Theoretical concentrations for solar furnaces

Theoretical Concentrations for Solar Furnaces B y M. H . C o b b l e * University of Delaware, Newark, Delaware

The cencentration for a paraboloid of revolution mirror and for a parabolic cylinder mirror are derived as a function of relative aperture for several target geometries. The expression for o p t i m u m concentration for each target-mirror combination is developed, and the equation for the ideal target shape for a target o f a p a r a b o l i c

l m = m a x i m u m distance from point on mirror to focus, ft

( n = relative aperture

Q,,~ = radiation rate to mirror, B t u hr Btu Qt = radiation rate to target, h r

CONCENTRATIONS

FOR SOLAR FURNACES

R~,~ = distance from the earth to the sun,

Several relationships arc derived for determining the concentration of solar furnaces, using either a singlecurved or a double-curved mirror, for several types of targets. The derivations are based on the following assumptions: (1) all the energy reflected b y an ideal mirror is used; (2) the target area is small c o m p a r e d to the mirror area, a n d so it is assumed t h a t the target does not shield the mirror; (3) all the radiation is parallel to the axis of the mirror,

(92.9 X 10 ~ mi) r = d, ft S = arc length, ft

S,¢ = diameter of sun, 8.66 X 105 mi S~ = .r = :r,,~ = # = g,~ = a = ¢~ =

NOMENCLATURE A = constant A ..... = a r e a o f m i r r o r n o r m a l t o r a d i a t i o n , At = area °f t'arget' ft2 a = vertex ot triangle

, dimensionless

Q = radiation rate, BtUhr

cylinder mirror having a relative aperture o f 4 is obtained, THEORETICAL

D) n =

radius of sun, 4.33 X 105 nfi rectangular coordinate, ft m a x i m u m rectangular coordinate, ft rectangular coordinate, ft m a x i m u m rectangular coordinate, ft angle subtended by the sun, dimensionless angle of deflection that. a ray makes when hitting mirror, dimensionless

2 ) , dimensionless 3'l = (90 ° - ~ ( a) 3,o = 90 ° + ~ , dimensionless

ft ~-

b = vertex of triangle B t u (target) / B t u (mirror) con(,.ent, ration - h r ft 2 C ~ hr ft.~

~i = angle, dimensionless p = polar radius, ft

c = vertex of a triangle D = mirror aperture, it d = t a r g e r width, ft d~1 d/]

INTRODUCTION~DEFINITION CONCENTRATION

Assuming an ideal mirror, the radiation rate falling on the mirror is equal to the radiation rate going

d~, t = target geometry, ft (See Figures)

t h r o u g h the target area, so that.

d~]

0m = 0t

d~)

dc/dn = rate of change of concentration with respect

OF

[11

Multiplying the left, side n u m e r a t o r and denonfinator b y the normal area of the mirror, and multiplying the right side n u m e r a t o r and d e n o m i n a t o r b y the area of the target results in Q~ A~,,,~ _ Qt A t [2]

to relative aperture, dimensionless ds = differential of arc length, ft f = focal length of mirror, fg 1 = distance from point on mirror to focus, ft * Associate Professor of Mechanical Engineering.

A,,,~ 61

At

Rearranging yields Q~the A t _c°ncentrati°nA .. - (7

[3]

"~I"

1

Art m

This concentration (C) will be used in the following derivations: I. C o n c e n t r a t i o n for a parabolic a small relative aperture

cylinder

j

/

having

The concentration is given from the introduction as c

.4 ......

f

"~¢~2

[~1

From the geometry of Fig. 1, it is seen t,hat D X L

I'"

D

'

;

[2]

tl

The definition of relative aperture (n) is n

=

D

[3]

f

~-

thus

Fro.

D = nf

2.

D

Paraboloid

aperture

of

revolution

d =f

2tan

with

small

relative

[41 so

The tangent of the solar half-angle is t a n a = d/2 = d 2 f 2f

[51

The concentration then follows -

__d2

C

-- d

k

f 2 tan

[7]

2 tan a/2 '~

[8]

C =~-.107.3n

[9]

Equations [8] and [9] are good approximations for n < .5, becoming more accurate as n ~ 0. ... ¢I !

N

,

~

~

II.

