5.17 Radiation View Factor
`
`137
`
`Table 5.6 Units Used with Radiation Exchange Equation
`
`Item
`
`Symbol
`
`English Units
`
`Metric Units
`
`Radiant heat exchange
`
`Q
`
`Stefan-Boltzmann constant
`
`View factor
`
`Emissivity
`
`Area
`
`Absolute temperature
`
`f
`
`e
`
`A
`
`T
`
`Btu
`hr
`
`cal
`sec
`
`0.1713 x 10-s Btu
`hr ft 2 0 R4
`
`1.355 x 10-12 cal
`sec cm2 °K 4
`
`Dimensionless
`
`Dimensionless
`
`Dimensionless
`
`Dimensionless
`
`ft 2
`
`cm2
`
`°F + 460 = 0 R
`
`°C + 273 = °K
`
`The finned surface increases the effective area for convection heat trans(cid:173)
`fer but has very little affect on the effective area for radiation heat transfer.
`Radiating surfaces must be able to "see" each other in order to transfer
`heat. The fins block the line of sight so that only the projected area bounded
`by the tight string is effective for the transfer of heat.
`
`5.17 RADIATION VIEW FACTOR
`
`The view factor f is defined as the fraction of the radiation leaving surface
`1 in all directions, which is intercepted by surface 2, as shown in Figure
`5.17.
`Heat transferred by radiation can be evaluated as taking place from
`surface 1 to surface 2, or from surface 2 to surface 1, if both are black
`bodies, whichever is more convenient. The product of the area times the
`
`"":-....
`
`\
`
`dA 1
`...... ....... ...... <',
`\'-
`', ............
`'
`...... ~ ',
`'
`, ............
`'
`'
`', \
`............ ',
`'
`' \ .................
`...,,._
`
`dA 2
`Figure 5.17 Areas involved in a radiation interchange.
`
`SEC et al. v. MRI
`SEC Exhibit 1020.151
`IPR 2023-00199
`
`

`

`138
`
`Practical Guides for Natural Convection and Radiation Cooling
`
`view factor works in both directions, as shown in Eq. 5.48 [15, 24].
`
`(5.48)
`
`Many complex equations have been derived, and many computer pro(cid:173)
`grams are available, which relate the view factor between two surfaces as
`a function of the geometry. Sometimes it is possible to determine the view
`factor between two rectangular plates with a piece of string, using the cross
`string method.
`For example, consider two long, narrow, rectangular parallel plates as in
`Figure 5.18, where only the narrow edges are shown. Using a piece of
`string, the dimensions across the free edges are shown by the dashed lines.
`As shown by Eq. 5.48, the product of the area times the view factor, per
`unit length of depth, is the sum of the lengths of the crossed strings stretched
`from the edges of the plates, minus the sum of the lengths of the noncrossed
`strings similarly stretched from the edges of the plates , all divided by 2.
`This is shown as ·follows:
`
`A f
`
`1 1-2
`
`= A f
`
`2 2-1
`
`f
`A 1 1
`-
`
`f
`=
`2 A 2 2
`
`-
`
`1
`
`= sum cross strings - sum noncross strings
`2
`(5.0 + 3.0) = 2 77
`·
`
`= (8.54 + 5.0) -
`2
`
`in2
`in of depth
`
`(5.49)
`
`The view factor from plates 1 to 2 (f1- 2) is
`
`f
`
`A 2f2 - 1 2.77 in 2/in
`.
`.
`.
`)(40) in 2 (40 m) = 0.692 (d1mens1onless)
`i - 2 = A
`= (
`4
`1
`
`(5.50)
`
`The view factor from plates 2 to 1 (f2_ 1) is
`
`A1f1-2
`!2-1 = A
`
`2
`
`2.77 in 2/in
`.
`.
`.
`)(40) in 2 (40 m) = 0.346 (d1mens1onless)
`= (
`8
`
`(5.51)
`
`Depth of both
`plates = 40 in
`
`I~ 8.0 in
`
`.I
`
`Figure S.18 End view of two long rectangular parallel plates.
`
`SEC et al. v. MRI
`SEC Exhibit 1020.152
`IPR 2023-00199
`
`

`

`5.17 Radiation View Factor
`
`139
`
`5 in
`
`'---. ,,,
`'---. 13 .
`.__ ____ ----...... 0
`'---.,
`r~12 in___j
`
`Depth of both
`plates = 40 in
`
`Figure 5.19 End view of two long rectang1,1lar perpendicular plates.
`
`The view factor for two long rectangular plates that intersect in a right
`angle, as shown in Figure 5.19, can be determined with Eq. 5.49. Here the
`cross strings are the width of each plate, and the noncross string is the
`hypotenuse of the triangle.
`
`A 1!1- 2 = A 2f2- 1 =
`
`5 + 12 - 13
`2
`
`in2
`h
`= 2 · d
`m ept
`
`The view factor from plates 1 to 2 (f1-~ is
`
`.
`.
`.
`2in 2/in
`A 2f 2-1
`= (5)(40) in 2 (40 m) = 0.40 (d1mens1onless)
`!1-2 = A
`1
`
`(5.52)
`
`The view factor from plates 2 to 1 (!2-1) is
`
`2
`A1f1- 2
`! 2- 1 =
`)( O) x 40 = 0 . 167
`= (
`Az
`12 4
`
`.
`(d.
`1mens1onless)
`
`(5.53)
`
`Figure 5.20 shows two infinite rectangular parallel plates of width b,
`directly above one another and separated by a distance h. The view factor
`
`cP
`
`/
`
`Figure 5.20 Two infinite parallel plates.
`
`SEC et al. v. MRI
`SEC Exhibit 1020.153
`IPR 2023-00199
`
`

