353 WK8TPCSWEB.TEX
ME M.M. Yovanovich Week 8 Lecture 1
Provide midterm results. Problem 3 of midterm will be re-submitted at start of next lecture. Outline of the solution procedure to be followed.
Lecture 2
Hand in Problem 3. Half-space solutions.
Neumann Solution:
x 2 p T x; t = Ti + q0 p t e,x2=4 t , x erfc p k 2 t Instantaneous surface temperature rise: T 0; t = Ts. 2 p Ts = Ti + p q0 t k
"
!
Robin Solution: 2 3 p ! ! ! T x; t , Ti = erfc p , exp 4 hx + h 2 t5 erfc p + h t x x Tf , Ti k k k 2 t 2 t
Instantaneous surface temperature rise: T 0; t = Ts.
Ts , Ti = 1 , exp 4 h Tf , Ti k
Instantaneous surface heat ux: qs x; qs = ,k @T@x t
!
x=0
2 !2
3 p 5 erfc h t
k
t
!
2 !2 = hTf , Ti exp 4 h
k
3 t5
erfc h k t
p
!
Approximations of error and complementary error functions from P. R. Greene, J. Fluids Engineering, Vol. 111, pp. 224-226.
erf x = 1 , A exp ,B x + C 2
and with coe cients:
h
h
i
erfcx = 1 , erf x = A exp ,B x + C 2 A = 1:5577; B = 0:7182;
i
C = 0:7856 Greene claims the approximations are accurate to 0:42. This is acceptable
accuracy for many engineering calculations.
s
Inverse of Complementary Error Function Inverse of y = erfcx is x = erfc,1y where 0 y 1 and x 0.
1 ln y x = ,C + , B A Accuracy of inverse is unknown. Commence overview of 1D transient solutions in plane wall, long circular cylinder, and solid sphere. See sections 5.4 through 5.6. Discuss temperature variation in plane wall during cooling.
Lecture 3
Return Midterm Exam at end of lecture. Provide new statistics. Internal transient conduction: plane wall T x; t, long solid circular cylinder T r; t and solid sphere T r; t; See text and Web site for Maple worksheets for Heisler cooling charts for one-dimensional conduction. Dimensionless temperature:
; Fo depends on dimensionless position:
= x=L for wall of thickness: 2L;
= r=a for cylinder and sphere of radius a; and dimensionless time: Fo = t=L2 where L = L or a. Dimensionless temperature is de ned as: = Tf , T
; Fo ; for heating
h
Tf , Ti
i f
and
c
= T
; FoT, Tf ; T, 2
for cooling
General Form of Temperature Solutions
; Fo =
1 X
where An are the temperature Fourier coe cients; S n
is the space-dependent 2 function, and exp, nFo is the time-dependent function. The eigenvalues: n are the positive, real roots of the characteristic equations: plane wall n sin n = Bi cos n ; and long cylinder n J1 n = BiJ0 n ; where J0x and J1x are Bessel functions of the rst kind order of 0 and 1, respectively; and sphere n cos n = 1 , Bi sin n ; where the Biot number: Bi = hL=k lies in the range 0 Bi 1.
n=1
2 An exp , nFo S n
Fo 0
Space-dependent Functions
and and
S n
= cos n
; S n
= J0 n
;
plane wall long cylinder
S n
= sin
n
; sphere n Fourier Temperature Coe cients An are obtained from: An =
and 2 sin n ; n + sin n cos n
2 0
plane wall long cylinder sphere
An =
and
n
2J1 n J n + J12 n ;
n n n
cos An = 2sin n , ncos n ; , sin
Heat Loss
The heat loss is de ned as
Qloss = Ei , E t = cP iV , cP V = cpV i , 3
where Ei and E t represent the internal energy within the body initially and at arbitrary time t 0, and V is the total volume of the body. The thermophysical properties are assumed to be constant during the cooling process. The volume average body temperature excess is de ned as ZZZ = 1 dV
Heat Loss Fraction
V
V
The heat loss fraction is de ned as which gives
Ei , E t ; Ei
t 0
are obtained from: and and
1 Q = 1 , = 1 , X B exp
, 2 Fo ; Fo 0 n n Qi i n=1 where Qi = Ei for convenience. The heat loss fraction Fourier coe cients Bn
Bn = An sin n ;
n
2
plane wall long cylinder sphere
Bn = 2An J1 n = 2 4Bi Bi2 ; 2 n n n+
6 2 Bn = 2 2 + Bi 2 , Bi ; n n Bi
For Fo Foc where Foc = 0:24; 0:21; 0:18 for plane wall, long cylinder and solid sphere, respectively, the general solution converges to the rst term of the summation, i.e., 2
; Fo = A1 exp, 1 FoS 1
and the rst eigenvalue can b...