# Solar Oven Equations by dffhrtcv3

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```									Approaches to Designing a Solar Oven
All good engineering design starts with a clear understanding of the project’s needs, goals,
budgetary constraints, material constraints, performance tolerances, and criteria (termed the
“Performance Index” or P.I.) for judging the effectiveness and success of the final product.
Once all of these design specifications are clearly understood, the process of actually
designing the product may begin. (Note that in the “real world” any or all of these
specifications usually change many times throughout the design process. For this project in
ENGR102, they are clearly defined and are not likely to change during your work on the
project.)

One possible approach to producing a solar oven would be to design and build an actual
oven, within the limits of the design criteria. You would then test the oven and measure the
Performance Index that you achieved with your first design. Unless you were extremely
lucky, it is highly unlikely that this first oven would achieve the desired P.I. You would then
have to redesign and build many other ovens, hoping with each cycle of design and test to
achieve the P.I. that the customer needs. Even with something as seemingly as simple as a
cardboard solar oven, this could take weeks, if not months!

Another approach would be to use Design of Experiments (DOE). Using DOE, one could
(hopefully) identify the most important factors in a solar oven, and then make good decisions
about the levels to test them at. Once these values were chosen, you could build all of the
ovens needed to test a full-factorial combination of these factors and levels. Then, using the
resulting DOE predictive model, you could build additional ovens and test them, until you
achieved the desired Performance Index. Again, this could take a very long time.

The solar oven project allows yet another approach to design, that of first building a
mathematical model used to predict a value for the Performance Index. Based on knowing
the physics of how a solar oven actually works, we are able to write down equations that
predict the flow of energy into and out of the oven cavity. For the solar oven, this works
because these equations are simple equations with closed-form solutions that are easily
solved. This allows you to calculate and predict the internal temperature of the oven cavity.
In turn, an equation for the Performance Index can then be written down and solved for, and
used as the basis for the actual design of a specific oven.

If a spreadsheet is used to solve these equations and to calculate the Performance Index, it is
easy to iterate solutions by changing the input variables (in effect, changing the oven design),
until the desired P.I. is achieved. (At this point, no time or money has yet been spent on
actually building any ovens.) Finally, the oven that achieves the highest calculated P.I. can
then be built and tested, using the values of the input variables in the spreadsheet. If the
model is accurate and complete, the P.I. that is actually measured should be close to the
desired value as predicted by the spreadsheet calculations.

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Modeling a Solar Oven
The approach we will use to design a solar oven is to create a mathematical model of the type
of oven that we actually desire to build. Basically, this will be an insulated “box-within-a-
box”, most likely covered by a transparent window. Figures 1 and 2 show cut-away views of
such a structure, placed in the sun.

A solar oven is a device that, regardless of its specific form or shape, converts solar energy
(in the form of visible and near-infrared light) into thermal energy (heat). (Appendix A
contains a more detailed description of this process.)

After placing the oven in the sun, the internal temperature of the oven cavity will begin to
rise, and will eventually reach a constant (steady-state) value. At this point in time, the oven
is said to be in thermal equilibrium. At thermal equilibrium, basic thermodynamics states
that the energy flowing into the oven must be equally balanced by the energy that flows out
of the oven (otherwise the oven would continue to heat up to an infinitely-high temperature,
which it does not do). On a per-time basis, this also says that the power absorbed by the
oven equals the power that leaves out of the oven.

It is perhaps more intuitive to think of this in “reverse,”—the energy that flows out of the
oven must equal the energy that flows into the oven. Intuitively, this seems like a more
direct ‘cause-and-effect’ relationship—the energy flowing out of the oven (the “dependent
variable”) comes from the energy that flows into the oven in the first place (the “independent
variable”). So, if you increase (or decrease) the energy flowing in, you also increase (or
decrease) the energy flowing out of the oven. After some time, the oven once again reaches
thermal equilibrium, with an associated increase (or decrease) in the internal oven
temperature. Equation 5.1 is a statement of thermal equilibrium for our solar oven:

Eout = Ein
(eq. 1)
Pout = Pabsorbed

Our goal is to write down equations for both Pout and Pabsorbed, as a function of the design and
material properties of our actual oven (the orientation of the oven to the sun, the size and
shape of the internal cavity, the color of the interior walls, the effects of using a window, the
type and amount of insulation used, the effects of using reflectors, etc.). Setting these
equations equal, we will be able to solve for the internal temperature of the oven in terms of
all of these material properties. This becomes the model for our oven, and can be used as a
design tool (especially if we do all of this using a spreadsheet, to iteratively change the
material properties and solve for a new, hopefully higher internal temperature).

Power Absorbed by the Oven—Finding the Equation for Pabsorbed

●       To calculate the power absorbed by the internal cavity of the oven, we first calculate
the solar power intercepted by the window of the cavity. This will be the solar power

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density, Io (watts/m2) at the surface of the earth times the area of the window “seen” by the
sun’s rays. This is the area of the window that is normal, or perpendicular, to the sun’s rays.

