zotero-db/storage/54ZKDP73/.zotero-ft-cache

CHAPTER 1
FIRST-ORDER DIFFERENTIAL EQUATIONS
SECTION 1.1 DIFFERENTIAL EQUATIONS AND MATHEMATICAL MODELING
The main purpose of Section 1.1 is simply to introduce the basic notation and terminology of differential equations, and to show the student what is meant by a solution of a differential equation. Also, the use of differential equations in the mathematical modeling of real-world phenomena is outlined. Problems 1-12 are routine verifications by direct substitution of the suggested solutions into the given differential equations. We include here just some typical examples of such verifications.
3. If y1 = cos Zx and y2 = sin 2x, then y; = -2sin 2x and y; = 2cos 2x so y1~ = -4 cos 2x = - 4 y, and y; = - 4sin 2x = - 4 y2.

and

5. If y=ex-e-x, then l=e... +e-x so y'-y = (ex+e-x)-(ex-e-i) = ze--'. Thus
y' = y+2e-x.

6 •

If

y1 = e-2x

an d

y 2 = x e-2x ,

then

y

,
1

=

-

2 e-2i ,

y,,. = 4 e-2x ,

y

,
2

=

e-2x -

2x e -2x ,

an d

Y,i. == - 4 e-2..- + 4x e -2.:r . H ence

and

8. If y, =cosx-cos2x and y2 =sinx-cos2x, then y; =-sinx+2sin2x, Yi'=-cosx+4cos2x, and y~ =:cosx+2sin2x, y;=-sinx+4cos2x. Hence

Section 1.1

1

y;+ y, = (-cosx+4cos2x)+(cosx-cos2x) ::;: 3cos2x
and
y;+ y2 = (-sinx+4cos2x)+(sinx-cos2x) = 3cos2x.

x 2 y.. + 5x y' +4 y = x2 (-5x-4 + 6x4 ln x )+5x(x-3 -2x-3 ln x )+4(£2 In x) = (-5[2 +sx-2 )+(6x-2 - lOx-2 +4x-2 )In X = 0.
13. Substitution of y =en into 3y' =2y gives the equation 3re'x = 2 erx that simplifies
to 3r=2. Thus r=213.
14. Substitution of y =e'x into 4y" = y gives the equation 4r2 e'' = e'x that simplifies to
4r2 =1. Thus r=±l/2.
15. Substitution of y =e'x into y..+ y' - 2y = 0 gives the equation r 2e'" + r en - 2 e'x =0 that simplifies to = r 2 +r-2 (r+2)(r-l) = 0. Thus r=-2 or r= 1.
16. Substitution of = y en into 3 y' + 3 y' - 4 y = 0 gives the equation 3r2e'1 +3re'x -4e'" =0 that simplifies to 3r2 +3r-4 = 0. The quadratic formula then
gives the solutions r = (-3± ✓57)16.
The verifications of the suggested solutions in Problems 17-36 are similar to those in Problems 1-12. We illustrate the determination of the value of C only in some typical cases.
17. C = 2
18. C = 3
19. If y(x) = Ce" -1 then y(O) = 5 gives C- I = 5, so C = 6.
20. If y(x) = Ce-x +x-1 then y(O) = 10 gives C-1 = 10, so C = 11.
21. C = 7

2

Chapter 1

22. If y(x) = ln(x + C) then y(O) = 0 gives In C = 0, so C = I.
23. If y(x) = ¼x5 + C x-2 then y(2) = 1 gives the equation ¼·32 + C ·½= I with solution
C:;:;-56.
24. C::: 17
25. lf y(x) = tan(x2 + C) then y(O) = 1 gives the equation tan C = 1. Hence one value
of C is C =n I 4 (as is this value plus any integral multiple of n).
26. Substitution of x = n and y =0 into y = (x+C)cosx yields the equation 0 = (Jr +C)(-1), so C = -n.
= 27. y I x+ y = = 28, Theslopeofthelinethrough (x,y) and (x/2,0) is y' (y-O)l(x-x/2) 2y/x,
so the differential equation is x y' = 2 y. 29. If m =y' is the slope of the tangent line and m' is the slope of the normal line at (x, y),
then the relation mm'= -1 yields m' = 1I y' = (y-1) /(x-0). Solution for y' then gives the differential equation (1- y) y' = x.
30. Here m =y' and = m' Dx (x2 + k) = 2x, so the orthogonality relation mm'= -1 gives the differential equation 2x y' = - I.
= 31. Theslopeofthelinethrough (x,y) and (-y,x) 1s y' (x-y)l(-y-x), so the differential equation is (x + y)y' = y - x.
In Problems 32-36 we get the desired differential equation when we replace the "time rate of
change" of the dependent variable with its derivative, the word "is" with the::; sign, the phrase "proportional to" with k, and finally translate the remainder of the given sentence into symbols.
32. dP/dt = k✓P
= 33. dvldt k v2
34. dv Idt = k(250-v)
35. dN I dt = k(P-N)
36. dN Idt = kN(P-N)

Section 1.1

3

37. y(x) = 1 or y(x) = x
38. y(x) = ex 39. y(x) = x2
= 40. y(x) = 1 or y(x) -1
42. y(x) = cos x or y(x) = sin x 43. (a) y(l0) = 10 yields 10 = I /(C-10), so C = 101/10.
(b) There is no such value of C, but the constant function y(x) == 0 satisfies the
conditions y' =y2 and y(O) :::: 0.
(c) It is obvious visually that one and only one solution curve passes through each point (a,b) of the .xy-plane, so it follows that there exists a unique solution to the initial
value problem y' =y2, y(a) =b.
44. (b) Obviously the functions u(x) = - x 4 and v(x) = + x4 both satisfy the differential
equation xy' = 4y. Buttheirderivatives u'(x)=-4x3 and v'(x)=+4x3 match at
x = 0, where both are zero. Hence the given piecewise-defined function y(x) is differentiable, and therefore satisfies the differential equation because u(x) and v(x) do so (for x $ 0 and x ~ 0, respectively). (c) If a~ 0 (for instance), chose C0 so that C0 a4 = b. Then the function
j[ X $ 0, if X 20
satisfies the given differential equation for every value of C.
SECTION 1.2 INTEGRALS AS GENERAL AND PARTICULAR SOLUTIONS

4

Chapter 1

This section introduces general solutions and particular solutions in the very simplest situation
= - a differential equation of the form y' f (x)- where only direct integration and evaluation
of the constant of integration are involved. Students should review carefully the elementary
concepts of velocity and acceleration, as well as the fps and mks unit systems.

1.

= f Integration of y' 2x + l yields y(x) = (2x + I) dx = x 2 + x+ C. Then substitution

of x=O, y=3 gives 3 = O+O+C = C, so y(x) = x 2 +x + 3.

2.

= f = Integration of y'=(x-2)2 yields y(x) (x-2)2dx t(x-2)3+C. Then

= = + substitution of x 2, y 1 gives 1 = 0 + C = C, so y(x) = (x - 2)3.

3.

J Integration of y' = ✓x yields y(x) = ✓x dx = fxm + C. Then substitution of

x=4, y=O gives O= 1~+C, so y(x) = f(x3I2 -8).

4.

f Integration of y' = x-2 yields y(x) = x-2 dx = -11 x + C. Then substitution of

= x=I, y=5 gives 5=-l+C, so y(x) -1/x+6.

5.

= = f Integration of y' (x+ 2r1I2 yields y(x)

(x+ 2)-112 dx = 2✓x+2 +c. Then

substitution of x = 2, y:::: -1 gives -1 = 2- 2 + C, so y(x) = 2.Jx+ 2 -5.

6.

= Integration of y'=x(x 2 +9l2 yields y(x) fx(x2 +9)112 dx- = f(x2 +9)312 +C.

Thensubstitutionof x=-4, y=Ogives 0=f(5)3+C, so
½[ J. y(x) :::: (x2 +9)3'2 - 125

7.

= J Integration of y~ 10/(x2 + 1) yields y(x) == IOl(x2 + l) dx- = lOtan-1 x + C. Then

= substitution of x=O, y=O gives O=lO·O+C, so y(x) 10tan-1 x.

8.

f Integration of y' = cos 2x yields y(x) = cos 2x dx = ½sin 2x + C. Then substitution

of x=O, y=1 gives l=O+C, so y(x) = ½sin2x+l.

9.

f = Integration of y'=1t.J1-x2 yields y(x) = 11 ✓1-x2 dx- sin-1 x+C. Then

substitution of x=O, y=O gives O=O+C, so y(x) = sin-1 x.

10. Integration of y' = x e-x yields

Section 1.2

5

(when we substitute u:::; -x and apply Fonnula #46 inside the back cover to the
textbook). Then substitution of x =0, y =1 gives 1=-1 +C, so y(x) = -(x+l)e-x +2.
= 11. If a(t) = 50 then v(t) = j50dt 50t+v0 = 50t+10. Hence

x(t) = Jcsot+JO)dt= 25t2 +10t+Xo = 25t1 +10t+10.
J = 12. If a(t) = -20 then v(t)::::: (-20)dt -20t+v0 = -20t-15. Hence

= x(t) j(-20t-15)dt= -10t2 -15t+x0 = -10t 2 -15t+5.

13. If a(t) === 3t then v(t) = J3tdt = ft 2 +v0 = ½t2 +5. Hence

Jc = x(t) --

2
2

,

2

+

5)dt

=

.21,J + 5 t + X 0

l2.t' + St•

14. If a(t) = 2t+l then v(t) = j(2t+l)dt = t2 +t+v0 = t2 +t-7. Hence

15. If a(t) = 2t+1 then v(t) = Jc21+l)dt = t2 +t+Vo = t2 +t-7. Hence

f 16. If a(t) = 1/ ✓t+4 then v(t) = 11 ✓t+4 dt = 2✓t+4+C = 2✓t+4-5 (taking
C =-5 so that v(O) =-1 ). Hence

(taking C = -29 I 3 so that x(O) =1).

J 17.

If a(t) = (t+lf' then v(t) =

(t+l)-3 dt

=

-½(t+1r2 +C

=

2
-½(t+Ir

+½

(taking

C =½ so that v(O) =0). Hence

6

Chapter 1

(taking C =-½ so that x(0) =0).
= 18. If a(t) 50sin5t then v(t) = Jsosin5t dt = -10cos5t+C = -10cos5t (taking C = 0 so that v(0) =-10). Hence
J x(t) = (-l0cos5t)dt::::: -2sin5t+C = -2sin5t+10
(taking C = -10 so that x(O) = 8 ).
19. v = -9.St + 49, so the ball reaches its maximum height (v = 0) after t = 5 seconds. Its
maximum height then is y(5) = --4.9(5)2 + 49(5) = 122.5 meters.
20. v = -32t and y = -16? + 400, so the ball hits the ground (y = 0) when
r = 5 sec, and then v = -32(5) =-160 ft/sec.
21. a = -10 rn/s2 and vo = 100 km/h = 27.78 mis, so v = -IOt + 27.78, and hence x(t) = -5t2 + 27.78t. The car stops when v = 0, t "" 2.78, and thus the distance
traveled before stopping is x(2.78) = 38.59 meters.
22. v = -9.St + 100 and y = --4.9t2 + lO0t + 20.
(a) v = 0 when t = 100/9.8 so the projectile's maximum height is
y(I00/9.8) = --4.9(100/9.8)2 + 100(100/9.8) + 20 = 530 meters. (b) It passes the top of the building when y(t) = --4.9? + lOOt + 20 = 20,
and hence after t = 100/4.9 = 20.41 seconds.
(c) The roots of the quadratic equation y(t) = --4.9t2 + 100t + 20 = 0 are t = -0.20, 20.6 l. Hence the projectile is in the air 20.61 seconds.
23. a = -9.8 rn/s2 so v = -9.8 t- 10 and

The ball hits the ground when y = 0 and

so t = 5.10 s. Hence

V = -9.8 t- JO:; -60,

Yo = 4.9(5.10)2 + 10(5.10) = 178.57 m.

24. v = -32t-40 and y = -16r2-40t+555. Theballhitstheground (y = 0)
when t = 4.77 sec, with velocity v = v(4.77) = -192.64 ft/sec, an impact
speed of about 131 mph.

Section 1.2

7

25. Integration of dvldt;;; 0.12 i3 + 0.6 t, v(O) ;;: 0 gives v(I) = 0.3 r2 + 0.04 t3. Hence v(10) = 70. Then integration of dxldt;;; 0.3 r2 + 0.04 t3, x(O) = 0 gives x(t) = 0.1 t3 + 0.04 t4, so x ( IO) = 200. Thus after 10 seconds the car has gone 200 ft and
is traveling at 70 ft/sec.

26. Taking .xo = 0 and vo = 60 mph = 88 ft/sec, we get

V ;: -at+ 88,

and v = 0 yields t = 88/a. Substituting this value of t and x = 176 in

x = -at2l2 + 88t,

= we solve for

a

=

22

ft/sec

2 .

Hence the car skids for

t

88/22 = 4 sec.

27. If a = - 20 m/sec2 and x0 = 0 then the car's velocity and position at time t are given by V = -20t+ Vo, X ;;; - 10 t2 + Vol.

= It stops when v 0 (so v0 = 20t), and hence when

X = 75 = -10 r2 + (20t)t ;;: 10 t2.

Thus t = ..fi5 sec so

v0 = 20..fi':s == 54.77 m/sec = 197 km/hr.
= 28. Starting with .xo = 0 and v0 50 km/h = 5x I04 rn/h, we find by the method of Problem 24 that the car's deceleration is a = (25/3)xl07 rn/h2. Then, starting with x0 = O and v0 = 100 km/h = l 05 m/h, we substitute t = vola into x = - at2 + vot
and find that x = 60 m when v = 0. Thus doubling the initial velocity quadruples the
distance the car skids.
= 29. If v0 :::; 0 and Yo 20 then
v = - at and y = -½at2+20.

Substitution of

t =

2,

y

= 0

yields a

=

IO

ft/sec

2 •

If

Vo

=

0

and

y0 == 200 then

= v = -10t and y -5t2 + 200.

8

Chapter 1

/40 Hence y = 0 when t =

= 2 ../w sec and v = -20Jto = - 63.25 ft/sec.

30. On Earth: v = -32t + v0, so t = vo/32 at maximum height (when v = 0).
= Substituting this value of t and y 144 in
y ;:; - I6t2 + vot,
v = we solve for 0 96 ft/sec as the initial speed with which the person can throw a baJI
straight upward.

On Planet Gzyx: From Problem 27, the surface gravitational acceleration on planet
= Gzyx is a 10 ft/sec2, so
= v -lOt + 96 and y = -5t2 + 96t. = Therefore v 0 yields t = 9.6 sec, and thence = Ymax y(9.6) = 460.8 ft is the
height a ball will reach if its initial velocity is 96 ft/sec.
= = 31. If v0 0 and Yo h then the stone's velocity and height are given by
= = v - gt, y --0.5 gt2 + h.

= Hence y = 0 when t ,J2h/ g so

32. The method of solution is precisely the same as that in Problem 30. We find first that, on
v = Earth, the woman must jump straight upward with initial velocity 0 12 ft/sec to
reach a maximum height of 2.25 ft. Then we find that, on the Moon, this initial velocity
yie lds a maximum height of about 13.58 ft.

33. We use units of miles and hours. If x0 = v0 = 0 then the car's velocity and position

after t hours are given by

= V

at,

X=

I
2

t2.

= Since v = 60 when t = 5/6, the velocity equation yields a 72 mi/hr2. Hence the
distance traveled by 12:50 pm is

x = (0.5)(72)(516)2 = 25 miles.

34. Again we have

V :: at, X = ½t2.

Section 1.2

9

= = But now v = 60 when x 35. Substitution of a 60/t (from the velocity equation)
into the position equation yields 35 = (0.5)(60/t)(t2) = 30,,
whence t = 7/6 hr, that is, l: JO p.m.
35. Integration of y' = (9/vs)( 1 - 4x2) yields

and the initial condition y(-1/2) = 0 gives C = 3/v5• Hence the swimmer's trajectory is
Substitution of y( 1/2) = 1 now gives Vs = 6 mph. 36. Integration of y' = 3(1 - 16x4) yields
= y 3x- (48/5)x5 + C, = = and the initial condition y(-1/2) 0 gives C 6/5. Hence the swimmer's trajectory
is y(x) = ( l/5)(15x - 48x5 + 6),
so his downstream drift is y(l/2) = 2.4 miles.

SECTION 1.3

SLOPE FIELDS AND SOLUTION CURVES

As pointed out in the textbook, the instructor may choose to delay covering Section 1.3 until later in Chapter l . However, before proceeding to Chapter 2, it is important that students come to grips at some point with the question of the existence of a unique solution of a differential equation - and realize that it makes no sense to look for the solution without knowing in advance that it exists. The instructor may prefer to combine existence and uniqueness by simplifying the statement of the existence-uniqueness theorem as follows:

Suppose that the function f (x, y) and the partial derivative ~f I ay are both

continuous in some neighborhood of the point (a, b). Then the initial value

problem

-dy = j(x,y), dx

y(a) = b

10

Chapter 1

has a unique solution in some neighborhood of the point a.

Slope fields and geometrical solution curves are introduced in this section as a concrete aid in visualizing solutions and existence-uniqueness questions. Solution curves corresponding to the slope fields in Problems 1-10 are shown in the answers section of the textbook and will not be duplicated here.

11. Each isocline x - I = C is a vertical straight line.

12. Each isocline x + y = C is a straight line with slope m = -1.

.Jc 13. Each isocline y2 = C ~ 0, that is, y =

or y = -Jc, is a horizontal straight

line.

14. Each isocline efy = C, that is, y = C', is a horizontal straight line.

15. Each isocline ylx = C, or y = Cx, is a straight line through the origin.
16. Each isocline x2- - y2 = C is a hyperbola that opens along the x-axis if C > 0, along the
y-axis if C < 0.
17. Each isocline xy = C is a rectangular hyperbola that opens along the line y = x if
= C > 0, along y -x if C < 0.
18. Each isocline x - y2 = C, or y2 = x - C, is a translated parabola that opens along the
x-axis.
19. Each isocline y - x2 = C, or x2 = y- C, is a translated parabola that opens along the
y-axis.
20. Each isocline is an exponential graph of the fonn y = Cex.
21. Because both /(x, y) = 2x2/ and iJJ I ay = 4x2y are continuous everywhere, the
existence-uniqueness theorem of Section 1.3 in the textbook guarantees the existence of a
unique solution in some neighborhood of x = 1.
= 22. Both / (x, y) x In y and iJ/ I c)y = xly are continuous in a neighborhood of
(1, 1), so the theorem guarantees the existence of a unique solution in some
neighborhood of x = I.
23. Both f (x, y) = y11-' and aJ I oy = (I/3)y-213 are continuous near (0, 1), so the
= theorem guarantees the existence of a unique solution in some neighborhood of x 0.

Section 1.3

11

24. f(x,y) = y113 iscontinuousinaneighborhoodof (0,0), but a11ay = (1/3)y-213 is
not, so the theorem guarantees existence but not uniqueness in some neighborhood of X = 0.
= 25. f (x, y) (x- y)112 is not continuous at (2, 2) because it is not even defined if y > x.
Hence the theorem guarantees neither existence nor uniqueness in any neighborhood of
the point x = 2.
= 26. J (x, y) = (x - y) 112 and aJ I dy -(1/2)(x - yf112 are continuous in a neighborhood
of (2, 1), so the theorem guarantees both existence and uniqueness of a solution in some neighborhood of x = 2.
27. Both f (x, y) = (x - 1/y and aJ I dy = -(x- 1)// are continuous near (0, 1), .so the
theorem guarantees both existence and uniqueness of a solution in some neighborhood of
X = 0.
= 28. Neither J (x, y) (x - I)/y nor aJ I Jy = -(x - 1)ly2 is continuous near (1, 0), so the
existence-uniqueness theorem guarantees nothing.
= 29. Both f (x, y) ln(l + /) and dj (i)y = 2y/(1 + y2) are continuous near (0, 0), so
the theorem guarantees the existence of a unique solution near x = 0.
30. Both f (x, y) = x2 - y2 and dj I dy = -2y are continuous near (0, 1), so the theorem
guarantees both existence and uniqueness of a solution in some neighborhood of x = 0.
i r = 31. If f(x,y) = -(J - y2)1'2 then aJ ldy = y(l - 112 is not continuous when y 1,
so the theorem does not guarantee uniqueness.
32. The two solutions are Y1(x) = 0 (constant) and y2(x) = x3.
= 35. The isoclines of y' == ylx are the straight lines y Cx through the origin, and
y' = C at points of y ;;: Cx, so it appears that these same straight lines are the solution curves of xy' = y. Then we observe that there is
(i) a unique one of these lines through any point not on the y-axis; (ii) no such line through any point on the y-axis other than the origin; and (iii) infinitely many such lines through the origin.
= 36. J(x, y) 4xy1n and dj I dy = 2.xy-112 are continuous if y > 0, so for all a and all
= = b > 0 there exists a unique solution near x = a such that y(a) b. If b 0 then the = theorem guarantees neither existence nor uniqueness. For any a, both y1(x) 0 and
y 2(x) = (x2 - c/)2 are solutions with y(a) = 0. Thus we have existence but not uniqueness near points on the x-axis.

12

Chapter1

SECTION 1.4

SEPARABLE EQUATIONS AND APPLICATIONS

Of course it should be emphasized to students that the possibility of separating the variables is the first one you look for. The general concept of natural growth and decay is important for all differential equations students, but the particular applications in this section are optional. Torricelli's law in the form of Equation (24) in the text leads to some nice concrete examples and problems.

1.

J; -f =

= 2xdx; in y = -x2 +c; y(x) ' e-x-'+c = C e-.r

2.

J:; J = = -

2x dx;

- -1 = -x2 -C; y

y(x)

x2 +C

3.

J; f= sinxdx;

in y = -cosx+c;

= = y(x)

e-cou+c

ce-COS_f

4.

f f dy =-- 4dx.,

y

l+x

lny = 4ln(1+x)+lnC;

y(x) = C(l+x)4

5.