Concentration having a small

I --

for a paraboloid relative aperture

of revolution

The concentration is

[~1

A ....

•~

C = At \

As seen from Fig. 2 this is equivalent to 7rD 2

-Fro.

1. P a r a b o l i c a p e r t u re

D cylinder

mirror

+J ~ith

small

C

relative

-

4

--

7rd24

62

-

D2 d2

[21

Equation I-[4] states

are of length L, Eq. [1] becomes D = nf

[3]

C -DXL-

D

~dL

rd

From Fig. 2

[2]

Using Eq. 1-[4] tan -a = d/2_ = d 2

f

[4] D = nf

[3]

sin • a =d / i2 d= -2 lm 21,,

[4]

2f

Thus as above

From Fig. 4 d = f

(

~)

[5]

2 t a n 7)_

Using the results of Eqs. [3] and [5] ill [2] results ill

or rearranged d

C-

D= -

(F

d = l,~ (2 sin a )

(nf)2

\

f 2 tan

[6] ~Y"

'~~

. 2 2 tan

l:rom Fig. 3

oy

tm=

2 tan ~

~ / { f - u,,~)~ + x ~ 2 :g

y [7] or

= 4fy

III. C o n e e n t r a t i o n for a parabolic c y l i n d e r h a v i n g cylinder

[8]

At the maximum position this becomes xm'-' = 4 fy.,

target

It is evident from the geometry of the problem, as shox~T1 in Fig. 3, that the diameter of the target that will just receive all the radiation is determined by the rays reflected from the extremity of the mirror (assuming the mirror is symmetric). Thus the concentration is C = A ....

[71]

4f

Equations [6] and [7] are good approximations for small n.

a circular

[(~]

The equation of the parabola with vertex at the origin is

or

C ~ (107.3) 2 n 2 ~ 11,550 n 2

,5]

[9]

Substitution in Eq. [6] yields 1,~, = %/(f -- !1,,)2 + 4fg,, this becomes with rearrangement lm = "v/ (f q- y,.,)2

[10] [11]

or

[1]

At t., = f +

y.,

[12]

From Fig. 3 it is obvious that if the target and mirror But 2

f ~ ] d.

I

Im

a

g" -

and

4f

D 2x., n = .f = f

nf

Ym

:~,,, = ,j ,

Xm = ~

-!

[14]

[151

t

X

Thus substitution of Eq. xm

I., =.l" + y., =.t" + i f

[13]

and [15] in Eq. [12] gives 1

= f q- ~f \ ~ ]

D

[l(~l ,~f ( n~ ) = .f-1-iti = f 1 + ] - 6

Fro. :3. Par,,bolic cylinder mirror with circular cylinder lrtH'get

63

Therefore using Eq. [31 and [16] in Eq. [2] yields C

D r

2 sin

lm =

nf

E(

7r f C=

[

nf

2 sill

1 + 16)

2)]

n

(

7r 1 + 16

2sin

'= [m Fro. 4. Cylinder target geometry

~1

OF

C ~-~

107.3n ~r 1 + n

of n = 4 into Eq. III-[17] gives the m a x i m u m eoneentration.

When n << 1, Eq. [18] simplifies to

C =

C --~ 107.3n ~ 34.2n -7r This relation is good for n < ½

[19]

7r 1 +

2 sill = 7r 1 q - ~ , .

IV. M a x i m u m concentration for a parabolic cylinder having a circular cylinder target The expression for concentration, n

C =

Eq.

C -

III-[17] is

7r 1 + ~ )

2sin

~. 2 X 107.:3

2 sm~

or

-

2 sin

2 ~

An -

[7[

4

1 q- ]~

C -----68.3

where the constant A is given b y

V. C o n c e n t r a t i o n having

[9]

for a paraboloid

a spherical

of revolution

target

1

A=

(

~) ~r 2 sin 2

[2]

Referring to Fig. :3, the diamet~er of the sphere t h a t will just receive all the radiation is determined b y the rays reflected from the e x t r e m i t y of the mirror. T h u s the concentration is

T h e m a x i m u m concentration is determined b y setting the derivative

dn

resulting equation,

equal to zero, and solving the

A,,:

which is

Looking at. Fig. :3, and evaluating for a sphere of diameter d, Eq. [1] becomes

/2n,- I

/

1

( - 1 )n ~ ) [ ~ It

l +~

1 +]~/

C -

rearranging gives

7rD 2 _ D 2 4

~-d'-' n~ 0 = 1 -4- 16

[1]

