLAYOUT The workstation for manual assembly of each Solar Stirling Charger is based on figure 52 below. Each part will have its own 15 in. by 20 in. bin laid out in a straight line on an 84 in. by 134 in. table. The bins will have three rows stacked on top of each other for a total of 33 bins for each unique part. This comes to a total of 200 sq. ft. /work station. This workstation layout will be used as an estimate in the cost analysis for floor space, as well as floor space for the welding station. Figure 53. Workstation layout. Component Bins Table Fixture Epoxy and Lubrication Bin
- 71. 71 7 TOLERANCING ANDCLOSURE Five locations on the design were selected for closure analysis. This will ensure all of the components have the correct clearance when being assembled together. A table of machining tolerances can be found in Section 11 Appendix. 7.1 SHAFT AND COUPLER An analysis is conducted to determine the radius of the shaft knowing that a loose fit is necessary. An analysis is also conducted to determine the max gap tolerance between the coupler and the shaft. Figure 54. Coupler and shaft clearance. 𝐶𝑜𝑢𝑝𝑙𝑒𝑟 − 𝑆ℎ𝑎𝑓𝑡 − 𝐺𝑎𝑝 = 0 Shaft: 𝛥𝑆 = 𝑆′ (1 ± 𝜌𝑠) 𝜌𝑠 = 0.06 (Silica Investment Casting) 𝑆 𝑚𝑖𝑛 = 0.94𝑆′ 𝑆 𝑚𝑎𝑥 = 1.06𝑆′ Coupler: (Fine Machining) 0.001 = 𝛥𝐶 0.4375/2 𝛥𝐶 = 0.00021875
- 72. 72 𝑔 𝑚𝑖𝑛 =(𝐶) 𝐿𝑀𝐶 − (𝑆) 𝑀𝑀𝐶 𝑔 𝑚𝑖𝑛 = (0.21875 − 0.00021875) − (1.06𝑆′) = 0 𝑆′ = 0.20616 𝑔 𝑚𝑎𝑥 = (𝐶) 𝑀𝑀𝐶 − (𝑆) 𝐿𝑀𝐶 𝑔 𝑚𝑎𝑥 = (0.21875 + 0.00021875) − (0.94(0.20616)) 𝑔 𝑚𝑎𝑥 = 0.025178 7.2 STARTER AND SLOT An analysis is conducted to determine the width of the starter knowing that a loose fit is necessary. An analysis is also conducted to determine the max gap tolerance between the slot and the starter. Figure 55. Starter and support clearance. 𝑆𝑙𝑜𝑡 − 𝑆𝑡𝑎𝑟𝑡𝑒𝑟 − 𝑔𝑎𝑝 = 0 Starter : 𝛥𝑆𝑡 = 𝑆𝑡′ (1 ± 𝜌𝑠𝑡) 𝜌𝑠𝑙 = 0.002 (Machining) 𝑆𝑡 𝑚𝑖𝑛 = 0.998𝑆𝑡′ 𝑆𝑡 𝑚𝑎𝑥 = 1.002𝑆𝑡′ Slot: (Water Glass Investment Casting) 0.09 = 𝛥𝑆𝑙 0.33 𝛥𝑆𝑙 = 0.0297 2𝑔 𝑚𝑖𝑛 = (𝑆𝑙) 𝐿𝑀𝐶 − (𝑆𝑡) 𝑀𝑀𝐶
- 73. 73 2𝑔 𝑚𝑖𝑛 =(0.33 − 0.0297) − (1.002𝑆𝑡′) = 0 𝑆𝑡′ = 0.2997 2𝑔 𝑚𝑎𝑥 = (𝑆𝑙) 𝑀𝑀𝐶 − (𝑆𝑡) 𝐿𝑀𝐶 2𝑔 𝑚𝑎𝑥 = (0.33 + 0.0297) − (0.998(0.2997)) 2𝑔 𝑚𝑎𝑥 = 0.0302997 7.3 KEY AND GEAR An analysis is conducted to find the min and max gap tolerance between the gear and the key on the shaft. This analysis concludes that both loose and press fits are functional fits for the gear and key. Figure 56. Key and gear clearance. 𝐺𝑒𝑎𝑟 − 𝐾𝑒𝑦 − 𝑔𝑎𝑝 = 0 Key: (Silica Investment Casting) 0.06 = 𝛥𝐾 0.125 𝛥𝐾 = 0.0075 𝐾′ + 𝛥𝐾 = 0.125 ± 0.0075 Gear: (Machining) 0.005 = 𝛥𝐺 0.13 𝛥𝐺 = 0.00075 𝐺′ + 𝛥𝐺 = 0.13 ± 0.00065 2𝑔 𝑚𝑎𝑥 = (𝐺) 𝑀𝑀𝐶 − (𝐾) 𝐿𝑀𝐶