`

`140
`
`Practical Guides for Natural Convection and Radiation Cooling
`
`t h
`{
`
`Figure 5.21 Two infinite perpendicular plates.
`
`!1-2 is shown by Eq. 5.54.
`
`h
`b
`
`(5.54)
`
`When the distance between the plates (h) is equal to the width (b) of the
`plates, the view factor f 1- 2 is shown by Eq. 5.55.
`
`f1-2 = v'l + (1)2 - 1 = 1.414 - 1.0 = 0.414
`
`(5 .55)
`
`Figure 5.21 shows two infinite rectangular plates of width b and h that
`intersect each other at an angle of 90°. The view factor f 1_ 2 is shown by
`Eq. 5.56.
`
`1[ h ~(h)2]
`!1-2 = 2 1 + b -y 1 + ~ "b}
`
`(5.56)
`
`1.0
`
`0.8
`
`.......
`.... - 0.6
`0 ....
`(..) "'
`'i: 0.4
`Cl> >
`
`'+-
`
`0.2
`
`0
`0
`
`Rectangle, high aspect ratio
`Rectangle, aspect ratio = 2
`
`'"'-.squares
`
`~Disks
`
`2
`
`3
`
`4
`
`5
`
`6
`
`7
`
`Ratio
`
`side or diameter
`distance between plates
`
`Figure 5.22 View factor for flat parallel plates directly opposed. (From Heat Transmission,
`by McAdams, McGraw Hill Book Co.)
`
`SEC et al. v. MRI
`SEC Exhibit 1020.154
`IPR 2023-00199
`
`

`

`5.17 Radiation View Factor
`
`141
`
`When the width of both plates are equal, the view factor f 1_ 2 is shown by
`Eq. 5.57.
`
`1
`f 1- 2 = 2 [I + 1 - VI + (1)2] = 0.293
`
`(5.57)
`
`Figure 5.22 show.s view factors for two recta ngular parallel plates of equal
`size but with different ratios of length to width , or aspect ratio [15].
`Figure 5.23 shows view factors for two rectangular plates of different
`sizes that intersect at an angle of 90° [15].
`The view factor f is not easily determined for radiation involving irregular
`shapes, unless the radiating surface (1) is completely enclosed by another
`surface (2), as shown in Figure 5.24. The view factor from surface 1 to
`surface 2 (!1-2) is 1.0 for this condition.
`
`0.4
`
`0.3
`
`0.2
`
`0.1
`
`.. :
`0 ...
`~ -
`
`0 ----r-
`
`0
`
`~---------- 8.0
`
`2
`
`3
`Ratio~
`c
`
`4
`
`5
`
`Figure 5.23 View factor for perpendicular and adjacent rectangles with a common side. (From
`Heat Transmission, by McAdams, McGraw Hill Book Co.)
`
`SEC et al. v. MRI
`SEC Exhibit 1020.155
`IPR 2023-00199
`
`

`

`142
`
`Practical Guides for Natural Convection and Radiation Cooling
`
`Surrounding surface
`and heat sink
`
`Figure 5.24 View factor for a completely enclosed irregular body.
`
`When partial surfaces are involved in the radiation interchange, the view
`factor must be calculated or estimated. In electronic systems, the view
`factor from a transistor on a printed circuit board (PCB) to a cooler wall
`may be extremely complex and involve lengthy mathematics. If there are
`50 transistors in different positions on different PCBs, the calculations may
`take several weeks. The amount of time required to perform these calcu(cid:173)
`lations may not be cost effective, so that the view factors for each transistor
`may simply be estimated.
`Radiation heat transfer depends upon the line of sight. Therefore, it is
`important to determine approximately what fraction of the radiation leaving
`the surface of each transistor in all directions is intercepted by the cooler
`wall. This is what determines the view factor from the transistor to the wall.
`The view factor from a hot transistor case on one PCB to the surface of
`an adjacent PCB is meaningless unless the adjacent PCB is much cooler
`than the transistor case. In general, there is very little heat lost by radiation
`between the hot components on adjacent PCBs. These hot components just
`"see" each other, so that there is a radiation interchange but no heat loss
`(see Figure 5.25).
`Electronic components mounted on the end PCBs will usually have a
`good view of the end walls. Radiation heat transfer for these components
`will be relatively good if the end walls are cool.
`
`View from transistor to cold plate
`
`View from
`transistor to - - -... 1
`end wall
`
`End PCB
`
`Little or no heat lost by
`radiation between PCBs
`Figure 5.25 Radiation heat transfer within an electronic box.
`
`SEC et al. v. MRI
`SEC Exhibit 1020.156
`IPR 2023-00199
`
`

