Designing a More Effective Car Radiator

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Designing a More Effective Car Radiator

Introduction

1. Original & Proposed Radiator Dimensions

2. Heat Transfer Performance of Proposed Radiator

3. Adjusting Heat Transfer Performance of Proposed Radiator

4. Export Optimized Radiator Dimensions to SolidWorks

Introduction

The demand for more powerful engines in smaller hood spaces has created a problem of insufficient rates of heat dissipation in automotive radiators. Upwards of 33% of the energy generated by the engine through combustion is lost in heat. Insufficient heat dissipation can result in the overheating of the engine, which leads to the breakdown of lubricating oil, metal weakening of engine parts, and significant wear between engine parts. To minimize the stress on the engine as a result of heat generation, automotive radiators must be redesigned to be more compact while still maintaining high levels of heat transfer performance.

There are several different approaches that one can take to reduce the size of automotive radiators while maintaining the current levels of heat transfer performance expected. These include: 1) changing the fin design, 2) increasing the core depth, 3) changing the tube type, 4) changing the flow arrangement, 5) changing the fin material, and 6) increasing the surface area to coolant ratio.

By increasing the surface area to coolant ratio, this application shows how one can minimize the design of a radiator and still have have the same heat dissipation as that of a larger system, given a set of operating conditions.

Figure 1: Components within an automotive cooling system

1. Original & Proposed Radiator Dimensions

The dimensions of our original radiator design can be extracted from the SolidWorks® drawing file (CurrentRadiatorDrawing.SLDPRT). The drawing is a scaled down version of the full radiator assembly which measures $24'' \times 17'' \times 1''$ . For the purpose of our analysis, the dimensions obtained from CAD are scaled up to reflect the radiator's actual dimensions.

Note: This application uses a SolidWorks design diagram to extract the dimensions of the original radiator. This design file can be found in the data directory of your Maple installation, under the subdirectory SolidWorks. If you have SolidWorks version 8.0 or above, save the design file, and then click the radio button below to tell Maple™ where to find the file. If you do not have SolidWorks installed on your computer, the values will be pre-populated.

Figure 2: CAD rendering of current Radiator Model

Original Radiator Model Dimensions

The table below summarizes the current radiator dimensions.

Current Radiator Dimensions
Radiator length $(rL__cur) &colon;$	$ft$ $m$
Radiator width $(rW__cur) &colon;$	$ft$ $m$
Radiator height $(rH__cur) &colon;$	$ft$ $m$
Tube width $(tW__cur) &colon;$	$ft$ $m$
Tube height $(tH__cur) &colon;$	$ft$ $m$
Fin width $(fW__cur) &colon;$	$ft$ $m$
Fin height $(fH__cur) &colon;$	$ft$ $m$
Fin thickness $(fT__cur) &colon;$	$ft$ $m$
Distance Between Fins $(fD__cur) &colon;$	$ft$ $m$
Number of tubes $(ntube__cur) &colon;$

Testing this radiator design under different coolant flow and air flow conditions yielded the following graph of heat transfer performance vs. coolant flow rate at different airflow speeds.

A heat transfer performance of $4025 \frac{Btu}{minute}$ was obtained using a coolant volumetric flow, air volumetric flow and air velocity of $30 gpm, 2349 \frac{{ft}^{3}}{minute}, 10 \frac{mi}{h},$ respectively.

These results are summarized in the table below.

Figure 3: Heat transfer performance vs. coolant flow rate at different airflow speeds

Radiator Operating Conditions
Coolant Volumetric Flow $(vf__c) &colon;$	$gpm$ $\frac{m^{3}}{s}$
Air Volumetric Flow $(vf__a) &colon;$	$\frac{{ft}^{3}}{minute}$ $\frac{m^{3}}{s}$
Air Velocity $(v__a) &colon;$	$\frac{mi}{h}$ $\frac{m}{s}$
Heat Transfer Performance $(q__cur) &colon;$	$\frac{Btu}{minute}$ $\frac{J}{s}$

Proposed Radiator Model Dimensions

Our proposed design has a radiator length that is 30% smaller than that of the original model. The dimensions of the radiator core (radiator length, radiator width and radiator height) can be adjusted to any dimension.