To calculate this area, consider Figure 2, which shows a cut-away side view of a solar oven
placed in the sun. The sun’s rays make an angle θs with respect to the ground, and the
window (therefore also the top surface of the oven) makes an angle β with respect to the
ground. The length of the window is L, and the width of the window (the dimension into the
paper) is W. The area of the window is given by the product of these two numbers:

Aw = W ⋅ L = area of the window                      (eq. 2)

However, the area of the window “seen” by the sun (the projected area that is perpendicular
to the sun’s rays) is equal to y×W. From Figure 2 and by considering the right triangle
having y as its base and L as its hypotenuse, it may be seen that:

y = L sin (θ s + β )                                 (eq. 3)

Combining these expressions gives:

Pincident to window = I oW ⋅ L sin (θ s + β )        (eq. 4)

where:         I o =incident solar power density, in watts/meter2
W = width of the window (into the paper), in meters
L = the length of the window, in meters

Pincident to window = I o Aw sin (θ s + β )          (eq. 5)

Note that θs is a variable that is out of our direct control. It depends on where the sun is in
the sky, which for any given location such as Tucson, AZ, depends on the day of the year and
the time of day that the oven is tested. On the other hand, the window angle β is a variable
over which you have direct control. It is said to be a design variable. To maximize the
power incident to the window of the oven, you want to mathematically maximize the sine
function by adjusting β so that sin (θ s + β ) = 1 . Physically, this is accomplished by
continually aligning the oven during testing so that θ s + β =90°. In other words, during
testing you must continually align the oven so that the sun’s rays strike perpendicular to the
window.

To actually do this, you need to tilt the top surface of the oven to the appropriate angle β,
using some kind of simple pointing device attached to the top surface of the oven (hint: think
“sundial”). This is made somewhat challenging, because angle β will change during the time
that it takes to test the oven (because during testing θs changes as the sun moves across the
sky). As an example, if at any particular moment in time the sun’s rays make an angle of 40°
with respect to the ground, then the oven must be tilted at 50° with respect to the ground so

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that θ s + β =90°. Figure 1 shows how the oven is aligned so that the incident rays of the sun
are perpendicular to the window.

●      Next, we need to calculate the power that is transmitted through the window, into the
oven cavity. This is simply the incident power times the optical transmission of the window
material:

Ptransmitted = τ ⋅ Pincident to window                         (eq. 6)

where:         τ = the optical transmission coefficient of the window material (0 ≤ τ ≥ 1)

Note that if more than one window is used, substitute τ=τn where n is the number of
windows.

●      Finally, we need to calculate the power that is actually absorbed by the walls of the
oven cavity. This is simply the power that has been transmitted through the window, times
the absorption of the cavity walls:

Pabsorbed = a ⋅ Ptransmitted                                   (eq. 7)

where:         a = the absorption coefficient of the cavity walls (0 ≤ a ≥ 1)

●       Combining these equations gives the power absorbed, in terms of the solar constant
Io, and design variables Aw, τ, a, and β.

Pabsorbed = I o Aw ⋅ τ ⋅ a ⋅ sin (θ s + β )                   (eq. 8)

From an engineering point of view, we want the power absorbed to be as large as possible.
To achieve this, we want both τ and a to be as close to 1 as possible. This implies the use of
a window material that is highly transparent across the visible and near-infrared spectrum
(see Appendix A). A typical value for τ is ≈ 0.92. It also implies that the walls should be as
black as possible. A typical value for a is ≈ 0.9-0.96.

Power That Leaves the Oven—Finding the Equation for Pout

In general, heat energy is transferred from a hot region to a cold region through the three
physical processes of heat transfer: convection, conduction, and radiation. This is expressed
by the following equation:

P = U ⋅ A ⋅ (Tsurface − Tregion INTO which power   flows   )   (eq. 9)
where:
U = overall heat transfer coefficient [units of Watts/(°C·m2)]
A = area through which the heat flows
T = temperature

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This equation implies that less power is transferred from a surface to another region if the
area of the surface is decreased, or if the heat transfer coefficient is somehow decreased.
This might be useful to remember when designing a solar oven!

Equation 9 is a simplified (but useful) “lumped-effect” model, in that the heat transfer
coefficient U accounts for all of the heat transfer processes: radiation, conduction, and
convection. A more complex model would include each of these effects separately.

An equation that is more suitable for our solar oven includes two separate heat-loss terms: a
term describing how much heat flows out of the sides and bottom of the oven, and a term that
describes how much heat is lost out of the window:

Pout = (U sb ⋅ Asb )(Tio − Tambient ) + (U w ⋅ Aw )(Tio − Tambient )   (eq. 10)

Pout = (U sb ⋅ Asb + U w ⋅ Aw )(Tio − Tambient )              (eq. 11)

where:
Usb = the overall heat transfer coefficient for the sides and bottom of the
interior oven chamber
Asb = the total area of the sides and bottom of the interior oven chamber
Uw = the overall heat transfer coefficient for the window
Aw = the area of the window
Tio = the temperature in the interior chamber of the oven
Tambient = the temperature of the air surrounding the oven
(the outdoor temperature on testing days)

It is worth noting that in this equation, Asb, Aw, Usb, Uw are all design variables:
▪ Asb is set by the dimensions you pick for the sides and bottom of the interior chamber.
▪ Aw is set by the dimensions you pick for the window.
▪ Usb is determined by the material you pick for the walls and insulation.
▪ Uw is determined by the material you pick for the window. It is also a function of Tio.

Values for Uw for a mylar window (Xerox transparency) have been measured experimentally
and are given in Table 4. Note that the value for Uw depends on whether 1 or 2 windows are
used (similar to using single-pane or double-pane windows in a home). Also according to the
data, Uw depends on Tio, the internal temperature of the oven cavity. As the numbers show,
more heat is lost from the window the hotter the internal oven chamber gets.

For this particular design of solar oven (a box within a box), it has been experimentally
verified that the major heat loss mechanism through the sides and bottom of the internal
chamber is conduction to the surrounding insulation. This allows us to develop a slightly
more sophisticated and generalized model for the Usb term, by including the effects of the
cardboard walls and insulation as separate terms. The result is that you are able to model
more design possibilities in how you insulate your oven.