- 74. 74 2𝑔 𝑚𝑎𝑥 =(0.13 + 0.00065) − (0.125 − 0.0075) 𝑔 𝑚𝑎𝑥 = 0.006575 2𝑔 𝑚𝑖𝑛 = (𝐺) 𝐿𝑀𝐶 − (𝑆𝐾) 𝑀𝑀𝐶 2𝑔 𝑚𝑖𝑛 = (0.13 − 0.00065) − (0.125 + 0.0075) 𝑔 𝑚𝑖𝑛 = −0.00315 (Press Fit) 7.4 HOT CYLINDER AND HOT SUPPORT An analysis is conducted to determine the radius of a hot cylinder knowing that a loose fit is necessary. An analysis is also conducted to determine the max gap tolerance between the hot support and the hot cylinder. Figure 57. Cylinder and support clearance. 𝑆𝑢𝑝𝑝𝑜𝑟𝑡 − 𝐶𝑦𝑙𝑖𝑛𝑑𝑒𝑟 − 𝐺𝑎𝑝 = 0 Cylinder: 𝛥𝐶 = 𝐶′ (1 ± 𝜌 𝐶) 𝜌 𝐶 = 0.06 (Silica Investment Casting) 𝐶 𝑚𝑖𝑛 = 0.94𝐶′ 𝐶 𝑚𝑎𝑥 = 1.06𝐶′ Support: (Water Glass Investment Casting) 0.09 = 𝛥𝐶 0.56 𝛥𝐶 = 0.0504 𝑔 𝑚𝑖𝑛 = (𝑆) 𝐿𝑀𝐶 − (𝐶) 𝑀𝑀𝐶 𝑔 𝑚𝑖𝑛 = (0.56 − 0.0504) − (1.06𝐶′) = 0
- 75. 75 𝐶′ = 0.48075 𝑔 𝑚𝑎𝑥= (𝑆) 𝑀𝑀𝐶 − (𝐶) 𝐿𝑀𝐶 𝑔 𝑚𝑎𝑥 = (0.56 + 0.0504) − (0.94(0.48075)) 𝑔 𝑚𝑎𝑥 = 0.1585 7.5 PARABOLOID AND ENGINE An analysis is conducted to determine the radius of the lower Paraboloid opening knowing that a particular amount of space must be between the Paraboloid and the engine in order to apply sealant. An analysis is also conducted to determine the max gap tolerance between the Paraboloid and the engine. Figure 58. Clearance between Paraboloid and engine. 𝑃𝑎𝑟𝑎𝑏𝑜𝑙𝑜𝑖𝑑 − 𝐸𝑛𝑔𝑖𝑛𝑒 − 𝐺𝑎𝑝 = 0 Paraboloid: 𝛥𝑃 = 𝑃′ (1 ± 𝜌 𝑃) 𝜌 𝑃 = 0.008 (Injection Molding) 𝑃 𝑚𝑖𝑛 = 0.992𝑃′ 𝑃𝑚𝑎𝑥 = 1.008𝑃′ Engine: (Silica Investment Casting) 0.06 = 𝛥𝐸 2.5 𝛥𝐸 = 0.15 𝑔 𝑚𝑖𝑛 = (𝑃) 𝐿𝑀𝐶 − (𝐸) 𝑀𝑀𝐶
- 76. 76 𝑔 𝑚𝑖𝑛 =(0.992𝑃′) − (2.5 + 0.15) = 0.06 𝑃′ = 2.7319 𝑔 𝑚𝑎𝑥 = (𝑃) 𝑀𝑀𝐶 − (𝐸) 𝐿𝑀𝐶 𝑔 𝑚𝑎𝑥 = (1.008(2.7319)) − (2.5 − 0.15) 𝑔 𝑚𝑎𝑥 = 0.40371
- 77. 77 8 COST ANALYSIS Thelanded cost is totaled to be $832.22 and can be used along with the direct labor, indirect labor, burden labor, and facility expenses to approximate the total cost per engine if 8,000 engines are manufactured per month. In order to ensure an even number of workstation production, the actual number of engines produced per month is 8,104. For this analysis, one shift that consists of 160 hours per month per workstation will be used. Based on the production layout discussed previously, the facility expenses are approximated. The calculated assembly time of fifteen minutes per engine is used to approximate the number of work stations necessary. The electricity cost is estimated based on the average commercial rate in Florida [7]. The cost per square foot is found to be $7.50 per sq. ft. per year [8] and the millage rate is found to be approximately $24.55/1k$/year [9]. The price for umbrella insurance is estimated to be $3,000. The product liability rate is determined based on the average for a new business [10]. All other values are decided based on experience. TABLE XIII FACILITY EXPENSES ITEM VALUE UNITS Sq. ft. per workstation 200 (ft2 /WS) Number of workstations 13 (WS) Sq. ft. for workstation 2600 (ft2 ) Indirect sq. ft. factor 100 (%) Total sq. ft. (raw) 5200 (ft2 ) Total sq. ft. 5700 (ft2 ) Sq. ft. cost 7.50 ($/ft2 /year) Electricity cost 0.08 ($/kwh) Power per workstation 300 (W/WS) Brazing power 10 (kW) AC power 35 (kW) Cost for lights 100.21 ($/month) Cost for AC/heat 449.68 ($/month) Brazing power 256.96 ($/month) Value of property 200000.00 ($) Millage rate 24.55 ($/1k$/year) Umbrella insurance rate 3000.00 ($/year) Product liability rate 17 (%/year) Estimated annual sales 0.19 ($M/year) Rent 3562.50 ($/month) Electricity 806.85 ($/month) Phone/internet 500.00 ($/month) Water 300.00 ($/month) Waste 500.00 ($/month) License 100.00 ($/month) Property text 409.17 ($/month) Security 200.00 ($/month) Insurance 2970.00 ($/month) Supplies 1000.00 ($/month) Total 10348.52 ($/month)