The table below summarizes the radiator dimensions for our proposed design.

Proposed Radiator Dimensions
Radiator length $(rL__new) &colon;$	$ft$ $m$
Radiator width $(rW__new) &colon;$	$ft$ $m$
Radiator height $(rH__new) &colon;$	$ft$ $m$
Tube width $(tW__new) &colon;$	$ft$ $m$
Tube height $(tH__new) &colon;$	$ft$ $m$
Fin width $(fW__new) &colon;$	$ft$ $m$
Fin height $(fH__new) &colon;$	$ft$ $m$
Fin thickness $(fT__new) &colon;$	$ft$ $m$
Distance Between Fins $(fD__new) &colon;$	$ft$ $m$
Number of tubes $(ntube__new) &colon;$

Coolant and Air Property Tables

The thermal fluid properties for the coolant and air are listed in the following two tables.

Coolant Properties: 50-50 Glycol-Water
Thermal conductivity $(k__c) &colon;$	$\frac{Btu}{h \cdot ft \cdot degF}$ $\frac{W}{m \cdot K}$
Specific Heat $(C__c) &colon;$	$\frac{Btu}{lb \cdot degF}$ $\frac{J}{kg \cdot K}$
Density $(ρ__c) &colon;$	$\frac{lb}{{ft}^{3}}$ $\frac{kg}{m^{3}}$
Dynamic Viscosity $(μ__c) &colon;$	$\frac{lb}{ft \cdot s}$ $Pa \cdot s$
Coolant Temperature $(T__c) &colon;$	$degF$ $K$

Air Properties:
Thermal conductivity $(k__a) &colon;$	$\frac{Btu}{h \cdot ft \cdot degF}$ $\frac{W}{m \cdot K}$
Specific Heat $(C__a) &colon;$	$\frac{Btu}{lb \cdot degF}$ $\frac{J}{kg \cdot K}$
Density $(ρ__a) &colon;$	$\frac{lb}{{ft}^{3}}$ $\frac{kg}{m^{3}}$
Dynamic Viscosity $(μ__a) &colon;$	$\frac{lb}{ft \cdot s}$ $Pa \cdot s$
Coolant Temperature $(T__a) &colon;$	$degF$ $K$

2. Heat Transfer Performance of Proposed Radiator Assembly

We expect the heat transfer performance of the smaller radiator assembly to be smaller than that of the original radiator model because we are reducing the surface area to coolant ratio. The question that we answer in this section is "How much smaller is the heat transfer performance?" If the heat transfer performance is only marginally smaller, we can take other approaches to increase the performance, for example, increase the number of fins per row, change the fin material, or change the flow arrangement.

The ε-Ntu (effectiveness-Ntu) method is used to predict the heat transfer performance of our new system.

The more common equations that are typically used in heat exchange design are listed below.