At

V dC

C-

2n 2 16

[21

4d 2

Using Eq. I-[4] [4]

or

D = ,f

[3]

sina _ d/2 _ d 2 /,, 2l,,

[4]

F r o m Fig. 4 n 2-

16 = 0

[5]

thus

solving for d yields n = ± 4

[6]

(9

T h e solution of n = -- 4 has no meaning. Substitution

d = lm 2 sin ~ 64

[5]

dO

db

~-

,.- .,~

2

~

)-

2 dc 1-.~]

-~

al~

,

2

db

-~"

~''--

X

]

!

d',

--?

~

,,~

Xm

Ym

)

D

I~

FIG. 5. Flat plate target geometry

Using Eq. IIl-[16]

Where the constant A is given by L = f

~ +

[6]

A -

Substitution of Eqs. [3], [5] and [6] in (2) gives (Y

O2

--

4d2

__

[(

4 f

n

C = 4

(1 +]~]

(n/) 2

1 + n 2 ~ ( 2 sin

16/\

.~

4 2 sin

,7,

Setting

2

dn

°Y

2sin ~

d(!

-- " C ~ (107.3)'~n 2 ~ ll,oo0n" -- 4 n'-''~" 4 ( 1 + ~ )/2~2 / (1 A- 16]

[,q]

0

4

n2~2( 2)2 1 + ~/ 2 sin

=

2n

=

(--2)n~

]6

+

[3] 1 -~- ~ i /

( r1~+) ~ i

16 4n:~

[4]

or

2n :~ = 32n

[5]

n = ±4 As before, the solution n = - 4 has no meaning. Substitutionof n = + 4 into E q . V - [ 8 ] , g i v e s t h e m a x i nmm concentration

An2

(

[

0 = 2n

The expression for concentration is 2

A

rearranging gives

C =~ 11,oo0n" ~ 2890n 2 [10] 4 ¥ I . ) l a x i n l u m c o n c e n t r a t i o n for a p a r a b o l o i d of revolution h a v i n g a spherical target

n

equal to zero and solving the resulting

1 -Jr- l(i/

When n < < 1, Eq. [9] may~. be., given as

(

[2]

equation yields

-

C =

1

nZy 1 + ~)/

[1]

65

C=

n

2

( Ill] 4

l +

2isin

_

90 ° -

c o s / 3 - - cos

2 sill

90°-- 2

sin/3

d:' e o s -

16/\

C =

)

(3/

a

db =

l

( 2 sin 2 ) 2

2

[( sill 90 ° cos ~o -- cos 90 ¢ sin 2

cos /3

[7] - - ( e o s 9 0 ° ( ' o s ~ . a + sin g0° sin 2 ) sin/3 ]

or o/

C ---~ 11,550

[8]

dl' e o s ~ de =

VII. Concentration ing a fiat plate

for a parabolic target

cylinder

[10l

2

hay-

cos ~ cos/3 -- sill ,~ sin/3

°

E °

1

But since

T h e starting point for the concentration is

O/

C -

,4 .....

cos ~ = 1.0000

[11]

a sin ~ = .00467

[121

[11

At From Fig. 5 it is seen t h a t for a mirror and t a r g e t of length L,

and

C = D X L = D d~ X L da F r o m Fig. 5 it is also a p p a r e n t t h a t

Eq. [10] m a y be written with good a p p r o x i m a t i o n for all values of /3 except, those approaching 90 ° as d(

[2]

[131

db --

tan a _ d~'/2 _

[3]

d(

2 cos B T h e value on the right of Eq. [13] is equal to d~ as seen

2 lm 2 X 1,o thus, upon rearranging this becomes

dl' = l m ( 2 t a n a )

from Fig. 5. B y syrnlnetry it is seen t.hat the largest image dimension

[4]

[14]

da = 2db

Using the law of sines on triangle abe, Fig. 5

d~

sin 72

F u r t h e r it is seen that

_ d?/2

[51

sin 5

rearranging yields

[151

_

l,,

or db-

d~'sin72 2 sin (3

[6]

dl' = 1,,

substitution of ~

= 90 ° +

~

a 2 tan 2

= f

1 +

2 tan

](i

[71

cos/3 - f - / y 1 8 0 ° - - 72 - - /3 =

180

--

(

90 ° +

0 ~

90 °

c~ _

2

_

cos/3,, -- 'f -- Ym _ f -- y,,, 1. . . .