- 78. 78 Next, the directlabor cost can be evaluated. This is the total cost associated with the employees at the workstations. The gross hourly rate is determined based on typical assembler pay. This is then multiplied by the total hours each direct laborer completes in one month. The taxes are found by multiplying the tax rate by the gross wages. The sum of the gross wages and taxes is equal to the total direct labor cost. This is equal to $0.40/min per workstation. TABLE XIV DIRECT LABOR Item Value Units Gross hourly rate 18.00 ($/hr) Gross wages 37440.00 ($/month) Taxes 7488.00 ($/month) Fringe 5000.00 ($/month) FICA/MC/UI/WC rate 20.00 (%) Total 49928.00 ($/month) Indirect labor costs will be analyzed next. This cost accounts for the employees that do not have direct contact with the product. The number of workers is found by assuming there are three runners, one maintenance person, and one shift manager. TABLE XV INDIRECT LABOR Item Value Units Number of workers 5 People Wage scale 1.25 Total 24003.85 ($/month) The burden labor is found by determining the salary for one engineer, one marketer, and one manager. TABLE XVI BURDEN LABOR Item Value Units Engineer 75000 ($/year) Marketing 80000 ($/year) Manager 90000 ($/year) Total 20416.67 ($/month) The use rate is then found to be $0.44/min/WS using the equation below where UR is the use rate, IDL is the indirect labor, BL is the burden labor, and FE is the facility expenses. 𝑈𝑅 $/𝑚𝑖𝑛 𝑊𝑆 = ( 𝐼𝐷𝐿($ 𝑚𝑜⁄ ) + 𝐵𝐿($ 𝑚𝑜⁄ ) + 𝐹𝐸($ 𝑚𝑜⁄ ) #𝑊𝑆 ) ( 1 𝑚𝑜 168 ℎ𝑟 ) ( 1 ℎ𝑟 60 𝑚𝑖𝑛 ) (70)
- 79. 79 The cost toassemble one engine is found using the equation below and yields $12.92. 𝐶𝑜𝑠𝑡 𝑡𝑜 𝑎𝑠𝑠𝑒𝑚𝑏𝑙𝑒 $ 𝑐𝑙𝑎𝑚𝑝 = 𝑇𝑖𝑚𝑒 𝑡𝑜 𝑎𝑠𝑠𝑒𝑚𝑏𝑙𝑒 𝑚𝑖𝑛 𝑐𝑙𝑎𝑚𝑝 ∗ (𝑈𝑅 $ 𝑚𝑖𝑛 + 𝐷𝐿 $ 𝑚𝑖𝑛 ) (71) The total cost per engine is found by adding the cost to assemble to the landed cost per engine. This totals to $845.14 per engine. The overhead is found to be 110% by dividing the use rate by the direct labor and multiplying by 100%. The fully burdened labor rate is then found to be $50.34/hour by summing the direct labor and the use rate. The total monthly budget is summarized below. TABLE XVII MONTHLY BUDGET Item Value Units Direct Labor 49928.00 ($/month) Indirect Labor 24003.85 ($/month) Burdened Labor 20416.67 ($/month) Facilities Expenses 10348.52 ($/month) Landed Cost 6744250.50 ($/month) Total 6848947.54 ($/month) The minimum sales per year in order to break even is approximately $82.19 million. This means that each engine would need to be sold for $845.14.