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This is done using the following equation, where each separate term describes conductive
heat flow through various parts of the oven: the cardboard walls of the inner chamber,
insulation material (newspaper, “R-19 pink fiberglass”, etc.), and the cardboard walls of the
outer box.

1   x x             x
R≡        = 1 + 2 + ..... + n                             (eq. 12)
U sb k1 k 2          kn

where xi is the thickness [meters] of the ith material and ki is the coefficient of thermal
conductivity [Watts/(m·°C)] of the ith material. The resultant Usb still has units of
[Watts/( m2·°C)].

Many engineering textbooks have tables of k-values. A website tabulation of k’s for many
substances is: http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html

The R in this equation is a measure of the thermal resistance to the flow of heat energy. Note
that it is the inverse of U, which describes how well a material (or different materials in
series) conducts heat. This R is equivalent to the “R-value” quoted for home insulation
materials (the “R-19 pink fiberglass” type of glass insulation commonly used in walls, for
example). For solar oven design, we desire to have a small value for U (or equivalently a
large R-value) in the material(s) we use to insulate the interior oven chamber. This will help
reduce Pout, the power that leaves the oven.

The Internal Temperature of the Oven—Finding the Equation for Tio

We are now in a position to solve for the internal temperature of the oven. To do this, we use
equation 1, and set Pout = Pabsorbed. As discussed earlier, this is an expression of the fact that
after some time in the sun, the oven will reach thermal equilibrium. At this point, the internal
temperature of the oven will reach a maximum, steady-state value Tio:

(U sb ⋅ Asb + U w ⋅ Aw )(Tio − Tambient ) = I o Aw ⋅τ ⋅ a ⋅ sin (θ s + β )   (eq. 13)

I o Aw ⋅τ ⋅ a ⋅ sin (θ s + β )
Tio = Tambient +                                                   (eq. 14)
(U sb ⋅ Asb + U w ⋅ Aw )

Constants in this equation:                Io, θs, and Tambient
Design variables in this equation:         Aw, τ, a, β, Usb, Asb, and Uw

At first glance, it seems simple enough to use this equation and solve for Tio. However, there
is a hidden problem—Uw depends on Tio, as was seen in Table 4. Said another way, Uw is a
function of Tio, written mathematically as Uw(Tio). In equation 14, Tio is actually still on both
sides of the “equals” sign!

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The Hidden Problem, Uw(Tio)—Solving for Tio

Equation 14 can be uniquely solved for Tio in one of 3 different ways: graphically,
algebraically, or numerically.

Solving for Tio graphically—Using this approach, you solve for Tio by finding the intersection
point of 2 curves on a single graph. It is easy and straight-forward, but does require that you
“manually” read the point of intersection, with some loss of accuracy in finding Tio (probably
not a major concern for the accuracies we need in our design process).

Solving for Tio algebraically—Using this approach, you solve for Tio by first curve-fitting the
Uw vs. Tio data in Table 4 (each column separately, for 1 window or 2 windows). A
spreadsheet such as Excel makes this easy, and a 2nd-order polynomial will do the job nicely.
Then, you substitute this polynomial equation into equation 14, in place of the Uw term. At
this point, equation 14 contains Tio explicitly on both sides of the “equals” sign—it is no
longer ‘hidden.’ Finally, after much algebraic manipulation, you can uniquely solve for Tio.
This approach has the advantage that using a spreadsheet, a numerical value for Tio can
automatically be solved for, and placed in a cell.

Example: The result of curve-fitting the Uw vs. Tio data in Table 4 using a 2nd-order
polynomial will give you an equation of the general form:

U w = aTio + bTio + c
2
(eq. 15)

Substituting this into equation 14 gives:

I o Aw ⋅ τ ⋅ a ⋅ sin(θ s + β )
Tio = Tambient +
(                     2
U sb ⋅ Asb + {aTio + bTio + c} ⋅ Aw    )   (eq. 16)

The only unknown is Tio, which can be solved for algebraically (after some work!).

Solving for Tio numerically—Using this approach, you solve for Tio by first curve-fitting the
Uw vs. Tio data in Table 4 (each column separately, for 1 window or 2 windows). A
spreadsheet such as Excel makes this easy, and a 2nd-order polynomial will do the job nicely.
Then, you substitute this polynomial equation into equation 14, in place of the Uw term. (So
far, this is exactly the same approach as for the algebraic approach, described above. And, as
for the algebraic approach, equation 14 will contain Tio explicitly on both sides of the
“equals” sign.) Now, however, you rearrange equation 14 and set all of the terms equal to 0:

I o Aw ⋅τ ⋅ a ⋅ sin(θ s + β )
Tio − Tambient −
(U                   2
sb ⋅ Asb + {aTio + bTio + c} ⋅ Aw   )=0           (eq. 17)

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To solve this equation numerically essentially means that you pick a value for Tio, substitute
it into the equation, and see how close the result is to 0. You keep picking new values for Tio
and substituting, until the left-hand side of the equation calculates to exactly 0. (There are
sophisticated ways to do this mathematically, and entire courses are taught on this topic,
called “Numerical Analysis”).

A very easy way to do this is to use the “Goal Seek” function in Excel. From the Excel help
“Goal Seek is part of a suite of commands sometimes called what-if analysis tools--(what-if analysis: A process
of changing the values in cells to see how those changes affect the outcome of formulas on the worksheet. For
example, varying the interest rate that is used in an amortization table to determine the amount of the
payments.) When you know the desired result of a single formula, but not the input value the formula needs to
determine the result, you can use the Goal Seek feature available by clicking Goal Seek on the Tools menu.”

“If you know the result you want from a formula, but not the input value the formula needs to get that result,
you can use the Goal Seek feature. For example, use Goal Seek to change the interest rate in cell B3 until the
payment value in B4 equals (\$900.00).”