/3

d~'sin 2 sin

Y=

( 9 0 ° + 0~

90 °

a2

f+

[18]

y.,

F r o m Eq. III-[13],

in Eq. [6] gives

& =

[171

oF

- - /3

[8] =

[16]

Dealing with the angle /3, it can be seen from Fig. 5 that

mad =

d,'/2

tan a 2

_

x4f ~

2

[19]

or [9] Y" =

/3

66

4f

16

[201

Substitution in Eq. [18] gives

f

where the constant A is given by

n~f

cos ~., -

1 - - - n~ --

16 _ ,n2f

f+~

16 2

l+n

A-

a [2] 2 The maximum concentration is determined by setting

[21]

2 tail

16 Using Eq. [14] in [13] used at, the mirror extremity gives 2 n 2 dl' f 1 -4- 16 tan ~ do

{dC'~

)(o)

(

1

\ ] d n equal to zero, and solving the resulting equation for n. Thus

e()s ~m

dC-

0 = A[(l

__

n 2

] ~ ) ( 1 +12(7))-2

~,2~ 2(__~) or f d,--~

( +Y(:) (+) 1 + 16)]

2

?l 2

it2~

[3] 3

tan

[23]

after some rearranging this results in

1 -- lii

n 4 -- 96n 2 + 256 = 0

As before

[41

which in turn yields D = nf

[24]

n = =1=4X/3-4- ,fig

Substitution of Eq. [23] and [24] in Eq. [2] yields C =

[51

Rejecting negative relative apertures, this gives n = 4~v/3 -4- x/g

nf

[6]

f ( 1 + ]~)/ 1 --

or

[ C = (

n2 \

n 2 y ( ~6 2) 1 q-~] 2tan

[25]

(n"=) 107.3n 1 -- 16

(

(/~ ~ -- [

I261

n

a - 107.3n 2 tan ~

[27]

1 -- T6

n2"~2(2) 1 -4- ~6/ 2 tan

[9]

C ='/' 107.3

[10]

2tan

C - A .....

("/?'2) = An (

) a

As before the starting point is

The expression for concentration Eq. VII-(25) is n

1

]1]

IX. Concentration for a paraboloid of revolution h a v i n g a flat p l a t e t a r g e t

VIII. Maximum concentration for a p a r a b o l i c cylinder having a flat plate target

(

' -- 16 rt2)2( a) 1 q-~ 2tan~

The value n = 9.64 is thrown out because it gives a negative eoncenm~tion.

which is Eq. I-[9].

C =

[8]

or

For n << 1

(7b2)

n _-- 9.64

n

C = (

1 -t- 1 6 ]

C '/'

/7]

Putting the solution [7] ill Eq. VII-[25] gives the value

or c =

n ~--- 1.65

1 -- ]~ n2~ 2 1 + 17J)/

[1]

At From Fig. 5 the target geometry ix determined by the largest image thrown on the plate from the extremities of the mirror. Due to symmetry this image will be a

[1-]

67

circle. Thus

The maximum concentration is determined by setting 7rD~

C -

D 2

4

-

~-d,, 2

de

[2]

d,"-

equal to zero and solving the resulting equation

for n. Thus

From Eq. III-[3]

dn

0

0 -- 2n

/,~~N]- 4

and from Eq. VII-[23] f

1 -t-~/\

do ~

tan

+ (-4)n 2 \

[4]

( n2)l-~

n4-

2 tan

(

96n 2 + 2 5 6 = 0

[4]

which in turn yields

(nf) 2

l + 16

\16/J

thisresults after some rearranging in

Substitution of Eq. [3] and [4] in Eq. [2] gives

f

1 + ~/

16 /

[5]

finally

i6/n2~2

or

with the applicable result ¢

~)

n2 l -

C ~---

[(

[6]

Using this value of n in Eq. [1] yields

~Y(~ tan ~

'

9;

1 + 16/\

n

or

~

16/ ( 1 -- ,
C =

(

11,550 n ~ 1 -C ~

n~ ~ ~]

----

C

[71

or

~/

+

[11

1 + Nn2~' ] (2 tan 2 ) ~

c

1

-

(:)

[8]

~

2 tan

When n << 1, Eq. [6] reduces to 2

~t

2 tan c) or

C ~ 11,550n ~-

I8]

These are the same as Equations II-[6] and II-[7] respectively. X. ~|aximum concentration for a paraboloid revolution having a flat plate target

of

The expression for concentration is n ~

.

