Heat Transfer Project 1 MECH 4315 Instructor: Evgeny Shafirovich 04/06/2015 Group 4 Irma Edith Duran Reyes Joao De Rezende Martins Lucas Da Silva Rigobello
Heat Transfer Project 1 MECH 4315 Instructor: Evgeny Shafirovich
04/06/2015 Group 4 Irma Edith Duran Reyes Joao De Rezende Martins
Lucas Da Silva Rigobello
Natural convection of the heat plate First, It is calculated the
heat transfer for the heater by itself. It is the rate of heat
transfer for a horizontal hot plate by natural convection. In all
calculations the gravity is constant = 9.81 / and the assumptions
are: room temperature at 24C, constant temperature at fins and
base, vertical positions of fins. Data Room temperature: = 24
Surface Temperature: = 85 Length: = 2.54 Width: = 7.62 Calculation
process Surface Area: = = 1.9354810 !! ! Perimeter: = 2 + = 0.2032
Film Temperature: = !"!!! = 54.5 Then using the film temperature,
the follow numbers were taken from the Table A-15 (Properties of
air at 1 atm pressure) from the textbook:
Heat Sink Design Now, to project the heat sink it is supposed
that the heat sink should be able to dissipate the same amount of
heat of the initial hot plate, but in a low temperature (70 ). It
is only possible because of the larger surface area than the actual
plate, which the heat sink is going to have. So, by calculating the
optimum space between each two fins, it is possible to have an
initial idea to choose some close or appropriate number desired for
the actual space between each two fins. As well, it can be assumed
some value for the fin high that goes from 2.54>H>0 (cm).
Then it is calculated the thickness of each fin and the needed
number of fins to dissipate the same amount of heat as the hot
plate. For this, a Matlab code was made
to help in the calculation and interaction to find the desired
parameters that would make sense for the project. Data Room
temperature: = 24 Contact Temperature: = 70 Length: = 2.5410!
Width: = 7.6210! High: = 2.5410! Parameters
Space between each two fins: = 5.2 High of the fin: = 13.0 Heat
Sink calculations Firs, to find the temperature at the base of the
fin, it is need to consider the conduction of heat, the total
dissipated by the plate, through the material from the contact with
the heater to base of the fin. Distance from the contact to the fin
base: = = 0.0124 Thermal conductivity of the material (Aluminum): =
173 !! ! Thermal Resistance: = !!"#!" = 0.037 !! Temperature at the
base of the fin: = = 69.9478 The second part of the calculation is
the part referred to the fins calculation. Film Temperature: =
!"!!! = 46.9739 Then using the film temperature, the follow numbers
are taken from the Table A-15 (Properties of air at 1 atm pressure)
from the textbook:
Kinematic viscosity: = 1.769210!! !!! Prandtl Number: = 0.72332
Thermal diffusivity: = 0.02735 Volume expansion coefficient: =
!!"!!!" = 0.0031 !! Calculating the Optimum Space between each two
fins: Rayleigh Number: ! = !! !"!! !!!! = 5.334110! Optimum Space
between each two fins: = 2.714 !!"!!.!" = 4.5361 (So the chosen for
the math is reasonable since it is bigger than ) Now considering
the from the parameters, the calculation can follows as:
Rayleigh Number: ! = !! !"!! !!!! = 457.7260 Nusselt Number: =
!"#!"!!! ! + !.!"#!"!!! !.! !!.! = 1.6612 Heat transfer
coefficient: = !"!! = 8.7371 !! ! Number of fins: = !! !!! !"!! =
5.3194 Thickness of each fin: = 9.125 Heat transfer in the side
faces of the base
Surface: = = 0.000944 ! Perimeter: = 2 + = 0.1772 Film
Temperature: = !"!!! = 46 Then using the film temperature, the
follow numbers were taken from the Table A-15 (Properties of air at
1 atm pressure) from the textbook: Kinematic viscosity: = 1.79810 5
!!! Prandtl Number: = 0.7228 Thermal diffusivity: = 0.02735 Volume
expansion coefficient: = !!"!!"# = 0.00313 !! Characteristic
Length: = !"! = 0.005327 Rayleigh Number: ! = !! !"!! !!!!! =
476.89 Nusselt Number: = 0.54!!! = 9.8775 Heat transfer
coefficient: = !"!!! = 2.52 !! ! Heat transfer rate: = = 0.109579 =
! = 1.3007
a) Figure a) CAD model b) final heat sink b)
Matlab Code g=9.81; %Hot
Plate------------------------------------------ Ts=85; T=24;
L=2.54e-2; W=7.62e-2; As=W*L Lc=As/(2*(W+L)) tf=(Ts+T)/2
kv=1.8421e-5; Pr=0.72163; k=0.027678; beta=1/(tf+273)
Ral=g*beta*(Ts-T)*Lc^3*Pr/kv^2 Nu=0.54*Ral^(1/4) h=Nu*k/Lc
Q=h*As*(Ts-T) %Heat
Sink-------------------------------------------- S=0.52e-2; % We
can change it H=1.3e-2; % We can change it (2.54e-2