For our problem, we know that the “formula” (the left-hand side of equation 17) must have a
result that equals 0. We don’t know the “input value”, Tio, but Goal Seek can be used to
numerically solve for it (almost instantaneously). Goal Seek will display the value for Tio
conveniently in a cell.

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Modeling a Solar Oven with Reflectors (Mirrors)
The use of reflectors can substantially increase Tio, the temperature of the internal cavity, by
delivering more sunlight to the window of the oven and into the oven cavity. For this
project, an important design constraint is that the reflectors must not have curved surfaces
(they must be flat). Curved surfaces actually concentrate (focus) sunlight at the window,
effectively increasing the value of Io to orders of magnitude above 1000 W/m2. The results
are impressive, but dangerous and disastrous—having an entire oven go up in flames is not
part of our desired Performance Index!

The use of flat reflectors technically doesn’t increase Io, but increases the amount of sunlight
at the window. We will model this using a “gain” term—a unit less number that multiplies
Pabsorbed, the power absorbed by the oven without a reflector:

Pabsorbed with a reflector = G ⋅ Pabsorbed without a reflector    (eq. 18)

Pabsorbed with a reflector = I o Aw ⋅ G ⋅τ ⋅ a ⋅ sin (θ s + β )   (eq. 19)

The gain term G, as we will see, is a number that depends solely on the geometry (size and
angular position) of the reflectors.

The Geometry of the Reflectors—How the “Lid Angle” Ω Relates to M/L

The following discussion and derivation are made with reference to Figures 3 and 4, which
show an oven tilted at an angle β such that the sun’s rays strike the window at normal
incidence. The following discussion and derivation assume this to be true—it is an
assumption to our model. The equation we will derive for G is based on this assumption, and
is valid only if this is true when you experimentally test the oven.

First consider Figure 3, to learn the nomenclature we will use in discussing the geometry of a
reflector. The length of the window is L, and the width (into the drawing) is W, as before.
The length of the reflector is M, and the sunlight makes an angle of incidence α with respect
to the surface of the reflector. Due to the optical Law of Reflection, this is the same angle
that the sunlight makes in reflection from the surface of the reflector (see “Ray 2”). Finally,
the reflector itself makes an angle Ω with respect to the oven window. It is easy to remember
this angle Ω if you think of it as the “lid angle.”

Once you decide on values for L and M and cut cardboard, the lengths of the window and
reflector are fixed. The remaining variable is Ω, the angle that you affix the reflector (the
“lid”) to the top surface of the oven. At first glance it might seem that its value doesn’t
matter all that much. In fact, it does matter, if you want to maximize the amount of extra
sunlight that the reflector delivers (reflects) to the window.

To understand this, consider Figure 3(b). Note that the upper ray (“Ray 2”, the ray that
reflects from the very top edge of the reflector) is reflected down towards the exact opposite

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edge of the window, and just enters the cavity. This is because, based on the Law of
Reflection, Ω has been calculated and set to the correct angle for this to occur. Also, “Ray 1”
strikes the very bottom edge of the reflector directly adjacent to the window, and also enters
the cavity. The cross-sectional width of the sunbeam defined by these 2 rays is labeled x. As
seen in the Figure, when angle Ω is correctly chosen, the sunbeam of width x completely fills
the reflector—all of this light enters the window. As we will see, the equation that describes
this turns out to be only a function of the M/L ratio—once you pick values for L and M, the
“lid angle” Ω is uniquely determined, if all of the light that reflects from the mirror enters the
oven cavity.

Figure 4 illustrates what happens if this condition is not met, if the “lid angle” Ω is set to
some other angle. In this Figure, M/L has been chosen to equal 2, and (as we will derive
shortly), Ω is therefore uniquely set at 111° (if M/L is chosen to be some other value, the
value for Ω will be different):

► If the reflector is stuck to the top of the oven at the correct angle Ω = 111° for M/L=2 (and
if the sun’s rays strike perpendicular to the top of the oven, our assumption in this model)
then all of the sunlight that reflects from the mirror enters the oven cavity. The internal
temperature Tio should then reach its maximum possible value.

► If the reflector is stuck to the top of the oven at an angle Ω < 111° (the Figure shows an
angle of 101°, only 10° less), the upper ray (“Ray 2”) reflects to about the midpoint of the
window. True, all of the sunlight that reflects from the mirror enters the cavity, but the new
beamwidth x′ is now smaller than our original beamwidth x. This means that less light enters
the cavity, and Tio will not increase to its maximum possible value.

► If the reflector is stuck to the top of the oven at a an angle Ω > 111° (the Figure shows an
angle of 121°, only 10° more), the upper ray (“Ray 2′ ”) reflects to a location that misses the
window entirely! The new beamwidth x′ is again smaller than our original beamwidth x, and
once again less light enters the cavity, and Tio will not increase to its maximum possible
value.

NOTE: The ratio M/L does not have to be an integer value. It can take on any value you
want to design to, and there will still be a unique solution for Ω. In a practical sense, though,
to keep the reflectors from becoming too large, you should keep M/L ≤ 3.

The Geometry of the Reflectors—Deriving an Equation for Ω in Terms of M/L

In practice, we first pick values for L and M, calculate the ratio M/L, and then desire to
calculate the “lid angle” Ω, the angle that each reflector makes with the top of the oven. This
angle can easily be measured, adjusted, and set using a simple protractor.