- 80. 80 9 INTELLECTUAL PROPERTY Theinvention claimed is a multiple-cylinder single-acting engine which utilizes a Stirling cycle wherein the hot and cold cylinders act on opposing sides of a swashplate in order to rotate the swashplate as shown in figure 59. This design allows the hot and cold cylinders to be separated in an effort to increase the temperature difference between the cylinders while still generating a large force on the swashplate. In addition, this engine is solar-powered wherein the devices employs a self-reflecting Paraboloid surface as shown in figure 60. This feature allows the system to be one assembly in order to decrease the size and increase the portability. Figure 59. Double-sided swashplate. Figure 60. Self-reflective Paraboloid Stirling engine.
- 81. 81 10 EVALUATION Based onthe analysis, the product meets the specifications determined in the preliminary phase of the design process, i.e., a portable 5 W Stirling engine powered USB charger. These specifications can be seen in Section 3 Performance Specifications. From the solar input (about 225 W based on the calculation above in Section 5), the solar to thermal efficiency is 16.2%. The thermal to mechanical efficiency is 44.6%, which results in 16.27 W of power from the engine. The mechanical to electrical efficiency is 50%, creating a final electrical output of 8.11 W. Based on these efficiencies, the design has a 3.6% overall efficiency. This exceeds the 5 W specification for the device. It is thought that the excess power can be used to charge additional devices, remain as a safety buffer (for charging in low light/cloudy conditions), or allow the design to be reduced/optimized in further prototypes. The final weight of the assembly is approximately 45 lbs., and it measures 25 in. in diameter. This puts the device within the parameters of man portable according to MIL-STD 1472D [4] and can be carried with two hands. This weight is considered to be fatiguing for a single person to carry, but the user would only have to physically carry it a short while, as the device is buoyant. Nevertheless, weight reduction is a point for future optimization. The stress analysis of the system shows that the factor of safety is higher than 4 on the critical components, which will provide a long fatigue life and leaves room for optimization. Fatigue life due to heating, solar radiation damage, salt water corrosion, and impact strength have not been considered and should be added in further evaluations. When producing 8 000 units, final cost of the device comes to $845.14 with an overhead of 110%. The sale of this product is primarily for high volume orders (e.g. FEMA or cruise lines) so this is not an unreasonable price range. The ideas used in this design are open to the possibility of patents which secure sales of this device from copycat devices on the market. Overall the device is a viable product and is ready for optimization and testing in further prototyping stages.
