Li James

Microprocessor heat sink design

problem # 1

13 October 2015ME 43300J. Gonzalez

Li James

Abstract: A Microprocessor is a computer chip that contains the Central Processing Unit (CPU) needed to run

computer, acting as the computer’s engine. However, the microprocessor can generate a lot of heat when

running the computer system and the excess heat can damage the processor and other parts of the computer.

This is when the microprocessor cooling system comes into play. A microprocessor’s cooling system contains a

combination of a heat sink with fins and a cooling fan to dissipate the processor’s heat. This design project will

primarily look at the heat sink design.

Project Description: The purpose of this project is that it asks for a new microprocessor cooling system design

to accommodate the Intel Pentium 4 processor due to the latter’s over heating problem. In order to simulate the

Microprocessor’s heat generation, a 20 Watt micro heater will be placed under the cooling system to test the

system’s heat dissipation. The following design restriction must be fulfilled, in order for the cooling system to

be successful. The Processor’s maximum temperature after the heat dissipation from the cooling system should

not exceed 55 degrees Celsius. The overall height of the cooling system should be 25 millimeters or less. The

budget for the cooling system is 25 dollars and the cooling system’s material must be compatible with the Intel

Pentium 4 processor. Detailed sketches of the design should be included as well.

Methodology:

Observations: 1: A type of heat removal system is required (Fans, Heat sink, etc.)

2: Material properties are required to define the system

3: Standard room temperature is recommended for ambient temperature

4: The system is air cooled

5: Analyze cost implementation

6: Compare the standards of the cooling system design to that of existing cooling system

7: Describe any potential human hazards from the system

8: Describe any additional energy required for the system in Watts.

9: Determine the system’s thermal performance.

Li JamesDesign Limitations:

- Maximum Surface Temperature is limited to 55 degrees Celsius

- Maximum height of the cooling system is limited to 25 millimeters

- Materials of the Cooling system must be compatible with the Intel Pentium 4 processor

- The budget of the cooling system is limited to 25 dollars.

Problem Definition:

The purpose of this problem is to demonstrate the performance of a microprocessor cooling system by

determining the removed heat flux, surface temperature of the chip and heat sink, efficiency, and effectiveness

using the heat equation.

Solution Plan:

- Select and define the material properties (2024 T6 aluminum in this case)

- Select and define the cooling system (Heat sink in this case)

- Find Biot’s number in order to determine whether or not to use the Fin analysis

- Use the heat equation for fins analysis (for Biot’s number less than 0.1)

- Determine the heat flux removal, surface temperature of the heat sink, efficiency, effectiveness using the heat equation

- Set the boundary conditions for the heat equation

Plan Execution:

Assumptions:

(i) Steady state heat transfer

(ii) Material is 2024 T6 Aluminum

(iii) Height of the cooling system includes heat sink and fan

(iv) Uniform thermal conductivity, surface temperature, ambient temperature and heat flux

(v) No contact resistance

(vi) No energy generated

(vii) No radiation

(viii) square heat sink

Analysis:

The Law of Energy conservation states the following:

Eq. 1) Ein - Eout + Egen = dE/dt

Based on Assumptions (i) and (vi) we can set Eq. 1 to the following:

Li JamesEin - Eout + 0 = 0 Eq. 2) Ein = Eout

Find Biot’s number to determine whether or not Fin analysis can be used.

Biot’s number = (h*t)/(2*k) < 0.1

If the Biot’s number condition is fulfilled, we can use fin analysis with an adiabatic tip boundary conditions

(dT/dx)] x = Lf = 0 and T(x)] x=0 = To with To < 55 °C

For fin analysis, convert Eq. 2 into the following:

Ein = P and Eout = qb + n*qf Eq. 3) P = qb + n*qf

Set the following formulas for each of the following variables:

qb = h*Ab*(To-Tif) ; qf = M*tanh(m*Lf)

Plug the formulas for qb and qf back into Eq. 3

Eq. 4) P = h * [w^2–n*(t*w)]*(To-Tif)+n*[(To-Tif)*sqrt(h*Pe*k*Ac)*tanh[(sqrt((h*Pe)/(k*Ac)))*Lf]] + Tin

Rearrange the Equation to solve for To

Eq. 5) To = [(P) / (h * [w^2 – n*(t*w)] + n*[sqrt(h*Pe*k*Ac) * tanh[(sqrt((h*Pe)/(k*Ac))*Lf)])] + Tin

After Solving for To, Plug the value back into Eq 4. to find the Energy removed and use that value to solve for the heat flux by using the following formula

qb” = qb/Ab ; qf” = qf/Ac ; qo” = qb” + n*qf” Eq 6) qr” = Pd - qo”

Solve for efficiency and effectiveness

Eq.7) ηf =(n* qf)/(h*As*(To-Tin))

Eq.8) ηo = 1 – (n*Af/At) * (1- ηf)

Eq.9) Ef = (n*qf)/(h*Ac*(To-Tin))

Engineering calculations: (the actual calculations will be done on Matlab).