C =

_ .an

1+~

2tan~

f

21 l

rt ~ -

-

1+Tg

Where ~he constant A is given by A

---

(

1 2 tan ~

[2]

x

i

~"

Fro. 6. Mirror geometry for optimum target shape for a parabolic mirror having a relative aperture of 4.

68

a n d finally

C~ 11,550

[9]

t40-

XI. Optimum target shape for a parabolic mirror having a relative aperture of 4

tan2

d

2X /

l

thus

d = l

ParabolicCylinder I ~-Plate Target (n << I)

120-

T h e target, should p r o v i d e a cross section j u s t wide enough to i n t e r c e p t each b e a m from e v e r y p o i n t on t h e mirror. T h e d e v e l o p m e n t follows. F r o m Fig. 6 it is seen t,hat~ o~ . . . . .d/'2 .

y<

Ioo-

-%~imum Targe! \

\

-~

.o_ =

FCylinder Target j

,
y--_

/ "~,<<1)

80-

o

g

2 tall ,~

Target

/ - - P l a t e

[l]

(o)

I

Mirror

60-

[2]

40-t

F r o m a p r e v i o u s development I = / + :]

[31

On t h e m i r r o r

2 o_

Y

z

2

N]

4.f

o

I 0

,

~

I

2

alld D -

f

2,c -

5

Relative Aperture

n-

n

4

3

FIG. 8 [5]

j'

F r o m Fig. 6 also

thus x

_

[6]

nf

c o t a n ¢~ - f -

3"

y -

f

./~

:~I = f 3"

X

x2

4f3;

2 ')

cot,a n N - 2 ---

~

,

-

-

S

~'t

~ ~

d

l

n

17]

8

m u l t i p l y i n g t h r o u g h b y 8n gives

/

8n c o t a n ~ = 16 -- n 2 -.

or

solving for n yields, finally -.

.

n = 4 (csc ~ -- cot, fl)

n = 4 tan 2 " Fro. 7. Target geometry for optinmm target shape with a par'd)olic mirror having a relative aperture of 4.

Eq. [9] is a c o n v e n i e n t r e l a t i o n b e t w e e n n a n d ¢~. S u b 69

L 14,000

Mirror Paraboloid of Revolution

Plate Target (n (< I) 12,000-

I

~.

/-Plate Target

/

,o,ooo_ , / Spherical Target

g 6,ooo0

4,000-

1

[

0

I

5

2

4

5

T

n- Relative Aperture

FI(~. 10. Ideal t a r g e t cross s e c t i o n for a p a r a b o l i c mirror with a r e l a t i v e a p e r t u r e of 4.

FIG. 9

stitution in Eq. [2] yields d = 1 2

=

(

f +

~

finally = (f+y)

O = ~f sec ~

2tan,~a

xf ) ( 2 ) 2 tan

ThelengthoiareSofl?ig. S =

2 tan 62

7isgivenby

dS =

v @ 4- p,2 d/9

[12I

1

d = f

1 -t- ~

2 tan ~

By symmetry

or

Where d =.f

(

I +~-~

2tan

[10]

o = A see2/9 2

From Fig. 7

P

and the constant. A is given by

2

2

f[ p =

[14]

1+

and p' is

(4 tan 2B-)21( 2 ) 16 2tan

A sin~2

O = ~f i 1 4- tan 2 ' ~ 1 ( 2~ t)a n ~

P'-- cos3-2 /9 7O

[16]

a

S o Earth Z

~

~

~r

_

_

~C. . . . . . . C .~7,~

_

~

x

_

~

The concentration then, for a mirror and target of length L, is

Sr ~

C - A,,,~ _ D X L _ D

SXL

At

Sd

C ~ 4.598f

-

Geometry

of earth and

=

2 tan 9.