From the geometry in Figure 3 and given our assumptions, it can be seen that:

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Ω = θs + β +α
(eq. 20)
Ω = 90° + α

Furthermore, using fundamental laws of trigonometry it can be seen in Figure 3 that:

(1)    If you apply the Law of Sines to the triangle defined by the reflector and the window:

M      L
=
sin ϕ sin α

⎛M    ⎞
or,                      ⎜     ⎟ sin α = sin ϕ                             (eq. 21)
⎝ L   ⎠

(2)     The sum of the interior angles of a triangle equals 180°. If you apply this to the
triangle that is defined by the window, the reflector, and the topmost ray (Ray 2) that hits the
window:
Ω + α + ϕ = 180°                               (eq. 22)

We now want to write equation 21 in terms of just M/L and Ω. Substituting equation 20 for
α, into equation 22 gives:

Ω + (Ω − 90°) + ϕ = 180°                    (eq. 23)

or,                             ϕ = 270° − 2Ω                              (eq. 24)

Then, substituting both this expression and equation 20 into equation 21 gives:

⎛M ⎞
⎜ ⎟ sin (Ω − 90°) = sin (270° − 2Ω )                 (eq. 25)
⎝ L⎠

We now use a law of trigonometry to simplify both of the sine terms:

sin(a − b) = sin(a) cos(b) − cos(a) sin(b)                   (eq. 26)

Therefore:
sin (Ω − 90°) = sin(Ω) cos(90°) − cos(Ω) sin(90°)
(eq. 27)
sin (Ω − 90°) = − cos(Ω )
and also:
sin (270° − 2Ω ) = sin(270°) cos(2Ω) − cos(270°) sin(2Ω)
(eq. 28)
sin (270° − 2Ω ) = − cos(2Ω )

This last expression may be further simplified using the following law of trigonometry:

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cos(2Ω ) = 2 cos 2 (Ω ) − 1                         (eq. 29)

Finally, substitute equations 27 and 29 into equation 25 to give:
⎛M ⎞
⎜ ⎟[− cos(Ω )] = 1 − 2 cos (Ω )
2
(eq. 30)
⎝ L⎠

Rewritten, this is recognized as being a quadratic equation in terms of cos(Ω):
⎛M ⎞
2 cos 2 (Ω ) − ⎜ ⎟ cos(Ω ) − 1 = 0                   (eq. 31)
⎝ L⎠

If we define cos(Ω ) as “x” this takes on the familiar form:
ax 2 + bx + c = 0                                   (eq. 32)
⎛M ⎞
where a=2, b= − ⎜ ⎟ , and c= -1. The solution is in the standard form:
⎝ L⎠
− b ± b 2 − 4ac
x=                                                  (eq. 33)
2a
Substituting for x, a, b, and c gives:
2
⎛M ⎞ ⎛M ⎞
⎜ ⎟± ⎜ ⎟ +8
⎝ L⎠ ⎝ L⎠
cos(Ω ) =                                           (eq. 34)
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Solving for the “lid angle” Ω alone gives:

⎧⎛ M ⎞ ⎛M ⎞
2  ⎫         ⎧⎛ M ⎞  ⎛M ⎞
2 ⎫
⎪⎜ ⎟ ± ⎜ ⎟ + 8 ⎪         ⎪⎜ ⎟ ± ⎜ ⎟ + 8 ⎪
⎪⎝ L ⎠ ⎝ L⎠    ⎪      −1 ⎪ ⎝ L ⎠ ⎝ L⎠   ⎪
Ω = arc cos⎨              ⎬ ≡ cos ⎨                ⎬                (eq. 35)
⎪      4       ⎪         ⎪       4      ⎪
⎪              ⎪         ⎪              ⎪
⎩              ⎭         ⎩              ⎭

As we will see in the following examples, physically-real solutions are found using the

⎧⎛ M ⎞ ⎛M ⎞
2  ⎫
⎪⎜ ⎟ − ⎜ ⎟ + 8 ⎪
⎪⎝ L ⎠ ⎝ L⎠    ⎪
Ω = arc cos⎨              ⎬                         (eq. 36)
⎪      4       ⎪
⎪              ⎪
⎩              ⎭

Finally, an expression for the “lid angle” Ω in terms only of the ratio M/L!! (Remember—
this equation is only true given our assumption that the oven is aligned so that θ s + β =90°)

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Figure 5 shows 3 different ovens, where M/L=1 (Ω=120°), M/L=2 (Ω=111°), and M/L=3
(Ω=106°). In each case, when the reflector is set to the corresponding value for Ω, the top
ray enters the exact opposite side of the window, and the full amount of light that is reflected
from the mirror enters the window to be absorbed the cavity. In each case, the reflectors are
used to maximize the value for the internal temperature, Tio.

Example (M/L=1):

Suppose we pick dimensions such that M/L=1. Then:
⎧1 ± 12 + 8 ⎫
⎪           ⎪          ⎧1 ± 9 ⎫
Ω = arc cos⎨           ⎬ = arc cos⎨      ⎬ so that either:
⎪
⎩    4      ⎪
⎭          ⎩ 4 ⎭
⎧1 + 9 ⎫
Ω = arc cos⎨        ⎬ = arc cos(1) = 0° which is not a physical possibility and therefore not a
⎩ 4 ⎭
valid solution, or
⎧1 − 9 ⎫
Ω = arc cos⎨        ⎬ = arc cos(− .5) = 120° which is the geometry shown in Figure 5.
⎩   4 ⎭

Example (M/L=2):

Suppose we pick dimensions such that M/L=2. Then:
⎧ 2 ± 22 + 8 ⎫
⎪            ⎪          ⎧ 2 ± 12 ⎫
Ω = arc cos⎨            ⎬ = arc cos⎨        ⎬ so that either:
⎪
⎩     4      ⎪
⎭          ⎩    4 ⎭
⎧ 2 + 12 ⎫
Ω = arc cos⎨        ⎬ = arc cos(1.366) which is undefined and not a solution, or
⎩ 4 ⎭
⎧ 2 − 12 ⎫
Ω = arc cos⎨        ⎬ = arc cos(− .366 ) = 111.47° which is the geometry shown in Figures 4
⎩ 4 ⎭
and 5.