JR - J dy - zd✓xx·. sin-1 y = ,J;+ C; y(x) = sin(Fx+c)

6.

f t = f 3--/xdx; 2,/Y = 2x"' +2C; y(x) = (x312 +cf

7.

f:~ J = f 1

113
4x dx;

y 213 = 3x413 +½ C;

3/2
= y(x) (2x413 +C)

f; f f(i:x 9.

= 1~d:2 =

+ l~x )dx

(partial fractions)

lny = ln(l+x)-ln(l-x)+lnC; y(x) = Cl+x
1-x

Section 1.3

13

10.

f dy f (1+y)2 -

dx (l+x)2'

l

1 - C = _ l +C(l+ x)

- - - = -1+ y l+x

l +x

= l +x

y(x)

l + y = -l +-C(-l + - .x)'

l+x _ 1 = x-C(l +x)

t +C(l+x)

l + C(l+ x)

11.

f dy = x d.x- _ _ 1_ = x2 - C_ y(x) = (c - xi )-112

f y3

' 2/ 2 2'

12.

13.

3
J Y

d Y

=

cos x dx;

¼In ( y4 + 1) = sin x + C

f y4 +I

y +l.y312 = x+l.x312 +C

14.

'.I

3

15.

-

-2 +
y

-31y3

=:

ln\xl+ -1 + C
X

16.

f sin y dy = xdx ; - ln(cosx) = ½ln (l+x2) + Jn C

f cosy

1+x2

17. y' = J+x+ y + xy = (l+ x)(l+ y)
I = dy = Jo+ x)dx; mll+ yl x +tx2 +C l+ y

f f ( - dy- = J+ y2

-1 - 1)dx; tan - 1 y = --1-x+C;

x2

X

f = = = 19.

d: Je" dx; In y e" + Jn C; y(x) Cexp(e-' )

y(O) = 2e implies C = 2 so y(x) = 2exp(e•')

14

Chapter 1

20.

f__EJ_ = f3x2 dx; tan-1 y = x 3 + C; y(x) = tan (x~+ C) l+y2

y(O) = 1 implies C = tan-' l = ,r/ 4 so = y(x) tan ( x3 + ,r / 4)

21.

J f= 2y dy

✓xx2d-x16 ;

y2

=

✓x1 -

I6 + C

y(5)=2 implies C= l so y2 = 1 + ✓x2 -16

22.

f; = f(4x3 -l)dx; lny = .x4 -x+JnC; y(x) = Cexp(x4 -x)

y(l) =- 3 implies C =-3 so y(x) = -3exp(x4 - x )

23.

f f _!!]_ = dx; ½ln (2y- l) = x+½InC; 2y- l = Ce2x

2y-1

t{ y(l) = l implies C = e-2 so y(x) = 1+ ei:<-z)

24.

J dy __ cos. xdx ·,

J y

s1nx

ln y = ln(sin x) + In C;

y(x) = Csinx

y(f) = 1} implies C = ½ so y(x) = tsinx

25.

fd; = f (~ +2x} ln y = ln x+x2 + lnC; y(x) = C xexp(x2)

y(l) =1 implies C = e-1 so y (x) = xexp(x2 -1)

26.

- -t = x2 + x3 + C;

y

y(l) =-1

implies

C =-1

so

1

y(x)

=

-l

-

x

2

-

x-3

27.

e Y = 3e2x +C·'

y (O) = 0 implies C = -2 so y(x) = In (3eix -2)

f Jj;; 28.

sec2 ydy =

tany = ✓x+ C; y(x) =tan-1 (✓x +c)

2

Section 1.4

t y(4) = implies C = -1 so y(x) = tan-'(-✓x -1)

29. The population growth rate is k = ln(30000/25000)/10 = 0.01823, so the population

of the city

t

years after 1960 is given by

P(t)

=

2

5

0

0

0

e001823 '

.

The expected year 2000

= population is then P(40) 25000eo.o,si3x4-0 >= 51840.

= 30. Thepopulationgrowthrateis k ln(6)/10"" 0.17918, sothepopulationafter t

= hours is given by P(t)

Po e017918 '.

To find how Jong it takes for the population to

double, we therefore need only solve the equation

2 P

= Po e0 179 81 •'

for

= t (In 2)/0.17918 = 3.87 hours.

31. As in the textbook discussion of radioactive decay, the number of 14C atoms after t

years is given by N (t) == N0 e-<>·0001216 '. Hence we need only solve the equation

= ¾N0

N 0001216

0 e--0

'

for

t

=

(ln6)/0.0001216

=

14735 years tofindtheageofthe

skull.

32. As in Problem 31, the number of 14C atoms after t years is given by

= N(t)

5.0x1010

0001216 e-0·

'

.

Hence we need only solve the equation

= = 4.6xl010 5.0xl0'0 e-<>·000121<>1 for the age t (ln(5.0/4.6))/0.0001216"" 686 years

of the relic. Thus it appears not to be a genuine relic of the time of Christ 2000 years ago.

= 33. The amount in the account after t years is given by A(t) 5000 e0081 . Hence the
= amount in the account after 18 years is given by A(20) 5000e0•0sx20 "' 2 1,103.48
dollars.

34. When the book has been overdue for t years, the fine owed is given in dollars by A(t) == 0.30e005' . Hence the amount owed after 100 years is given by
= A(l 00) 0.30 eo.oSxtoo ,,. 44.52 dollars.

35. To find the decay rate of this drug in the dog's blood stream, we solve the equation
½:::: e-sk (half-life 5 hours) for k= On 2)/5"'0.13863. Thus the amount in the dog's
= bloodstream after t hours is given by A(t) Ao e--0.n1163'. We therefore solve the

equation

A(l) =

Ao 3 e---0. nSG

= 50x45 = 2250 for

Ao = 2585 mg, the amount to

anesthetize the dog properly.

36. To find the decay rate of radioactive cobalt, we solve the equation ½= e-s.m (half-life
5.27 years) for k == (]n 2)/5.27 = 0.13153. Thus the amount of radioactive cobalt left
after t years is given by Ao A(t) == e--0·131531. We therefore solve the equation

16

Chap~r1

A(t) =

Aoe-0

13153 '

= O.OlAo for

t= (In 100)/0.13153 ""35.01 and find that it will be

about 35 years until the region is again inhabitable.

37. Taking t = 0 when the body was formed and t = T now, the amount Q(t) of 238U in
= = the body at time t (in years) is given by Q(t) Q0e-k1, where k (In 2)/(4.5Jx109).
The given information tells us that

Q(T)
:::: 0.9. (J.o-Q(T)

After substituting Q(T) = Q0e-kT, we solve readily for ekT = 19/9, so T = ( l/k)ln(l 9/9) = 4.86x109• Thus the body was formed approximately 4.86 billion years ago.
= = 38. Taking t 0 when the rock contained only potassium and t T now, the amount
= Q(t) of potassium in the rock at time t (in years) is given by Q(t) Q0e-k1, where
k = (In 2)/(l.28xl09). The given information tells us that the amount A(t) of argon at time t is
A(t) == ¼[(1- Q(t)]

and also that A(]) = Q(T). Thus

{!o - Q(T) ::c 9 Q(T).

After substituting Q(T) = {1e-kT we readily solve for

T = (In IO/ In 2)(1.28 x 109 ) = 4.25 x 109 .

Thus the age of the rock is about 1.25 billion years.
= = 39. Because A = 0 the differential equation reduces to T' kT, so T(t) 25e-k1. The = fact that T(20) =:: 15 yields k (l/20)1n(5/3), and finally we solve
= 5 25e-1c1 for t = (In 5)/k °" 63 min.
40. The amount of sugar remaining undissolved after t minutes is given by A(t) = .4ae-k1 ;
we find the value of k by solving the equation A(l) = Aoe-k = 0.15Ao for k =-ln 0.75 = 0.28768. To find how long it takes for half the sugar to dissolve, we solve
the equation A(t) = Aoe-kr:::: ½Ao for t = (In 2) /0.28768,,, 2.41 minutes.

41. (a) The light intensity at a depth of x meters is given by l(x) = / 0e-1.4~. We solve
= the equation / (x) = /0 e-14" ½/ 0 for x = (In 2) / 1.4 = 0.495 meters.

Section 1.4

17

(b) At depth 10 meters the intensity is /(10) = /0e-1•4"10 = (8.32xl0-7 )/0 .

(c) We solve the equation = /(x) 10e-i.4x. =0.01/0 for x =(In 100)/1.4 = 3.29 meters.

42.

(a)

= The pressure at an altitude of x miles is given by p(x)

29.92e--0

2 ..r.

Hence the

pressure at altitude 10000 ft is p(I0000/5280) ""20.49 inches, and the pressure at

altitude 30000 ft is p(30000/5280) ""9.60 inches.

(h) To find the altitude where p = 15 in., we solve the equation 29.92e--0 2, =15 for
x = (In 29.92/15)/ 0.2 =3.452 miles= 18,200 ft.

43. (a) A' = rA + Q

(b) The solution of the differential equation with A(O) = 0 is given by

rA+Q = Qe".

When we substitute A = 40 (thousand), r = 0.11, and t = 18, we find that
Q = 0.70482, that is, $704.82 per year.

44.

Let

N 8

(t)

and

N5 (t) be the numbers of 238U and 235U atoms, respectively, at time t (in

billions of years after the creation of the universe). Then = N8 (t) N0e-k' and

= N 5(t) N0e-<1, where N0 is the initial number of atoms of each isotope. Also,

k = (In 2)/ 4.51 and c = (In 2)/0.71 from the given half-lives. We divide the equations

for

N 8

and

N5 and find that when

t

has the value corresponding to "now",

= -N 8 := 137.7.
N5

Finally we solve this last equation for t = (In 137.7)/(c-k) ~ 5.99. Thus we get an estimate of about 6 billion years for the age of the universe.
45. The cake's temperature will be 100° after 66 min 40 sec; this problem is just like Example 6 in the text.
46. (b) By separating the variables we solve the differential equation for
= c-r P(t) (c- r P0 ) err_
With P(t) = 0 this yields

18

Chapter 1

= c rPoer 1 /(ert- l ). = With Po = 10,800, t = 60, and r 0.010 we get $239.37 for the monthly payment at 12% annual interest. With r = 0.015 we get $272.99 for the monthly payment at
18% annual interest.
47. If N(t) denotes the number of people (in thousands) who have heard the rumor after t days, then the initial value problem is
= N/ k(lOO-N), N(O) = 0
and we are given that N(7) = 10. When we separate variables ( dN 1(100- N) = k dt ) and integrate, we get ln(I00-N):;:: - kt +C, and the initial condition N(0):::: 0 gives
C =ln lO0. Then 100-N = l00e-.u, so N(t)=l00(1-e-kl). Wesubstitutet=7,
N= 10 and solve for the value k =ln(l00/90)/7 =0.01505. Finally, 50 thousand people have heard the rumor after t =(In 2) / k = 46.05 days.
48. With A (y) constant, Equation (19) in the text takes the form

We readily solve this equation for 2 Jy = kt+ C. The condition y(O) = 9 yields
= = C 6, and then y( I) 4 yields k = 2. Thus the depth at time t (in hours) is
y(t) == (3 - t/, and hence it takes 3 hours for the tank to empty.
49. With A = n-(3)2 and a == n-(1 I 12)2 , and taking g == 32 ft/sec2, Equation (20) reduces
= = to 162 y' = - Jy. The solution such that y 9 when t 0 is given by
324 jy = -t + 972. Hence y = 0 when t = 972 sec = 16 min 12 sec.
50. The radius of the cross-section of the cone at height y is proportional to y, so A(y) is proportional to /. Therefore Equation (20) takes the form
y2y' = -k.Jy,
and a general solution is given by
2/ '2 = -Skt + C.
= The initial condition y(O) = 16 yields C = 2048, and then y(]) 9 implies that = 5k 1562. Hence y = 0 when
= = t C/5k 2048/1562 = 1.3 I hr.

Section 1.4

19

= 51. The solution of y' -k.Jy is given by
2Jy = -kt+C.
The initial condition y(O) =h (the height of the cylinder) yields C =2 ✓h . Then substitution of t = T, y =0 gives k =(2 ✓h )IT. It follows that
= y h(l - t!T/.
If r denotes the radius of the cylinder, then

52. Since x = y3'4, the cross-sectional area is A(y) = nx2 = re y312• Hence the

general equation A(y) y' = - a,J2gy reduces to the differential equation yy' = -k

with general solution

(l/2)l = -kt+ C.

The initial condition y(O) = 12 gives C = 72, and then y(l) = 6 yields k = 54.
Upon separating variables and integrating, we find that the the depth at time t is

y(t) = .J144-108ty(t).
= Hence the tank is empty after t 144/108 hr, that is, at 1:20 p.rn.
53. (a) Since x2 = by, the cross-sectional area is = A(y) 'f!x2 = nby. Hence the
equation A( y) y' = - a.J2gy reduces to the differential equation

with the general solution

(2l3)y312 = -kt+ C.

The initial condition y(O) = 4 gives C = 16/3, and then y(l) = 1 yields k = 14/3. It follows that the depth at time t is
y(t) = (8-7t)213.

(b) The tank is empty after t = 817 hr, that is, at 1:08:34 p.m.

20

Chapter 1

(c) We see above that k = (altrb) jii = 14/3. Substitution of = a 1rr2• b = I,
g = (32)(3600)2ftlhr2 yields r = (l/6D)-J7112 ft""0.15in fortheradiusofthc
bottom-hole.

54. With g = 32 ft/sec2 and a = tr(l / 12)2, Equation (24) simplifies to

A(yd)y- = --tvr Yr:: ·

dt

18

If z denotes the distance from the center of the cylinder down to the fluid surface, then

y = 3 - z and A(v)

=

10(9 -

2 112 2) .

Hence the equation above becomes

= 10(9-z2 )1'2 dz ..::._(3-z}1'2 ,
dt 18
180(3+ z)1'2 dz = trdt,

and integration yields

= 120(3+z)112 m+ C.

Now z ;;: 0 when t = 0, so C = 120(3)312. The tank is empty when z = 3 (that is, when y = 0) and thus after

t ; (120ltr)(6312 - 3312) "" 362.90 sec.

It therefore takes about 6 min 3 sec for the fluid to drain completely.

55. A(y) = n(8y-y2) as in Example 7 in the text, but now a = ,r / 144 in Equation (24),
so the initial value problem is

-Ji, 18(8y- y2)y' =

y(O) = 8.

We seek the value of t when y = 0. The answer is t = 869 sec = 14 min 29 sec.
56. The cross-sectional area function for the tank is A = tr(l- y2) and the area of the bottom-hole is a = I0-41r, so Eq. (24) in the text gives the initial value problem

tr(l-y2)dy = -10--4rr:.J2x9.8y, = y(O) 1.
dt Simplification gives

so integration yields

Section 1.4

21

The initial condition y(O) = 1 implies that C = 2 - 2/5 = 8/5, so y =0 after
t = (8/5)/(1.4xJ0---4,Ji'o) ""'3614 seconds. Thusthetankisemptyatabout 14 seconds after 2 pm.

57. (a) As in Example 8, the initial value problem is

dt = 2 dv

r:;

rr(8y-y )

-1rkvY,

= y(O) 4

where k = 0.6 r 2..fig = 4.8 r 2 . Integrating and applying the initial condition just in the
Example 8 solution in the text, we find that

When we substitute y = 2 (ft) and t = 1800 (sec, that is, 30 min), we find that
k "" 0.009469. Finally, y = 0 when
t = 448 "" 3154 sec == 53 min 34 sec. 15k
Thus the tank is empty at 1:53:34 pm. (b) The radius of the bottom-hole is
r = ,Jk / 4.8 '°" 0.04442 ft "" 0.53 rn, thus about a half inch. 58. The given rate of fall of the water level is dy/dt = -4 in/hr = -(1/10800) ft/sec. With
A = m 2 and a = m-2, Equation (24) is
Hence the curve is of the form y = kx4, and in order that it pass through (1, 4) we must have k = 4. Comparing ✓Y = 2x2 with the equation above, we see that
(8r2)(10800) = 1/2,
so the radius of the bottom hole is r = 11 (240✓3) ft ""' 1/ 35 in.
59. Let t = 0 at the time of death. Then the solution of the initial value problem

22

Chapter 1

T' ::: k(70 - T),

T(O) = 98.6

is

If t = a at 12 noon, then we know that
T(t) = 70+28.6e-w = 80,

Hence

T(a+l) = 70+28.6e-k(a+ll = 75.

= It follows that e-k = 1/2, so k In 2. Finally the first of the previous two equations
yields

a = (ln 2.86)/(ln 2) "" 1.516 hr "" 1 hr 31 min,
so the death occurred at 10:29 a.m.
60. Let t = 0 when it began to snow, and t = to at 7:00 a.m. Let x denote distance along the road, with x = 0 where the snowplow begins at 7:00 a.rn. If y :;: ct is the snow depth at time t, w is the width of the road, and v = dxldt is the plow's velocity, then
"plowing at a constant rate" means that the product wyv is constant. Hence our differential equation is of the form

k dx = !_
dt t
The solution with x = 0 when t = to is

t = toekx.

We are given that x = 2 when t = to+ 1 and x = 4 when t = to+ 3, so it follows
that

= = to+ 1 toe2k

and

to+ 3

t

oe4

k
4

Elimination of to yields the equation

e4k - 3e2k + 2 = (e2k - I)(e2k - 2) = 0,
so it follows (since k> 0) that e2k = 2. Hence t0 + 1 = 2t0, so to = 1. Thus it began
to snow at 6 a.m.

Section 1.4

23

61. We still have t = to e1o=, but now the given infonnation yields the conditions
to+ 1 = toe4k and to+ 2 = toe7k
at 8 a.m. and 9 a.m., respectively. Elimination of to gives the equation 2e4k -e7k -1 = Q,
which we solve numerically for k = 0.08276. Using this value, we finally solve one of the preceding pair of equations for t0 = 2.5483 hr = 2 hr 33 min. Thus it began to
snow at 4:27 a.m.

SECTION 1.5

LINEAR FIRST-ORDER EQUATIONS

1.

= p=exp(Jtdx)=e"; D .. (y·ex)=2e-'; y•e., =2e'+C; y(x) 2+Ce-x

= y(O)=O implies C=-2 so y(x) 2-2e__,

= y·e-2.t = 3x+ C; y(x) (3x+ C)e2.r

= y(O) = 0 implies C = 0 so y(x)

3

x

e

2 -

'

5.

= = p

=

exp(f (2/

x)dx)

=

e

21 "-

'

x2;

D.. (y· x 2 ) = 3x2;

2
y•x

3
x

+

C

y(x) = x+Clx2 ;

y(1)=5

implies C==4

so

y(x)

=

2
x+4/x

= = = v(x) x2 +CI x 5 ; y(2) 5 implies C 32 so y(x) = x 2 + 321 x5

24

Chapter 1

y(x) = 5✓x+Cl ✓x

8.

p =exp(J(ll3x)dx )=e(lnx)/l =¼; DX (y ·¼)= 4½; y .if; =3x413 +c

= y(x) 3x+cx-113

9.

p=exp(J(-l/x)dx)=e-10.,.=l/x; D,(y·l!x)=llx; y·llx=lnx+C

y(x) = xlnx+Cx; y(l)=7 implies C=7 so y(x) = xlnx+7x

= y(x) 3 .X3 + CX 312

11. p=exp{J(11x-3}dx)=e1u-3.x=xe-3'; Dx(y·xe-h)=O; y·xe-:ix=C

= = y(x)

C

x

-1e

3 '

;

y(l) = 0 implies C 0 so y(x) = 0 (constant)

(J (y · = = = = = 14. p exp (-3/ x)dx) e-3lnx x-3; D,, x-3 ) x-1; y·x-3 In x+C
= = y(x) x 3 1nx+Cx'; y(l)=lO implies C==lO so y(x) x3 Inx+10x3

16.

= p = exp(fcos xdx) = e'ini; DX (y · esinx) e•inx cos x; y. e•inx = esinx + C

y(x) = I+Ce-sinx; y(n)=2 implies C=l so y(x) = I+e-,;•.,

(J 11.

p = exp

1/(1

+

x)

dx)

=

e

1"<

1 +xJ

=

l

+

x;

Dx ( y- (1 +x)) = cos x;

y-(I +x) = sin x + C

Section 1.5

25

= C+sinx
y(x)
l+x

y(O) = 1 implies C = l so y(x) = l +sin x l+x

18.

= = = p exp (f (-2 Ix) dx) e-21"' :::; x-2 ; Dx( y •x-1 ) ~ cos x; y •x-2 sin x +C

y(x) = x 2 (sin x+C)

19.

(J = p = exp cot xdx) = eln(•inx) = sin x; D, (y ·sin x) sin xcos x

y-sinx=½sin2 x+C; y(x) = ½sinx+Ccscx

y(O) = 0 implies C = I so y(x) = -1 +e-x-•112

y(x) = x 3 sinx+Cx3 ; y(2.n-)=0 implies C=O so y(x) = .x3 sinx

22. p =exp(j(-2x)dx)=e-x'; D... (y·e-x")=3x2; y-e-..-' =x3 +C

y(x) = (x3 +C)e-x\

y(0)=5

implies C=5

so

y(x)

=

(x3

1
+5)e-x

y·(x2 +4)312 =½(x2 +4)312 +C; y(x) = ½+C(x2 +4r312
y(O) = 1 implies C:::::: 1: so y(x) = ½+ \6 (x2 +4r3' 2
25. First we calculate
23 [x 2 -ln(x2 +l)].

26

Chapter 1

= It follows that p (x2 + 1r312 exp(3x2 /2) and thence that

D.,(Y•(x2 +Jr312 exp(3x2 /2)) = 6x(x2 +4r5' 2 ,
= y·(x2 +It3' 2 exp(3x2 /2) -2(x2 +4r312 +C,
= y(x) -2exp(3x2 / 2) +C (x2 + 1)312 exp(-3x2 / 2).
= Finaily, y(O) 1 implies that C = 3 so the desired particular solution is
= y(x) -2exp(3x2 /2)+3(x2 +1)312 exp(-3x2 /2).