[20]

n

4.598

2 tan

Since n = 4, the concentration is

sun

T h u s continuing

4

+{

J°'/24/

(~)2

fit 2

\Atan6/

Thiscomparestothevah,

cos- ~

= 68.3foracircular

x = p cos fl

[22]

p sin fl

[23]

and

.~/~ see ~ ,~; = 4A J0

eofC

To obtain the equation of the target in rectangular coordinates let

1 + t a n 2 a dfl

cos - - 2 fl 2

[21]

C_~93.8

dN

./~ -0

4 X 107.3 4.598

C

/ =

°)

Res

FIG. 11.

S

S

nf

/

[19]

dfl

,/2

~ d ~ = 4A f~o eos~ 97

[17]

-

Y =

fl

3"he equation i or O is

cos '~ p = Asee ~ fl9

~/2 8A ~ dfl/2

-

A

~

_

2A 1 + cos fl

COS2 ~

[24]

A cos' ~

but also

integration yields

,s = 8.4

~

p = ~v/.c~ + y-~

ft + 21

tan

COS2 _

l] i]

+

2

I

J

acos 13 - . p v T . ~ - -+ y2 setting p = p, and rearranging

_]0

7F

2A -

sm~

tan( 4

i

[25]

8)

-

1 +cosfl

i

[26]

V/~2 _~_ y2

2A 1 -+-.x

I_

- V / ~ + y2 +

cos 2 0 + In t a n

+ 0

= % / ~ + g~

V':r~ + y2 + .~. finally N = 4A [ 2 ! . v/ ~ + In tan 3~--] k (½~v/~)2 ~ - ,I S--4A

[ 1~7071 +

In 2.42

]

%/x~-+ q.2 + x = 2A

[18]

[27]

this can be modified t()

v~ + u._,=

= 4A[1.414 + . 8 8 5 ]

2 d --

.,.

:c2 + y2 = 4A 2 _ 4Ax + x 2 S----~4A[2.299] ~ 4

2tan

= 4A 2 -- 4Ax = 4A 2 1 -

[2.299]

S ~=4 . 5 9 8 f " 2 tan

Y = ±2A 7l

~1

A:r

x

[28]

b u t since Se

Substitution of the value of A gives y = 4.2

f

V=&,

2tan~

sin '2 f

2 tan

2R~ - - 2 X 92.9 X 106

the equation for one-half of the surface when

[3al l

~ .00467

214.6 [0
< f (2 tan 2) ]

also

T h e equation for the other half of the target is y = 4-2

[( :)1,/ ( : ) f

2 tan

1 -t-

x

f

•

-f

'2 sin a ~'~ 1 2 - - 107.3

2tan

2 tan

[3t)]

The following trigonometric series are useful in solar furnace relationships :~ a 7

[30]

< x < 0

The slope is given b y

it;

it:

.F

~

5!

7!

X3

9 5 ~:r

6,~2 X 9

17:C 7

tan x = x -~ ~ + ] 5 + 315 + 2 8 ~ -t- " "

[51 y, _

4- 1

T h u s when a is a small angle, it is true t h a t A

y' = -4

1

[6a]

2 sin ,~ ~ 2 t a n ~ ~ a

[(ib]

and a

f

sin ~ ~-- t a n ,~ _ ~

2 tan ~

On the earth, the solar half angle is

[ 0 --<

x < f

(2

a -- Sill •--1 (.004()i~. ")

tan 2)]

[7]

'2

The points of interest are at x = f

2 tan

, and x = 0. •

For the first the slope is

x

3

1.3x ~

sIn-~x = x -I- 2 . 3 -t- 2 ~

1 . 3 "5 x

7

-/- 2 - 4 . 6 . ~ + "'"

Y' = 4- ~ . Forx

[8]

E

= 0, t h e s l o p e i s

if" < 1 , -

y'=

4-1

APPENDIX

~
x

'1

Thus I

a

~'~ .00467 radians ---~.2675 degrees ~ 16.08~mimites. [.9] 2 --

F r o m Fig. 1 sin a _

S~

2

R~,

[1]

I t follows then t h a t the solar angle is =~ 32.16 minutes.

72

[10]