`

`5.18 Sample Problem-Radiation Heat Transfer from a Hybrid
`
`143
`
`Electronic
`chassis
`
`1.0 X 1.0 in hybrid G)
`surface temperature
`212°F (100°cJ
`
`End wall Q)
`temperature
`122° F (50°C)
`
`Figure 5.26 Hybrid mounted on a PCB that faces a chassis end wall .
`
`5.18 SAMPLE PROBLEM-RADIATION HEAT TRANSFER FROM A
`HYBRID
`
`A 1.0 in. square (2.54 cm) x 0.180 in. high (0.457 cm) flat pack hybrid is
`mounted on the end printed circuit board so that it faces the end wall of an
`electronic chassis, which has a temperature of I 22°F (50°C), as shown in
`Figure 5.26. The hybrid is about 0.40 in (1.02 cm) from the wall, so that
`natural convection and conduction heat transfer will be negligible. The
`maximum allowable case temperature of the hybrid is 212°F (I00°C). De(cid:173)
`termine the maximum allowable power dissipation for the hybrid with and
`without a conformal coating. The inside surfaces of the aluminum chassis
`are painted light blue.
`
`SOLUTION
`
`The top surface of the hybrid and its electrical lead wires have a good view
`of the chassis end wall. The view from the side walls of the hybrid to the
`side walls of the chassis is blocked by other electronic components around
`the hybrid. The bottom surface of the hybrid has no view of any wall.
`Ignoring radiation from the lead wires, the view factor from the top surface
`of the hybrid to the end wall will be very high, because the end wall is so
`close to the hybrid and the end wall is so much larger than the hybrid. The
`view factor f from the hybrid case to the chassis wall in this case would be
`about 0.95.
`
`SEC et al. v. MRI
`SEC Exhibit 1020.157
`IPR 2023-00199
`
`

`

`144
`
`Practical Guides for Natural Convection and Radiation Cooling
`
`Equation 5.47 can be used to determine the amount of heat the hybrid is
`permitted to dissipate, in English and metric units.
`
`Given
`
`<r = 0. 1713 x 10-a Btu/hr ft 2 0R 4 = I.355 x 10- 12 cal/sec cm 2 °K 4
`f = 0.95 estimated, based on geometry
`e 1 = 0.066 emissivity of bare hybrid steel case
`e 2 = 0.90 emissivity of painted aluminum wall
`I
`I
`I = 0·0655
`e = l /e1 + l/e 2 - 1 = 1/0.066 + 110.90 -
`(combined emissivity)
`
`A 1 = 1.0 in 2 = 0.00694 ft 2 = 6.45 cm2 (area)
`t 1 = 212°F = l00°C (hybrid case temperature)
`t 2 = 122°F = 50°C (chassis wall temperature)
`
`Substitute into Eq. 5.47 for heat transferred from the hybrid by radiation,
`without a conformal coating, using English units.
`
`Q = ( 0.1713 x 10-s hr :~UOR4) (0.95)(0.0655)(0.00694 ft2)
`(460 + 122)4)
`((460 + 212)4 -
`
`0R4
`
`Q = (0.1713)(0.95)(0.0655)(0. 00694) [ ( ~~~) • -e~~) ·J
`
`watt
`Btu
`Q = 0.065 ~ x0.293 Btu/hr = 0.019 watt
`
`In metric units for the hybrid with no coating:
`
`(5.58)
`
`Q = ( 1.355 x 10- 12
`
`((273 + 100)4
`
`-
`
`cal2 0 K4
`sec cm
`(273 + 50)4
`
`] °K 4
`
`) (0.95)(0.0655)(6.45 cm2
`
`)
`
`Q = (1.355)(0.95)(0.065)(6.45)
`
`373 )
`1000
`
`[ (
`
`4
`
`( 323 )
`1000
`
`-
`
`4
`
`]
`
`(5.58a)
`
`Q = 0.00457 cal x4.187 watts sec = 0.019 watt
`sec
`cal
`When a thin conformal coating is applied to the hybrid case, the emissivity
`will jump to about 0.80. The combined emissivity will jump to
`1
`1/ 0.80 + 1 /0.90 -
`An examination of Eqs. 5.58 and 5.58a shows that the amount of heat
`transferred by radiation is directly proportional to the emissivity. Therefore,
`
`e =
`
`= 0.734
`
`I
`
`SEC et al. v. MRI
`SEC Exhibit 1020.158
`IPR 2023-00199
`
`