= = 26. With x' dx I dy, the differential equation is /x' + 4 y 2x 1. Then with y as the
independent variable we calculate

= = = p(y)

exp(J (4/ y)dy)

=

e41 ny

y4; DY (x• y 4)

y

= = x-y4 -1y2 +C; x(y)

-l+ C-

2

Zy2 y4

= 27. With x' dx I dy, the differential equation is x' -x = y eY. Then with y as the
independent variable we calculate

28. With x'=dxldy, thedifferentialequationis (]+y2)x'-2yx=l. Then with y as the
independent variable we calculate

= = p(y) = exp(f (-2yl(l+y2)dy) e-ln(y2+1) (1+y2rl

l r l r = D_,, (x-(I +

1 )

(1 +

2

An integral table (or trigonometric substitution) now yields

I f 2 I+Xy2 = (I+dyy2

= }(

J
1+ Yi

+

-1
tan

y

+ C

]

x(y) = ½[Y+(t+ /)(tan-1 y+c)]

Section 1.5

27

29.

30. After divjsion of the given equation by 2x, multip1ication by the integrating factor
P = [ 112 yields
x - 112y'-½x-312y = x-112 cosx,
Dx(x- 112y) = x-112 cosx,
f x -112y = C+ 1- 112 cost dt.

The initial condition y(l) = 0 implies that C = 0, so the desired particular solution is

f ,- = y(x) x 112

112 cost dt.

31. (a)

32. (a) If y =A cos x + B sin x then
= y'+ y (A + B)cos x+ (B-A)sinx = 2sinx
provided that A ;; - 1 and B = 1. These coefficient values give the particular solution Yp(X) = sin X - cos X.
(b) The general solution of the equation y' + y = 0 is y(x) = ce-x so addition to the particular solution found in part (a) gives y(x) = Ce-x + sin x- cos x.
= (c) The initial condition y(O) 1 implies that C = 2, so the desired particular
solution is y(x) = 2e-x + sin x - cos x.
33. The amount x(t) of salt (in kg) after t seconds satisfies the differential equation
= x' = - x/200, so x(t) IOOe-11200• Henceweneedonlysolvetheequation
10 = IO◊e-''200 fort :::461 sec= 7 min41 sec (approximately).
34. Let x(t) denote the amount of pollutants in the Jake after t days, measured in millions of cubic feet. Then x(t) satisfies the linear differential equation dx I dt = 114- x I 16 with
solution = x(t) 4 + 16 e-,n6 satisfying x(O) =20. The value of t such that x;; 8 is

28

Chapter 1

t = I6 Jn 4 ""22.2 days. For a complete solution see Example 4 in Section 7.6 of Edwards and Penney, Calculus with Analytic Geometry (5th edition, Prentice-Hall, 1998).

35. The only difference from the Example 4 solution in the textbook is that V == 1640 km3 and r == 410 km3/yr for Lake Ontario, so the time required is

t

=

V - In 4

= 4 In 4

""

5.5452

years.

r

36. (a) The volume of brine in the tank after t min is V(t) = 60 - t gal, so the initial
value problem is

The solution is

dx = 2-~,

dt

60-t

x(O) = 0.

x(t) = (60-t)- (60-t•/ 3600

(b) The maximum amount ever in the tank is 40 / ./3"" 23.09 lb. This occurs afler r = 60-20./3'"" 25/ 36 min.

37. The volume of brine in the tank after t min is V(t) = 100 + 2t gal, so the initial value

problem is

dx = 5_ 3x ,

dt

100+2t

x(O) = 50.

= The integrating factor p(t) (100 + 2t)312 leads to the solution

x(t) = (100 +2t)- 50000
(I00+2t)·312

such that x(O) = 50. The tank is full after t = 150 min, at which time x(150) = 393.75 lb.

= soe- 38. (a) dxI dt == - X 120 and x(0).:::: 50 so x(t)

1120 •

(b) The solution of the linear differential equation

with y(O) = 50 is

-dy - -5x- -S-y - -5e-,120 - -1y

dt IO0 200 2

40 •

= y(t) 1SOe-1140 - IOoe-1120.

(c) The maximum value of y occurs when

Section 1.5

29

PROBLEM 2.1
KNOWN: Steady-state, one-dimcDllional heat conduction through an axi•ynunetri< shape.
FIND, Sketch temperature dis\ribution and c,rplalo shape or curv,. SCHEMATIC:

T,
E,.

·--Lx - -f"-/l- I -1;

------

-
--

-

-·

l'•'

T, >T,
t.., Tr•)

L '

T,
T/x)
r.: 0

~TL

---

)(

L

-Sl.,p.oi
curve

ASSUMPTIONS, I1) Sleady-stai.,,, one-dimensional conductloo, (2) Constant
prop<•rties, (3) No internal heal genProtion.
~ALYSIS: Performing an onorg, h•Jance on the obj<·cl accordlog to l::q 1 It•· E,. - E ,, - 0, it follnws that
= f:lh - Eg,,H Cb
= •nd that q. 'h(x). That is, Ibo beat rate within the object i• ,,v,rywhcre ron,tnnl
From Fourier'• law. Cb. :a --k.A. -dT .
dx and •inr• I\, and k nre both con•I.Aat.s, it follow• <hat
dT A, - - Constant.
dx
Th!ll i.!;1 tbP product of tbt• cro'-•H1t"Niona_l artn. normal to the heat rat<' and tf'm('.-rntlJrt!
grn.dieot t'"mnin~ 11 constant and iadepcodcnL of di!ll3nt"t x. It follow~ that since .-\ 1
incre~ with x1 then dT/dx mu.st decrease with x. Hroce, the Lt"mperature- distrlbullon
appears as shown Above. Notri the- gradic.nt Jrcrease.s with increasing ,c.
CO~IENTS: (!) De suro lo rc,ognlto Lbal dT/d,c I, tbo slop• of ihe temperature
di,trlbution. (2) \\'bat would lhe dbtribution be whoa T, > T1 7 (3) Show on th• Ahove
plot how tho boat Du,, q:, varic, wilh disULat<.

PROBLEM 2.2
KNOW!\": Hot waler pipe covered with thick la.yer or insulation FIND: Sketch temperature distributioo. and give brief explanation lo justify shape. SOREMATIC,

Hor
woter p,

r,

7;

Tz
Tir)

Insulat,on

-'-.'

.
--

,

.,

, I

r;

T,
>1i.

~
r,

r, r

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional (radial)
conduction, (3) No ln1,eroal beat generation, (4) losulntlon bas uniform prnpcrties
indrpcndent of l.cmperature IUld position

A."'sALYSIS, Fourier's law, Eq. 2.1, for this one-dimensional (cylindrkal) radial sy•lem has the form
q, - -kA, ~; - k (2s-rt) ~~

where A, - 2r.rf and ( is the a.xial length of the pipe-iMulation oy,,tem. Reco1,nlzr that ror !ILt-ady-staLe t"ODdiLioos with no internal hea.c. generation, an energy balance on the
!)'!item requires E,n = £011\ since £, = f:-' = 0 and hence
q, = Con,tnnL

Thal I•, q, is Independent of radlUl! (r). Since the thermal conductivity is iuso constant,

It f<>llows tbaL

I!~ I r

= Const1rnl.

Tbb rr.lation requir<.. that th• produ<L of the radial ~perature gradient, dT/dr, and thr rndiw., r, remains coll!ta.nl Lhrnughout the insulation. For our situation, th<· t•mp..mtun• tli>tributioa mu5t appear as shown in the :1bove, righL •ketch.

COMMENTS: <on.slant. Bo"

(I) does

['q;o•,t(cr)thvaatrywhwiliethq,

Is r1

a constant and (2) Rccogniic

independent of r, q; is not a that the radial tcmpcrnturr

grndicn1, dT,dr, decreases with iocrea.sin~ radius.

PROBLEM2.3
KNOWN: A ,pberlcal shell with pr..cribed geometry &nd surface t.emperatur...
FIND, Sketch temperature distribution and explain shape of the curve.
SCHEMATIC:

T,
Spher,cal

Shell

Tjr}

T,

T,
, 4 , -~ T, T, >T,

r,

r

ASSUMPTIONS: ( I) Steady-state condillom, (2) One-dimension<>! conductioo la radial (spherical coordiaateJ) direction, (3) No iDtera,1 geaor•tioo, (4) Coostaat properti~.

ANALYSIS: Fourier'• law, Eq. 2.1, for this ooe-dlmemional, rodi:11 (spherical <oordlnac.e) system bas the form

q, = -k ,\,

dT dr

= -k (4;;-r')

dT dr

wberP :\r Is the- surface area or a sphere gh~c.n u Ar - 4 :'fl''. For stc..dy-st.n.Le condition_,,

n.n ,~nt,qp· balanct- on the system require5 that s-lnce Ee- = En - O. E ~ = Eoul and thu!!I 11

'lu, - q,,., = q, ~ q,(r) ,

1/o'.\,,.,9-;";",

\ qr \

.,,,-:: .. 'l.rn

,...

~,~

~\
re-I

That is, q, it a consLanL, i.ndepCDde.nt. or the radia.f coordiaate.
I conductivity Is coo.suwt, il follow!I that r•· [cdiT;"' -. Consl::u1t.

Since thf! tberma.J

Tbl• relatioo requlret tbftl the produ<t of lhe radial tempcralure gradient, dT/dr, and
thfl radlu! squared. r~, remaioQ ron.~tant throughout. tbe shell. H.e.oce the lempe.ralurc 1
di1trlbutioa app4'art a.• shown In lhe above, right sketch.

COMMENTS: Noto everywbe,rf' constant, B

that ut how

fdoore!!lhq•e,

abov& vary as

coodiLiom, a ruact1•00 or

q, ,,. q,(r); radius!

that,

i•,

q,

is

PROBLEM 2.4
KNOWN: .ruruiymmetric shape with prescribed cross-sectional area, temperature dirtribution aod beat rate.
FIND: E'<))rcssloo Cor the thermal conductivity, k. SCHEMATIC:

---1I ----

Un, t.:;

--A(xJ=(l·x)

JT·K

T(x)=300(7-2x-x3) x-m

x

A-m4

ASSUMPTIONS, (1) Steady-state coodltion.s, (2) One-dlmenslooal cooductloo io x-
direcllon, (3) No iotcroal heal generation.
ANALYSIS: Applicati':'n or the energy balance relation, Eq. l.lla, to the ,ystem, it
Collows that since E,0 = E0 .,,
q,, - Constant"" C(x) .
Using Fourier's law, Eq. 2.1, with appropriate expressiollll Cor A1 and T, yields
q,, = -k Ax ddJTt

6000\V - - k • {I -x)m: • j_ "300(1 - 2x - x3 )j K .

dx

m

Solving fork and recognizing it.s uoits are W/ m • K,

k -

-GODO

= - - -20- - -

(I -x)l300(-2 - 3x:) (1 - x)(2 + 3x')

<I

COMMENTS: (1) Note that nt x=O, k=IOW/ m·T< and LbaL indeed tbc unil.5 are corrcetly obtained.
(2) RPcognizc thRt the 1-D assumption is an approximation which is moro appropriate as the area c,hange with distance x is less.

PROBLEM 2.5
KNOWN: End-face temperature., and temperature dependence of k for a trunc.lted cone.
FIND: Variation wilh axial distance along the cone or q., q•., k, and dTJdx.
SCHEMATIC:

~ ~ , c : - -7 ; , A l

X

~~'--'H:-;:-k •k0 -llT

ASSUMPTIONS: ( I) One-dimensional conducLlon in x (negligible Lcm111·raiure
(tradicnt.s along y), (2) St•ady-state condiUons, (3) Adiabatic sides, ( I) ;s;o i11t1•r1ml hut generation.

AJ'sALYSIS: For the prc,icribeJ conditions,.it foll~w• from const·l"·ation of cnc·rg), Eq.
l.l l~ lhat for n. differential control \'olum1·1 F.111 - = Eo-ut or q1 qx-+dr• Hcucc
•t. i> indcp1•ndeot of x.

Slnc<- .A(x) 1rrrrta.$t'.C with int"rta!ting x. il folluw!!I Lhat q; - q./A(x) drr.rr4H'3 whh
mcrr,u1ng x. ,:::;inrt- T dtcrca..'-tl with ,ntrtc.-,,ug x, k urrrta.u1 with 1ncr-ca.uug J.. fl('Of'f', rrom F'ourit•r'!l la~, Eq. :!.:.!.

= •
·n11

-k -dd-Tx •

h foflo.,., lhllt I dT/dx I dtcrtaus with increasing x.

PROBLEM 2.6

KNOWN: Temperature dependence of the thermal conducti,·ity, k(T), for bta.l transfer lbrougb a plnn• wall.

FIND: EITcct of k(T) on tcmporature distribution, T(x).

ASSUMPTIONS: I I) One-dimen•ional conduction, (2) Stcady-,t•t• rondition,, (3)
f\'o mlerual heat genera.Lion.

A."IALYSIS: From Fourier'• lnw and the form of k(T),

• q, -

-k

dT dx

= -

(ko + aT)

dT dx

•

( l)

The sbnp< of the t1•mpcralure distribution n1ay Le Inferred frnrn kuo"h•dK<' of
d:T/dx: e d(dT /a)/ux. Sin« q: i• lndcpendcnl of x for lh• pn•vrilwd condition•,

~ -

d,

[(k0

+ aT)

dT ]- dq:

dx

dx

•

o

I ' -

(k0

+ nT)

d'T
-

dx'

-

dT
n -
dx

-o.

lien co.

from "h,eh 11 follow• that for
a '> Cl: .J'T/dx' < 0
a• O: d'T/dx' =O
a < ll: d 2T /d• ' > O.

.__ _ _ _ x

\\Lt-re lht- run:tture for th~ tem~u·raturt• 1fo,trHt11liou T{x) is ftl"l,:lili\'t•• U:ro, and p1,,ri1tin•, r,•spt.-(tl\·cly.

CO\NE!':TS: The ,h•p•• of the distribution could nl"" ht• inf1·rr1•d from C~ (1).
~ioct' T dL-c rp:1,..-,~ with inf'rt•:s...t11iog x,

I I n. • tt k drcr,•::t.-•H-s wu.L int"rc~qing x => dT/dx iuut::isr~ with irJtrt;L\illg x

a .- CJ: k a=. k0 --> dT/d~ I:. ronstan\

-=> I < n O• k lnn,•nJIL':I v.·itL ir1l"rn1..;iog x

dT/dx J dt·crtast~ v. ith lnc-rta.siug ~.

PROBLEM 2.7

KNOWN: Thermal conductivit.y nod thickness of a one---dimt.ns-ion:tl syst.-m wlth no
int.e:rna.l bi,at i;tincratioc aad steady•state condit.ion.s.

FtND: Unknown lllr(Act' t.e.mpe.rat.ures, i.emp~raturc gradic.ni. or beu Rux,

SCHEMATIC:

f i r. ---~

L,as.

• 9~

9!, Tompor4turo 9r•d•ont

k

.

zs

w.'., f,

K-.

.

-X -.-.i.r,

ASSUMJ'TJONS, (1) On...cl!men.,lonal h•at Oow, (2) No internal beat geoontloo, (3) Stea.dy-sLale c.onditinn1, (·l) Constant propcrUea.

ANALYSIS: The raLc equation and umperaLure 1r1dlent for thlt !l}'Slem an

•

dT

C'lc ----k -dx

and

dT T,-T, -dx= L

( 1.2)

U1iog Eq>. (l) ond (2), the unknown qullntllles can be delormlned.

C•)

dT _ dx

(~00-300)K 0.5m

= 200 K/m

q; = -25 ..Y!.._ x 200~ = -5000 W/m1

m·K

m

fb) q: = -25 -m\•\K' " I-260mK-l =8250 \V/m'

~ I dT
T,-T,-L[ dx

= 1000 • C - O,Sm 1-250

T: = 226 • C.

(•) q; - -16 --W-:- > 200K- - -5000 \\'/m2

m·K

ID

' 'tOOK
cz;
JOO"[
aot

~
300K

q·• •<J

x-l

T,

dixI.- -25iomK

x-i

<J

dslx.L,+200mK

T 2 - liU r: -o.~,n[2oo~j--2o·c.

9; <J

I j I I .. (d) dT • ~• =

dx

k

4000 W/m' 25 \V/m·K

_ _ 160~m

9,:

4000ri,_ Tr = I. ddTx + T, • 0.Sm -160-;K:;. + (-~ C)

><-l
-s·c

T1 - -as· c.

! l {e)

dT -dx

-

•
'h - k- •

_"..-,_32_S0=0\-0=I'.\,.mI.'..·.mK.=7-~) -

K 120m-

T, =JO· C - O.f>m [ 120 = -30 • C:,

T,
><-i

<J

.

PROBLEM 2.8
KJ'JOWN: Oo~dlmt.M1onn.1 sysl.Ha wlt.b prescribed Lhctm::U eoaUudil·hy .and
thickoell'!I FlN02 Unknown•i for •·nr-1ou. tN11perlltUrl· eondillon• 11nd skt·tt'II di.,lrilrntlnn.
SCHEMATIC,

ASSUMPTIONS: ll) Steady•sl.nlr. c:oadillont, (2} Onl1..ftlmt-u.1ionnl conduction, (3) No int;-rnal hcL grncrn1.ion1 (-1) Com;Lft.U1. proJwrt.it:1.

ANALYSIS: Tb.- uh' NJUation and lrmfK'rnturn gra,lit•uL ror thi~ !>y1t"m arn

•

dT

r\1 - -K -dx

:a.ad

dT -dx

=

T,•-I.T-,

•

I 1,2)

fn)

JT •
th.

1-20 - bOl K 0.:?5u1

= _2~0 K/iu

IL

· ". I I •
q, -

.,u - - I(
m·K

'-1Im0(-

=

50'C
·2.0'C'

r-•

<J

q·•

·ro·c

~

<J

1.•. =rLd,•hT-. ,.T1

lt.1K)- t70"f'
Ill

,., - 110'(",

~ ;,n m·K , [-Ml mK] - 1.0 kll'/rn'

l..·-ddT •

tu' C -

0.'!:irn

K
-kO_;_

•

rn

[too

K
Ill

]

=

-10.0 k11

/m'

q; __,,, T, . i K

70'( .-,

dx •/60 m

1-X
<J

<J

9,

30"C

r,

dT,zooK

d•

m

<J

PROBLEM 2.9

KNOWN, Pla.oe wall with prescribed Uiermal conductivity, thickness, and surface tempcrat.urc::s.

FlND: He&'- flux, q;, and tempere.ture gradienL, dT/ch~ for Lhe three dlffcrent. coordlnau; ,ystems shown.

SCHEMATIC,
. T,•400K r t.-~

7;,600K T;

k.; 100 W/m•K.

L=IOOmm 0

L-x

tr

tT

"G
r;
x-L
(b) 0

~

Yi

x+-L

0

(c)

ASSUMPTIONS: (1) Ooe-dimensional beat flow, (2) Steady-stote coodltions. (3) No internal genoratloo, {◄) Coostanl properties.

ANALYSTS: The rate equation for conduction heal Lransrer ls

q; -k dT

(1)

dx '

where the t"mper3Luro gradient i• constant lbroui;bouL Lbe wall and or tho form

dT = T(I.) - T(O)

()

dx

L

2

Substhutiog numoricAI vnlues, find the l<!mperaLure gradioois.

- - (a) dT T 2 -T1 = (600-400JK - 2000 K/m

dx

L

O.JOOm

<l

(b) - dT= T,-T, = (·100-600)K = -2000 K/m

dx

L

0.JOOm

<l

(c)

-dT= T,-T,

dx

L

=

(600-IOO)K 0.IOOm

_

2000

K/m

<]

The h•al rates, usln~ Eq. (1) with k = 100 V;/m·K, are

(a) 'I", - -100\\'- x 2000 K/m -200 k\\'/m2

<]

m·K

(b) q•, • -10m0w·K~ (-2000 K/m) - +200 kW/m'

<]

.

w

(c) q, = -100 nrK 2000 K/m - -200 kW/m' .

<]

P ROBLEM 2.10
KNOWN, TemperaLure distribution in solid cylinder and convectioo cocfficieol at cylinder surface. F rND, Expressioos for bent r11le at cylinder surface and Ouid temperature, S C I I E M A TI C :
L

- -----<t>

h,

T00

I>
---!>

ASSUMPTIONS : ( I) Ooi,-dlmen.sional, radial eonductian, (2) SLcad)'·•Lale conditions, (3) CoasLant propcrlics,

ANALYSJS, The beat rlllo from f'ourier'• I•" for lhe rndi•I (eyllodrical) 5)'11Lem bas Lbe form
dT
q,--kA•-;j;-·

U•ing lhe expression for Lhe lemperaLure dislributioo, T( r) - a + hr' 1 evo.lunte lbo
L•mp1•rnlure gradirot, dT/dr, lo 6nd 1hr heal rntr,

q, - -k(2:rrl..) 2hr = -I rrkbLr'

At

the

outc_r

surface,

(r

=

r 0

)1

tbr

couduction

be:lt

rt1.lr

is

'1,. r_ = -I :rkbl.r; •

<J

Prom

a

surfnee

energy

bnlunce

bl

r

=

r 0

,

q,.,, - 'la,0 , = h(2,r0 L) jT(r0 ) - T 10 J,

a.ad suh·ing for T ~ \J11in~ Lhe hc:.L rau: ;LC. r - r0 , flrnJ

T,,

-T(r0 )+

2

kb b

r0

. . I..,

= n +

., br;;

+

-2k-br-0

h

T" - n + br0 Ir• + ~k I·

<J

PitOBLE:M 2.11

KNOWN: Twc,.dimenaion&J body with 1peciJled thermal cooduclivlty a.ad two isotherm&! surfaces or prescribed t.empera-1.ure&; ooe surracr, A, bas a prescribed lemporature g-radioal.
FIND: T•mp,ratur• gradient■, tn'/{'x and ilT/cty, at lhr turf•<• fl.

SCHEMATIC:

B, ~=IOO"C

w,._=Zm
ASSUMPTIONS: (I) Twc,.dlmensional coaduction, (2) Steady-st.ate conditions, (3) 'so beat generation, (◄) Con,,\a.nt propcrli...

