Two-Dimensional Fin Analysis Project for Heat Transfer (AME 3173)

This project focuses on the analysis of heat transfer within a two-dimensional fin, addressing the problem outlined in the AME 3173 Heat Transfer course. The fin, made of aluminum, is subjected to a convective environment, and the project requires the development of a program to solve the two-dimensional temperature distribution for elapsed times of 3, 30, and 300 seconds, across three different heat transfer coefficients (h = 400, 4000, and 40000 W/m2K), totaling nine configurations. The analysis includes the determination of steady-state temperature distributions for all three h values and the time required to reach that state. A grid size of 0.25 cm is specified for the computational analysis. The results are presented using contour plots and line graphs, and a comparison is made between the centerline steady-state data and the temperature profiles derived from fin theory, as covered in the textbook. The document details the assumptions made, the governing equations, and the boundary conditions used to model the heat flow through the fin. The project's objective is to provide a comprehensive understanding of heat transfer principles and their application in the analysis of a finned heat sink.

Running head: HEAT TRANSFER
HEAT TRANSFER
Name of Student
Institution Affiliation

HEAT TRANSFER 2
HEAT FLOW THROUGH RECTANGULAR PLATE HAVING UNIFORM CROSS
SECTIONAL AREA.
Assumptions
i. K is constant throughout the material.
ii. HT is unidirectional.
iii. There is no heat generation in the field
t0= temperature at the base of the fin
ta= ambient temperature
l= is the length of the fin
Acs= is the crossectional area of the beam
As = surface area of the beam
P= perimeter of the fin
And these can be illustrated using the following diagram

HEAT TRANSFER 3
Consider an element at a distance x having thickness dx and temp develop is dt and
temperature of the surface is T.
According to energy balance equation
Qx=Qx+dx +Qconv …………………………………………….1
Equation of heat flow ks
Qx= - KAcs dt
dx …………………………………………………..2
Outward of the heat transfer Qx+dx = Qx+ d
dx (Qx) dx …………..3
Heat loss by convection from element
Qconv= hAsource(t- ta)
= h. p. dx (t- ta)
Q/x + Q/x+ d/dx (Qx)dx + QConv

HEAT TRANSFER 4
d (1−KA cs )
dx . dt d / x
dx = - hpdx (t – ta)
KAcs d2 t
d x2 = ph (t-ta)
θ= t-ta
dθ
dx = dt
dx ( t is variant )
d2 θ
d x2 = ph
KAcs , Consider ph
KAcs = m2
d2 θ
d x2 – m2 θ=0
Second order differential equation
D=±m
PI= = C.F = Ce mx + C2e –mx
θ= Ce mx + C2e –mx
Boundary conditions,
X= 0 , θ−θ 0=to - ta
FIN LOSING HEAT AT TIP
Let t1 be the temperature at the tip
i. at x=0

HEAT TRANSFER 5
θ−θ 0=to - ta
ii. Qconv ( at x=u ) = Qconv ( at x = u)
−KAcs ¿ ) x= u =hA(tl−¿ta)
dt
dx )x=u = −h
k . Qu
From the above equation we obtain
(tl−¿ta) =Ce mx + C2e –mx
dt
dx = 0+ mCe mx −¿ mC2e –mx
−h
k . Qu=m(Ce mx −¿ C2e –mx)
−h
k . Qu=m(Ce ml + C2e –ml)= (Ce ml −¿ C2e –ml)
Where θl= (Ce mx +C2e –mx)= (Ce mx −¿ C2e –mx) and θ 0=C1+C2
−h
k .(Ce ml + C2e –ml)=(Ce mx −¿ θo e−ml+ Ce−ml)
θ e−ml( 1−h
km ) = αC 1 [coshml+ h
km sin hml ]
C1= θ e−ml (1− h
km )
2 [ coshml +hkm . sinhml ]
C2= θo−C 1
= θ ¿

HEAT TRANSFER 6
C2=
θ eml [ 1+h /km ]
2 (coshml + h
km sin hml )
Rate of heat transfer
Qfin= - (dt/dx) for x=0
Where θ= θ e−ml
( 1− h
km ) emx
2 [ coshml +hkm . sinhml ]
. θ eml
(1+ h
km )emx
2 [ coshml +hkm . sinhml ]
θ
Q0 = coshm(l−x )
coshml +
h
km .sin h m(l−x )
h
km . sinhml
t= ta+(t 0−ta)¿ ¿
dt
dx where x = 0 = to- (t 0−ta)¿ ¿
At x=0
( dt
dx )x=0 = -m ( to- ta) [sinhm(1-x)+h/km coshm(l-x]
Q = - KA dt/dx
Q= -
√ phKAcs(¿−ta) [sin h m (l−x )+ h
km cos θ ]
cos hml+ h
km sin hml
At x=L
( dt
dx ) x=L = −m(¿−ta)¿ ¿

HEAT TRANSFER 7
=
−h (¿−ta )
k [ cosh ( ml ) + h
km . sinh ( ml) ]
Q = −KAcl ¿) x=L
= KAcs × ( −h
K ¿
Q=
−hAcs (¿−ta)
cosh ( ml ) + h
km . sinh (ml)
Heat transfer at tip due 2 sides exposes to convection.

HEAT TRANSFER 8
References
Naterer, G. (2018). Advanced Heat Transfer. Hull: Taylor & Francis, CRC Press.
Thomas, G. (2012). Heat Transfer: A Problem Solving Approach. Chicago: Routledge.