HW #3 /Tutorial # 3 WWWR Chapter 18 ID Chapter 5

HW #3 /Tutorial # 3WRF Chapter 17; WWWR Chapter 18

ID Chapter 5

• Tutorial # 3• WWWR #18.12

(additional data: h = 6W/m2-K); WRF#17.1; WWWR#18.4; WRF#17.10 ; WRF#17.14.

• To be discussed during the week 1-5 Feb., 2016.

• By either volunteer or class list.

• Homework # 3 (Self practice)

• WRF #17.9; WRF#17.16.

• ID # 5.6, 5.9.

Unsteady-State Conduction

Transient Conduction Analysis

pC

qT

t

T

2

Spherical metallic specimen, initially at uniform temperature, T0

Energy balance requires

Large value of Bi •Indicates that the conductive resistance controls•There is more capacity for heat to leave the surface by convection than to reach it by conductionSmall value of Bi•Internal resistance is negligibly small•More capacity to transfer heat by conduction than by convection

Example 1 (WWWR Page 266)

• A long copper wire, 0.635cm in diameter, is exposed to an air stream at a temperature of 310K. After 30 s, the average temperature of the wire increased from 280K to 297K. Using this information, estimate the average surface conductance, h.

Example 1

Heating a Body Under Conditions of Negligible Surface Resistance

BC (1) -> C1=0BC (2) -> = n/LFo = t/(L/2)2

IC -> Fourier expansion of Yo(x) …..> Equation (18-12) Engineering Mathematics: PDE

BC(1)

BC(2)

IC

V/A = (WHL)/(2WH)=L/2

x

Detailed Derivation for Equations 18-12, 18-13

Courtesy by all CN5 students, presented by Lim Zhi Hua, 2003-2004

Detailed Derivation for Equations 18-12, 18-13

Courtesy by all CN5 students, presented by Lim Zhi Hua, 2003-2004

Example 2

Heating a Body with Finite Surface and Internal Resistance

Heat Transfer to a Semi-Infinite Wall

Temperature-Time Charts for Simple Geometric Shapes

Example 3

or Figure F.4

Example 4

WWWR 18-12; 18-13

WWWR 18-16

(a) T=Ts @ x =0 WWWR 18-20

(b) -k dT/dx = h (T-T∞) @ x =0 WWWR 18-21

Transient (Unsteady – State) Conduction Summary

i) Calculate Biot Modulus (Bi)

kA

VhBi

if Bi ≤ 0.1 → Lumped Parameter Analysis

TT

TT

tA

Vch op ln

if Bi ≥ 100 → There is temperature variation within the object. If the geometry of the solid objects falls into the 6 shapes given in Fig. 18.3 → Use figure 18.3 to calculate the temperature at the specific time.

Calculate

TT

TTo or 21x

t

And read off 21x

t or

TT

TTo

To find t or T if 0.1 ≤ Bi ≤ 100 → Use appendix F of W3R ( refer to examples 18.3 and 18.4)

Using Y =

oTT

TT X =

21x

t

n = 1x

x m =

1hx

k

Courtesy contribution by ChBE Year Representative, 2006.

ii) Slab Heating Heating of Body under negligible surface resistance. Check Bi no. and let m = 0. Heating a body with finite surface and internal resistance

0x

T (At centerline) and

TTk

h

x

T (At surface)

iii) Heat transfer into a semi – infinite wall Different from (ii) because there is no defined length scale Use Appendix L

For Heat transfer into a semi – infinite medium with negligible surface resistance

t

xerf

TT

TT

oS

S

2 or

t

xerf

TT

TT

oS

o

21

For Heat transfer into a semi – infinite medium with finite surface resistance

t

x

k

therf

k

th

k

hx

t

xerf

TT

TT

o

21exp

2 2

2

(18 – 21)