ANALYSJS: AL Lbe s-urface A. lbe t.empcra.t.ure gra.die.at io the x-direcliou must be tcro. That b, (ilT/clx)A - 0. Thi» follows from the requirement that lh• h,at fiux vect.or mun be normal to An b1ar.hermal surra.ce. Tbc- bC.15it rate a.t. the rurfl\cl'! A i! ghen b~• Fourier', law wriLlen u

ur ] \,. q,." = -k·wA ~ - - 10-- x 2m < 30K- - -600\V/m

"

• 'Y

m·K

m

A

On the surface B, it follow, lhat

(,rr/•'Y)e =O

<J

in ordor to utbfy the requiremrol that th, heat flux vc,ctor be normal lo the l,othcrmal

1urfa<0 B Uting tbr consen·ation of cooriy roquircmcnt, Eq. I.I IA, on the body, find

•

•

<ly,A-q,9=-0

or

l'sot'" that,
L ~~ q~ n - -k·wn

;1' and heoct-

•

(,'f/•'x)9 =

·weA -

-(~Oil \V m !Of\\' m·K x Im

= 60 K/m,

<I

C•OMMENT: No1f' tbaL in ustn, the tonscr-vatioo ~ulr~me.nt, '11•.11 - ¾,•,A and
qout ,_ --Hl:i R·

PRODLEM 2.12

KNOWN: Leogtb and thermal conductivity or a sbarL. Tempernturc dls1ribuLlon along shart.

FINO: Temperatures and heal rat.cs at ends or sbaft.

SCHEMATIC:

- -'1(0)
... ...

Supporf1n9 .shd+,

. ; ,', .

k=ZSW/m·K,L:lm,

:

0
:2,

A:0.005m<- ~-.,,_·.x.-=-,L---,------J!'•"">-c:::::::·_1T:JOO-JSOx,-10x2.

G-ro~.,;d

'l(L)

ASSUMPTIONS: (1) 5tendy-stnle condlUons, (2) One-dimensional condutlion in x, (3) Con•L"r·t propcrtiL·s.
ANALYSIS: Temperature, at the lop Rad hot tun or Lhr. sh:ll't .,n•, r<-.J>ecti,·cly,

T(O) 100 • C

T(L) .- --JO. C.

<J

Ar>plyin& Fourirr's law, Eq. 2.1,
q, = -kA ddT - -:!5 \\'/urK(0.005 1112 )(-150 + 20x)"C/m :(
= q, U. I :!fi( 150 ~lh )\\.
lln,n·,

•t.[O)- ll!.75 \\'

•1.(l.) = 16.25 \\'.

<l

The diffcrt·nri• in h~;iL rate,, q,(O) > q,(L). is dur lo lu,at losses q, fror'l tlw air!,· ur the
•,haJI

CQt,,Cl,fENTS: llrat lnss from Llw side n,quir,-. ti,~ 1,xi,tt'l1c,, or lcmpcrn\urc gTAdicots
ov«:r tL, sh.aft rrn~~-:,t•«-Lion, H•·nc~, 5perification or T n.s oi. funcLlon of only " i~ an nppro\: imnt 1011

PROBLI.M l.l J
h.1\0\\ , : A n-,J CJf ..-:on,IJflt th«:nn.11 i:onJu,uvtt} l .anJ \Jn,1hle uoo.-..•sci;lion11I il/~J ,.\,1,):: •\ r .. wht'rr A. Jnd I ilN! c:on~anu
f l \ l): .;1 f,ph!\.'lhln lut tht Clm.Ju.d1,1n hc.•;ir rall: q 1t ,,~ u,e Lhi.. t1.pr~1.,uin ht JL·lcmu~ fh1,· trmr~rutur.._ J1 ..m'1ut1Pn, Ti>: I, .tnJ ,Lc1h:h uf 1he lt:m)1t'r-,1lurt" Jhlnhuuon. 1b, Con1.1J.tnnl! thit r,~,t'n..~ ,,1 \ ,1Jumc1n, h<.11 t.!cnetJllun rah~, q q.,t,pl- .1,1 ol!laJn W1 cxph."\\Hlh for q,,, 1 \lht:n lht• ld1 l.1..:L".,:; 1.J. 1. "'.-11 rn,ul1J1c:J
Sl' lit:\l ,\TI!':
T(tJ

0
\S.''il , IJJ I I(>,s: 1I 1 ( 1t1l•.Jur1c:n1r,1on.1I ~1u1,lu1,,:11c-r, 1n thC' roJ. 1: , C,1n,t11nt pu:,~n1~~
111Ju111n

l '
11 Ste,1J~ -.1.11::

l.,.., .. F,'-""'11 q y,., •<t ~(,) J,=U
lli~ r ,nJui..•11111 hc-~11 r.11,.- t(nw, cu.n he c:-,p1e""C'J :J-' .1. Tit~h11 ~ne, .mJ ,11lh111u1m~ c\rti:\~1-.,n, IPI 'I
.111d ,\1,
-~J1l1.t•q_t,pl ..,, ., C\f"IJ.\"l-11
dT
l Al'q J,

J ,J\

l•1,

I

n

1n1J1,.,111111o 1h,J1 lhc hf.II l-lk' I~.; \t1,1.u,1 v.11h, El)· ~l1mtlm1n~ b.J'- 1I. ,mJ ,21

< ',, <

PROBLE~t 2. 13 (Cont.I

That 1;, 1hc produi..1 or the cro~-.~~ccttun.,I dn:a anJ the temperuturc gradient i, n i.:.t1nMant. indi:rr:ndt'nt r1J
\ , H~nt,;c. wnh Tt01 > Tc LI. the tcmp!!rarure d1,1nbut1on 1s t"'tponenual. ;ind i.l, ...hov.n III the .,~i=h.:h Jh,w,: St>raraung , ,mahlt!, ;anJ 1ntcgra11111;! F.:q I~ 1. 1hi.: geneml fonn lor 1he ttmpcr3ture d1,Lnbu1 '"'n ~..an
he Jd<rmmcd.

A e,p(a., I -JJT, " C'
JT=C1,-\. ' e,pj - a.,Jd•
<

\\'c ,,1uld u,c 1he l\Hl tt!rnpera,urt!' houmJ.a~- &:ond1uon~. T.. =TIO) and TL= Tl LI. 10 t-valu,11e C, ,mJ C.
•1mJ , htn,1..· oh, Jtn lhl.' t~mper:nurr: J1,1nbuuon m L~nn.!'I oi l ;md Tl•

,t,, \\'uh Ill< m1<mal ~•nerauun from l:4 11,.

"'

<

( ·o\l ~LF !\TS: In pun Ch), ~uu ,ould Je1cnmnc 1h, tempcru1urc ,h,1nhu11on u\lng Founcr', IJll "11u kn,,\\ kJ~i: ..1t thi: ht:JI rJh: J~r~mJ~n1,;t• upnn the ,-i:oordmati:. GJ\'C II a try'

PROBLEM 2.14
KNOWN: Dimensions and end temperatures of n cylindrieo.J rod which 1s rnsulntcd on its side. FIND: Rate of heat transfer associnted with dlffen:nt rod matcrio.Js. SCHEMATIC:

D•ZSmm

-x

L =0.1m

ASSUMPTIONS: (I) One-dimensional conducrion along cylinder axh. (2) S1cady-srn1c conditions, (3) Constanr properu.:s.

PROPERTIES : The properucs muy be evaluated from Tables A-1 to A,J :u n mean tempcrorwc of 50"C-= 323K and arc summarized below.

ANALYSIS: The bent rransfer T'Jte may be obtained from Fourier's law. Smee the ax1nJ temperature gradient ,s linear, this e.xpression reduces to

Tt - T2 n(0.025rnl

q=kA

-= k

L

4

Cl OCH)l°C -= 0.49 l(m·°C)·k
0.lm

Cu

Al St.St SiN Oak Magnesia Pyrex

lpurel (20241 (30!)

(~5%)

k(W/m Kl .I()]

q(W)

197

177 16.1 1-1.9 0.19 0.052

87

llO 7.3 O.O'JJ 0026

1.4
0.69 <J

COtl,ll\tENTS: The k values of Cu and t\l were ob1runed b) linear tntcrpoluuon; the k value of S1.S1 was obtamcd by linear cxtrnpolaunn, a, W11' the value ror SiN, the value for mugncsu w;" c,h1a1ncd h) lincru- mtcrpol,11ion and the ,,uues for Oitk ,lDd pyn:x an: for 300K.

PROBLEl\12.15

KNOWN: Onc-d,mcn<ional sy<icm wuh prescribed surface 1empcra1wcs and 1h1cl..nc,s

FIND: Heat flux through system cons11Uc1cd of these materials: <•I pun: aluminum. (bl plain carbon steel, (c) AISI 316, swnless steel, (dl pyrocenun, (e} 1cflcm and ffl conc-n:ie

SCHEMATIC:

i+•l- - -•-1 +- - L •2.0mm

Material of_l--__,._ known k

Tz.=Z75K

ASSUMPTIONS: (I) Steady->tatc condJuoru, {2) Onc-dimc,a,ionaJ coadU<:tion, 01 Su he•I genenwon, (.1J Conslllllt thermal propcmcs.

PROPERTIES: The thermal conducnvt1)' 1s evaluated a, the average 1empcm1urc of 1hc
system. T= IT1+T1 l/2= (32S+275)Kf.!= 300K Property value, and table idcn11tica1111n .ire sho"' n bclo"'.

ANALYS IS: For llus sysicm, Fonner'< l•v. c::u, be wnucn 1li
dT Tz-1'1 q, =-k dll =-k L

Sub,•11u1tng numcnc3l value,. the heal flu, in 1enn, of the ,y,1em !hernial conducu,i1y "

q, =· k (275-325JK =+l500~ k

20,10 1m

m

v.here q; v.111 have uniis W/m2 tf k hu, un11, W/m K The heat flu,es tor tuch >}Stem follo"'

Ma1cnal

Thermal conduc11vny Hca1 flu, Tahlc k!W/m·K) q, (kW/m2)

{it) Pure aluminum

i\· l

(h) PJ;i,n carbon SICCI A•I

(cl AISI ""• S.S

A-1

(di Pyro<cram

A-1

(cl Teflon

Al

11) Concrcir

,,.1

217 611.5 IJJ
19~ 035 I.J

593 I~ I 33.5
995 O.~!i 35

COMMENTS: Rccogni,c 1hc r:tn~< of thermal tonducnviry for the"' ,-oltd nia1enals "ncarl~ two decades

PROBLEM 2.16

KNOWN: Different thicknesses of three ma1enals, rock, 18 ft: wood. IS an; nnd tibcrglas<
1nsula1ion, 6 an.

FIN D: The insulating quality of 1he matenals with given thicknesses n, mea,ured by 1he R·
value.

PROPERTlES: Table A-J (3{IOKJ:

t.1a1erial
Limestone: Sofrwood BlankeI (glass, 6ber IO k[!/m1)

Thcnnal conductivuy. W/m·K
2.15 0.12 0.048

ANALYSIS : ·lbc R-value, • quannty commonly used an the consuucuon industry and bu1ldmg 1echnology. is defined ns

-R =

Uin)
k(B1u·1n/h·f'r

'F)

The R ·,alue can be in1erpretcd a, 1he thcnnnl rcsa<UIOce of a I ra= cro" section ot the matcriill .

=· U~ing lhe convcr..ion rac1or for 1hcnn:!l conducuvi1y between the SI and Cn!!ll>h ,ysicm,, the
R•value<

Rock, Limestone, IR ft.

18 fl• 12-'"

R = _________r_i_ _ _ _ _ : 15.51B1u/h fl:·°Ft· 1

W

Btu/h·ft 'F

tn

2.1.SmK x(l5778 W/mK l'.!fl

Wood, Sortwood. 15 m:

R = - - - - - - . . :1.5::..:tn:.:__ _ _ _ _ = 18 (Bm/h fl1 ·0 Fl-1

0. 12~ • 0.5778 Btu/h ft·'F ·' 12 in

mK

W/m·K

fl

lnsuln11on, Blunke1, I\ in:

K =

.

(l ,n

. ,,

= IK (Btu/h·fr·' F)-1

O(l48~•0.,778 Btu/hit F. 12.!!!.

m K

\V/m·K

fa

C0\11'>I E 'IT: The R-valuc of JQ j?I\Cn m 1he ntlvcniscmcni is reasonable.

PROBLEM 2.17

KNOWN: Electncll hca1cr ,nndw,chcd between rwn 1denncal cyhndncnl 130 mm d1,. , 1,(1 mm length) samples whose opposite ends com.ua plates mtUntaincd 11t T0 .
FIND: Ca) Thennal conductiVJly 01 SS3lb ,nmples lor the preo;cnbed cond,uons CAI nnd us avenge tcmpcnnurc. (bl Thermal condix:11v11y o( Annco imn s11mple rnr 1he p«"'4·nhcd condillons IB), (c) Commen1 on advantllgc, or c:qx:nmcntal um.npcmcn1, l.icral hell! '"""'· condition when t.T1 '¢~Ti ,

SCHEMATI C:

To= 7 1 • c
_1,..,._ t>.x=lSmm

Heater,} ..,_ ,r-.ic.7j=2.S.0°C

lOOY, 1--b---SS 316

0.'353A

ATz. =zs.o·c

c.x=lSmm

To= 77•c
t>.1i,•15.0"C
Armco ,"ron
t,.Tz:150-C T.=11·c
Case B

,\SSUMP'flONS: (I J One•dJlllt'.n"onaJ heal tnn.,fcr 1n wnplcs, (2) S1eady-'1u1,: comltth>rt,,
n \ Ne~hg1ble co~tact rc4-an,nce t>etl.\t"CO materi1th,
PROPERTfES: Tobi, A.2, Slllinkss Sled 31h IT~ 400 Ki k,. E 15.2 W/m K: Annco 1ton (T
= 380 Kl k.,,.,,. = 71.6 W/m K

•\ 'I/AL YSIS: ca) Rccogn11e 1hn1 hall Lhc hc:11cr power w,U pas; Lhrough c.irh of the >'llllple,

wh"h MC pmsumcd Jdc.mh:a.l \CC C'.a.,c A 01~.we Apply F~unc..r·, la" ta a sample.

,1T

q=k/\,··

"' l = qt.x = 0.51 IOOV>il.353,\J 0.CH5 ".?, ~ IS.U W/m K

<J

/\1~.: T

rn0.0.,0 m1lt.•,•.~__O_ '·'C_

·c. n,c re>tnl 1cmpcr:11urc drop •<ross die lcn~•h c,f the ,:unple is .\T11Ut.x) = 25'C Cl>O mmil5
mm)= HIO Hence. the hc.11ertcmpc,ru111rc 1, Tb= 177'C. Thu,, the a,·cn1i;c tcrnpcratun: ul the 11i3mplc ,~

<J
We 1:ompa.re this r,:sull w11h 1he t:1huh11t:J \·a.luc (Sl-C above) JI ..100 K and nocc lhc l!nod t1[rrccmcm,

(b) F'ar 1he Caar.c B nmi.n~emcnl, v.c M,ume I.hat thi: 1henn1.1l conducuvh) of the SS~ ltt 1s the <amc o, 1ha1 round In Pan f•I The hc•t ruie thr"u~h the Armco iron ~•mple"

Conunucd ~

PROBLEM 2.17 (Cont.)

= quoo = qh..,., - q,. IOOV..0.60IA - 15.0 W/m·Kx 11(0.030m)2 )< 1.s.o•c

4

0.015 m

q..,. = (60.1- 10.6)W =J9.5 W

where
q., = k,.A,;6T2/Ax2

Applying Fourier"s low 10 the 110n s.1mplc.

k.,.., =

quonAX2 A,;AT1

=

49.5 W.0.015 m 11(0.030 m)2/4~ I 5.0°C

= 70_0 W/m·K.

<]

The 10Ull drop across lhc iron sample ,~ 15°C(60/15) = 60°C; the heater 1empcrature 1s (77 +
60)°C = I 37°C. Hence the average 1cmpcra1urc of the iron sample ,s

T= (137 + 77)0 C/2 = I 07°C = 3110 K

<J

We compare this resuh w,lh the 1abula1ed value (see above) 111 380 K. Note good agreement

(cl The principle adv11magc of having 1wo identical samples is the assurnncc 1h01 all lhc electncu.l power dissipated m 1hc healer will appear ns hcu1 Row through the samples. W11h only one sample. he:u can flow from the bac_ks,de of 1he heater even though msulaied.

Hca1 leakage ou1 the lacer.ti surfaces of lhe cylindrically shaped snmplcs wtll become sigmfican1 when the ~ample 1hcrmnl conduclivny is only slightly higher than lhat of the insulating material used on those surfaces. That i,. the method 1s suimblc for metnllics. but must be used with caution on non-meutlhc ma1erials.

For any combination oi matenals 1n the upper and lower position, we cxpcc1 .lT1 = .lT2 Ilowe,cr. 1f the insulauon 1>crc improperly applied along 1hc la1eral rurfoccs, 11 1s possible thnt heac lc.tknl,!c will occur. Thi- mav cause uT1 ~ uT2.

PROBLEM 2.18

KNOWN: Comparanve method for measunng lhennal conductiv11y involving 1wo 1denncal samples ~!Acked wilh a reference mn1erial.

FIND : Ca) Thermal conductivny of 1es1 matenal and average temperature,. (bl Condlnon~ when AT1 .,_AT2,

SCH EMAT IC:

T,, =4001<-~-

c.x=lOmm
A'tJ: 3.32•c

Reference mi,/eri;,f,-...., Armco iron
Test Si!mp/e (2) --+-;'
Tc=~OOK-~ _ _ ,

ASSU MPTIONS: (I) Steady-state conditions. (2) One-d1mcnsionnl heat transfer through ~mples and reference material. (3) Seglig,ble thermal contact rcsisuincc between material,

PROPERTIES: Tablt! A 2 Armco iron CT= 350 K): k, = 69.2 W/m K

ANAL VSIS: (a) Rccogmnng that the heat ra1e through the samples and reference material , nil of the snmc dtnmcter, is the snmc, 11 follows from Fourier's ln-. Llrnt

k

t.
,

T-,-1

=

<iT, k,-=

<'IT,.:: ~,--

.lx

.:ix

A~

,•c oT,

2 49•c

,

k1 =

k,-T
_:i I

=

69.2

W/m·K

3

·1-

= 51.9 W/m K.

<J

We ,hould n.,si~n this vnluc a temperature of 1~0 K

<J

lbJ If 1he 1c,1 ,ample, arc 1dcnu.:al in every rcspcc1, .:iT ,. t.Ti only when 1he thermal conducuvuy •• highly dcpcndcn1 upon tempcrarure Also, if there is heat leakage ou1 1hc lateral
,urfacc, we can expcc1 .!\T:: < AT, Thi< would occur when lhe thtnnal conducovi1y of the 1cs1
material were onl)· an order of magnitude above that of 1hc insulnung matcnal tmployed.

PROBLEM 2.1 9

KNO\\•N: lden,lcal samples of pre,cnbed dlome1cr length and den,11~ 1nmall) •• a unoic,m,
tcmpertnures T,. sandY. 1ch l.U1 clrcmt:: heater which provide~ a uniform hc:st ftux q~~ for a pcnetd
of umc :',I,.,. Condmon~ shttnly ufu:r cn~r~zmi; and a lonj! tune after dc-cne:rg11.1ng hc-mi:r "'""
prc!l(:nlxd

Fl.NU: Specifu: hl!.lt and thcrm.U tondue1i..,11y uf the 1c,1 "';smplc ma1crinl. fn11n 1hc.,;,c:
pmpcmc~, ldcnnfy ,\-pc- of m.alcn.n.l u~inJ;; T,1bli: A 1or A.~

SCIIE:l,IATIC:

. . +-I n5u/,,fion about the
L=IOmmr~D=60mm ~ , enfire block Sample I,_p=3965"9/m3

7;,(t)

Heater, P(W)
i......,...,,..- - ..-,-,.~ _;,J.. Sample l.;,,T;=l.3.00"C
-~ .. .. •. .._.. t

Ca<e A forll <f Is" .\I, = 120 ,. P =I~\\ T0 00" = 25.~3 C
CJM!' 8 lor t > ,\l"(J. P =Cl\\. fort> .,t1., Tpl-1 = 33.50t"-C

\SSl \rlPTIO"-'S: Cll Om:-d1mcns1nn.tl hcul rrum,rer m \llmpln. (J1 \Jmfnm, prnpcn1c.... (.'\J Per1t:L·I m,ulam.,n n,, '"'''"' nf hc-J1c:r J'(1'-"<t rn 1n1-ulr111t1n. 1--l I I fca1cr ha\ nej!lip;1blc ma!,o;

.\,-\I \ ''-;-IS: Cn111.u.kr 4 control , 1tlu1nc ahnut ihc .,;tmpk, .md hi:.il~r lot Jn ,ntcrv.;tl uf um~ t a ti tv,..... und v.ritc the ~•11JhCl"\'.ltlUl1 llf cnrr~~ tl"lfUlttffll'nt.
re' Euw-; .\14 -= f., - F-.1
l'.'.IJ,, -ti= M,0 11 ,_, - T, I

,T~)-.r,--z, oo•c

I-- -T-,..-,1-=-3-s-.s-o-·c-.

' E=.-n+III '

-'•'I

~oh·mi tnr "v• "iuh,tnuun~ numc.rk.11 \·3lu~. and rccngm11ng lhl" energy rn •., prr,c:nh~d Ii)·

C1!ie :\ P4-'.,,.l!I 1:1.1ndl1ion untl chc rin.tJ 1cmpamun: T1b~ Ca.'<' B. lind

<J

C'onunui:d

PROBLEM 2.19 (Cont.)

k =

30 s

]l [ 2><2653 W/m1 = 36.0 W/m·K

<J

nx3%5 lcg/m3x765 J/kg·K [24.57 - 23.00J°C

where

10560W2

,

=

l 2653 W/m

.