Example (M/L=3):

Suppose we pick dimensions such that M/L=3. Then:
⎧ 3 ± 32 + 8 ⎫
⎪            ⎪          ⎧ 3 ± 17 ⎫
Ω = arc cos⎨            ⎬ = arc cos⎨        ⎬ so that either:
⎪
⎩     4      ⎪
⎭          ⎩    4 ⎭
⎧ 3 + 17 ⎫
Ω = arc cos⎨        ⎬ = arc cos(1.780) which is undefined and not a solution, or
⎩ 4 ⎭
⎧ 3 − 17 ⎫
Ω = arc cos⎨        ⎬ = arc cos(− .281) = 106.3° which is the geometry shown in Figure 5.
⎩ 4 ⎭

13
The Geometry of the Reflectors—Deriving an Equation for G

We want to derive an equation for G, the gain of the reflectors. To begin, again consider
Figure 3. As just described, the assumption is made that the oven is tilted so that the sun’s
rays strike the window at normal incidence, and that the reflector is tilted to the correct “lid
angle” Ω.

From equation 3, we have already seen that:

y = L sin (θ s + β )                         (eq. 37)

From the geometry of the oven, it can be seen that:

x = M sin α                                  (eq. 38)

Without reflectors, we had seen that the area of the window “seen” by the sun (the projected
area perpendicular to the sun’s rays) is equal to y·W. Now, the additional projected area of
light “seen” by the window is x·W. Therefore, the total power incident on the window is
given by:

Pincident to window = I oW ( y + r ⋅ x)                 (eq. 39)

where:          r = the reflectivity of the mirror

This is just an extension of equation 4 to include additional light from a reflector.
Substituting expressions for y and x gives:

Pincident to window = I oW [L sin (θ s + β ) + r ⋅ M sin α ]
⎡                     ⎛M ⎞      ⎤
= I oW ⋅ L ⎢sin (θ s + β ) + r ⋅ ⎜ ⎟ sin α ⎥               (eq. 40)
⎣                     ⎝ L⎠      ⎦
⎡                     ⎛M ⎞      ⎤
= I o Aw ⎢sin (θ s + β ) + r ⋅ ⎜ ⎟ sin α ⎥
⎣                     ⎝ L⎠      ⎦

We next apply equations 6 and 7 to arrive at the following expression for the total power
absorbed using a reflector:

⎡                     ⎛M    ⎞       ⎤
Pabsorbed with a reflector = I o Aw ⋅ a ⋅ τ ⋅ ⎢sin (θ s + β ) + r ⋅ ⎜     ⎟ sin α ⎥    (eq. 41)
⎣                     ⎝ L   ⎠       ⎦

Recall from equation 8 that the total power absorbed with no reflector was given as:

Pabsorbed without a reflector = I o Aw ⋅ τ ⋅ a ⋅ sin (θ s + β )

14
Finally (!) we use our model for the gain term G, from equation 18,
Pabsorbed with a reflector = G ⋅ Pabsorbed without a reflector

and substituting we arrive at an equation for G:

⎡                     ⎛M       ⎞       ⎤
⎢sin (θ s + β ) + r ⋅ ⎜ L      ⎟ sin α ⎥
⎝        ⎠
G = ⎣                                      ⎦
sin (θ s + β )
⎡ ⎛M ⎞            ⎤                                       (eq. 42)
⎢ r ⋅ ⎜ L ⎟ sin α ⎥
=1+ ⎢ ⎝ ⎠             ⎥
⎢ sin (θ s + β ) ⎥
⎢
⎣                 ⎥
⎦

Given our assumptions and substituting from equation 20, this is simplified to:

⎛M ⎞
G = 1 + r ⋅ ⎜ ⎟ sin (Ω − 90°)                                   (eq. 43)
⎝ L⎠

If N reflectors are used (typically N=4), the general expression for G becomes:

⎛M       ⎞
G = 1+ N ⋅ r ⋅⎜        ⎟ sin (Ω − 90°)                           (eq. 44)
⎝ L      ⎠

Finally, by setting Pabsorbed with a reflector = G ⋅ Pabsorbed without a reflector = Pout it may be seen that:

I o Aw ⋅ G ⋅τ ⋅ a
Tio = Tambient +
(U sb ⋅ Asb + U w ⋅ Aw )                      (eq. 45)

This is our final (!), overall design equation for the solar oven.

___________________________________________________________________________

The Geometry of the Reflectors—Cutting Cardboard

Figure 6 shows the shape and dimensions of a trapezoid-shaped reflector, the geometry
typically used when using 4 reflectors. This Figure defines the various angles involved in
finding a general expression for T, the top dimension of the reflector. The final result is
written in terms only of L (the window length), N (the number of reflectors), and Ω (the “lid
angle”):

T = L[1 + 2 N sin (Ω − 90°)]                                      (eq. 46)

15
Tables of Useful Data:
The solar altitude θs can be calculated at: http://susdesign.com/sunangle/
This site needs reasonably accurate location, date and time zone data. For the solar oven
project use the following information. You will also need to input the test days.
Location-Harshbarger bldg            32.23 (32°13')-110.95 (-110°57')
Latitude                         32.23N
Longitude                        110.95W
Elevation, ft                        2548
Time Zone                            T
Daylight Savings                     No
Table 1. Location of the University of Arizona.