Given values:

Material Properties for 2024-T6 Aluminum at 20 °C

Density (ρ) 2770 [Kg/(m^3)]

Constant pressure specific heat (Cp) 875 [J/(Kg*°C)]

Thermal Conductivity (K) 177 [W/(m*°C)]

Reflectivity (α) 7.3 * 10^-5 [m^2/s]

Ambient temperature (Tin) = 20 °C

Power generated by the micro heater (P) = 20 W

Power Density (Pd) = 7750 W/m^2

Li JamesThermal coefficient (h) = 30 [W/(m^2*°C)]

Number of Fins (N) = 24

Thickness of the fins (t) = 0.02 m

Length of the fins (Lf) = 0.08 m

To find the Heat flux removed (qr”), Base temperature of heat sink and chip (To), thermal efficiency (ηf and

ηo) and effectiveness (Ef) solve for equations 5, 6, 7, 8, 9.

Eq. 5) To=[(P)/(h*[w^2 – n*(t*w)]+n*[sqrt(h*Pe*k*Ac)* tanh[(sqrt((h*Pe)/(k*Ac))*Lf)])]+Tin = 52.7736 °C

Eq 6) qr” = Pd - qo”= 802.4191 W/m^2

Eq. 7) ηf =(N* qf)/(h*As*(To-Tin)) = 23.8828 %

Eq.8) ηo = 1 – (N*Af/At) * (1- ηf) = 24.4338%

Eq.9) Ef = (N*qf)/(h*Ac*(To-Tin)) = 6.0662 (Effectiveness at h = 30)

Additional energy provided by the fan

V = 12; I = 0.15 A; P = V * I = 1.8 W

Height of the system:

Fan height + Lf + y = 15mm + 8mm + 1.5 mm = 24.5 mm < 25

Cost analysis

2024 T6 Aluminum (0.5 Kilograms) $ 2.00

Socket 478 / Pentium 4 CPU Cooler (with Fan) $14.99

Cooler Master High Performance Thermal Compound $4.29

Total cost $21.28

Summary and Discussion of results

Results and concluding remarks: The heat sink proved effective at cooling the microprocessor. After solving all

the equations in the methodology and engineering calculations section, the following results were provided. The

heat sink managed to lower the base temperature of the Processor chip to 52.7736 °C, which is less than 55 °C.

The Height of the entire system is also acceptable at 24.5 mm, which is less than 25mm. The block of aluminum

can be machined in the CCNY manufacturing lab and the cooling fan is compatible with the Intel Pentium 4

processor. The height of the thermal compound paste is negligible and not included in the overall height. The

Li Jamestotal cost of the cooling system tops put at $21.28, which is under the $25 budget. Overall, the cooling system is

well within the given constraints and the design can be used to manufacture the cooling system.

Human Health Hazards: Potential hazards associated with the cooling system includes: the fan, the thermal

paste and the heat sink. The fan’s spinning blades can cause potential laceration or dismemberment or hazards if

the cover is either removed or damaged. The Thermal paste can be harmful if ingested. The Heat sink is what

absorbs and dissipates the heat generated by the processor and as such, it can be hazardous to touch it if it is too

hot.

Benchmark of the cooling system: The system design is within a similar price range as existing cooling systems

($21.28 versus $15 - $30). There was no reliable information for the thermal performance and endurance for

existing cooling system for Intel Pentium 4 processor and as such the design’s thermal performance and

endurance cannot be compared to existing systems.

System originality: The heat sink was designed from scratch from the given constraints and thus, it is a personal

design not based on existing heat sinks. The fan and thermal compound paste has to be purchased online and is

not part of the original design.

