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
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
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''×17''×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:
ft m
Radiator width rW__cur:
Radiator height rH__cur:
Tube width tW__cur:
Tube height tH__cur:
Fin width fW__cur:
Fin height fH__cur:
Fin thickness fT__cur:
Distance Between Fins fD__cur:
Number of tubes ntube__cur:
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 Btuminute was obtained using a coolant volumetric flow, air volumetric flow and air velocity of 30 gpm, 2349 ft3minute, 10 mih,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:
gpm m3s
Air Volumetric Flow vf__a:
ft3minute m3s
Air Velocity v__a:
mih ms
Heat Transfer Performance q__cur:
Btuminute Js
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:
Radiator width rW__new:
Radiator height rH__new:
Tube width tW__new:
Tube height tH__new:
Fin width fW__new:
Fin height fH__new:
Fin thickness fT__new:
Distance Between Fins fD__new:
Number of tubes ntube__new:
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:
Btuh⋅ft⋅degF Wm⋅K
Specific Heat C__c:
Btulb⋅degF Jkg⋅K
Density ρ__c:
lbft3 kgm3
Dynamic Viscosity μ__c:
lbft⋅s Pa⋅s
Coolant Temperature T__c:
degF K
Air Properties:
Thermal conductivity k__a:
Specific Heat C__a:
Density ρ__a:
Dynamic Viscosity μ__a:
Coolant Temperature T__a:
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⋅Cmin⋅ITD:
The rate of conductive heat transfer
UniversalHeatTransferEquation≔1UA=1hc⋅Ac+1nfha⋅Aa:
The overall thermal resistance present in the system
ReynoldsEquation≔ReynoldsNum=ρ⋅v⋅D__Hμ:
A dimensionless modulus that represents fluid flow conditions
HydraulicDiameter≔D__H= 4⋅AminWP:
Parameter used to equate any flow geometry to that of a round pipe
DittusBoelterEquation≔ NusseltNum= 0.023⋅ReynoldsNum0.8⋅PrandtlNum13:
An equation used to calculate the surface coefficient of heat transfer for fluids in turbulent flow
PrandtlEquation≔PrandtlNum= C⋅μk:
A dimensionless modulus that relates fluid viscosity to the thermal conductivity, a low number indicates high convection
NusseltEquation≔NusseltNum=hc⋅DHk:
A dimensionless modulus that relates surface convection heat transfer to fluid conduction heat transfer
NtuEquation≔Ntu=UACmin:
A dimensionless modulus that defines the number of transferred units
ϵNtuEquation≔ϵ=1−ⅇ−CmaxCmin⋅1−ⅇ−Cratio⋅Ntu:
A mathematical expression of heat exchange effectiveness vs. the number of heat transfer units
ITDEquation≔ITD = CoolantTemperature − AirTemperature:
Measure of the initial temperature difference
Perform calculations in FPS unitsPerform calculations in SI units
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
1UA=1hcAc+1nfhaAa
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⋅2⋅TubeHeight⋅RadiatorLength +2⋅TubeWidth⋅RadiatorLength:
AirSurfaceArea≔Aa= TotalNumberOfAirPassages⋅2⋅FinDistance⋅FinHeight+2⋅FinHeight⋅FinWidth:
where
TotNumAirPassages≔ TotalNumberOfAirPassages=NumRowsOfFins⋅RadiatorLengthFinDistance:
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≔simplifysolvesubsNumRowsOfFins=ntube__new−1,RadiatorLength=rL__new,FinDistance= fD__new,TotNumAirPassages,TotalNumberOfAirPassages
9216.
Ac__new≔simplifysolvesubsTubeHeight=tH__new, TubeWidth=tW__new, RadiatorLength=rL__new, NumberOfTubes=ntube__new, CoolantSurfaceArea,Ac
8.49968ft2
Aa__new≔simplifysolvesubsTotalNumberOfAirPassages=TotalNumberOfAirPassages__new,FinDistance=fD__new,FinHeight=fH__new,FinWidth=fW__new,AirSurfaceArea,Aa
61.7456ft2
Atotalnew≔simplifyAc__new + Aa__new;
70.2453ft2
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=ρvD__Hμ
The value D__H is found from the HydraulicDiameter equation:
HydraulicDiameter
D__H=4AminWP
Amin__c≔simplifytW__new⋅tH__new;
0.000413872ft2
WP__c≔2⋅ simplifytW__new+tH__new;
0.171712ft
D__Hc≔simplifysolvesubsAmin=Amin__c, WP=WP__c,HydraulicDiameter, D__H
0.00964108ft
The velocity of the coolant as it flows through the tubes is:
v__c≔simplifyvf__cntube__new⋅Amin__c
4.89391fts
ReynoldsNum__c≔simplifysolvesubsD__H=D__Hc, v=v__c, ρ=ρ__c, μ=μ__c, ReynoldsEquation,ReynoldsNum
5982.76
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=0.023ReynoldsNum0.8PrandtlNum1/3
PrandtlEquation
PrandtlNum=Cμk
NusseltEquation
NusseltNum=hcDHk
PrandtlNum__c≔simplifysolvesubsC=C__c, μ=μ__c, k=k__c, PrandtlEquation,PrandtlNum
6.59999
NusseltNumc≔solvesubsReynoldsNum=ReynoldsNum__c, PrandtlNum=PrandtlNum__c,DittusBoelterEquation, NusseltNum
45.3346
Knowing the NusseltNumber we can now solve for hc__new
hc__new≔simplifysolvesubsk=k__c, DH=D__Hc, NusseltNum=NusseltNumc, NusseltEquation, hc
1128.53Btuhft2degF
Determine nfha__new
We solve for ha__new in a similar manner as we did for hc__new (by determining the ReynoldsNum for air)
Amina≔simplifyfH__new⋅fD__new
0.000203025ft2
WP__a≔2⋅ simplifyfD__new+8⋅fD__new
0.0937500ft
D__Ha≔simplifysolvesubsAmin=Amina, WP=WP__a,HydraulicDiameter, D__H
0.00866240ft
ReynoldsNum__a≔simplifysolvesubsD__H=D__Ha, v=v__a, ρ=ρ__a, μ=μ__a, ReynoldsEquation,ReynoldsNum
701.978
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 = ha__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=UACmin
Cmin is obtained by comparing the thermal capacity rate CR for the coolant and air.