2(1txO. /4)m·

With the following pmpcrncs now known,

p = 3965 kg/m3

',, = 765 J/kg K

k = 36 W/m·K

SCllrCh now in Tobie A.2 fint 10 ,cc whether values arc typicnl of mctalhc or non-metallic
mntcrinls. Consider the following.

• metallics with low p genemlly have higher thcrnw.l cooductivirics,

• specific heats of both type., of mntcnnls arc, of ~imilar magnitude,

• the low k vnJue of the sample i~ typicnJ of poor metnlhc conduc10n which ~•ncrnlly have much higher specific hcalS.

• more thD.ll likely, the matcnnl is non-mcllllhc

Begin <enn: h through Table A.2, and rmd the ,ccnnd cnu-y, polycrysutllinc alummum o"dc. hu,

propenics at 300 K corresponding w those found for the samples

<J

PROBLEM 2.20
KNOWN: Temperature distnbuuon. TC~,t,z). withm an infinnc, homogeneous body •ta gl'cn 1nsum1 of lime,
Fl',D: Region, where the tempernturc changes with time ,
SCHF.l'tlATIC:

c--T(x,y,1.)• x•-2y2 +z'-xy +2yz Z~Infinite med,um

X

\SStl\1PTIONS: 11) Con,rant prnpcnic, of infinite medium and r2) No internal heat
i;tncr.umn

\ 'lfAL\'SIS: The temperature d1<tnbu11nn throughout the medium. at any ln<tant nf umc, mu,1 ,.n,fy the heat equauon. For 1hc thrce•dtmenslonal canes,an eoordlna1c system, wtth con.\lam

pr<>pcn10, and no 1n1emal hcJI !!•ncranon, 1hc h•a• cquauon. Eq 2. 15, ha< the ronn

,iT i12T ci2T I ill

-,-'l"-----,+7"'"'T=--::-- .

(I)

r),• cl)· dr u di

When T1,.y.ll <au,tics 1h1, relation. then con-..:r-anon 01 .-ncrgy at eve') point in the mew um is sall'licJ Sub,111ut1nj! T(x,y,t) 1nto the Eq l I l. lirst tind the gradient>, clT/ih, .n-rii), ct~.•

d
-
u,

(.!~-))-

cl
-dv•

(~,·-)(+~11

+

d
-d1z:?1+2.v)=

ldT
-u -ilt

Performing the d,ffcrcntiaul'.ln. tind

,
-

_

4

,,=-1
-n

iJT
ii,

That i,,

i,~,, = fl

,-.h1ch unphe~ 1hal. rur the: g1,·c:n tnMani (lf ume, 1he ,cmpcnuurt: \\-1II e1-:erywherr not chanie.

C0\1Mf'\TS: Smee v.e do not ,now th< 1n1llal ond boundary conchllon,, we cannot detcrmmc the 1cmpcrnturi tlJ!ltt1hu1mn. Tr,.y.1). :n an} future umc. We only can dctemunc that. lor tJ11, 1-pccrnJ msrn.nt uf llmt, tin: 1i:mpcr,.uurc wJll not change.

PROBLEM 2.21
KNOWN: Steady-state temptrntwc dullibuuon in • cyltndricdl rod having uniform heal gcnera11on o( (}I:; Sxt07 W/m3
FIND: Cal Stendy-siate centerline nnd surf.Ice heat 1mnsfer rJtc, per un11 length, q; (bt lnmal umc nuc of change or 1he cemcrhne and surfoct 1em1'1CrJtu~~ 1n rcspon,-c: 10 01 change 1n the
generation rn1c [T(Jm <it 10 <ii= 10" W/m'
'iCHEMATI C:

ASSUMPTIONS: Cll One-dlmcn,ional conducuon ,n the r d!n:cuon, (11 llmform gcncr.mnn,
and (3) S1cady•St•te for q1 = 5•107 W/m1

Al'A(, YSIS: (al From tl1c rate cquauon, fi,r crltndncal coordlna1c,,

q, =-kil-l
i)r

q = - k Ad,T- .
ilr

Hcm.:c:.

"'

•

i)T

'Ir= -1nlrr ilr

(I)

where iJT/ilr mn) be evolwued from 1hc p«wnbcd tcmpcrnturc d1>U1buuon. T(r)

,\t r=O. the gr•dlent I> (;rf/iJrl = II. I l<nce, fr<'m F.q (I) the htllt mtc i,

~.,o, =o

<]

Al r=fuo thC' tcmpcr-.uurc gradient I:,

"1 i<JirT]n.

=-2

[ ~

167,JO-'

, 01

(r,.)=-2tJ lb7vlo'l!002Sm)

~; ] = -0~0&, In' Kim
""·

Conunucd ....

PROBLEM 2.21 (Cont.)
Hence. the beat rate at the outer surface (t=r0 ) per unu length as q,(r0 ) = -2Jt (30 W/m K) (0 025mJ [-<>.208> 10' Kim)

q,(r0 ) = 0.980 x IOS W/m

<J

(b) Transient (nme-depcndent) cond11Jon• wi LI cx1s1 when the genr:rution as chani;cd. and for the prescribed assumptions, the 1cmpera1urc is dc1ennmcd by the following fonn of the heat cquution. Eq. 2.20
aT) . -rI - ii)Jr (kr..u._r +q1=P<i;dl,Ti-

llcnce

Howr•er. muinlly (at 1=01, the temperature disuibuuon as given by the prescribed fom,.
Trrl = K(Xl- ,1, 167• tn5r'. and

r r I iiiir [la iill'rT] = k ~i) (r(-83~4,llf·r))

= -k (-16h6~•h,.r, n
r

= 30 \V/m K [-It, ('68, ta5 K/m~)
= = -5x107 Wtm' (the ong,nnl q q1 J,

llencc. evci; where an the wnll.

ilT

1
11Cl0/k!!lm·1 x ~no

J/kg

K

(-5,to'

+

10•1

\Vim'

or

<J

COI\I \1E','TS: 1I I The value of (c/T/ii1) 11.111 decrease with 111creasing nmc beyond t=CJ. unul n new ,1ead~-,rntc condition is reached and nnc-.: again fc/T/01) = 0.
. r2J By :,ppl.ying t.he energy conser.auon requirement. Eq .. I I l.a. . 10 the rod for the s1cady-sin1c
conwum1, E,, + E,,,, -E.,u, =O llenccq,(01-4,cr0 1=-q1cirr;1.

PROBLEM 2.22
KNOWN: Tcmper.1turc dismbuoon In • one-dimcns,onal wall wuh pn:;cnbcd 1h1ckne,, and thermal conducuvuy.
l'CIID: ro) The heot generation rnu:. q, in the wall, rbl Hc.11 Huxes at the woll foce, and rclaut>n
1oq.
SCHEMATIC:

IL,SOmm ASSU~1PTIO!IIS: (I) S1ead)·\Ul1c t(>ndl1ion,, (?J Qnc,dimcns,onlll hcot flo..,. (3J Con<tnnt propcrtiC>.
ANALYSIS: laJ 1ltc npprop.riatc form nf tht hcor cquorjan for .;1e11dy~m1e. nne-dlmcn,mnal
condnions wi1h con51unt pro:pcrucs is Eq. 2.15 tt•wmtcn as

4=-k d~ [ : : ]
Using llu: fnrm for the. cempcmturr dnmbuuon. cvalwuc lhc grucbcm giving.
l' 4 - -k ..d.!x!... _d!i!t,_(a-;-bx21] = -l _d!<!,_ f2bx f = - 2bk

~1.:. :c !OOO<?CJ-m2l - 50 W/m·K= !.0...10-" V,'/n,J .

<l

1h1 ·1nc bc:ul Hu,cl'I at Lhc wall fat:cs. \:Jn t,c:: t:\.'alu1utd from F,,uncr'i Liv.,
!!' ], q:(,l= - k

Uc;m~ the 1cmpcrarurc dl~mbuuon Tix, to t:V31Ul'H~ the gradJenl, find

• ,1- ,(,1=-k -dd fo-;-b•\"J= -21.b,
The' flux iU the- fa.1,:c x::O. is thc.n

4:!01 = 0

<l

lllld Jt, = L, 4:1L1 • -2kbL -~ • ~OW/m K 1-1000°C/m2J• 0.050m

q:(Ll = 10,000 Wlm'

<l

CClM'.IE.'ITS: From 1n o-cr,11 energy ba!w1ce on the wall, it follow> 1hot

E, E,,,,+E1 =0

q,COJ-q,(LJ+qL=U

o.~;o::· - q·,(LJ - q,•tO) JO ()()("'/

q =

I,

= '

'

ti =~.a. J01W/m1

PROBLEJ\.1 2.23

KNOWN: Wall Ilttclmcu. Iltcmlll conducuvny. 1cmpcnrurr dmnhuoon. ond nu,d 1cmpenuun:

FIND: (it} Surface hca1 nuc1 And nuc of change or wall energy s1oragt pc:r unit area, a.nd (b) Con-.•cction cocfhctfflL
SCHE\fATIC:

k=IW/m·k

T(>t) •200 · ZOOx + .30"'

I I

142..7°(

- , - --+ ,
q•in •

,---➔
, q•out

I

I

11 - - -...,

f--x IL•0..3111

ii{&;) ,;,, •1oo·c,h

,\SSL MP'TIO'\S: I 11 One-d11nen<1onal condur:uon to •• fl) Con•tanl k.
A.'-IALYSIS: (a) From Founcr', I•-..
q, =-k -~ ih = f2(Xl - 60xl·•

<J

q:,, = '] uL E (lOO- fiO • 0 3fC/m • 1 W/m K = 18! WJm' .

<J

r\prh·ms Un energy balani:c to .i concml "·t1lume .1.hou1 the wall, Eq I. 1ta.
• •

- t,1 =-q"111 -q<IU,; IK W/m.~.

<J

H>l AppJylnJ! .t \url.tce c-ncri:~ h;,.lnm.-1:! .11 \-=I .. q:,, ~ h[T(LJ - T.. I

h =

tfuu1

i:;

182 W/m'

T(LI •T.. flJ2.7- IU01°C

<J
C'CIM'l'1F'\TS: 1ll Fmm the h~t cqu•11011.
!olJilll = (k/pcpJ iJ2T/nx' = 60!k/pe,. I .
II rnUo..., sth.11 chc 1c-mpcn.1urt •~ in<:~)IOJ wnh ume 3.C C\.'tf)' po1m 1n the w1U
<cl The value (lf h ",mall :ind" typical or tree ccwwe.1100 10 ag.u

PROBLEM 2.24
KNOWN: Tetnpcrarurc distnbuuon and di,mbuaon of heat gcncraaon ,n contral l•ycr of • solar pond FIND: (a) Beat fluxes at lower and upper 1iurfaccs of the centr.tl layer. (b) Whc1her c.'.t"l1ldmonJ
arc steady or IJ'lltlsicnt, (c) R3tc of thenn111 energy gcncroaon for th< <nun, central i•)'cr SCIIEMATIC:
M i,ced layer

ASSUMPTIONS: ( ll Central l•)'rr 1, ""gnant, (2) Onc-<limcn>1onl1I conJu,uon. ni ('"'""'"'
propcn,c,.
A ,,__Al\'SIS: ht) The de\tttd Ou."Cc, 1.:orrcspood to conduction fluxes 111 the central layer a:u 1he lnwcr and upper 1;ur1uceS- A icncral fonn far the conduction ttu, 1.5

• q,_, :

-l

oiJT.

5

- k [ •A•

-u C

+

BJ'

<J

,i1T • I oT
--+~::--
ih2 k a ilt
I Jenee condh1on!i arc 1rca,/\· "lim.:c.

-A- e~u +/-\ e ., =l -in-

'

k

ttih

<JT/dt -0

tlor 1111 U S, S LI

<J

::J (c-> For tb~ ccnrnl lt1yt:r. the cnrrav ~enernuon t,

f.1

l l4d,=AJl e·uru'"

1

1

E ;-" e•U IL:-~(c"--11:~(l-c aL)

<J

'

.i

n .i

:i

Ahcm.tt1\'ely. from in D\'er.tll cnerg)· b.1.Ltnu,

COMME1.' S: Coodu,uon ,, In the nega.uve •·dircc1lon, ncccst.1t.1ung u~ of m1nu, ~1gn!- 1n chc above energy balanc,.

PROBLEM 2.25
KNOWN: Temperature dtsaibuoon 1n I scm1-1rnnsparcn1 medium subjected 10 rndtonvc flux.
FIND: l•l F.ll]mssions for the hco1 Rux OI lhc fro01 ond rear surfaces, (b) Hca, gcncrnaon 1111c q(x), (cl Ex11rcss1on for absorbed nsd1a11on per uDll surface "'"" m terms of A. •· B. C, L, and k SCU£MATIC:

ASSUMl'TIONS: (II S1c:id~-<1~1c cond.111ons, 1!1 Onc-<luucnsionnl conduc11t1n m mcdmm, (3) Con'1IDII propcn,cs, 141 All la.scr 1rrnd.1111ton is •bsorbcd and can be choracitnzcd bi· an 1nu:mal
,.-(,lumemc heal gcne.ranon tem'I {l(x).
ANA.LYSIS: fa) JVmwin~ &he tc::mpcr,uurc dhtribution. 1hc surface hea1 ftu.;\Cl 11te found using
Fourier·~ la~.

<I

<I
1b11l1e he.at diffo~icm cquilllon for the medium Ii
.Ju_(dd.,T)•~k II "' q
<I
(ct Periom11ni; 11n cncrg)' ba.J.111\cc on the mcJ1um tl\ !tihrtwn abo\.·c.

n::~·n,gniz.e 1ha1 E~ rep~scnis thr :sh,cir~ UTt'ld1nuun On u unil an::i b:u.1s

<I e·~ /\ltrmaU\·cly. cvaluatr hy rntetrauon fl\•er 1hc \·olumc of the medium.

E"1

= f
""

q'f).Jdl •

ri. Ae -"d.tc
.le

-~~ 1e-0
.1

Jl = ~{I
u .:i

e•..i,

PROBLEM 2.26

KNOWN: Stcady-sm1e tcmrrature dtsaibuoon in • onc-dllnens1onal wall of thermal conductivity, T(x) c Ax3 + 8,r + Cx + D.

FIND: Expressions for the heat gcncrnnon m1c m the wall ond the hem ffuxes at the two wall
faces (x • O,L).

ASSUl\fPTI ONS: (I) Stctldy-suuc conditions. (2J One-dimcns,onal heal flow. (3)
Homogencou~ medium.

ANALYSJS: The appropriate fonn of the heal diffusion cquntion for these condition• L<

d1T •

.

d2T

-+_g_=O
dx2 k

or

q = -dk,c1-.

Hence. the genernnon rn1c b

q• =-kd- [ -dT ] =-k-dl3Ax1 +2Bx+C+OI

dx dx

dx

ti= -kf6Ax.,. 281

<l

which ,i; linear with the coordma1c, , The heat fluxes nt 1he wall faces CilJl be evaluated from
Founer•s law.

q~ =-k : : =-k[3Ax1 +!Bx+ CJ

using 1hc c,prcSMOO for the tt:mper•1ure gmd1cn1 dcnvcd nbove. Hence, the heal fluxes an: ·

Surfarr ,=O,

4';10)-=-kC

<J

Surface f:L:

<J

CO/',,ll\1ENTS:
. .

CI) From on nvcrnll energy b:dnncc nn "1e wall, .

find

E., - F~•• + E, =O

<i':,01 -q'; Cl .J = (-kC1 - <-kll ~AL1 +181.. +CJ+ £1 = 0

E'~ =-3AkLl - 2BkL.

From tnlC(!T'JUon of the volumec.nc he:11 mtc, we can ;tlso find E'i n,

F.'~ = J;"iil•Jdx = iL-klliA, + 2B Id, = -kl JA,2 .,. 1B~I~

F.'g = - 3Ak.Ll - !BkL.

l\.,ow,~ Pl..mc "'.ill '41th no m1em.lJ encrg~ gencr.-.liun
► 1"11)1 lklL"Trnmc Y.llc1her lhc: P"=,;c-nhcJ tcn1p<rJ.1ure d1\tr1hu11cm h J'K•"1bl1!' r,plJm ~our ,e.a~omn~ Wnh 1hto lt"mpcr.uun=.f. T,n, a; O'C 41nJ T_ : ~q-C li1t"tl.. c::t1mputc ilnd plv11he 1empcr.uun- T1L1 "'-' ,1 1un..:1111n ~,1 lh( u,n,c~lllln c,1C(i1~lt'DI tvr th,: ri1n1c- IO! h S l(IO \\'.Im"." K
SC"HLM \ TIC·

12D r1•q

q•, (LI

-T(LJ q•cv

D ' - - - - - ' -• •

=JI .... T(OJ ......

:t-c.-==--...... q•O - -
•= • 5 W1m-K

o I

T/1.)
I
l • O 18m

r.= 20 "C
h •30W/m2K

L

~~l \JPTIO,'i: (II On,-d1men1imnJI ,11nJl.k't11•n 1~t l\o 1m!.'rnJI cni,g~ iene~u100. C' 1C'llfl,Lmt
pr, n,·ni. .11 I\, r.i.J1.11u,n tl,h..tnj.:t" J.l the wr1.Kt,:: L. ,md t .5) .~lu"1)•'1..ttc c::nnd1uom

"*"• \'AI \

h rhe rre" nhc.-J tc-mJ't'r.uurc- dht111',cut11111 rM.,.>1ble" If ,\,. 1~ cni:r~~ h;alJn•e at 1h-c-

•u1r1':

... r-i .,~ th,1wn illl-0,'C' an 1hc Sdl('m.,11~ 010,1 be 1;msf1eJ
f • '= ''II

•(Ill• lJT' •-~ fll.J TIO)• .J~\1/m l.:x(l~fl -Oi"C/Ol km a-10001\'/m'

J, /,,,

I

,j"' h[Tfl.l l,.);-lf1\\/m" ~,,.11.:!0 1ofr.aiOflflW/m!

'\ut,,11111ru1g the he.JI nu.-. ,·.ilue, mri, l:~ f :,,, lmJ ,..,uoo, .. i ll'.Jf11.J1 • Ulllld thcrdorc. ,h~ 1cm~r.t1ur..-
ll1\lt1t,ur1~•11 1, n,,f rlb,1hli:

\\.11h 110 _ ,, c· ,,nJ r_ :i :u·<...1hc IC"1('\.'J,t1Utc JI the \Url.11.t" '.: L TtL1, C.Ul l'C tli:1t·t11UMli lmn1 Jh
o,('1.11! L'1Wf~) l,.1.lJ.110.i: c,n lh1..• Yollll ,I\ •hl)"4n JJ"(}\e 1n lh~ ~1i:ht"mJf1;;

q;1111 q"' .o

-~T-IL)L-T(-O-i - h[TtLJ-1._] ;;:0

-.J<\\lm i-.[r1L1 11 r)lob-m-111\\/111· "-[TU J-~o-C]:u

111 .-w: 10•1 C
I 1mr 1111, ,;amt" .1nJlu11, ltU ,u .1 lu~IIL•n tJI 1hc .;cmc1i 1,...,n tt1cfrl,.1cm c.sn ti~ Jc1crnuonJ .an.I pl11t1t\J \\ c J~,n•I C':\rc-c-1 T•l. 110 PC l111C"wrl~ J~rcn.L·t11 Uj"'\•11 h ~ote thJl ,.,.. h Hl,.R:"J-.Ci. 11,
t. '.,r,cr \ dw:,., Trl , .1prro.1.:h~s l'n "hJl
\ .a!ui: \~ 111 l1 l I Jprrna.:h ,I'> h Jc1.1c.~,~~ I

<

•

~-

', •
i •

-,

- I
••• •

~
•

' • • ·--•····•·-·· • 1

PROllLt.M ?.lH

h.NO\\ '\: Coa1 p1lt' ol rreK.nbcd tJ\!pth r,peneni."1n1 umlotm \ 1.'tlumctn, icnerauo11 "1th i;on\'c-.;111,n. Jb~1rhc!d 1rr.uhu1tun 1.111J enu~i,on on 11, upper ,urfac.c

nl'\O: ;,) The ,,ppwpmuc. fom1 ,,r th~ lit~I di..lfU.'-ll'\A equiJIIOO i HOEi 1tnd \\ hc1hcr the prt'~nb<d 1cm~ra1ufl:- Jhmhuunn ,.iu!.f1c1, th1, HUE.: tt"ruht1011'- .u tht hc,m,m Circhc rile. ,-. = ll, -.l.c11.:h or 1111:

ti:m~rillu~ J,..mhutil,n with JJ.bd1ng of L..c~ ft.ttur~,-; tbJ E,r,c~,1nn fur 1h.: ,ondu~lton he.JI mh! 111 1hc

L. ~.-,r~'"'"

uron

b.ll1m.:c: L

1uc:st111n

.\ : -

for

lht"

'-urfa,.;c

tc.mp,c_rJ.lurc

T i b~c,ed

.i.

s:urt11,c

cocrg~

111, --

t:"'llluau: T i.md T101 ror tht pr!!.1;cr1bcd ccind1t1CtR\. !cl B.1Jcd upcm l)pu:.tl d,ui) a"crn.@l-'""' forG" .anJ h,

1.·11mpu1r :mJ r,lm T~ w,d rf0.1 for I J• h = ~ Wlm' I\ w11h 50 S. G\ ~ ~00 W/m·. t!l Ci, = 4tM'I \\'/m· "1d1 ~

< h "4J \\ 1m' K

f
•

\~'\LI\IP'JlO"''.'\: 1I I One,duncn,h,naJ 1.nnduc.11.nn. 11) l'nuorm ,,aluml!mt hca1 ~cncrJU(tn. \ l, C ,11 IJnl P"''I"'"'"'· .J, ~'-"J?hl!Jhl(' m.&J111Hrin frt•rn the. .,urruundlllt!-"'• 11nd t,1St<IW,-,t.th: ",,0J111nn,
l'ROrf RTlf-"; Tuhl~ \ {, Cl"•al ,ooK, Si..,.,. U16 \Wm K

\"·\I \ SIS: ,1, F11r Llnc-J1mt'm1tm.1I, ,.tc-.W;,-....,Lu~ conJui:um, 1,1,uh umform rnlumctn1; hc.11 ~cncr.iuon
•1,I ,in,.1.1n1 pr~'f't'm1,,·s 1hc hl•,11 Jiltui.mn cqu.-1u1n tHOF> follo"' rmm Eq : 11,,

J '!f)• •I •U

q I<

Ju J\ ~

tl [ a.-q·-l' ( o
d'< ~k

I z_,\J

<

h,1111 E1.1 •~L 11,He lh.11 th,: h:mrc,-rnturi: 1,h,1nhu111m nu,t-.1 hcijUJJJJU.;. "•th nw.'lmum \.Llm: .i.1 \ ; I.I \I \ -0 1hc ht-..11 i1u, Iii

,.. II
,io 1hJ1 th,: ~r,1,hrn1 .11 , : fl I" /t"hl Hcm:c. thl· h,1t1r1rn
1-. in~ut.ucJ

1

/ f>arabohc shape

Zero gradlenl
al boUom

0 '--:----'--

r,

T!OJ

q',llt

qL

<
Cun11nucJ

PRORI.F.:\I 2.18 IConLt

From ii ,url.K'C cners> hal.ancc ptr unn orl!J ~hown 111 lh~ S1.'.hcm.n11.: .abc.nc.
E,. - 1.:c::. • l', = II

yl-li(T, -T. J • OY~G, «:rT,' = O

1ow·m·,<1n1-~\\'/m K(T, ~•l~I\J •0••~,4001\"/m' -u~5.-:\1>7•11l '\\"jm "'l ~o

T ~ ~'15 7 K:!! 1 C

<

from Eq 1:.1 \\Uh,:. o. find

.,..her" 1ht lh~rru1d cnndu1:.t1\'II\ Jett C\JJI 'o'·J, 11!,tJmc:d Hom I ,1hk ,\ :i.