Charts and Drawings:

Figure 1: Effectiveness chart

Effectiveness versus Heat transfer coefficien t

1.01742.0322

3.04454.0542

5.06156.0662

01234567

0 5 10 15 20 25 30 35

Heat Transfer coefficient (W/(m^2*°C))"

Effe

ctiv

enes

s

Li James

Figure 2: Heat sink design (Dimensions are in Millimeters)

Reference and citations

http://www.51aluminum.com/product_367753/2024-aluminum-tube-price-per-kg.html

http://www.accessotronik.com/product_p/701853_m119.htm

http://www.newegg.com/Product/Product.aspx?Item=N82E16835103027&nm_mc=KNC- GoogleAdwords-PC&cm_mmc=KNC-GoogleAdwords-PC-_-pla-_-Thermal+Compound+%2f+Grease-

_-N82E16835103027&gclid=CIqYgNGItsgCFcEXHwode0cA2Q&gclsrc=aw.ds

Latif M. Jiji.(2013). Heat Transfer Essential, 3rd edition. New York, N.Y. : Quadratic Press

Gonzalez Jorge.(2015). Class Lecture Notes. New York, N.Y. : The City College of New York.

Li James

Appendix

Matlab codeclcclear all % Variables to inputh = input('What is the thermal coefficient? '); %W/(m^2*°C)n = input('What is the number of fins? '); %unitlesst = input('What is the fin thickness? '); % mLf = input('What is the length of the fin '); % m % Given valuesk = 177; % Thermal conductivity (W/(m*°C))tin = 20; % Ambient temperature (°C)w = 0.0508; % length of the sides (m)P = 20; % Power generated by the microheater (W)f = 7750; % Power density (W/(m^2)) % General EquationsAb = w.^2 - (n.*t.*w); % Area of the base (m^2)J = Ab.*h;Pe = 2.*(t+w); % perimeter of the fins (m)Ac = (t.*w); % tip area of one fin (m^2)M = (h.*Pe.*k.*Ac).^(1/2);m = ((h.*Pe)./(k.*Ac)).^(1/2); % base temperature equation To = ((P)./ (J + n.*(M.*tanh(m.*Lf)))) + tin; %Base Temperature (°C)disp('Base temperature (°C) = ');disp(To); % heat flux equations qb = h.*Ab.*(To-tin); % Heat transfer rate of the base (W)qb_f = qb./Ab; % Heat flux of the base (W/m^2)qf = M.*tanh(m.*Lf); % Heat transfer rate of a fin (W)qf_f = qf./Ac; % Heat flux of the fin (W/m^2)qt = qb + n.*qf; % Total Heat transfer (W)qt_f = qb_f + n.*qf_f; % Total Heat flux (W/m^2)removed_flux = f - qt_f; % Heat Flux removed from the system (W/m^2)disp('Heat flux removed (W/(m^2))')disp(removed_flux) % Efficiency equation Af = 2.*w.*Lf; %surface area of one fin (m^2)As = w.^2; % area of the heat sink without the fins (m^2)At = n.*Af + Ab; % Total Area of the heat sink (m^2)nf = (n.*qf)./(h.*As.*(To-tin)); no = (1 - ((n.*Af./At) .* (1-nf))); nfp = nf.*100; % Efficiency of a finnop = no.*100; % Overall Efficiencydisp('fin efficiency percentage')disp(nfp);disp('total efficiency percentage')disp(nop); % Effectiveness equation

Li JamesEffectiveness = (n.*qf)./(h.*Ac*(To-tin));disp('Effectiveness of the heat sink')disp(Effectiveness);

Index:

Ein = Energy rate entering the system (W);

Eout = Energy rate exiting the system (W)Egen = Energy rate Generated by the system (W)

dE/dt = total Energy rate change (W)h = thermal coefficient (W/(m^2 * °C))

t = fin thickness (m)k = thermal coefficient (W/(m * °C))dT/dx = temperature gradient (°C/m)

x = x-direction distance (m)Lf = Length of the fin (m)

T = temperature (°C)To = base temperature (°C)

P = Power generated by the microheater (W)qb = Heat transfer rate of the heat sink base (W)

qf = Heat transfer rate of the fins (W)

n = Number of finsAb = Area of the base (m^2) = (w^2 – n*(t*w))

Tin= Ambient Temperature (°C)w = sides (m)

M = sqrt(h*Pe*k*Ac) m = sqrt((h*Pe)/(k*Ac))

Pe = Perimeter of the fins (m) = 2*(t+Lf)Ac = tip area of one fin (m^2) = (t*w)

qb” = flux of the base (W/(m^2))qf” = flux of a fin (W/(m^2))

qo” =heat flux removed (W/(m^2))qr” = heat flux removed from the system (W/m^2)

Pd = Power density (W/m^2)ηf = fin efficiency

ηo = total efficiencyEf = Effectiveness

Af = total surface area of the fin = 2*w*LfAs = Total surface area of the heat sink (m^2)

At = total surface area of the heat sink (m^2) = Ab + N*AfV = Voltage (J)

Li James I = Current (1/s)

P = Power of the fan (J/s = W)y = heat sink base height (m)