ThermalCapacityRate≔CR=C⋅mfr;MassFlowRate≔mfr = FluidViscosity⋅ρ;
CR=Cmfr
The Mass Flow Rate for the coolant and air are:
mf__a≔simplifysolvesubsFluidViscosity=vf__a,ρ=ρ__a,MassFlowRate,mfr
2.77964lbs
mf__c≔simplifysolvesubsFluidViscosity=vf__c,ρ=ρ__c,MassFlowRate,mfr
4.23765lbs
The Thermal Capacity Rates for the coolant and air are:
CR__a≔simplifysolvesubsmfr=mf__a,C=C__a,ThermalCapacityRate,CR
40.0268BtumindegF
CR__c≔simplifysolvesubsmfr=mf__c,C=C__c,ThermalCapacityRate,CR
223.748BtumindegF
Since CR__a<CR__c:
Cmin__cur≔CR__a;Cmax__cur≔CR__c;
and
Cratio__cur≔CR__aCR__c
0.178892
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
ITD__value≔combinesimplifyevalfT__c−T__a,units = 150.degF
ϵNtuEquation
ϵ=1−ⅇ−Cmax1−ⅇ−CratioNtuCmin
ϵ__cur≔simplifysolvesubsq=q__cur, Cmin=Cmin__cur, ITD=ITD__value,HeatTransferEquation,ϵ = 0.670384
Using the HeatTransferEquation, the value of Ntu__curcan be found.
HeatTransferEquation
q=ϵCminITD
Ntu__cur≔solvesubsCmax=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:
UA__cur≔solvesubsNtu=Ntu__cur, Cmin=Cmin__cur,q=q__cur,NtuEquation, UA = 49.5199BtumindegF
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≔simplifysolvesubsNumRowsOfFins=ntube__cur−1,RadiatorLength=rL__cur,FinDistance= fD__cur,TotNumAirPassages,TotalNumberOfAirPassages
12288.
Aa__cur≔simplifysolvesubsTotalNumberOfAirPassages=TotalNumberOfAirPassages__cur,FinDistance=fD__cur,FinHeight=fH__cur,FinWidth=fW__cur,AirSurfaceArea,Aa
82.3274ft2
Ac__cur≔simplifysolvesubsTubeHeight=tH__cur, TubeWidth=tW__cur, RadiatorLength=rL__cur, NumberOfTubes=ntube__cur, CoolantSurfaceArea,Ac
11.3329ft2
Finally at this point we can solve for the value of nfha__cur by substituting the values for Aa__cur Ac__cur, and UA__cur into the UniversalHeatTransferEquation:
nfha__cur≔simplifysolvesubsUA=UA__cur, hc=hc__new, Ac=Ac__cur, Aa=Aa__cur,UniversalHeatTransferEquation,nfha
47.0113Btuhft2degF
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
Solve for q__new
We can determine the heat transfer performance q__new of the new radiator assembly by using the HeatTransferEquation:
We can determine the value of ϵ__new from the ϵNtuEquation.
The unknown value for Ntu__new can be determined using the NtuEquation:
Finally the value of UA__new can be determined from the UniversalHeatTransferEquation
Solving for UA__new, Ntu__new, and ϵ__new yields the following:
UA__new≔simplifysolvesubshc=hc__new, nfha=nfha__new, Aa=Aa__new, Ac=Ac__new, UniversalHeatTransferEquation, UA = 37.1398BtumindegF
Ntu__new≔solvesubsUA=UA__new, Cmin=Cmin__cur,NtuEquation,Ntu = 0.927873
ϵ__new≔solvesubsCmax=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 Cmin into the HeatTransferEquation:
q__new≔simplifysolvesubsϵ=ϵ__new, Cmin=Cmin__cur, ITD=ITD__value,HeatTransferEquation,q
3450.48Btumin
As expected the heat transfer performance of our proposed radiator design is smaller than that of the original.
q
q__diff≔simplifyq__cur−q__new
574.52Btumin
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
1560.09Btumin 27414.7 Js
1.0ft 0.304800 m
2661.09Btumin 46762.1 Js
1.5ft 0.457200 m
3450.50Btumin 60633.9 Js
2.0ft 0.609600 m
4025.01Btumin 70729.5 Js
Radiator Length vs. Heat Transfer Performance
m ft
Js Btuminute
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 Btuminute ≈70729.3Js), we must increase the number of fins per row. The procedure called DetermineNumberOfFinsq__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.
simplifyDetermineNumberOfFinsq__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 Btuminute ≈70729.3 Js. 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
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:
Distance Between Fins fD__SolidWorks:
* 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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