Many engineering textbooks have tables of k-values. A website tabulation of k’s for many
substances is http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html
Specific heat thermal conductivity k-values applicable to the solar oven are presented in the
following table:
Material                                                        k-value
Cardboard (Holman, 1981)                                    0.064 W/(m·°C)
Table 2. Thermal Conductivity Coefficients of Cardboard and Newspaper.

Foil Condition            Reflectivity r
Smooth                    .75
Almost smooth Al Foil     .70
Crinkled                  .45-.6
Table 3. Aluminum Foil Reflectivity (bright-side values).

Tio (°C)      Uw, single W/(m2·°C)  Uw, double W/(m2·°C)*
66            10.10                 4.88
93            13.90                 6.69
121           18.66                 8.96
149           24.34                 11.74
177           31.60                 15.20
204           40.11                 19.35
*based on 1-2 cm gap between mylar sheets for a double window.
Table 4. Temperature Dependent Heat Transfer Coefficient for Mylar Window.

Material                                      Density [gm/cm3] Density [lb/ft3]
Newspaper (the Wildcat), “stacked solid”           0.5631                35.15
Newspaper (the Wildcat), “wadded up”               0.0128                  .80
Shredded white paper                               0.0540                 3.37
Corrugated Cardboard (1/8” thick)                  0.1869               11.67
Pink Fiberglas® Insulation R-19
Table 5. Measured Densities of Various Insulation Materials.

16
Appendix A—Planck’s Law, Solar Radiation, and Solar Ovens
All objects at a temperature above absolute zero emit electromagnetic radiation, according to
the Planck equation:

−1
⎛ 2hc 2 ⎞ ⎛ λhc   ⎞
P(λ , T ) = ⎜ 5 ⎟ ⋅ ⎜ e kT − 1⎟
⎜ λ ⎟ ⎜           ⎟                            (A.1)
⎝       ⎠ ⎝       ⎠
where:
P (λ , T ) is the spectral radiance of the object              [W/(m2·sr·m)]
(Physically, the spectral radiance P (λ , T ) is the amount of power that an object at
temperature T emits per unit area, into a solid angle of one steradian, per unit of
wavelength.)

h = Planck constant                                                  [6.626x10-34 J·s]
c = speed of light                                                   [3x108 m/s]
k = Boltzmann constant                                               [1.380x10-23 J/K]
T = the temperature of the emitting object, in Kelvins               [K=°C+273]

Note that the Planck equation describes the power emitted by objects such as the sun,
incandescent light bulbs, or the walls of a solar oven.

Fig. A.1. Plot of P(λ,T) vs. λ, for various temperatures.

● The total power per unit area emitted by an object is found by integrating the Planck
equation over all wavelengths:
∞
M tot = ∫ P (λ , T ) ⋅ dλ = σT 4                            (A.2)
0

where:
M tot is the total power emitted per unit area             [W/m2]
σ = Stefan-Boltzmann constant                              [5.67x10-8 W/(m2·K4)]

17
Note that this result is only a function of the temperature, T, of the object. (Equation A.2 just
indicates the final result—no details of how to actually solve this integral are shown here.)

The total power per unit area that the sun delivers to the surface of the earth is found using
this equation (also taking into account the area of the sun, the distance between the sun and
the earth, and the spectral transmission of the earth’s atmosphere). It can be shown that the
total power that the sun delivers to each square meter of the earth’s surface is approximately
equal to 1000 W!!

● The peak wavelength, where the object emits the most power, is found by differentiating
the Planck equation with respect to wavelength:

dP(λ , T )
= λmax ⋅ T = 3000 μ ⋅ K                           (A.3)
dλ

If the temperature of an object (in Kelvins) is known, the wavelength (in microns) where the
most power is emitted is given by:

3000
λmax =      in microns                             (A.4)
T
The following table shows a variety of sources at different temperatures, and the wavelength
where their output power is maximum:

Source                    Temperature (K)                Peak Wavelength (μm)

Sun                           ≈6000                            ≈0.5

Tungsten Lamp
≈3000                            ≈1
(100W light bulb)
Toaster
≈600                            ≈5
(nichrome wire)
Solar Oven Cavity
≈500                            ≈6
(at 400°F)
Objects at Room
≈300                            ≈10
Temperature
Table A.1. Objects at different temperatures emit infrared energy at different wavelengths.

Note that as the temperature of an object decreases, the wavelength where the most power is
emitted increases, and vice versa.

18
Solar Radiation Delivered Into the Oven Chamber

The surface temperature of our sun is about 6000K, and as shown in Fig. A.1 and from
equation A.4, its peak output power is at 0.5 microns (500nm). This wavelength corresponds
to green light, right in the middle of the visible spectrum.

Further study of Fig. A.1 shows that most of the sun’s energy falls between ultraviolet (very
short) wavelengths, and near-infrared wavelengths of about 1.5-2.0 μm. In fact, about 60%
of the total output of the sun falls between these wavelengths! Mathematically, this is
calculated as:

1.5                        1.5

∫ P(λ ,6000K ) ⋅ dλ ∫ P(λ ,6000 K ) ⋅ dλ
0
=   0
≈ 0.6                     (A.5)
∞
σT 4
∫ P(λ ,6000 K ) ⋅ dλ
0

For purposes of building a solar oven, this implies what we already know—the window
covering the inner chamber should transmit all of the visible and near-infrared energy
delivered by the sun. In other words, the window should be made of some highly
transparent, clear material (Xerox transparency film, glass, plexiglass, etc.).

Role of the Oven Chamber

The role of the oven chamber is to (a) absorb as much of this visible light/near-infrared
energy as possible, thereby heating up the walls of the chamber, (b) transfer this heat to the
air in the chamber in order to cook food, and (c) keep as much of this heat in the chamber as
possible, in order to achieve the highest internal temperature, Tio.