Heat Exchange Equations:	Definitions:
$HeatTransferEquation ≔ q = ϵ \cdot Cmin \cdot ITD &colon;$	The rate of conductive heat transfer
$UniversalHeatTransferEquation ≔ \frac{1}{UA} = \frac{1}{hc \cdot Ac} + \frac{1}{nfha \cdot Aa} &colon;$	The overall thermal resistance present in the system
$ReynoldsEquation ≔ ReynoldsNum = (\frac{ρ \cdot v \cdot D__H}{μ}) &colon;$	A dimensionless modulus that represents fluid flow conditions
$HydraulicDiameter ≔ D__H = \frac{4 \cdot Amin}{WP} &colon;$	Parameter used to equate any flow geometry to that of a round pipe
$DittusBoelterEquation ≔ NusseltNum = 0.023 \cdot {ReynoldsNum}^{0.8} \cdot {PrandtlNum}^{\frac{1}{3}} &colon;$	An equation used to calculate the surface coefficient of heat transfer for fluids in turbulent flow
$PrandtlEquation ≔ PrandtlNum = \frac{C \cdot μ}{k} &colon;$	A dimensionless modulus that relates fluid viscosity to the thermal conductivity, a low number indicates high convection
$NusseltEquation ≔ NusseltNum = \frac{hc \cdot D_{H}}{k} &colon;$	A dimensionless modulus that relates surface convection heat transfer to fluid conduction heat transfer
$NtuEquation ≔ Ntu = \frac{UA}{Cmin} &colon;$	A dimensionless modulus that defines the number of transferred units
$ϵNtuEquation ≔ ϵ = 1 - {&ExponentialE;}^{- (\frac{Cmax}{Cmin}) \cdot (1 - {&ExponentialE;}^{- Cratio \cdot Ntu})} &colon;$	A mathematical expression of heat exchange effectiveness vs. the number of heat transfer units
$ITDEquation ≔ ITD = CoolantTemperature - AirTemperature &colon;$	Measure of the initial temperature difference

We must first calculate the overall heat transfer coefficient $UA__new$ of the smaller radiator before we can determine its heat transfer performance, $q__new$ .

Solve for $UA__new$

The Universal Heat Transfer Equation is defined in (1)

$UniversalHeatTransferEquation$

$\frac{1}{UA} = \frac{1}{hc Ac} + \frac{1}{nfha Aa}$

(1)

The next several steps will take us through the process for solving for the unknown values of $Ac, Aa, hc$ and $nfha$

Solve for $Ac__new$ & $Aa__new$

$CoolantSurfaceArea ≔ Ac = NumberOfTubes \cdot (2 \cdot (TubeHeight \cdot RadiatorLength) + 2 \cdot (TubeWidth \cdot RadiatorLength)) &colon;$

$AirSurfaceArea ≔ Aa = TotalNumberOfAirPassages \cdot (2 \cdot (FinDistance \cdot FinHeight) + 2 \cdot (FinHeight \cdot FinWidth)) &colon;$

where

$TotNumAirPassages ≔ TotalNumberOfAirPassages = NumRowsOfFins \cdot (\frac{RadiatorLength}{FinDistance}) &colon;$

Figure 4: Expanded view of tubes

Figure 5: Expanded view of fins

Solving the unknown values leads to the following values for the $TotalNumberOfAirPassages$ , $Ac$ , and $Aa$ .

$TotalNumberOfAirPassages__new ≔ simplify (solve (subs ((NumRowsOfFins = (ntube__new - 1), RadiatorLength = rL__new, FinDistance = fD__new), TotNumAirPassages), TotalNumberOfAirPassages))$

$9216.$

(2)

$Ac__new ≔ simplify (solve (subs (TubeHeight = tH__new, TubeWidth = tW__new, RadiatorLength = rL__new, NumberOfTubes = ntube__new, CoolantSurfaceArea), Ac))$

$8.49968 ⟦{ft}^{2}⟧$

(3)

$Aa__new ≔ simplify (solve (subs ((TotalNumberOfAirPassages = TotalNumberOfAirPassages__new, FinDistance = fD__new, FinHeight = fH__new, FinWidth = fW__new), AirSurfaceArea), Aa))$

$61.7456 ⟦{ft}^{2}⟧$

(4)

${Atotal}_{new} ≔ simplify (Ac__new + Aa__new) &semi;$

$70.2453 ⟦{ft}^{2}⟧$

(5)

Solve for $hc__new$

The value of hc depends on the physical and thermal fluid properties, fluid velocity and fluid geometry.

The ReynoldsEquation defined below can be used to determine the flow characteristics of the coolant as it passes through the tubes.