- 82. 82 11 APPENDIX 11.1 TABLESFOR SHAFT FAILURE ANALYSIS
- 83.
- 84.
- 85. 85 11.2 MANUAL ASSEMBLYTIME TABLE Notes Step Number Step Name Step Description Repetitions Tool Access Alpha Beta Sum Handling Code Handling Time Insertion Code Insertion Time Total Time 1 Handle hot cylinder support Initial handling of hot cylinder support 1 0 360 180 540 20 1.8 0 1.8 2 Handle and insert hot cylinders Insert hot cylinders into hot cylinder support 1 0 180 90 270 00 1.13 00 1.5 2.63 3 Weld hot cylinder support Weld hot cylinders to hot cylinder support 1 120 NA NA NA 0 96 12 132 4 Handle and insert hot cylinder seal Place hot cylinder seal into hot cylinder support 4 0 180 0 180 01 1.43 31 5 25.72 5 Handle hot cylinder assembly Place hot cylinder assembly aside 1 0 360 180 540 20 1.8 0 1.8
- 86. 86 6 Handle generator support Initial handling of generator support 1 0360 90 450 10 1.5 0 1.5 7 Application of epoxy Apply epoxy below dowel pin of generator supports 4 2.9 NA NA NA 0 97 12 50.9 8 Handle and insert starter cover Insert starter cover until stopped by generator support pins 1 0 180 90 270 00 1.13 00 1.5 2.63 9 Handle generator support Place generator support assembly aside to cure 1 0 360 180 540 20 1.8 0 1.8 10 Handle dry bag Initial handling of dry bag 1 0 360 180 540 20 1.8 0 1.8 11 Application of epoxy Apply epoxy on bottom of dry bag 1 2.9 NA NA NA 0 97 12 14.9
- 87. 87 12 Handle floatation device and attach dry bag Initial handlingof floatation device and insertion of dry bag 1 0 360 0 360 10 1.5 06 5.5 7 Pre- Assmbly Above 13 Handle floatation device assembly Place floatation device assembly aside 1 0 360 360 720 30 1.95 0 1.95 14 Handle cold piston Initial handling of cold pistons 4 0 360 0 360 10 1.5 0 6 15 Handle and insert of O Rings Initial handling of O ring and place O rings in groove of cold pistons 8 0 180 0 180 03 1.69 31 5 53.52 16 Handle cold piston assembly Place cold pistons aside 4 0 360 0 360 10 1.5 0 6 17 Handle cold cylinder support Initial handling of cold cylinder support 1 0 360 180 540 20 1.8 0 1.8
- 88. 88 18 Handle and insert cold seals Placecold selas into cold cylinder support 4 0 180 0 180 01 1.43 31 5 25.72 19 Insert cold piston assembly Handle and insert cold piston assembly into cold cylinder 4 0 360 0 360 10 1.5 31 5 26 Using tooling support to raise and hold assembly 20 Handle cold cylinder assembly Place cold cylinder assebly on tooling fixture 1 0 360 180 540 20 1.8 0 1.8 Piston joints are press fit 21 Handle and insert swashplate Place swashplate on ball joint of cylinder support 1 0 360 0 360 10 1.5 31 5 6.5 Dowl pins help to align the regenera tor 22 Handle and insert regenerator Place regenerator on cold cylinder support and align 1 0 360 180 540 20 1.8 00 1.5 3.3
- 89. 89 Screw gun does not interfere due to fixture 23 Handleand insert screws Secure regenerator to cold cylinder 4 2.9 360 0 360 10 1.5 38 6 32.9 Dowl pins help to align the regenera tor 24 Handle and insert hot cylinder support Place hot cylinder support on top of regerator and align 1 0 360 180 540 20 1.8 00 1.5 3.3 25 Handle and insert screws Secure regenerator to hot cylinder 4 2.9 360 0 360 10 1.5 38 6 32.9 26 Handle hot piston Initial handling of hot pistons 4 0 360 0 360 10 1.5 0 6 27 Handle and insert of O rings Initial handling of O ring and place O rings in groove of hot pistons 8 0 180 0 180 03 1.69 31 5 53.52 28 Handle hot piston assembly Place hot pistons aside 4 0 360 0 360 10 1.5 0 6