`

`5.19 Sample Problem-Junction Temperature of n Dual FET Switch
`
`145
`
`a direct ratio of the combined emissivities can be used to determine the heat
`transferred when a conformal coating is added to the hybrid case.
`
`Q = 0.019 watt (
`
`e coat )
`eno coat
`
`0.734)
`= 0·019 0.0655
`(
`
`Q = 0.213 watt
`
`(5.59)
`
`This shows that just adding a conformal coating to the hybrid will increase
`its heat transfer capability by a factor of 11.
`
`5.19 SAMPLE PROBLEM-JUNCTION TEMPERATURE OF A DUAL
`FET SWITCH
`
`Figure 5.27 shows a dual field effect transistor (FET) switch in a T0-8 size
`case that dissipates 0.150 watt. The switch is mounted on a PCB within an
`electronic chassis that has painted walls at a temperature of I85°F (85°C).
`The maximum allowable junction temperature is 302°F ( 150°C) and the
`thermal impedance (0jc) from the junction of the switch to the case is
`49.2°F/watt (27.3°C/watt) as rated by the manufacturer. The case of the
`switch is made of polished iron and the view factor f from the case to the
`chassis walls is estimated to be about 0.80. The chassis is enclosed, so that
`natural convection and conduction heat transfer from the switch are negli(cid:173)
`gible. Determine if the design is satisfactory.
`
`SOLUTION
`
`The following information is required for Eq. 5.47.
`
`f = 0.80 view factor, estimated from geometry
`e 1 = 0.24 emissivity of switch with bare iron case
`e 2 = 0.84 emissivity of painted chassis walls
`
`I
`1
`-
`e = - -- --
`1/ 0.24 + I/0.84 -
`Ile1 + I /e2 -
`I
`(combined emissivity)
`
`= 0.23
`
`I
`
`0.460 in
`
`0190 in~ ~
`
`Figure 5.27 Dual FET switch mounted on a PCB.
`
`SEC et al. v. MRI
`SEC Exhibit 1020.159
`IPR 2023-00199
`
`

`

`146
`
`Practical Guides for Natural Convection and Radiation Cooling
`
`The bottom surface of the switch faces the PCB, which is a poor heat
`conductor, so that its area is ignored. Also, radiation from the lead wires
`is ignored. The switch surface area is :
`A = 7r(0.460)(0.190) + (7r/4)(0.500)2 in 2 = 0.00327
`144 in 2/ft 2
`
`f 2
`t
`
`Q = 0.150 watt x 3.413 Btu/hr = 0.512 Btu (heat dissipation)
`hr
`watt
`
`Substitute into Eq. 5.47 for the case temperature.
`
`0.512 = 0.1713 X 10-s (0.8)(0.23)(0.00327)((460 + t 1) 4 -
`0.512 = 1.03 I x 10-12 (460 + t 1)4 - 0.1784
`
`(460 + 185) 4]
`
`460 + t = (0.512 + 0.1784)
`1.031 x 10- 12
`
`l
`
`1/4
`
`t 1 = 444.6°F (229.2°C) (case temperature)
`
`(5.60)
`
`The temperature rise from the switch case to the junction is determined
`from the manufacturer's thermal impedance from junction to case (0jc),
`49.2°F/watt.
`
`op
`Atjc = 49.2 - - x 0.150 watt = 7.4°F
`watt
`
`The junction temperature becomes
`ti= 444.6 + 7.4 = 452°F (233.3°C)
`
`(5.6 1)
`
`(5.62)
`
`This is substantially above the maximum allowable junction temperature of
`302°F (150°C), so th.at the design is not satisfactory.
`To improve1 the radiation heat transfer, a black oxide finish is added to
`the surface of the switch, which raises its emissivity to a value of 0.95. The
`combined emissivity is now
`
`1
`e = 110.95 + l/0.84 _ 1 = 0.804 (combined emissivity)
`
`Substitute back into Eq. 5.47 for the case temperature.
`
`0.512 = 0.1713 x 10-8 (0.8)(0.804)(0.00327)[(460 + t 1 )
`
`4
`
`-
`
`(460 + 185)4
`
`]
`
`+
`
`460
`
`- (0.512 + 0.62·3)
`3.60 x 10- 12
`t 1 -
`
`114
`
`t 1 = 289°F (142.7°C) (case temperature)
`
`(5.63)
`
`SEC et al. v. MRI
`SEC Exhibit 1020.160
`IPR 2023-00199
`
`

`

`5.20 Radiation Heat Transfer in Space
`
`147
`
`The thermal impedance 8jc from the junction to the case is still the same.
`The junction temperature · of the switch with the black case is shown as
`follows:
`
`t J = 289 + 7.4 = 296.4°F (146.9°C)
`
`(5. 64)
`
`Since the junction temperature is less than 302°F (150°C), the design is
`satisfactory.
`
`5.20 RADIATION HEAT TRANSFER IN SPACE
`
`In the hard vacuum environment of space, the amount of radient energy
`that a body receives will depend upon its distance from the sun, the view
`the body has of the sun, and its proximity to other planets or satellites. A
`spacecraft in close orbit around the earth or around the moon can receive
`radiation from the sun, the earth, and the moon at the same time. Most of
`this radiation is in the ultraviolet and visible light frequency range, which
`has a wavelength up to about J µ.m, which is 1 x
`l0- 6 m, as shown in
`Figure 5.15. In a near earth orbit the direct radiation intensity is about 444
`Btu/hr ft 2 (130 watts/ft2
`) (0.0334 cal/sec cm2
`) [4, 5].
`In addition to the direct solar radiation energy, there is reflected radiation
`energy, or albedo. Albedo is defined as the fraction of solar radiation that
`is returned to space due to reflections of solar energy from the atmosphere,
`clouds, and surface of a planet. For the earth, the albedo is about 0.40,
`depending upon the amount of cloud cover in the sky. For the moon, the
`albedo is very low, about 0.07, because the moon has virtually no atmos(cid:173)
`phere.
`When the earth is viewed from a distance, its average temperature appears
`to be about - 20°F (- 29°C). The mean surface temperature appears to have
`a value of about 57°F (l4°C), and the mean temperature of the upper
`atmosphere, at about 150,000 ft, is about - 85°F (-65°C). The sun, when
`viewed from a distance , appears to have a temperature of about 10,400°F
`(5760°C).
`The amount of radiant energy that a body in space will receive will depend
`upon the proximity and the view the body has of the sun and, or the moon,
`or perhaps some other planet. Table 5.7 shows how much radiant energy
`various planets receive from the sun. This is the energy that reaches the
`planet's upper atmosphere, not the energy that reaches the surface (except
`for the moon).
`The upper atmosphere of the earth attenuates the radiation energy re(cid:173)
`ceived from the sun, and this reduces the amount of solar radiation that
`actually reaches the surface of the earth. The amount of radiation energy
`that actually reaches the surface of the earth depends upon the month,
`hour, parallel of latitude, and cloud cover. Some typical maximum values
`
`SEC et al. v. MRI
`SEC Exhibit 1020.161
`IPR 2023-00199
`
`