(CJ Tv,o plul"' .uc i;c:nl.'r.tte<l u,ing E4 u:nJ 1j1 lur T. Tio,.

I 1 \\Uh t: - ~ \\j m K f1.•r

(--l

I

ornJ

r t ! 1 p c 1 . : " 1 1 \· l " I }

1

~II S: r;, S" IIKI l\1m' and 1!) wath G, = 41XI \\"1m· for~~ h !, 511 W/m· ii;

.,

~11o,,t.'ll'~ f'l•l'f'l.~IC

.,

.•••· .,

-- f IQ
t •

... - ""' r;:.

.,,

• '"'

:~,..- i
.,. .. "' - · ~ 1llfl..- .. ,. w ....-."
- -- t,.

I-rum thl." T \,- h pl1.1r v.nh (11, c .tOIJ \\",111 •• nl11t..• 1h.111hi: i:uML',:lluH lu<llktc:nt U1'4.:'~ not h.1vc 1l m.1.Jut
rnnu..·m:1.- I'll the )U:n,1-.--e 11r l'i1•U1tll1 "·r,.11 ri le l~mpt:rJIUrt'\ Frum th\:" l ,~. ("j._ plut '>Atllh h _.;_" \\ 1m I\,
IWI\.' th.11th(' ~l'l;.ir nr.aUt.11111 11 h,h J \Cl} !ttfntli.;.rnl 1.•fll.'1.:111n thr: temp1!r.1turc~ Th!! , ~cl th.11 l ,,. 11.•,,. lhJr.. tlu: ,l:llhlrlll ;11r •~mpt'rJtUJC. T ••mJ i11 lhc" (J.\C" \'11 \Cl} !11" \",tlllt'" 01 Ch, hdo,,_, fr'-·c11n!,;-- 1," " 1r1.,c4uc"n1.t· -ii the: IJ.C~1.• m..i.g111tudr Qt the: c::111,,,,c p11\\1.·r 1~
C-0\1\JF, n~: 111 IUI 1U1,,J~,,.. 1g11u,i.:J nnJ1:11111u lf1•111 lht ..1,,l ,In l.'n\trnnrn~ni.il r.:1J1,1T11111i.:hi:,1 \I.C'
, 1u·11 :1•n• 1Jt • m ChJph·r I~ ltcJh:J :.1\ l,ug"" 1,111hcm1J.J ,urruu11Jmy•, r:loL, ~ ur;:~ ,\ht:ri: T,, =- to· C
IM \Cf~ ... le.JI ~-•11J1111111, .m,I n..:.trl~ Jtr k111r,a.·1.111m.• lt1r ..:.k1ud~ c11mJ111on, four l.,m G:i 1.'.011J111tin, Vil." ,tii,ulJ t:<•n..,JL•r G"'" lht: d lc~t ol \\hii.h ¼Ill t,." Ill rredJ\1 l11ghl'I ,.itue~ to, r ••mJ 1,01.

PROBLEl\1 2.29
KNOWN: Cylindncol ,y~m wilh negligible temperature vananOfl m lhe r,1 dlrccnons.
Fll',D: (a) Heat cquauon beginning w11h • properly defined contr0l ,·olume, lb) Tempcra1un: dt<1nbu1ion T{ti>l ror s1eady-sia1e condmons wilh no u11cmal heo1 ~•nerauon >nd con"on' propcrues. Ccl Hcoi ra1c far Pan (bl condltton,
SCHEMATIC:

\SSIJMM'IONS: (11 T " tndepcnden1 or r.z. !lJ ru = cr0 -r, l < r,.

\ 'I .\J YSl5: la) Deline lhe control volum,: n1 V = r1d,;, ru L where L 11 lcn~th nDlllllll lo p•f•·
Apply 1h• ·nn<ervauon nr energy requ1remcn1. Eq I I la.

• . F.., - E,,., + E1 = E,1

q0

- q0 , J o

.
+qV

=

pVc

aclTt

( 1.21

where

q0

=-k(Ar

Ll

-ill
r,iio

'lo-<IO = q0 +

,.i.i_
""'

(q0) dQ

(3,,11

Fqs. n1 .,,,1 C4J Follov. from Founcr', 1•11.. [q 2.1. ond lrom f.q 2.7, re,pccavcly. Comb1n1ng
F.q, ()Jan~ (4) wah F.q I~) .tnd canceling like 1cnn,. hnd

r/I
-

<l
-

lk -iY--fJ + q = pc: -a,·

JQ dO

t)l

<J
1.\1

Cun!.tdl!nnl! that r:t. art con,i.ant, th1~ lonn -igrt..-cs wuh Eq 2 20

q (b) f-ur 'lit.C.idy·\UllC C'OndlllOR\ \f..'llh -::0, lht: heal equ;mon, (~J. ~come~

~ ll dT)=n
do dO

161

' \\'1th c-01u1.an1 pmpcnit), ii rollnw). 1ho11 JT/dQ •~ con,1.t.tu wh1th unpbn Tt4n a hntJI m o

Th.1.11 ...

'"

fc) The- htat r:ue for the C(•l\d111an" nf Pan <b) folh\'-A>·~ Imm Fount..--r·, h1w, r:41. n,. u~ln~ the

l [ l 1empcrornr, !(r.ldJcn1 of Eq C7 ), Thai "· t]o"-ki,\rl.lJ- [~-1(l,-111 =-~ -r.-r, L1T,-T,>

r, it

ITT,

t:O\tMEJ\TS: No1e !he e,pre'""" for lhe 1cmpcr>1urc grad1cm tn Founcr's lov., Eq. (3), 1< al'tr,,1,:, not ilT/ilo For the condJ11on, <>f Pan (b) and le), no,. 1h,11 q(¢l 1> 1ndepcnden1 of¢, th" 1, lir<t nntcd m Eq. C6J and hnally conlirmcd in Eq. 19),

PJ'lOBl,EM :I. I
KNOWN: Onc--diD'tf.lD!ionn.l, planr· wall tteparallnl; bot and cold !luid1 at T ,;- 1 :1.ntl
T . 2, r~pedh·clr. FIND: Tcmprrat.urt disLr-lbutlon 1 T(xl, and heat Hux, •h•.• i11 terms or T ..-. 1, T-.: J• ba, h2, k and L.
SCHEMATIC:

Hor f/u,d
r...,1, ,,, /

L-,

L

ASSUMPTIONS, (1) One-dimen,ioool ,oaduc<ion, (2) Stcady-,mw eonditions, 13) Constant properties. (·I) NC11H~lbl• r:uliatioo, (5) No R•nrration,

A.t'l.ALYSlS: For th~ for('gnintt conditions, th<- ~encral i:;,olution to th1• lw111 dilfw;ioo equ..a.tion U, or Lhe form, Equ:Hioo 3.2,

T(x) =C1x +c,.

(I)

The coo,l.L\Ilr.s or int.egrat.1on, C 1 :md c,.. are drt.rnninrd by u~lng '!lurr.ac, enrrgy ha.lance conditions M x=-D :t..nd -x L, Equation ?.23, :1od as Hlustr.al.C'd abo,·t•,

-kddTxI •h,lr•<.1-T(O), .-u

-kddxTI =h,IT(L)-Tx,I•
,-t

For the BC aL x =U. Equation (2), u,~ Equation (J) to find

-k (C, -"- 0) = b, ITx,) - (C1 ·0 ..._ c,JI

and for tho RC at x= I, to llnd

Mu!Uply Eq. (·I) by h2 •nd Eq. (5) by h1 , •dd the equ,t1nns In obtoln r,

1Ubstilutc C1 into Eq. (-t) to obtnin C2 , Tho rcsuhs .a.re

c,-=

rr, , -T~,,l
x[h-1• - _b,I+~k ,

(T.., 1 -T ,.,)
c, s=- - - - - - - - - - ---Tt'IC' I
~ hI 1h~, ~h,.;. ~k ,
I l kX .•. bIi° +T. I

(2,3) (4) (S)
Then
<J

<J

PRORII \I .I.?

K~CH\ '\. J ~rl.'.'ratu1e, .111J t.:nO\ i:.:tiun -.:oc11 IL 11:nh .1,,1.)(1J1eJ w11h air .ir lhc 11111cr ..ind l,u1r:1 ,m1 .1.:-I!"'
•I .a I •Jr '-' 10d11"

Flt\ 0 : .a Inner Jnd outt-r "inJL1" ~ur1.a-.:~ t.:mpL·rnturt1. T .mtl T .inc.I •1', 1 •.ind T, ,h J 1u111.111111 ,,1
1h l!Ut 1~ Jir tr:mpr:r.1rnh.• T J.nd tur ,c-lci:1~•u \ .1iul.'.'ti. ,,1 C\Uh:r i::C1m-ci;1n,11 c,.,dfl.:-11.·m . h
\CHI \I \fl ,: Glass -

•••

Li1I

q'

L - 0.004m

I T,. I= 40°C
- h1 ~ 30Wlm2 • K

rrro,s: \SSt \l

1 I '-;(cJd\•\l.&k ~-unJuwn,. 12 (Jnt-•Jlln('ll',l!J11al cc,nlluct10n. 1 11 ~~gllg1hl~· fJJJ.itl<ln

·th: t ➔ C 1n.,t.10l pn)rcr11cs

q

T

-T•
L

I

J1rc 1-1u·c1
- ---,------,l""J1'-11s-1-1.:,=:n---.---

h ~ • h. o5\\' m i.; IJ \\ mi.; 10\\ m· i.;

,n- ("
flJIJJ5J•UIJ0~9-11llJ3JJm K \\

ll::r1i..r \\llh •I h (T. - r.

<

r

q
h, -Ill ( t,5 \\' Ill Is - -PJ C'

<

1b1 l "",111~ tr.c !J,lmc .mJh,1,, T JnJ T, ha\·c hL"cn 1.:1Jmpu1cJ ~1nJ pln11ed J...' j lunctmn ol the oul,1Jc ~ur

r ti:111pcr~lUl'c.'.'

lnr OUh:r l.:'00\Cl..'.111111 t:11.:ff1~·1cnt, C•I Ji e 2. 05•.md JOO \\/m· K A, C''tlX'dcJ, TI .snJ

r T iilrl'.' hn~r \\Ith ~hJng~.. in ,he ,m1•1.1J~ .,1r h:·mpcr..1turc The U1lferen-:i: ht.·l\\~~n [ J.nd mtrra.5-c.,

11

\.\Hh 11hr1•.1"tn~ iii:'Oll\'~~lh1n C\~111,,:11:nt. ,.,nee lhl" hL·.11 clu\ 1hwu~h 111~ \\lnJ1,"- lt~c\\hf! 1n...rtJ\l":S nu..

=: hrh.'rL""n~e 11,, l.n1!cr ,ll h,v,t'r om,1di: ..11r 1l"fll)'c.:rJtutt.•., h>r thi: -..Jtnt: rc.:tM,n "\,',Jtt: lh..11 v. uh h

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T f l'I tnn ...m.1l111, • hll\\ un fill' rlu1

PROBLT\I J,.! CC11111,1

• ►
t
l
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. . coM,tE,TS: 'I The large.s.1 rc-11.i.unce 1t that :btoc1.ued wuh ,on,.·cL'.UOn ill thr: mnl.'r ~un.1...::c Th•
\';aluc ll T .md T ~nuld h: ini;rcas.cd b,· mc.tt.n•un~ th~ value nl h.
, ~ The JIil Tri, rtr1ul R~ uHa,ru .\irnwrl. \JoJi1I "'" u,e,l t" 1.n:att .1 modd N 1hc "'inJm, .md r~•n<n.11t 1he .it1i1,,c rlut The- Wor!i.~pJ~c 1, ,-hown belo,..,

HTtl«m.11 R•1i•i.nn N•«worli. Modal ,
Tf'I• t'#TNO(.

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rle"1 •a1ttt. l'!O •'OGll 1,;..1 tllTl)i.-;p'I ll'!•lfflaJ ,. .111ar,1:• ~1,
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l"'CH.S•~

·• Au,Qt!ed va11.at11•1 ~ d•Ml•,:I 111• q R.4i a!"IO ' 1~~ A'lt '-"U\0¥,n1 Ml 01 ._ O •or • ' " o ~ l'IO(la' poinra

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Tl• 1'11

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ft H•■t rale W. nao. 2. ni: ad•tr\lr n1twi 101.1rc.
'r; I ~ &ii l~ptt'Jh.Jtll C

04 •

11 Heat ta! t \1'1

,,Tharmat AnlltJIICHt

R21

no·••I

Rl2• 1. ,._.,,_ ,

P~1s, ,,.. ~•

Co,wet:t,ol\ tr>"'m.l '•'"•t.lfll:e. KJW, o...iei turl•e•
'ConoUClll;l't lnt,tffllll lHIS.larll:·• tf./w ljli.u
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hi.a 65

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1,. • 'J OOol
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' Thlc1t~na ~ vi.u
Tho-t'Tl•I condu.:tnt•tv w.m I( Q;..LH I tnsiae 1111 ll!lfCMll.l'lllt C

I\> • 30

CottW"aeliol'I cottt,c1on1 w.m '-2 I( 1nr,1t1 lufTae■

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PRORLC\J _l.l
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• Odlf:

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Film-type heater q;
Inside au
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h, & 65 WJm2 • I<

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-I Ambient t'Ut
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rttJ"- lh" r1.tcr11n 1.rm, c.:111,n v. ill 1n-.r-eJ<tC;. 1r11uir1nt 1n~·1r.i..t:tJ hc.n f'\11.\er 10 m,un141n tilt' I 4 t
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PROUL L\I 3.4
f\.'\j()\ \ , ( unnp- ,.,1 J tr,m!<irarent cdm b\ r..sJ1JJ11 11c;.11m,; ~ 11h ,ub,1r,11~ JntJ film \url,1.;( ,uh1C\.·,~J 1, ._"\'"" .h1.·rm.L1-:,,nd1uon-. I· 11' 0. 1.11 Thc-nnAI ~n,ull 111t th1-. ,11u.1u,1n.1t,1 k.uhJ111 hL""JI tlu\, '-I 1\\1m l;1 m:unum t.H,md ..1.1
urmi· 11·mrcr.1..1u1c. T, h, •Comrucc- .mJ plot ll. J~ .. tun\.u,111 ,.11 chi: f11m 1lud.~"'!> k1 n "'L ,. I mm .aml ,J, II 1hl! him a:-.. n,:ii cr.m..pan:nl ,kt~tmm:: 1-1.. rt~um:d 1,, a.. htt''tl" ho"Jmt r,h.,, u·,ull'- ;J, .11tm\11 111
,,1 l
'iCllt.M .\ I It':
L1: 0 25 mm

r,: ao 0 c
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lmt"u•

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.... .. ""~ t?•·~r.1.1.e the abtwe ploc The Wnrl,.,p:k;c is )ho'l4n ht-11\'14 II fl'l.,,,.l ~i•tan-c• *•wot~
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~. o~

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Tti,Cll",_H,_ l"I' ,_.,.
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PROBLE~1 3.5

K..~O\\"N: ~1aximum alluwa.h,1• tc-mpNahHt• ttntl ope-rating cooditioo1 of A rode.et ooulc "all.

FIND: PrcfurPd m:llt1ri1.I: Cu or 30-1 Stainless.

SCIIEMATJC,

Ts.,

,I.,•__,...,....,.._•.I. L

,

.....-,-(opper

•

T.
.,

"

8 1.3K

i i i Tw•.3023K
h•2.xJO•W/m• K. q ;.,-.,.-,-

,1 '

s,,.,,,/ess:

: :

1
j

1 t

9"

:i:.;-sw._d,--,rfs•1 =4Z3K.

7;_ 1 ~125:JK..

~9-"

-•

T~T.~ , 7,.
1/h L/k

ASSUMPTIONS: (1) One-dimcnsion•I condutLlon Lhrouah • plane wall, (2) Stradyst.atc mdilion.s, 13.l Con5L.L-Dl propt-rth.'-5.
= = = PROPERTIES: Tahl, ,1.1, Cu (T (423+~13),r2 618K): r ,033 kg/m'. k ~ = = 37'< \\ m·K; St.St. (30·11 IT 123+1:?!>J /2 83RK): p - 700() •c/m', k - 13.2
W,m·K.

ANALYSJS: Tiu• dtcision <"onceruiug which rnnluial to us.: rn:1y hr madt by fiHt

t"Dmpulin1t lhe thi~knCS5 I.. rcquirrd to in1urc LhAt T,. 1 nmah1!! ~ ltbln thr acctptablr

limit. Tht> !igbtn waU (rorrt!lponding lo tho smallf'r ~:1Jue of L0 1l) would Lheo bt· lht

qbv1ou1 choice..

Appl)•in,t an

f'Ut>r~r balaoct·

Lo Lbe ioaer wurfaee.

•
q,;i,o•

•

Qc• a.ad"

JIN1cr,

b(T., - T, I) - k T,.1 L-T•.•

L -lb-e = TT,.,-I --~TT,~ .,1:

For tht! topper:

L _

376 \\'/m· K
:hd04 \\'fu~'·K

1~13--12311, (:l02J-,l3)K

-_ ...,.31· mm

1I, - t•VJa kg/m' , 0.00134 m) = 20, kg/m'
For tb~ 1-io1t1ltu 1tttl·
L • :?3. :! \\ /ud, 1253-l23)K Do 2< 10' \1'/m, I{ {:m23- I:!Z.3JK • • ·I mm

1L • 17000 kg/m' >< 0.000511 m) • -1.3 kg/m: .

CO.\n..tE~TS: 'fbt' abO\.'l' C'ansidu:itions i~nort' 1ttcngLh f!•quir,·trwhls. y, hieh determint• l ht n inimurn v.a:J thicknen tll'l•dt•d to su~L:i.:n th" nuulr lo:.ds. ~uch re:quirr·menu wo·Jld ;la,·l' to lu- conside:f'd to cornpl!!U'! Llil" design c:ilcuhitions.

l'ROl!U\I .,_.,

rl'wl):

T ., ( ,,n t,uun coe111i.:1cnt lt1r v. Jlcr 0,1\4

-= 17"C, .inJ c-rr•-'t J.(.11.0\: .11cJ v. 11!1 M<!~h.•\ 1inf:

•1nl.tlLI ,111 111 1hc 1n,uluu,m lb ('\,n\·e~1u,n ~fhcicm Im ,ut lln" 1T = 1.!~· ( 1,tnJ ~nor J•"1\.w1,J

~ 11h t,o=i;:li:i:Hn~ cunJuc111 ,n 110J nt,J1at1lln, IL J 1:lle~I CII ,(ln\c:i.:llt '" i.~111.:ll"lll ''" error ,l\.,l.11;.I.U(d \\ Uh.
nct!I ;11n~ct•11Ju.. u1,111nr I • 2",C

, c lll \t.\TII :
(~:r) ___

--------- --- -.-°c h -.;_.- 25 ""T'I"'- - - - - - - - -

L • 10 mtT!