$ReynoldsEquation$

$ReynoldsNum = \frac{ρ v D__H}{μ}$

(6)

The value $D__H$ is found from the HydraulicDiameter equation:

$HydraulicDiameter$

$D__H = \frac{4 Amin}{WP}$

(7)

where

$Amin__c ≔ simplify (tW__new \cdot tH__new) &semi;$

$0.000413872 ⟦{ft}^{2}⟧$

(8)

$WP__c ≔ 2 \cdot simplify (tW__new + tH__new) &semi;$

$0.171712 ⟦ft⟧$

(9)

$D__Hc ≔ simplify (solve (subs ([Amin = Amin__c, WP = WP__c], HydraulicDiameter), D__H))$

$0.00964108 ⟦ft⟧$

(10)

The velocity of the coolant as it flows through the tubes is:

$v__c ≔ simplify (\frac{vf__c}{ntube__new \cdot Amin__c})$

$4.89391 ⟦\frac{ft}{s}⟧$

(11)

$ReynoldsNum__c ≔ simplify (solve (subs ([D__H = D__Hc, v = v__c, ρ = ρ__c, μ = μ__c], ReynoldsEquation), ReynoldsNum))$

$5982.76$

(12)

For fluids that are in turbulent flow (that is, ReynoldsNum $> 5000$ ), we can use the DittusBoelterEquation to relate the ReynoldsNum wwith the NusseltNum. The NusseltNum is dependent upon the fluid flow conditions and can generally be correlated with the ReynoldsNum. Solving for the NusseltNum will enable us to determine the value of hc.

$DittusBoelterEquation$

$NusseltNum = 0.023 {ReynoldsNum}^{0.8} {PrandtlNum}^{1 / 3}$

(13)

$PrandtlEquation$

$PrandtlNum = \frac{C μ}{k}$

(14)

$NusseltEquation$

$NusseltNum = \frac{hc D_{H}}{k}$

(15)

$PrandtlNum__c ≔ simplify (solve (subs ([C = C__c, μ = μ__c, k = k__c], PrandtlEquation), PrandtlNum))$

$6.59999$

(16)

${NusseltNum}_{c} ≔ solve (subs ([ReynoldsNum = ReynoldsNum__c, PrandtlNum = PrandtlNum__c], DittusBoelterEquation), NusseltNum)$

$45.3346$

(17)

Knowing the NusseltNumber we can now solve for $hc__new$

$hc__new ≔ simplify (solve (subs ([k = k__c, D_{H} = D__Hc, NusseltNum = {NusseltNum}_{c}], NusseltEquation), hc))$

$1128.53 ⟦\frac{Btu}{h {ft}^{2} degF}⟧$

(18)

Determine $nfha__new$

We solve for $ha__new$ in a similar manner as we did for $hc__new$ (by determining the ReynoldsNum for air)

${Amin}_{a} ≔ simplify (fH__new \cdot fD__new)$

$0.000203025 ⟦{ft}^{2}⟧$

(19)

$WP__a ≔ 2 \cdot simplify (fD__new + 8 \cdot fD__new)$

$0.0937500 ⟦ft⟧$

(20)

$D__Ha ≔ simplify (solve (subs ([Amin = {Amin}_{a}, WP = WP__a], HydraulicDiameter), D__H))$

$0.00866240 ⟦ft⟧$

(21)

$ReynoldsNum__a ≔ simplify (solve (subs ([D__H = D__Ha, v = v__a, ρ = ρ__a, μ = μ__a], ReynoldsEquation), ReynoldsNum))$

$701.978$

(22)

The ReynoldsNum for air indicates that the air flow is laminar $(that is, ReynoldsNum__a < 2100$ -- LaminarFlow). As a result, we cannot use the DittusBoelterEquation to relate the ReynoldsNum to the NusseltNum and hence determine the value for $ha__new$ . Another approach to determining the value of $ha__new$ is to solve for the value of $ha__cur$ since the value of $ha__new = h a__cur$ . In the next section, we will show how the value of $ha__cur$ is calculated by first obtaining the heat transfer coefficient for the original radiator $(UA__cur)$ .