`

`148
`
`Practical Guides for Natural Convection and Radiation Cooling
`
`Table 5.7 Average Radiation Characteristics of the Planets
`
`Incident
`Radiation
`Intensity
`(Btu/hr ft 2)
`
`Equivalent
`Reflected Black Body
`Radiation Temperature
`(Albedo)
`(oF)
`
`2922
`860
`444
`434
`192
`16.4
`4.9
`1.2
`0.5
`0.3
`
`0.06
`0.61
`0.40
`0.072
`0.15
`0.41
`0.42
`0.45
`0.54
`0. 15
`
`342
`127
`-5
`51
`-56
`-242
`-297
`-346
`-366
`- 380
`
`Planet
`
`Mercury
`Venus
`Earth
`(Moon)
`Mars
`Jupiter
`Saturn
`Uranus
`Neptune
`Pluto
`
`Night sky = - 40°F
`
`on a clear day are as follows: [37, 38]:
`
`40° north latitude USA: 300 Btu/hr ft 2
`
`32° north latitude USA: 3 IO Btu/hr ft 2
`
`24° north latitude USA: 320 Btu/hr ft 2
`
`Deep space has the ability to absorb virtually an infinite amount of heat
`that makes it a very useful heat sink with a temperature of-460°F (-273°C),
`absolute zero. The view factor of a body in deep space is 1.0, because the
`body can radiate to deep space in all directions.
`If two aluminum spheres are placed in deep space and if one sphere has
`a polished aluminum surface while the other sphere has a surface that is
`painted white, the white sphere will show a steady state surface temperature
`of about 80°F (26.6°C) and the polished sphere will show a surface temper(cid:173)
`ature of about 450°F (232°C), as shown in Figure 5.28.
`
`5.21 EFFECTS OF ale ON TEMPERATURES IN SPACE
`
`When thermal radiation energy strikes a body, the fraction of the incident
`radiation energy that is absorbed is called the absorptivity a. In the infrared,
`or long wavelength range, the color has no effect on the absorptivity, and
`a black smface will absorb about the same amount of thermal radiation
`energy as a white surface. In the visible, or short wavelength range (see
`Figure 5.15), the amount of radiation energy absorbed by a body is sharply
`
`SEC et al. v. MRI
`SEC Exhibit 1020.162
`IPR 2023-00199
`
`

`

`5.21 Effects of ale on Temperatures in Space
`
`149
`
`t = 450°F
`
`c = so°F
`
`.
`.
`Whi te paint
`
`O~ona: aluminum
`
`;- = 0. 16
`
`Figure 5.28 Temperatures of painted and unpainted aluminum spheres in space.
`
`affected by its color. The stabilized temperature a body reaches in space
`will therefore be related to the ratio of the so]ar absorptivity (as) to the
`infrared emissivity (e), which is shown as the ratio a 8/e. For the polished
`aluminum sphere this ratio is shown in Eq. 5.65 and for the white-painted
`aluminum sphere, this ratio is shown in Eq. 5.66.
`Polished aluminum sphere in space:
`
`_as _ solar absorptivity = 0.10 = 4
`e
`infrared emissivity
`0.025
`
`White-painted aluminum sphere in space:
`a s = solar absorptivity = 0.15 = 0 16
`0.94
`·
`infrared emissivity
`e
`
`(5.65)
`
`(5.66)
`
`A high ratio of ale means thal more solar radiation energy will be ab(cid:173)
`sorbed, so that the resulting temperature of a body in space will be higher.
`Conversely, a low al e means that less solar radiation energy will be ab(cid:173)
`sorbed, so that the resulting temperature will be lower. The equilibrium
`temperature of a body in space can be controlled, to a great extent, by
`controlling the ratio of a sfe.
`A body in space will receive radiation from the sun and emit radiation to
`space at - 460°F, which is absolute zero. If the amount of solar energy
`absorbed by a body (including albedo) is shown as a 8Q8A 8 , the general
`radiation reJation shown by Eq. 5.47 can be modified as shown by Eq. 5.67.
`
`(5.67)
`
`where a s = absorptivity of surface receiving solar radiation
`Q s = solar irradiation on absorbing surf ace
`Q i = internal power dissipation
`As = projected area of body normal to sun
`(T = Stefan-Boltzmann constant
`e = emissivity of surface radiating to space
`A = surface area of body radiating to space
`T = absolute temperature of body in space
`
`When the body in space is away from all the planets, with no internal power
`
`SEC et al. v. MRI
`SEC Exhibit 1020.163
`IPR 2023-00199
`
`