Part (a): High absorption ( a ≥ 0.9 ) is accomplished by painting the walls of the oven
chamber black. In turn, this absorbed energy heats up the walls of the chamber to
temperatures of potentially around 400°F (at least on a clear, sunny day!). (As Table A.1
shows, walls at a temperature of around 400°F radiate at a peak wavelength of about 6 μm.
In other words, the oven chamber acts as an energy-converter, capturing visible light/near-
infrared solar energy and converting it to longer-wavelength infrared heat energy.)

Part (b): Heat is transferred to the interior of the chamber mainly by convection with air
molecules in the chamber, and also by long-wavelength thermal radiation from the hot walls.

Part (c): The window keeps most of the heat in the chamber, predominately by limiting
convective losses to the surrounding outside air. In addition, the (ideal) window has very
low transmission of the infrared wavelengths (> 6 µm) emitted by the hot walls, which also
helps keep heat in the chamber. Figures A.2-A.7 show the infrared transmission of various
possible solar oven window materials, each from 2-18 µm. (Note that all of these windows
are also highly transparent to the incoming visible and near-infrared solar energy.)

19
In effect, the window acts as a “one-way light valve,” allowing visible light in, but not
allowing the longer infrared energy to radiate out of the oven chamber. This action,
combined with limiting convective losses, gives rise to a “greenhouse effect” within the
chamber. As a result, the internal temperature of the oven chamber increases to a steady-
state value that is much higher than the temperature of the ambient air.

Heat Energy Lost From the Oven Chamber

In general, heat energy is lost from the oven chamber due to the mechanisms of conduction,

Convective Losses
Convective losses occur wherever air molecules come in contact with the hot oven chamber.
This occurs at the exterior surface of the window, where outside air carries heat away from
the hot window. This convective loss is especially pronounced if the oven is tested on a
windy day! This may be substantially reduced by using two windows, separated by a “dead-
air” gap. The trapped air greatly reduces convective losses to the outside air, helping
maintain a higher internal oven temperature. This is clearly evident in the table of data
showing Uw vs. Tio, for one and two window layers.

Similar convective losses occur between the outside surfaces of the interior oven chamber,
and air molecules trapped inside of the oven box itself. Proper choice of insulation material
greatly reduces these convective losses. Ideally, the insulation should have many tiny
“closed-air spaces” to help trap air molecules, limiting their motion inside of the oven box.
Also, the insulation should be made of a material that is not conductive to the flow of heat.
Fiberglass insulation used in houses meets both of these requirements. Shredded newspaper
does also, and is a cheaper alternative.

Conductive Losses
Conductive losses occur wherever the hot interior oven chamber is in contact with other parts
of the oven. This happens where the interior chamber is held in place (at the point of
connection to the oven box), and where the insulation comes in contact with the outside
surfaces of the interior oven chamber.

Radiation occurs from the hot window itself, but little can be done experimentally to reduce
this loss. Radiation also occurs from the outside surfaces of the interior oven chamber, into
the oven box. One possible way to reduce this loss would be to make the outside surfaces of
the interior oven chamber highly reflective. This would greatly reduce the thermal radiation
from these hot walls, helping keep heat inside the chamber. (It is for this same reason that
spacecraft are wrapped in highly-reflective gold foil.) Aluminum foil would be a much
cheaper alternative for a solar oven, but note that this effect is not included in the spreadsheet
model for our solar oven.

20
Infrared Transmission of Various Window Materials
(measured in the Optical Detection Laboratory, College of Optical Sciences, University of Arizona)

Xerox Overhead Transparency Transmission Spectra (8 cm-1)

0.9

0.8

0.7

0.6
Transmission (%)

0.5

0.4

0.3

0.2

0.1

0

-0.1
2       4         6             8            10             12            14    16   18
Wavelength (μm)

Fig. A.2. One layer of Xerox Overhead Transparency Sheet (mylar)

Xerox Overhead Transparency (2x) Transmission Spectra (8 cm -1)

0.9

0.8

0.7

0.6
Transmission (%)

0.5

0.4

0.3

0.2

0.1

0

-0.1
2       4         6             8            10             12            14    16   18
Wavelength (μm)

Fig. A.3. Two layers of Xerox Overhead Transparency Sheets (mylar)

21
Gladwrap IR Transmission Spectra (8 cm-1)

0.9

0.8

0.7

0.6
Transmission (%)

0.5

0.4

0.3

0.2

0.1

0

-0.1
2   4        6             8            10             12           14   16   18
Wavelength (μm)

Fig. A.4. One Layer of Gladwrap plastic sheet

1/8" Plexiglass IR Transmission Spectra (8 cm-1)

0.9

0.8

0.7

0.6
Transmission (%)

0.5

0.4

0.3

0.2

0.1

0

-0.1
2   4        6             8            10             12           14   16   18
Wavelength (μm)

Fig. A.5. Plexiglass, 1/8” thick

22
Float Glass (Thin) IR Transmission Spectra (8 cm-1)

0.9

0.8

0.7

0.6
Transmission (%)

0.5

0.4

0.3

0.2

0.1

0

-0.1
2   4      6               8            10             12            14   16   18
Wavelength (μm)

Fig. A.6. Window glass, 1/8” thick

CD Jewel Case IR Transmission Spectra (8 cm-1)

0.9

0.8

0.7

0.6
Transmission (%)

0.5

0.4

0.3

0.2

0.1

0

-0.1
2   4      6               8            10             12            14   16   18
Wavelength (μm)

Fig. A.7. CD jewel case

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