• q cond

lFo11cPOioc.-= 2000 w,m2)

T1 : is;:

21 °c rWaten 126 °c tAin

l

lntutauon

°c Surroumhngs
Tsu,.: 25

(k : o 040 w,m Kt

<---Tb• 2~ac
\ !,' l ,1Pn o , ~: l • S1~J.d~•\l!dC ~It )nc:--d1tnl:!n..111n..il i.:on JW.:tll.)0. CJ C.:in~li.lnl ~

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,.-.. -t. • ~t--... .I; 111 T, T - HT -T,J/L
lkm.c,

r-
h

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T r

.,

"

<

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hi I

r. 111

t(>!T' -·1~ j-L1T T I L
T T
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:1101-IJt, -ff!Ul\\·m IIMJ ~

K 1 j)4~1
l(ij11(

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<

l'ROlll.l \I J.I, rl 11111.1

II conducuon, ud111uon. ur t.:.1.1nJu,uon .,nd rad1iluon o.1u nt!plc.:tcJ the conc)ponJ,n~ \·i.1.Jut'.'> 111 h .1.nJ the pc:rc,ntJ!,!C crrt1r<\ JR' IS !i \Vtm· K 111 tVY> J(i Wlm; K r!O.;\'i 1, .1.nd ~O \\'/m K tJ":' Q~ i

l~' rur J fl)(.Cd \J.lue oi T, = ~7°C'. the condut:UOO I""' rcmJIR'- Ill t:1:~ -: s \\'1m· "hu.;h ,...d-.,1 1hc- l°l\t•J
J,f(ereni:c between P:::r.: ,md q71'M -\hhough 1h1, d1htr~n1.I! ,... fhll di:Jrl\· ,lrn\l.n m th1..· r1ol 11,, lfl ~ h ....
1000 \\'/m· K. 111, rcve,1lc:J 1n 1hc: ,ubrlo, for In$ IOl'l \\"1m: K

J,:·JQ,,")

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f

'•• '""
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/ ., ••••&• "'
i

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:7
- ,7
,. /
'
•• "' "' '"'

Errou ,b\0( 1111ed w 11h ncglccung l'.ttnducuon dt:t:rea~c. wnh 1n1:.re.t-,1n1; h from vuluc:) wh11.•h au , 1grufo:un1 fnr ,m..ill h th< 100 \V/m· KJ u, \"UJu~, wh1,h 1Ji:t ncglip1hlc for l.tr~c h

CO~l~lENTS: In llqutd.s I lurge h>. 11 1s an ext:cl1c;n1 ,1ppru,1.1rm,111on 10 nep-k~l condu.cuon und J'.!111,un·1~ lh,u 311 of lht d1~)1p:ued ro\lr,·Cr 1:11: tmnsierrcd 10 lh( nu1d

PRO l l L E \1 3.7

K.."t0\\1'; \ Ja~'tr of f111ty U.sJiUc wuh taxed 1n~1d~ Lempt'rarurc LJ.n c,pc:ni:ncc dltl~~n1 n111'.\14k convccoon cond.Juon'li

FIND: faJ Rauo of hc:u lo~, (or dlffcn:n1 ..:nn\'1.:cuon condll.ion,. rhl Outc.i ,urt1u:c temJ)<'r-.ttwt' for d1ffc.ren1 l:.Onvecdttn cnndmnn-.. .md ft..• i Tcmper.uure ril ,1111 tur y, h1ch ,11;h1c\'c~ "1111ll"
coohn~ .a, mnqn~ O.Jt c,wnd chW d'fec1 t

SCHEMATIC:

~ L•Q00.3m--l

r..tty

t I T.
I t $t

I I

t 1 s 5 u e - -~

I
~·

I,~9~-➔►

1,.,_,,;i,:c_M::::d::,al con,

Ta.=-ls•c
fif 1,.zsWlm•· ·c,or
fo : 65 W/m• •c

'\SSL \ll'TIO'iS: 1 I I Onc-d,mcnsionJI comJu,uon lhmuph • plane ""ll. , ~ S1catl~ 1.11,
condu1on~. 0) I (omogenenus. medJUrn wuh cunsmn1 prop~mcs. ,-, 1 l\'o lnLcmaJ heat l!(n~rJthin
cmct•boh, effects ,re nc~Lwblcl, c5l .'-cghg1blc nidtntlon cifcc"
PROPERTlES: TahltA •J , T,.,uc. !'at lu)·ct ~:IJ.2 W/rn K
ANALY~lS: The che:rmal c1n·u11 for 1h1~ (IIU3lion J\

Hcncc.1hc heal r,1tc al

=T,, -T. T,1 -T_

<I

R,., .c: l.lkA + In,,\

4.

t\pplytO£ face energy bat.in~,: in 1hc ouier ,urfocc u al!io tolloY•\ 1ha1

Conuout"d ...

PROBLEM 3.7 (Cont. I

Hence.
Lk <T, t - T,.z>: h rr,.= - T_ >

I.
T_ + iii:" T,.t

Ts..l =

1 +I.-
1\L

= To dctc:nnme the wind chill effect, we must dc1cnmnc the heru los, for the wmd,· da,· and u,-e 11
10 cvnlu3tc the hypothcucal ambient temperature. T'.... which would pm,·1d~ thi same heat

loss on a calm dny Hence.

f'mm these rclanons. we can now llnd the result, <ought·

q~ o~~;:K + (15W;m1 K 0.015+0.0154

'tt)
q'"-Ulth

0.003 m
o.~ W/m K

0015--0.0,l

•

<1_::a1m = U.S5~

<l

qwmct,,·

- I5°C ,.

0•2, W/m· K

36°C

(bl

= o.~ T,.:

)

··•-
_..

~~.1•c = !25 W/m· KH0.003 ml
-----....c.,;_;_,;__;_c..:...c..:..;,;__;_.:....._ _

I ...

W/m K

<l

125 W/m• K) CU.003 1111

-15'C +

O 1,\V/m·K

36°C

T, •'

~ _ _ _..t.6.:5.;;W;;/._m.·...K:.t;t;0;;._0;0.;3.;m.=l=='---- = l0.8°C

<l

• )wu,d,

I

0.2 W/m K

+

,

165 W/m· K) !0.CJ0.1m1

(C)

T_ = 36'C-t36+15l'C i0.00}/0.'.!+ l/2Sl =-56.3'C (0 003/0 1 + 1/651

<l

C0\1~1 El\1S; The wand chill effect t> cqut,alcnt 10 a uccrca,c ofT,.1 by I I.Y'C and 1n,Tca-,
m the heat los, t,, a foctorofl0.5S3>"t = I.SI.

PROBLE~I 3,8
5'.."o"·,: Suri1.1cc-mount cnnu~1nr "'-llh rrc-'-tnhed ptt"'t'r 01-.,1pJt1L111 anJ con,,t\.1100 1.:uol!uc
condmon!l,
FNU: l"\1n~ thcnna.J rcusum:t' cm:u1t. an c,rm:,i1on tor the: .:a~ tcmr,er.uurc T'". c.-\ .1lu~11l· IN ~ll~n:uu illf and condutU\·c p.urc- tilJcd gar

PROP~RTIFS: 1Gi,en1 -\,r, , 1., = 0021,~ W/m·K; 1'3,tt ~,.r a 0.1~ W/m K. \lct.,l kads. k. - 25 \V/m·K

!>'P
~h~rc the ahc:nnal rc~1st.ru1cci .lR:. with .A1 =. L,AL: and ..\, : ,.1,1,.
R1 •..,. I !h,\, "" l/ffl W,m! ~I~ •JW; 1411 •m' • h!S.OIv\\'
R - .. I f.111..A...-\. ) = fl1(0 l"l'IJ m/'...S WIm Kn •-0 '15 • 10-- m,' • "'I J ~ ftv'V,

R,•1'-• • IA.1_, A. • Ol • ICC 111/IJ/J~J \\Om M~•J ..1rr•1m1 a ~J7.6 Kl\\

R,.,., • lll..,A. c Of' (1: m/o I! "-·tm Kdu.;,-. U,...,rn:: - 'i~ I K,M,'

from the chcmu.1 cm:uu .and 1hr thcnna.J n:1it\utnce cxpn:won.\, rind

£.,..aq •r.t1-=lT, T...)l~•IT Ti.,,fMt,_, PR,.._.,

T • ' E,.R.,0 + T.. • T" --'R'-.'.-, ,~-- -l/1 I • :-;-:,---R--:_-:::--;c T I

11,1<-,..• ..- IIR.,,. r

(l,R..,.ia • ll'R,""S

Subr.111u1.rng ,.&Jun tor 1hc ~Ul~nant Glf•RcJfl ,ondm('ln, nnd

I ~fl\, ,n-' w,.tt:?5 '1 KJ\\ • :n C. !~C.C-

'115•0 -

T, • - - - - - - I ,..

.

llf.!ll.l- 1/217t,H
L

,. .:1 n r·

tt~Hl

<J

I lf.!ll.l • ••~, nr

<J

PROBLEM 3.9
K., OWN: Length. ,wtocc lhcnnal condrnon,. Jnd rhcmul condu,11,11, 111 u plarc PIJI< m1dpo1n1 1cmpera1urc. FN O: Surface convecuon coefhc1cn1 . SCHEMATIC:

\ SSL"\1PTJO"-S: cI J One-dirnen,ional. ,ready conducuon "ilh no gcncr:11100, c:!I CnnstJOI
pmpcmes

4NAL\'S IS: For prcscnbed conduion,. q'" i\ con)1an1 I fence.

" q wnJ

=

Tr -T, •
I Lr.! )/1..

l~_•C
= ~0 ~ 5 ~ m/~50-c\\c1/-m-,K-

=

J 5(XI

W/m1

"= T,-r_ =

3o•c

=1500\\'/m=

q (L/1;1 • <1/h) (0.0~ + 1/hirnl·K/\\'

<I

n,e COI\I\IE., TS:

cunrnbuuon, <II conJucuun anJ convc,uon m 1hc rmal thcm1al rC\l>IJncc

JfC

R,1,• con- JLk----1).0..' m ~ ,r•v,uHc>
f R';.,..,d = =0.033 m~ Kl\\'.

PROBU:\I J.111

f\1\0\\ "\ ·1micn11on\ "'t .a rhcnn,,J1J111." "ui.tn,, l~Llom .mJ .,mt-111."nl JJJ (t>11d1l1un.,
n,u: .11 He.ii 1i-,.,, 1hrr•ur!h "'111Jl'"' 1l,1 E11~1 N ,.m.1111,n m l•Ut,.,Jc ._1,11\·c,:.11i•n t.lltU1,1.·n1 11,1 .J,,ohl,.
.1.r:1J 1rl1 p.mr 11'>1rn.;u,m

"'\l'll.E\I \ TIC UoubJr Pant:t:

W1noow 08mx05n,

:c~-~i,:· EHJ . . L,0007m '

hT-;:·•0

:.10°C 80 Wlm2•1<

Tw l
..L

-q

h,A

ltHCW~RflES: fohl,--t._t.Ul..uti1iflOt-.1 l =IJ\\',mh. 7111•,',,l.J AtrH=:..S:1\.1 ~-: 111~..;l
\\,111 f,.,

•

r.. ,- l~ ..

·•-- • I f

L I. L

\

h , \ I

• l,

h

I

.:.,fr 1-111 C'1

'I

l "' n -1111 w,,

110,r:m
" IJ\\' Ill I(
10' C'

UUOi'm

•) 1.11rm

-----

" "' " 1JO~ IH\' 1'11

I a I\

lJI" f

1,.1 I\\ '11

"

➔ W1~ 1111(~C-tJ '"'I;;; .. tluJ~~ .. 11 u,1 !_< 11' I\ : I U!l 1-. \\ - ,:,, J \\

<

h f th(' tnrl'-" p.i.n~ \\ 1t1Juy, the ..JJ11l:111.al r,.i.hc- .mJ .au,r:a..c '"'n:J,i: the luWI re 1,1.m,.·c Ho'lu I 11. I N\\ 11> l .,J•t ..._;"' lllc:n-h, rcJu.::1n~ the hi:,1' I,!\, tr0111 ~•J -l t, 1- l W Tltt> tlle"~i -.11 J1 ~,n 1r,c lu~.i.J l~•n
t, r1~111:J .1:o lcillo"',

< •

,",
..

I

/

!•I ,,

"

.ri,-•~

•
u

•
..

.. • V
u

0..... - - - ~«c ......,...

PRORI.I \I J. lfl 1( 11nt.1
1"h.1ll\.!O 1h h 1nrlu1;n..:l' th~ ta,Jt IL1,, JI ,nmll \.1lu~, 1•1 h it,r \,h11:h 1h1,,• 1.•Uhok '-·,,01,a.1n111 rl•..,.,i.mci.• 1,
1h11 lll."L'lh11hl,· rcl.1lf\l! h.1 lhl" t111i.1J r.:w,L.1nu.; Jhmt\·t't th~· rl':,1..1.in..eo ~~c,~~ f1l'!1lt5?1hlt:' \\llh ;11.. rc.a,111t•
h. rJrll,ulart, l•·r th~ mrlt! r~m..- "'"J1•\\ .1nl1.;h.mgc~ 111 la liJ\~ ltttk ~1h.-i..:I c1n the h~.11 k ..-.
t ·o.\l\ll'\ IS: Th1.: lur~-=~• (ontnhutt1.111 ,,~ th~ 1hcm1al rl"~I ,1,m.:e ,.. Jue r, t.:l•hl.lUi.:ll<•n J-.,11._111!1 lhi.' ~n.. l, td ur .'\I.ill" thJ.1 1h1:- Jlr ,.:;,ulJ h< 1n Oh)lh1111ful!' 11 lfi!'~ 1:,111\'l!dll,n -:uucuh, If tht.· ...t,mo•,rL11'hJm:=
..,,ir•.·c,l!iln" ·~·11;.. 1~01 c,..:~l".il!J 1 -( \\'mr ~ thr thi:tn1JI (\"\l'-l,m,.:: \\OUIJ t'il: Ii!!<>"' th.in th.II r1~•.h.:ll"J t"I\
,1uum101 -...'nJui..11cn .i.:tl.•~' 11tJ_i!n.1n1 .11t

PROllLE\1 J.11
KNO\\'N: Willi ('()nurucnon tor p.1.S!,IH~ sol.tr collec1or. Set i.idiauon Huit 10 nnc ,,.1.rh11.:e Amb1cn1 tcmpcrnturcs and com,ccuon t('IC:ffidcnt.s for opposuc surilK'.cs. ~1eh1ng roin1, 1i~1u1eJ
convccunn coeffic1cn1. illld solid lhcnn.U conducm·tt)· of pha.sc chance m:ucnll
FlNP: \.1ch ~pon thidnc,s and sun11ec temperatures..
SCIIE\IA TIC:

\SSlJ!\(P'TJO!liJS: ( I> Onc-d1mcn~1on.tl. 1tcady-,1111e heat tr.Uhler rhrouin the \4.i1U. 1:.1 Vcnic.J '-Clhd- liquid 1mc.n.1ec: 1n thc PC\1. l 3) Nc~hgiblc conducHon re~a,t.1ni:c- m the
,uppomnt;, ,1urfiu,:c,., f,tJ ConM.1111 k,

r • .;':., _._ h, T.1 • h..,T. • ( lr.QJ • :?o,CO • ,lOOO)W'lffl: • "' \'C

'

1h • !\.ii

•20 • IU!Whn..,--K

<I

lien«. from

.nd L, (1U"C/JB\\:m:, lll1m1 tVW •UOl7Mm l

L..=L-L.=0062:m

<I

\ho.

T--T

<I

COM\1E\rS: 01 ~ntc 1hc lov. cncr!!) 1;c,Uccuon dric1cnq u1•q'iA(;a:1=01:t.l1 The
dtic1cnC\ ma~ he in1.71!il!icd b~· JncreJ..,1ng hm u.nd/or dt1.n.:hin~ Trn
t2> The g;c1ual Johd-lit.1urd rnicn::u:c \'-lll nn1 bi( vcruc.11. bu1 \\OUlcJ illtll c.iu•.n1wJ.rtl to 1hc !ell

PROBLEl\-1 3.12

KNO\\'I'•,; Mrucnlll th1ck.nesses an a compos11e wllll con,i,lini or bnck. gloss fit>cr, •emuculltc nnd pane pt111cl IMcr 3nd ouaer convccoon coefficicntS

FIND: Total lhcrmal rCSJStancc and uveraU heat ll1lnsfcr cocffic1cn1.

S CIIE l\,l A T IC:

Glass fiber (Z.Bk9/m3)
Br,,c-;k...,..~1'11"G1ypsum, kgy
Pine penel, kp

1
h;

ASSUJl,IPTIONS: (I l One-dimensional conducuon, (~) Con>1nJ11 propcnits, l ~l l\eghiiblc
con1nc-1 TCSUCQllCC

PROPERTIES: Table A-3, T = 300K Brick. kb= I ~ W/m K: GlJt.~• tibt,r 12& J.gim11.
= k11 = tl.031\ W/m K: Gypsum, k1> ll.17 V.'/m K. Pane panel. J.p = 0.12 W/m·K

.\.NAL YSIS: Constdcnni a unJ1 surface :ire.i. !he 10ml 1hennn.J rcs1&tMce h
R.., =I - +Lh- +L1-t " " L~v- -L-i, + -I
h., kh k11 k,y kp h,

• = [-'- + .!!:!_..,

0.1

.., 0.01

.._ 0.006

+ ...!.,_]

1
mK

Run 70 1.3 0.038 0. 17 0.12 10 W

-R101 = I0.01-11 + fUl71\9.., z.6316 + 0.0588 + 0 05fl0 + 0.1) m,· ·K/\V

- . R,,,. =2.93 m··KJW .

<]

fmm Eq. ~-18 the o•trdU htm tramicr coeffickna is

l! =O.J-11 W/m1·>,. ,

<]

t:O \IM ENTS: A, amic1pa1cd. ahc domanum contnbuuon to ahc 101ul rcsistnnet ,., mur.k hy aht
1osuJa11on

PRORLF.!\1 J.13
"'1'/0\YN: Tiuc,nc~es 01 thn:c m.,u:nah v.hoch lllnn • cnmp<'<Uc 1<,tll and 1hcrmJ, ,:onducuv1Lic~ of two of the. m;ucnah Inner and ou1cr ,urfocc tcmp<"rJlu.rr!- ◄-•I the compm,Hc,
1ho. crmperaru.re and con\Jtcuon rneH1C1cn1 a,,tic1a1cd u. 1th ~d.101nin1 gas.

SCIIE~1AT IC:
T.•, =6oo·c ,;,,,aoo·c
h,25W{mi-K

L., ~0.3m
L,, =L, ~ OJSm
k_. .z.oW/m•K k,Lsow/m·K

\SSmrPTtO"IS: rt) S1eady-,i;i1c condiuon,. (?I Onc-dlmen<ional cnndu,unn. f3l Consuun propcmc, f41 '1c~bj!oblc con1a,1 "'"'tancc. (51 Ncgbg,blc rndmuon dlecis.

\NAL \'SIS: Referring 10 1hc 1hcnnill cm:uu. 1hc hc,u llu, may he •~prcS\ed :i,

T,., -T,_.,

1600-JOl"C

fl3m

O.ISm 015m

~,-- ~ -=-,,.,-.,-

kn

50 W/m h.

4 =

580

\V/m~

1101~+0 15/kn

The hca1 llu, may be obwncd from

q• ~ 51X~l \V/m•·

Suh,111u11ni; for the heat tlu11rom Eq 11) in10 F.<j, 11 J. hnd

Ol.a(S

=

S~O q

-ll.018=

5811
:;ooo

-0.01~=0098

<I

CO.\I\IE:\TS: Rad1011on cff<c1< .ire hi.di 10 ha,c, \l~noriconi tnlluence "" tho ncl he•• ttu, :u the innc-r ~urface flf the O\'C.n

l'HOIILL\I J. IJ

H NU: H.:..ahl

',l"III \I \TI C:

lea••••• I

fc-,.,,

L.p L, L•v
~ I- 5'I """ -I ~ 10 """

lg

+·~ l-o "'• ·o
3 mm -1---i-'5 mm

© ~F,\fLll ~~

u,ettian• to.am di

WOoa,\..... 1'1 : 5 Wtrn2. K
r., •10CIC

Glau11;p

G•u• 1g1

1r.10.

=

, swrm.'.-K •• ,sec

0

• •
Ci1a.u (iii+

1-'KUl'l RTIL'· , I :i JOO I\ f.;bJ,, \ f rJ;htL"r l'>\l~HI. I. -= o 1., \\ 'm 1' ilf t'ID.&nc ~. - 11 UZt, \\ Iii I'\

"" ~ J ~

, ... \\ ·rn K a:iau. L.1 • I ~ \\ 1m ~ f.lf,J,· \.: .,,, ~.: ll u:ti.\ \\ 1m "'-

\ , \ L, !,~ : 1J Th~ he.ii Im~ ttu\ I'<' l•h1.11n\'J hir J1\ Idint 1hc- ••,~r.tll tcmrculurr \J1lli:rcn..~ 1,\ thi

•t;

c- h m-.il ri , , ..,.m..

f-Ortnc comr,,~11.: wull cil umt ,,u1J.:c m.•., . .\-;: I m·

T. T. [,1h1 (L,l,,f•<L lj-11.. I l•tihlj\
10 C.: I 1, C/
1 •1,·,-,.-,-.-,-,,-,-,,-,-.-,-,-,~-.•HlSl .. 1JtJt,f111,· K wJ Im

q- !

"r
BK\\

:

1~1·1\\

<

•t· I w 1h1 un~1- r:1ne 11( i;11,~

T -T
,;·,-h-,-,L. I. ~ II h •I \

~,·c
'I

1'0 "'

<

•~ 1 b1r 1t,-t J·1ut,1i: p.tnt' \' 1ml11\l,
r. - r.
·[,-,-h---=-11-.-,., I Il., l I· 11 h 1] ,\

T :J.41i
IH_IIJ◄ ..,lll•lU-!'!lh~~ \\'rim·

if, J \\

<

l '0\1\H-' 1,; "'he,;, mpMUt W..JJl 111, h:.arl:, ~ur-.·1101 uorn 1hc ,.1.1nJp-,1n1 I 1~u1. 111!" M..Jr 1~--,,1 ,inJ th!"
J,,,n,111,, "mnt,u11an tu 11411'1,II therm.al re,h1,1n..C' 111'.~'".>1 li..Jn1."-IJll!'d ,1.·nh lh~ h,.an m,ul,1,l11 1 f '"1"0
••Iii 1J1 Iii' 1 ,n • n~tru~ur111. heal hh!t 1hrc,111:;h tl1i• • ln,tP" 1~ f\Jl!mltt:.i.nlh ~~i•t lh,Ut 1h.,111nr ·hr
..cmpoi.1h, " .i. ll

PROBLE\I 3. 1.5

K.,"'0\VN: Cornpo,1cc wall of a hOUloC w1th pre~bcd co"vecno,, {'rtX:t'.,iCl ilt mncr 1111d nu1cr ,urt'ilcc,
FL\ u~ 1ill Expre-s.s1on for thcrm.;al rc~1s1.mcc of houc\e. wall~. R.._,1: lb\ Tom.J hcou lo,,. 1.11 \\",. 1c1
Effi~ct M he.at Joss due 10 HK'l'l:A\C In out~1dc hc-.1t transfer con'-·c..:unn ntt:ftkJ!!nt, hJJ . .and 1d,
Cont.n,llinp rc:.s1sta.ncc tor htat l(h\ Irum house

SCHEMATIC:

A 350m•
A,,30W/m• K
r; ,io·c t t t

F.ber9l8S$ blan kc t (2.BJ..gim~), k1,
r-,.--,e,-"f~ .......-- --Plywood 51d,n9, ks
t t tlio• 60W/m--K, T., -15°C

Lp•lOmm I• ,l,Li,,JOO..,~• 1

t Onc-d.lmcn.-.1on2J tc:1nducunn. Su:ady~'.'-liifC ,.-,mdH101v•• 1',:c~httblc

\SSL:MPTt o

,

s:

1

\

12l

r

'\

PROPERTIES: Tublc A J, 11' • 1T, • T0 l/l = 1lO-IS1"C-:! : :.S'C = •110K1· !'1bcrit•»
blankc1. l8 kg1m· k, ~ 0.0,8 W/m· K: Plywood <idmg. k, • 0 11 \\'/m K: Plo\lorbn.1rd •r = 0.17 W/mK

A'IAL \'SIS: 1.11 The c,pna.!imn for the 101.al thermal rc~t\1.Uwc ot the hnu~ wjtl, tollo~·.s from E4 i 18

II,. • _I • _Ii • ~ . ~ __I__
h,A L,A "-" I., A h,,A

<J

lbJ The toulJ heat IO!<,.S ctuouµh the hou,c v.alh; i,.