`

`150
`
`PracticaJ Guides for Natural Convection and Radiation Cooling
`
`dissipation and no heat transferred into or out of the back side of the body,
`and when the projected area of the body is normal to the sun. The absolute
`temperature of the body in space is shown by Eq. 5.68.
`
`( Q) 0.25
`
`T= as
`s
`<re
`
`(5 .68)
`
`In a deep space orbit similar to that of Venus, but away from all planets,
`the direct solar radiation intensity Qs is about 860 Btu/hr ft2 (252 watts)
`(0.0648 cal/sec cm2
`). Substitute this value into Eq. 5.68 to obtain the relation
`for the absolute temperature of a body in a deep space orbit similar to that
`of Venus.
`
`T = (
`
`860as
`0.1713 x 10-se
`
`) o.25 = 842 (as) o.25
`
`e
`
`(5.69)
`
`Since the solar absorptivity (as) and the infrared emissivity ( e) are both
`characteristics of the body surface, the control of these values will help to
`control the body temperature.
`
`5.22 SAMPLE PROBLEM- TEMPERATURES OF AN ELECTRONIC
`BOX IN SPACE
`
`• It is
`An electronic box has an insulated panel with a surface area of 1.0 ft 2
`in deep space approximately the same distance from the sun as the earth.
`The panel is painted white and always faces the sun. Determine the stabi(cid:173)
`lized temperature of the panel with no internal power dissipation and with
`an internal power dissipation of 50 watts.
`
`SOLUTION
`
`Given Q8 = 444 Btu/hr ft 2 solar radiation on panel in an orbit similar to
`that of Earth
`as = 0.15 solar absorptivity of white paint (ref. Table 5.4)
`<J = 0.1713 x 10-s Btu/hr ft 2 0 R4 constant
`e = 0.9 emissivity of white paint (ref. Table 5.4)
`
`Substitute into Eq. 5.68 for the stabilized temperature of the panel with no
`internal power dissipation.
`
`J 0
`(0.15)(444)
`(0.1713 x 10- 8)(0.9)
`
`·
`
`25
`
`= 455 ·
`
`90
`
`R
`
`T =
`
`[
`
`t = 455.9 - 460 = -4. l°F
`
`(5.70)
`
`SEC et al. v. MRI
`SEC Exhibit 1020.164
`IPR 2023-00199
`
`

`

`5.23 Simplified Radiation Heat Transfer Equation
`
`151
`
`The stabilized temperature of the panel with power on can be obtained with
`Eqs. 5.67 and 5.68, wh ich requires adding the internal power of 50 watts
`(170.6 Btu/hr).
`
`T = [ (0.15)(444) + 170.6 ]
`(0.1713 X 10-8)(0.9)
`
`0
`
`25
`
`·
`
`= 2
`6 6
`
`OR
`3
`
`'
`
`t = 626.3 - 460 = 166.3°F
`
`(5. 7 1)
`
`If the electronic box was in an orbit around the earth , where it was
`possible for the panel to receive reflecte.(I albedo radiation (40%) from the
`earth, the panel would receive an additional incident radiation of 444 x 0.40
`or 177.6 Btu/hr ft. Substituting into Eq. 5.68 for the panel temperature, we
`obtain
`
`T = [ (0. 15)(444 + 177.6) + 170.6] o.z:s = 643
`(0.1713 x 10- 8 )(0.9)
`
`.20R
`
`t = 643.2 - 460 = 183°P
`
`(5.72)
`
`There wou ld also be a n interchange of radiation energy between the
`electronic box and the earth, which wou ld depend upon the orbit of the
`electronic box around the planet. This energy interchange was ignored in
`the problem.
`
`5.23 SIMPLIFIED RADIATION HEAT TRANSFER EQUATION
`
`It is often convenient to express the radiation heat loss in a form similar to
`the convecti on heat loss, as shown by Eq. 5.7. This can be accomplished
`by setting Eq. 5.7 equal to Eq. 5.47. The radiation heat transfer equation is
`then simplified as shown by Eq. (5.73) [15, 17, 24].,
`
`(5.73)
`
`In this expression, hr represents. the radiation heat transfer coefficient in
`terms of Btu/hr ft 2 0 P (cal/sec cm2 °C). The radiation coefficient is then
`defined as shown in Eq. 5.74 for English units and in Eq. 5.74a for metric
`units.
`
`h r = 0 .1713 f e {[ ( t 1 + 460) / 100 ]4
`t 1 -
`12
`
`-
`
`[ (t 2 + 460) / 100 )4
`
`}
`
`Btu
`= - --
`hr ft 2 0 P
`
`(5.74)
`
`SEC et al. v. MRI
`SEC Exhibit 1020.165
`IPR 2023-00199
`
`