Solve for $nfha__cur$

The equation, which relates Number of Transferred Units $(Ntu)$ to Universal Heat Transfer, will be used to determine the Universal Heat Transfer Coefficient $(UA__cur)$ of the current model.

$NtuEquation$

$Ntu = \frac{UA}{Cmin}$

(23)

$Cmin$ is obtained by comparing the thermal capacity rate $(CR)$ for the coolant and air.

$ThermalCapacityRate ≔ CR = C \cdot mfr &semi; MassFlowRate ≔ mfr = FluidViscosity \cdot ρ &semi;$

$CR = C mfr$

(24)

The Mass Flow Rate for the coolant and air are:

$mf__a ≔ simplify (solve (subs ([FluidViscosity = vf__a, ρ = ρ__a], MassFlowRate), mfr))$

$2.77964 ⟦\frac{lb}{s}⟧$

(25)

$mf__c ≔ simplify (solve (subs ([FluidViscosity = vf__c, ρ = ρ__c], MassFlowRate), mfr))$

$4.23765 ⟦\frac{lb}{s}⟧$

(26)

The Thermal Capacity Rates for the coolant and air are:

$CR__a ≔ simplify (solve (subs ([mfr = mf__a, C = C__a], ThermalCapacityRate), CR))$

$40.0268 ⟦\frac{Btu}{\min degF}⟧$

(27)

$CR__c ≔ simplify (solve (subs ([mfr = mf__c, C = C__c], ThermalCapacityRate), CR))$

$223.748 ⟦\frac{Btu}{\min degF}⟧$

(28)

Since $CR__a < CR__c &colon;$

$Cmin__cur ≔ CR__a &semi; Cmax__cur ≔ CR__c &semi;$

$40.0268 ⟦\frac{Btu}{\min degF}⟧$

(29)

and

$Cratio__cur ≔ \frac{CR__a}{CR__c}$

$0.178892$

(30)

Next, we need to calculate the Number of Transfer Units $(Ntu)$ of the original radiator assembly. To do this we for $ITD__cur$ , $ϵ__cur$ and $q__cur$ from the $ϵNtuEquation$ $HeatTransferEquation$ , and $ITDEquation$ , respectively.

$ITDEquation$

$ITD = CoolantTemperature - AirTemperature$

(31)

$ITD__value ≔ combine (simplify (evalf (T__c - T__a)), units)$ = $150. ⟦degF⟧$

$ϵNtuEquation$

$ϵ = 1 - {&ExponentialE;}^{- \frac{Cmax (1 - {&ExponentialE;}^{- Cratio Ntu})}{Cmin}}$

(32)

$ϵ__cur ≔ simplify (solve (subs ([q = q__cur, Cmin = Cmin__cur, ITD = ITD__value], HeatTransferEquation), ϵ))$ = $0.670384$

Using the $HeatTransferEquation$ , the value of $Ntu__cur$ can be found.

$HeatTransferEquation$

$q = ϵ Cmin ITD$

(33)

$Ntu__cur ≔ solve (subs ([Cmax = Cmax__cur, Cmin = Cmin__cur, Cratio = Cratio__cur, ϵ = ϵ__cur], ϵNtuEquation), Ntu)$ = $1.23717$

We can finally solve for $UA__cur$ by substituting the values of $Ntu__cur$ and $Cmin$ in to the $NtuEquation &colon;$

$NtuEquation$

$Ntu = \frac{UA}{Cmin}$

(34)

$UA__cur ≔ solve (subs ([Ntu = Ntu__cur, Cmin = Cmin__cur, q = q__cur], NtuEquation), UA)$ = $49.5199 ⟦\frac{Btu}{\min degF}⟧$

Now that we have the value of $UA__cur$ , we can use the $UniversalHeatTransferEquation$ to determine the value for $nfha__cur$ , which is equal to the value of $nfha__new$ .