- 90. 90 29 Insert hot piston assembly Insert hot piston assemblyinto hot cylinder 4 0 360 0 360 10 1.5 31 5 26 30 Handle and insert insert snap ring Insert snap ring onto shaft 1 0 180 0 180 03 1.69 31 5 6.69 31 Handle and insert key Insert key into keyway of shaft 1 0 180 90 270 00 1.13 00 1.5 2.63 32 Handle and insert gear Insert gear onto keyway of shaft on top of snap ring 1 0 180 360 540 20 1.8 02 2.5 4.3 Tightenin g of set screw 33 Handle and insert bottom coupler Insert and secure bottom coupler 1 2.9 360 0 360 10 1.5 38 6 10.4 34 Handle sub assembly B Set aside sub assembly A 1 0 360 90 450 10 1.5 0 1.5 35 Handle DC motor Initial handling of DC motor 1 0 360 180 540 20 1.8 0 1.8
- 91. 91 36 Handle and solder buck converter Handlingof buck converter and sodering wires of DC motor to buck converter 1 32.9 360 360 720 30 1.95 95 8 42.85 37 Handle and solder USB converter Handling of USB converter and sodering wires of USB to buck converter 1 2.9 360 360 720 30 1.95 95 8 12.85 38 Handle generator support assembly Initial handling of generator support assembly 1 0 360 90 450 10 1.5 0 1.5
- 92. 92 39 Handle and insert electronics assembly on generator support assembly Insertmotor into hole of generator support and place electronics next to motor 1 0 360 360 720 30 1.95 00 1.5 3.45 M3x10m m screws 40 Handle and insert screws Insert and secure bottom of motor with screws 4 2.9 360 0 360 11 1.8 38 6 34.1 Using Press Fit 41 Handle and insert USB cover Insert and secure USB cover onto generator assembly 1 0 360 360 720 30 1.95 02 2.5 4.45 42 Handle and insert top coupler Insert coupler onto top shaft and tighten set screw 1 2.9 360 0 360 10 1.5 38 6 10.4 43 Handle sub assembly C Set aside sub assembly B 1 0 360 360 720 30 1.95 0 1.95
- 93. 93 44 Handle paraboloid Initial handling of paraboloid 1 0360 120 480 10 1.5 0 1.5 45 Handle and insert clip strap Handling of strap and insertion into clip of paraboloid 3 0 180 180 360 10 1.5 02 2.5 12 Make sure floating device is dry bag up 46 Handle and insert floatation device assembly Handling of floatation device and insertion of paraboloid into floatation device 1 0 360 360 720 30 1.95 00 1.5 3.45 47 Place and tighten straps around floatation device Handling of straps around floatation device and tightening 3 0 360 120 480 10 1.5 32 4 16.5 48 Handle sub assembly A Set aside sub assembly C 1 0 360 360 720 91 3 0 3 49 Handle sub assembly B Initial handling of sub assembly A 1 0 360 90 450 10 1.5 0 1.5
- 94. 94 50 Handle and insert sub assemblyC Place sub assembly C into the grooves of sub assembly B 1 0 360 90 450 10 1.5 00 1.5 3 51 Weld legs of sub assembly C Weld legs of sub assembly C onto the hot cylinder support of sub assembly B 4 120 NA NA NA 0 96 12 168 52 Handle sub assembly A Initial handling of sub assembly A 1 0 360 360 720 91 3 0 3 53 Application of epoxy Epoxy inside of paraboloid of assembly A 1 2.9 NA NA NA 0 97 12 14.9
- 95. 95 54 Handle and Insert sub assembly B/C Handlingof sub assembly B/C and insertion into sub assembly A 1 0 360 360 720 95 4 0 4 55 Handle and insert water seal place water seal on top of paraboloid 1 2.9 180 0 180 03 1.69 97 12 16.59 Total Time (s) 924
- 96.
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