`

`152
`
`Practical Guides for Natural Convection and Radiation Cooling
`
`[(t 2 + 273)/1000]4 }
`1.355fe {[(t 1 + 273)/ 1000]4 -
`hr=--- - - - - -- - - - - - - - - - - -
`f 2
`!I -
`
`cal
`= -- - -
`sec cm2 °C
`
`(5.74a)
`
`A plot of Eq. 5.74 is shown in Figure 5.29 for various heat receiver
`temperatures and various temperature differences between the emitting a nd
`receiving surfaces [70].
`
`5.24 SAMPLE PROBLEM-RADIATION HEAT LOSS FROM AN
`ELECTRONIC BOX
`
`An electronic box, for a space application, has hybrid circuits mounted on
`the cover facing the inside of the box. The backside of the cover can "see"
`the back side of a space radiator panel within the spacecraft, as shown in
`Figure 5.30. The maximum allowable hybrid case temperature is 212°F
`(100°C). The temperature rise from the component case to the cover is 12°F
`(6.7°C), which means that the maximum allowable cover temperature is
`200°F (93.3°C). The radiator panel has a temperature of 20°F (- 6.7°C) and
`an emissivity e2 of 0.75. The electronic box cover has a painted finish with
`an emissivity e, of 0.85 . Determine the maximum allowable power the
`electronic components on the cover can dissipate.
`
`Temperature difference, t.t (°C)
`50
`80100
`20
`30
`
`10
`
`200
`
`300
`
`500
`2.0
`
`These curves for
`f = 1.0
`L' = 0.9
`
`1.0
`0.8
`
`14.0
`
`10.0
`8.0
`6.0
`5.0
`
`3.0
`
`2.0
`
`1.4
`
`1.0
`0.8
`
`. 4.0
`u:
`-
`"' ....
`...
`......
`.&:.
`....
`-
`:i
`Cl)
`..
`
`~
`
`Temperature of
`surrounding wall
`(o F)
`(also temperature of
`heat receiver)
`300
`
`200
`150
`100
`
`50
`0
`
`M
`I
`0
`
`-
`.
`
`'::;
`
`~
`
`x
`0.5 u
`0.4
`"'
`E
`0
`0 "'
`"'
`"'
`2. ..
`
`0.3
`
`0.2
`
`0.1
`
`0.6
`10
`
`200 300
`20
`100
`60
`30
`Temperature difference between surface and
`surround walls (° F)
`Figure 5.29 Radiation heat transfer as a function of temperature. (From General Electric Co.
`Heat Transfer/Fluid Flow Data Book)
`
`600
`
`SEC et al. v. MRI
`SEC Exhibit 1020.166
`IPR 2023-00199
`
`

`

`5.24 Sample Problem-Radiation Heat Loss from an Electronic Box
`
`153
`
`Space radiator, depth = 48 in
`
`t 2 = 20°F
`e2 = 0.75
`
`Electronics box cover
`Depth = 48 in
`i, = 200°F
`e1 = 0.85
`
`Components
`mounted on
`cover
`
`Figure 5.30 Electronic box mounted in a spacecraft.
`
`SOLUTION
`
`The cross string method shown in Eq. 5.49 is used to determine the view
`factors between the radiating surfaces .
`f = 2(12.04) - 2(8.54) =
`f =
`in2
`3 5
`·
`. d
`h
`A 1 1-2 A 2 2-1
`2
`m ept
`f _ = A2f2- 1 = 3.5 in2/in x 48 . = 0. 583
`A 1
`(6)(48) in 2
`m
`3.5
`.
`A if1-2
`f 2_ 1 = A
`= (12)(48) X 48 m = 0.292
`
`1 2
`
`2
`
`The combined emissivity is obtained from Table 5.5.
`
`e =
`
`1
`l/e 1 + 1/e 2 - 1
`
`=
`
`1
`1/0.85 + 1/0.75 -
`
`I
`
`= 0.662
`
`The view factor from area 1 to area 2 (/1- 2 = 0.583) will be used for the
`radiation heat transfer, so area I must be computed.
`
`A = (6)(48) in~ = 2.0 ft 2 = 1858 cm2
`144 in 2/ft 2
`
`1
`
`(5.75)
`
`Substitute into Eq. 5.74 to obtain the radiation coefficient in English Units.
`
`h = 0.1713(0.583)(0.662) {[(200 + 460)/100]4 -
`200 - 20
`r
`
`[(20 + 460)/100]4}
`
`Btu
`hr = 0.502 hr ft2 op
`
`(5.76)
`
`SEC et al. v. MRI
`SEC Exhibit 1020.167
`IPR 2023-00199
`
`