Solving for the unknown values of $Aa__cur$ and $Ac__cur$ yields the following:

$TotalNumberOfAirPassages__cur ≔ simplify (solve (subs ((NumRowsOfFins = (ntube__cur - 1), RadiatorLength = rL__cur, FinDistance = fD__cur), TotNumAirPassages), TotalNumberOfAirPassages))$

$12288.$

(35)

$Aa__cur ≔ simplify (solve (subs ((TotalNumberOfAirPassages = TotalNumberOfAirPassages__cur, FinDistance = fD__cur, FinHeight = fH__cur, FinWidth = fW__cur), AirSurfaceArea), Aa))$

$82.3274 ⟦{ft}^{2}⟧$

(36)

$Ac__cur ≔ simplify (solve (subs (TubeHeight = tH__cur, TubeWidth = tW__cur, RadiatorLength = rL__cur, NumberOfTubes = ntube__cur, CoolantSurfaceArea), Ac))$

$11.3329 ⟦{ft}^{2}⟧$

(37)

Finally at this point we can solve for the value of $nfha__cur$ by substituting the values for $Aa__cur A c__cur$ , and $UA__cur$ into the $UniversalHeatTransferEquation &colon;$

$UniversalHeatTransferEquation$

$\frac{1}{UA} = \frac{1}{hc Ac} + \frac{1}{nfha Aa}$

(38)

$nfha__cur ≔ simplify (solve (subs ([UA = UA__cur, hc = hc__new, Ac = Ac__cur, Aa = Aa__cur], UniversalHeatTransferEquation), nfha))$

$47.0113 ⟦\frac{Btu}{h {ft}^{2} degF}⟧$

(39)

Since the value of $nf$ is the same for both the original and proposed radiator models we can determine the value of $nfha__new$ directly from the value of $nfha__cur$

$nfha__new ≔ nfha__cur$

$47.0113 ⟦\frac{Btu}{h {ft}^{2} degF}⟧$

(40)

Solve for $q__new$

We can determine the heat transfer performance $(q__new)$ of the new radiator assembly by using the $HeatTransferEquation &colon;$

$HeatTransferEquation$

$q = ϵ Cmin ITD$

(41)

We can determine the value of $ϵ__new$ from the $ϵNtuEquation$ .

$ϵNtuEquation$

$ϵ = 1 - {&ExponentialE;}^{- \frac{Cmax (1 - {&ExponentialE;}^{- Cratio Ntu})}{Cmin}}$

(42)

The unknown value for $Ntu__new$ can be determined using the $NtuEquation &colon;$

$NtuEquation$

$Ntu = \frac{UA}{Cmin}$

(43)

Finally the value of $UA__new$ can be determined from the $UniversalHeatTransferEquation$

$UniversalHeatTransferEquation$

$\frac{1}{UA} = \frac{1}{hc Ac} + \frac{1}{nfha Aa}$

(44)

Solving for $UA__new, Ntu__new$ , and $ϵ__new$ yields the following:

$UA__new ≔ simplify (solve (subs ([hc = hc__new, nfha = nfha__new, Aa = Aa__new, Ac = Ac__new], UniversalHeatTransferEquation), UA))$ = $37.1398 ⟦\frac{Btu}{\min degF}⟧$

$Ntu__new ≔ solve (subs ([UA = UA__new, Cmin = Cmin__cur], NtuEquation), Ntu)$ = $0.927873$

$ϵ__new ≔ solve (subs ([Cmax = Cmax__cur, Cmin = Cmin__cur, Ntu = Ntu__new, Cratio = Cratio__cur], ϵNtuEquation),)$ = $0.574696$

The heat transfer performance $q__new$ of our smaller radiator design can be found by substituting the value of value of $ϵ__new$ and $C m i n$ into the $HeatTransferEquation &colon;$

$q__new ≔ simplify (solve (subs ([ϵ = ϵ__new, Cmin = Cmin__cur, ITD = ITD__value], HeatTransferEquation), q))$

$3450.48 ⟦\frac{Btu}{\min}⟧$

(45)

As expected the heat transfer performance of our proposed radiator design is smaller than that of the original.

