Solutions Manual for Heat Conduction (ALL Chapters ) by Latif M. Jiji – Step-by-Step Problem Solutions

SOLUTIONS MANUAL

,
,SOLUTIONS MANUAL FOR HEAT CONDUCTION
CHAPTER 1-23



A rectangular plate of length L and height H slides
down an inclined surface with a velocity U. Sliding x
T
friction results in surface heat flux qo . The front and h W
y U
top sides of the plate exchange heat by convection.
L
The heat transfer coefficient is h and the ambient h
T 0 g
temperature is T . Neglect heat loss from the back
side and assume that no frictional heat is conducted
through the inclined surface. Write the two- 
dimensional steady state heat equation and boundary
conditions.

(1) Observations. (i) This is a two-dimensional steady state problem. (ii) Four boundary
conditions are needed. (iii) Rectangular geometry. (iv) Specified flux at surface (x,0) points
inwards. (v) Plate velocity has no effect other than generating frictional heat at the inclined
plane. (vi) Use Section 1.6 as a guide to writing boundary conditions.
(2) Origin and Coordinates. The origin and Cartesian coordinates x,y are as shown. The
coordinates move with the plate.
(3) Formulation.
(i) Assumptions. (1) Steady state, (2) two-dimensional, (3) no energy generation, (4) all
frictional heat is added to plate (inclined surface is insulated) and (5) stationary material
(plate does not move relative to coordinates).

(ii) Governing Equations. The heat equation in Cartesian coordinates is given by

(k T )   (k T )   (k T )  q  c (T  U T  V T W T ) (1.7)
x  x y y z z p
t x y z
where c p is specific heat and  is density. The above assumptions give
 
U V W   q  0
z t
Eq. (1.7) simplifies to

(k T )  
(k T )  0 (a)
x x y y
If we further assume constant conductivity, we obtain
2T 2T
 0 (b)
x 2
y 2


(iii) Boundary Conditions. Four boundary conditions are needed. They are:

(1) Specified flux at boundary (x,0)

, PROBLEM 1.11 (continued)
(2)