`

`154
`
`Practical Guides for Natural Convection and Radiation Cooling
`
`Substitute into Eq. 5.74a to obtain the radiation coefficient in metric units.
`1.355(0.583)(0.662){[(93.33 + 273)/ 1000]4 - [(-6.7 + 273)/ 1000]4
`93.3 -
`(-6.7)
`
`h r=
`
`}
`
`hr = 0.0000679
`
`cal
`2 0 C
`sec cm
`
`(5.76a)
`
`Substitute Eqs . 5.75 and 5.76 into Eq. 5.73 for the maximum heat that
`can be transferred by radiation, using English units.
`-
`
`Q = ( 0.502 hr~:~ 0F) (2.0 ft 2)(200 - 20)°F
`
`Btu
`Q = 180.7 -
`hr
`
`x 0.293
`
`watt
`Btu/hr
`
`= 52.9 watts
`
`(5.77)
`
`Substitute into Eq. 5.73 for the maximum heat that can be transferred by
`radiation using metric units.
`
`Q = (o.0000679
`sec cm
`
`cal 2 oc) (1858 cm 2)[(93.3 -
`
`( -6.7)]
`
`Q = 12.61 cal x 4. 187 watts = 52.8 watts
`sec
`cal/sec
`
`(5.77a)
`
`The heat transfer coefficient h,. could also have been obtained from Figure
`5.29 for English units and metric units . Notice that Figure 5.29 is based
`upon a view factor of I and a combined emissivity of 0.9. If the geometry
`under consideration has a different view factor and a different emissivity,
`corrections must be applied to obtain the true radiation coefficient.
`For the previous problem, using a heat sink temperature of 20°F and a
`~T of 200 - 20 = 180°F, h,. = 1.18 Btu/hr ft2 °F for English units. Correc(cid:173)
`tions must be applied for view factors and emissivity.
`
`0.662)
`h,. = 1.18 x 0.583 x - 0 9
`(
`.
`
`Btu
`= 0.506
`2 0
`hr ft F
`
`(5.78)
`
`E quation 5.78 compares well with Eq. 5.76.
`
`5.25 COMBINING CONVECTION AND RADIATION HEAT
`TRANSFER
`
`In many electronic boxes, radiation heat transfer and convection (both
`natural and forced) occur simultaneously. Therefore, it is convenient to
`write the combined heat transfer equation in a form similar to the general
`
`SEC et al. v. MRI
`SEC Exhibit 1020.168
`IPR 2023-00199
`
`

`

`S.26 Sample Problem-Electronic Box in an Airplane Cockpit Area
`
`155
`
`relation shown by Eqs. 5.7 and 5.73. The combined relation is shown in
`Eq. 5.79 [15, 17, 24].
`
`(5.79)
`
`where he = Btu/hr ft2 °F or cal/sec cm2 °C (convection coefficient)
`hr = Btu/hr ft 2 °F or cal/sec cm2 °C (radiation coefficient)
`A = ft 2 or cm2 (area)
`At = 0P or °C (temperature difference, surface to ambient)
`
`For most small electronic boxes, typical values of he are about 0.8 Btu/
`hr ft 2 0P (0.000108 cal/sec cm 2 °C) and typical values of hr are about 1.2
`Btu/hr ft 2 0P (0.000162 cal/sec cm2 °C), which results in a combined heat
`transfer coefficient, he + h n of about 2.0 Btu/hr ft 2 °F (0.00027 cal/sec cm 2
`oc).
`
`5.26 SAMPLE PROBLEM-ELECTRONIC BOX IN AN AIRPLANE
`COCKPIT AREA
`
`An electronic box is mounted in the cockpit area of an airplane that must
`operate at 50,000 ft without a pressurized cabin. Plug-in PCBs within the
`box are cooled by conducting their heat to the side walls of the electronic
`box. Natural convection and radiation are used to cool the external surfaces
`of the box. The maximum expected cockpit temperature is 100°F and the
`maximum allowable box surface temperature is 140°F. Because the top and
`bottom covers are fasten~d to the box with a few small screws, the covers
`make poor contact with the box and will not be effective heat transfer
`surfaces (see Figure 5.31). Determine the maximum amount of heat' that can
`
`Bolted top and
`bottom covers
`
`Ambient
`t0 = 100°F
`
`Pilots
`seats
`
`Mounting shell
`
`Section through cockpit
`
`Figure S.31 Electronic box mounted on a shelf in an airplane.
`
`Box surface
`t = 140° F
`
`$
`
`SEC et al. v. MRI
`SEC Exhibit 1020.169
`IPR 2023-00199
`
`

`

`156
`
`Practical Guides for Natural Convection and Radiation Cooling
`
`be transferred from the box to the surrounding ambient at sea level and at
`50,000 ft. The box has painted ex ternal surfaces.
`
`SOLUTION
`
`Onl y the vertical side a nd end walls will be effective for heat transfer by
`natura l convection and radiation. The top and bottom covers are ignored
`here because the covers make poor contact with the chassis. The natural
`convection coefficient for the vertical side and end walls is determined first
`for the sea level condition.
`
`lit) o.2s
`h e= 0.29 T
`(
`
`(ref. Eq. 5.4)
`
`Given
`
`li t = 140 - 100 = 40°F (temperature difference)
`L = :~ i; =0.583 ft (vertical height)
`I m
`t
`
`40 ) 0
`h e = 0.29 -
`-
`(
`0.583
`
`·
`
`25
`
`= 0.83
`
`Bt
`ft~ F (convection coefficient)
`hr