$q$

(46)

$q__diff ≔ simplify (q__cur - q__new)$

$574.52 ⟦\frac{Btu}{\min}⟧$

(47)

3. Adjusting Heat Transfer Performance of Proposed Radiator Design

Effects of Radiator Length on Heat Transfer Performance

The effects of radiator length on heat transfer performance (while keeping all other parameters the same as in the proposed design) can be examined by changing the adjacent dial.

The heat transfer performance values for four different radiator lengths are summarized in the table below.

Radiator Length	Heat Transfer Performance
$0.5 ft$ $0.152400 m$	$1560.09 \frac{Btu}{\min}$ $27414.7 \frac{J}{s}$
$1.0 ft$ $0.304800 m$	$2661.09 \frac{Btu}{\min}$ $46762.1 \frac{J}{s}$
$1.5 ft$ $0.457200 m$	$3450.50 \frac{Btu}{\min}$ $60633.9 \frac{J}{s}$
$2.0 ft$ $0.609600 m$	$4025.01 \frac{Btu}{\min}$ $70729.5 \frac{J}{s}$

Radiator Length vs. Heat Transfer Performance

$m$ $ft$

$\frac{J}{s}$ $\frac{Btu}{minute}$

From the table, we can confirm our hypothesis that changing radiator length alone will not be sufficient to generated the desired heat transfer performance. As mentioned in the previous section, there are several methods available to increase the heat transfer performance of a radiator assembly. For our proposed design, we have chosen to increase the metal-to-air surface area by increasing the number of fins per row.

Effects of Surface Area on Heat Transfer Performance

To achieve a heat transfer performance for our proposed design equal to that of the current design ( that is $, 4025 ⟦\frac{Btu}{minute}⟧ \approx 70729.3 ⟦\frac{J}{s}⟧$ ), we must increase the number of fins per row. The procedure called $DetermineNumberOfFins (q__cur)$ , defined within the Code Edit Region, calculates the number of fins per row needed to achieve the desired heat transfer performance for our assembly.

Calculate number of fins per row needed to achieve threshold heat transfer performance

$simplify (DetermineNumberOfFins (q__cur))$ = $NumFinsPerRow = 436.056$

Thus, the number of fins per row must be increased from 384 to 437 to achieve a heat transfer performance of $4025 \frac{Btu}{minute} \approx 70729.3 \frac{J}{s}$ . The graph in Figure 6 shows the effects of changing the number of fins per row on the heat transfer performance for our smaller radiator design.

Figure 6: Effects of surface area on heat transfer performance

The application below allows you to compare the effects of changing the number of fins per row on the heat transfer performance for two different radiator lengths based on a given reference radiator length. The two different radiator lengths can be defined in the terms of the percent or absolute change of the reference.

Maple Application -- Effects of heat transfer performance vs. number of fins per row for different radiator lengths
Reference Radiator Length	Radiator Length 1	Radiator Length 2


$m$ $ft$	$m$ $ft$	$m$ $ft$

4. Export Optimized Radiator Dimensions to SolidWorks

We can create a CAD rendering of our smaller radiator assembly. The design parameters of our new design are the same as the original, except it is smaller in length and has more fins per row.

The parameters of our new radiator model are listed in the table below. It is important to note that the number of fins per row is actually a measure of the distance between the fins (that is, how the fins are spread out within a row).

Export Radiator Dimensions to SolidWorks
Number of Fins Per Row
Radiator Length $(rL__SolidWorks) &colon;$	$ft$ $m$
Distance Between Fins $(fD__SolidWorks) &colon;$	$ft$ $m$

* Note: For consistency, we are creating a scaled CAD rendering model of the new optimized radiator assembly similar to that of the original CAD rendering

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