.. •~TIR..,•IT1 r.li"R,.• Sub~tuuung numcncal value-), hnd

. - - - - ~ 0.02m -

I

(U'!Wlm 1"11350m 6'.1\\-tm' K~1~r,m'

ll.., I~ ~! •l6 It •i~ --11 h ....: 76 t'l"lffhCI'\,\ ..,JI le w ➔ ~r!\\

The hc.u loss 1, 1hr:n.

q •r.t' i-15_11•otn1-.1w1 c-i.v. J.11 ~w

<)

ti:) lf ho tU;ln~e, from tiO tu JOO \\Jim: K. then R..,-= 1/lt.,-\ cha.n&;c."~ rn.1m ..t /6, 111 ' Cf\\ 1,1
=0.9.S•Hr~-CJ\',' Thi, ttduccs R,m 11, li2tl"'H1_.,.Ci\\' \l,.·hu,:h i\ -0.S'r tli!(tto-.r m R1o11 or O5'i- mcrc3~ in q

tdJ Frnm 1hc R1t1, nulTlC'n~aJ c'C.plb)tOn ln p.in ,b1, ni-,1c 1ha1 the an:tuldllun rc!>l\t.J.nc:c. Li,/l.1-.:\. 1,;
752/830 !!': 9ni;;. of the l.0141 ru,i.-.r.inc~ Ht:ni:t, thi\ tt1J1cnal layer comml\ the ru1~11;1nn· of thr wall. From pan 11.:i note rh.11 .1 ~-fold Ucc:.rt'.L\C 1n tC\1\l.ut("c due tu wmd vclL-1'.'UY 1nt.1c-J..,c h.i, .1
ncgl11,?1hh: c:.ffc,.,, on the hcJt Im,.

PRORLE~1 3.16
KNO \\'N: Composu~ wall of A house with prcscnbcd con\.ec.:Uon ptl).;c~.., at rnni:r .uu.l 11Ull'f 1wfacc1.
FIND: Daily heat tou for prescnbed dtum:tJ vanaoon in amb1cn1 .ur lcmpcr:1turt:

ASSUMPTIONS: (I) Onc-d1rncnsional. stcady-,...tc conduc11on (ncth£1hlc change 1n s.all
1.hcnnn.l energy uorngc over 2-ih pc.nodl, 12) l\egli,ibJc conU'lct rc!'.utance

PROPERTIES: Tobt, A·J T = JOOK F1bcrgl•"· blanket tlS k!!lm1 . •• ~ o .n,, Wim .:.
Ply"-ood. i., = 0 12 W/m K; Pla.1crbo>rd. •e • 0 17 W/m·O:

M A L \ ' SIS :

"Inc heat loss may

be

•ppro,muucd

Ill

Q

=

:areti
6

T -

,R-~T, ...

_I) 1 ~ ~ R"' = _I [-I +

..'::!?_ "' +

' A h ; kp kh k , h ,

I[ I

O.Olm

O.lm

0.02m

R,.. • 200m: 30W/m' K OJ7WlmK 0038W/m·K O i2W/m·K

R.., = O.OIJS-1 K/W . Hence: the heat r11tc lS

Wh

Q= 16.18 kW•h = I )Ol,Hi"J

<l

C0\1M"E1'TS: From k.oo,,,lcdgc nf the fuel M:"it , rhe 1nt.1I d~ll~ hc,i1Ung hill l.°tiuld h(
,:lc1crmu,cd for c<Mlplc. •• • CO<t or0 .1OS/I.Wcdoth. the he;111ng bill would be S1 ~2/dJ\

PROBLEM 3.17

KNOWN: Dimcrmons ilnd mntcnah aS..\OCl!Utd \vnh a compn\11.t. w:.tU f~...Sm), o..5m. 10 ,twh each 2.Sm high).

FND: Wall lhcnn:u n:sc,uncc

.. . SCI-IE.MATIC:

1:·• -

-0-.&-!-i - I

Insulst,on

X I

I

' GIBss Fiber,

I

- ptJper faced (DJ I

t

_IJO,,,m•L•
J

L0

-- - . (28kg/mJ)

--

I
I I
'

I,,,.-,HBrdwood s,dmg (A)

:E--8,,,m =L,.
-"1 0mm
J..lardwood (8)

•

f-/:.-12.mm• L,
Gyp•um (C)

.\SSU,1l'TIONS: t I J S1ead•-'1••• condmon,. (2) Tcmperamn, nf <nmpo,uc depend< only on
x hurf11ccs n('lnrull co ,.: are I\Olhtnn.tll, 0) Consu:mt propemc,:,, t·H Kc~Jg.Jblc 1-:ont.a~t
~Sl$f,1Jh.;C

PROl'ERllES: Titbit ,\..i IT= :100K ): Hanlv.ood >id1np. , , ~Otl'l-1 \\"/rn f,; : IIJUd1Anod.
ke = 0.16 W/m K. Gvpsum. kc= 0 17 \V/m·K. lnwl•tinn Lglu, llbcr paper faced . ~R k(1/111' },
ko c 00~8 W/m K

ANALVS IS: U1,mg the i.!othcrmill surface a.ssumpoon, Ult lhcnnaJ cu·cun :U.)QClatcd wnh a
,,n~Je um1 Ccnclo<cd by d"'hcd bnc,i of1hc v.llll 1>
L. fAaA•

The cqu1v,llcn1 rri,i.11.mtc of the cort' 1,
= R., rllR11 - I/Rol"1 = rl/R 1:; + 1/2.14)1 I = l.75~ K,,v
and the ,oul um1 rt._'iismnce 1~
R.,, 1 - R, R,q ~ R,- = 1.115J 1(/\1'
Wi1h IO ..uch unit~ m p,.1.nt.llrl. ihc LutuJ w.1ll re"i!-t;mcc h
<J
C:OM)IEr-.'TS: If ,urbt·cs parlllkl 101hr heal ftov. dnecuon .ue a,,um•d a<habam· 1hc 1hcrm.tl
l.:LKUH Md the v'1luc ol Rint will chtft':r

l'ltllllLf.\1 J. LK

k. '\U\\ , i. onJ111un-. .1...., ,,.:li.lttJ " uh m.un1.un1nl-! hc.lt~.J .111J c~1icd ,.('11Juwn ~ \I. ulun
11mp1M1 Ill
n , o : L ~'lfl..:i:nl 1.1J pL·r11.1mu11~:( CUI' J

rl"lfl!!t't 11 {"It

,CIIFM \ rlC":

-
f'I

.. a

=2 !JO

0w°,,c.,,..,.

• ~

'
- l'
Ele:::t11e • heaier

Case 1a,

(

,_, • 90 °c
••
T _ 0 ■ 25°C
C

Ca-: IOI

-

• -

-
•

Pluggec
W,n = 125 000 J
.it-= 1Z h

\SSL \IJYflO:"\S: Sti:.1d\ -~l..1tc- t'f!Crallm.· cc11J1111nh. , :! I \.:rph_t,'.1bl~ r;uhJtton l I C..\.1mp,mm 1,, ,,mrkt.el· "-·~1ieJ ltom .,mb,eni :.an

.\, \I \ \lS: l 11 ~ C.J,r t .& ft' \pt:nlnl.'nl 1,. pn'lonn~ 1, ► Ji:ta:nmnc.- the f.l\ c.-rall 1hem1.d ,c.-,,,.t.~n-.~· 10 h~.Jt r.w,-1,·r h '1'41 ,r11h.: 1ntc1 lllf ill 1hc rdtlF,c."tJIN Jnd lhc< .atnbt~nl ,UI ·\prl~ in~· tn l.'RC'fL\' t,.11,nk.L" l.> l ,,mi ,1 un.1 Jt,.:11111~ t.:U•~t.aror II h,llc•\\I 1rcim bl I 11.t tlut .al Jn\ m,1.ant

r, f. - (I

~ft'n.;t
" 'I" 'I~
"'h~tl!'~ .. IT. T. 1/R.

Jt toll\l\\-. 1lul

:nv. R T

T

l l)fJ ~ii C
:

l:._"i'(,\\

q..,

l or L.i.1.C' •~! hC'at 1r.1n.,11:1 Jr••m lhe .1mh1cnt ,IH 101111." ~ ,mr,1mm•n1 uh,.~ h,:~1 lu.1J• "h.Jl.10.:~J t-i\ hc.11 u.m~Ji:r 11 11~ r.!rn~..-r.1n1 •q c •1•-.i' IJcn1.t" thC' thC'ntul en~,~~ u.1,n,hm~·J huin 1hc rclriJl'i::r.tWt n;r lhc 11 h•mr rt!n11J 1"'
T -T
q ..J.1;;;:;q,11.1,= ~•R • ~\'
'
iJ.,.,,

:t,bJ>OII I IJJ
I!~ IN.ill

<

t'UI\C\ll \l'\: The 1dcJI 1C.irn •II (.t;f' 1..

nP

;7K~ ~ JlO
r,. - r 121}ij :~~11'

1', u " '\. 11.1.i.l il(k'!f ,i,p,,.,.c :.1nit \(t11l·.1.I ,h..,tJn..c f:\etv.e>t>n llo1:1n r,:,r .t hlUJ.te~ 11.at h~I ~11Wmt

•• , n

r,rr~''"'"

l..if v.:1J1h

\'I

bullJinr: "'hlth

m1n1m1te) htJl lt1'~. tfil

.anJ W1J1h

11urn~r (II

lh.1tlr,,

~hr ·h nut!-•1111:e hc.JJ los, tor a rt,e'°'nh.:J f 1'Yl•1 ..r,1..'t' .and J1•1.1ncc hctwttn 1looJ1 C(lni:~1"-111J111c 11('.1'
roar" r ,._ 11t h.:..tt I~ rt·Ju.. t,on ll'QfTI :

,c llf \1 \ 1H':

At• 32. 768 m2

.q
•

• ••

•

•A -
w2

1a1

b)

\..;Sl \JP110,,; '"lhi;1hlc h<Jl lu), 10 prnunJ
\'-\I, Sl1<i: 1.11 rl nm11m1.r:e 1h.:- h.:-4t lu,, l.j 1ht'c-,1en.., t-utiJ..:t' jrt., ~. mu.., ti( m1mm11cJ I 1,>m h,r ,.. ,

JA
,J \\
' ,,

\\

1!:\ H 11

<

n~ ,· mri.·1111S' 1.•ltcc:1\ ,,1 \\ \;'n thl' .ir~il.." f thl: roc,1 ..mJ ..1Jt\l.JII\, ,mJ hrru.;~• the t--..01, h•r ,m '-'Jlllm 1111 1•
11h,1~n k.ht1n.,11~·.1th III Fit hJ

lq I •1 \ 1 - ~., °'(J~ Ill JIil.i fl : J ffl

.' \\ 1! • ,, -,,~m J m) f\.J In

<

Hnic.._

'

\. II

. 1~ "Tr,l'I 111

~

fn-l m 1

<

1 I \ 1T IWf1T1 I( ({!.1 m I

- :~·r _ J • 1! -~,:-.m • •l In •

lu7 ~00\\

t'l.ltn

<

PROBLE:\13.19 IC'unt.!

'

c~ reJuc11011 In y :

12.IJ<K)

J07,2UOll5 12,l)(Kl : 41l'l\

512.00()\\'

<

PROBLE\>1 3.20
KNOW": Mntcnah ~i.nd d1mcns1on" 01 1. compo,su~ Wi.lll ~epi!T3tmg a cornbu-.uon ~.u- Imm J liquid Ca<tl::tnL
FlNO: (al llc.u 1Di5 per unn area. and tb) Tcmpcralurc tfo,.mhuucm
SCfli,:MATIC:

ASSL \1PTIOSS: Cl l Qnc.duncn...innal heal IJ11.D,ikr. t'.!) Stc-'<Sy-tiGUe i:ondJt1on,, t~) Cun1itan1 propcruci t4) ~cghgiblc r:adi:uion cff~cts.

PROPERT1t:S: Tuhk ~. I Sa S1. 131\J1(t = lllOOl.1· k: ?.5.J W/m K. Tahir A!, BcryUiwn
O.ide rT = IIOOK); l - !1.5 W/m K.

ANAL\'SIS: fa) The dc~lred hc.u. flux m.:iy ht- c<tprc,\ed a-.

q

: - ~T-• -1--T.~ : - -

-~ h1

L-~•~-,.R

. '

-

~
kg

-

'h:

~'.-.!_-0-01-~r~ 2O60S0-+-IHI-OJ0!-1•c-

~-
m'

Sil !1.5 • ~SA ll~l(I \\

-
K

'

q = .14,1,(JO W/m'

<J

lhl The ct)mposne ~url.tL-e t~mpcr.uu:rc!a- nu~, be obi.uncd h} appl~·mg w.ppropru11.c rtll\: equ.:mnni r-rum the ta.ct 1h1u q :: h 1 (T-..1 - T~. 1). u toJ111w\ 1h.11

T 1=T. - q" =:!600'C- .lJ,(,Ofl WJm' a loO~'C-

•

·' h,

SIi Wim' K

\\,.11.h q.. -= 1L."'1LAl lT,.1 -T«.1).11 =u,ofoUow\ thtn

~ i•=••c- Tr.l =T t.l -

L,.q" i...

11..,,

~,n

•

-

_11_111_..,m.,.•..,J-1.., ..,(l()()-'-.,.W>-'/_m1_ ?l.,.Ci:\V/m~

-_

l"ft-'•C' fP)'.

S1mJIMI~·. w1t.h q = ITr., l -Tc.~ 1/R,~<•

T,.: = T, 1 - Ru q = l892'C - 0.05 "'_~ K •3J,60:J 11'. = 1111'('

n

m-

Conunucd ..

PROBLEM J.20 (Con1.J

= nnd wnh q• l ke/Le )(7',.z - T,.z ,.

•

T 2 =T . - L9q "162•c- 11.02mx34.600 W/m· "I 34.6'C .

'·

•• J..8

25.4 W/m K

The temperatun: distribution i~ therefore of the followin~ form:

Tm_, ,2600:C_/
I / Ts.i •1908 C
r;, · =1892. c-

r - ~-• slbZ °C / ~.• : /31-.6 ••C T..,, =100 C

<J

COl'-11\1Ei'<TS: ( 11 The calculations may be checked by rccompuung q from

q

= h2(T,.? -

T-.?) =

,

,

IOOOW/m· K(l J:J 6-100)°C = ~4.tiOOW/m·

12) The mtcial es11ma1es of the me-.m mntenal tempcrntures are ,n error. pamcularly ior the
slllllllcss steel. For improved accuracy the calculnuons should be n:pcatecl u,mg k v,tlucs
corrcspondtng to T = I 900°C for the ox,de and T :c I 15°C for the steel

. . 13) The mnjor eontnbuuon, to the total rcs,stnncc nrc made by the rombu~ttnn l!.!\ hnumL.tn
layer antl the conmtt. where the tempcromrc drop, arc largest.

PROBLEM 3.21
f<"0W1'': Th1d..ni:,~. C')\ientlJ 1c.mpcr:1wre Wfftrc.m:c, 1md prcs,urc tor two ,tamlc\, ~1crl plates.
FIND: I.ti lleJt Rux nnd lb) Com.1c1 pl.me 1<mpcr:11un: drop.
<;CJIEM \ T I C:

0 Olm -ij•-~-j'l-•~-.j1-o OJ m

T,,
L
k

J... ~9.,,.•.,
k

!}--t ---Cantacl pr~ssure 1 bar
T, I -T. 4 :/Oo•c
T,•,
'--'-Stain/~:.• steel

ASSUMPTI ONS: 10 One-dnn<:n<aC>nol hca11rnn,icr (lJ S1<ady-<1a1c cand111on<.1'.11 Con,1.1n1 rropcmes

PROPERTCES: Tah/c A-I, S1:unle.s S1cel tT= -lOOK): k = lo.b W/m K.

ANALYSJS: 1>1 Wilh R., = I 5xlll-' m' ',;JV,' fmrn Tobk 3.1 and

iL:-=

O.Olm
16.6\V/m·K

=

6

Q?
• ~,

JO-'

:,KJW

,n

'

II tollow, 1ha1

he.nee
<]
(b) Frr>m 1he 1hcnnal rucu,1

C0~1T\-fE1'TS: The conmct rcs.islilncr Js sign1ric:i.n1 rcllm\'c 10 the condur:u()n n:si,1:am.a.·~ The vt\luc or RL -.ould d1mm1.sh. hOWC\'er. wnh mcre-:t\ing pn:~sure.

PROBLEM 3.22
KNOiVN: Tcmpc:ra1urcs nnd convecuon coctfictents assoc101cd wuh t1wd, ,. mner 3nd ,,u1er surfates of ::i composite wa.lJ Contac1 rcsi~1nnce dimensions. and thcmmJ condueu\'1ll~t. 11Ssocia1ed with wall mrucnals.
FO-.D: en) Ra.1e of heat 1ransfer u,rou1th the wall. (bl Ttmper.uure d1<1I1huuo11
SCHE\1ATIC:

ASSUMPTIONS: (1) S1eod)•-s1me condmon,. r::?) One-dimcns1onal hea1 ITTJnstcr. ,,, Ne~ligible rndsauon. (4) Cons1nm properucs.

ANALYSIS: (nJ Calcula1c tJ,c 101:ll rcsis1nnc:,: 10 hnd 1J1e heat rn1e.

I

L,.

Rtot = - - + -

➔• Rt.c: •

Le
-

+ -I -

h1A k,.A

k9A h2A

R101=

I 0.0 I 0.3

[

-+..
10><5

.

;

..
O.

.;.
lxS

.

_

+5 -

+

0.02 0.04,5

...

I !Ox5

] K
-
W

R101 = (0.02 + U.02 + O.Oo + 0.10 + 0.0 llwK = 0.21 wK

= q

T•.,-T• .z
Rio,

=

1200-10)°C 0.21 K/W

=

76,-

\\'

rb) It lollow, tha1

T'· 1 =T-.I - _h19A_ = 200"C - 5706W2 W/K -- I.",.•.,8°C

TA= T, 1- -qL,. = 18-1.s•c- -7=62-W=-,U=.0=1:rn:.= 169.6°c

• kAA

H 1~ , ~m:

m·K

Ta =TA - 4Ri_, = l69.6"C- 762W,ll 06¾ = 1!3.8°C

·•c T '

•
·

=

T 8

-

qL9 = I'3 s•c - 762W,O.O!m = J7

ks A - '

0 . 0 4W- , S m ·,

·"

•

m·K

T., =T,, - _9_ = 47 .6°C - 762W = 41l"C

•• ·• h1A

IOOW/K

<J
A B
r,

l' RllllLI \I .1..!J
~ NU\\ , . ( Jucrr unJ 1nnrr ,uri.l:"r 1,,nn,'-"~hl'II ..·,111Jn1cn, a,..,111.·1.11~J \\dh 1m;on11.1 i.c.•.ak'J ln,;,,nd
'1lrtnr,. bbJ" n11d.11c,,c" 1ticmu.l "t'Otlu.:uvmt"~...mJ 1111-:n.1...-w1 f l•,1,tJ11~t' ,,1 1hc hlJJ,: mJ.t~n.1I-.
\t.u1mum .1J11m.1blt u~mrcr.uurc ,,I ln1..c.,11d
Flt\ I>: ,\·hrthi.-r hlo1Jc LipaJrt", h<IL'" m:l\1mum h:tnp~t.nu,~ l ~mrt'ra1urt Jh1nt,u11un 111 hl.uk. "1111
J ltd ",1huu1 fhl" ( BC

111

i i I

,h • 500 Wlm2. K
T, 1• 4D0K

Zreon,a

R·,c'-'10...¢m~ K W

• ~ 1 l WrrrH<

ln.col"iel tt:25Wlml<
Tm••= ,isoK

\SSl \IP11 0 , s. , , Unt•Jtrm'fhlURJ;I. ·•tc.uh •"lf,HC t1,nJuu1on .tll J ~0111r\holl.: pl.lilt.! ..., JU, 1.:1 C ,m~um rt•'f'\; ·11 • ' "lit, thtiblc r.a.d1.a11un

R ,. h ~ILl~•,,-R," •(I.J~I, ·h 1 R". (1<1•-,~;.111-•.1n_J-::.ur•-:!• IO"')m E-::\\'=.,.t,9 .. Jfl"'m• K' \\

T •

W11hrn11 :h:: fBC

R,~ •

ah· ·ll/~I1, , h

1l~Ull1',1 •:O.•lfl m Iv\\"; -l l>h• Iii" \\'Im

,lit" lhc-11

-1

T

r. r

T,.J/R: . ;

PROIILl\I J.13 1(11111.1

'"'°

!

,.;:~

~ •

ti!:~.'0

-i.
t

I !ll()

... .... .... 11"0

• 110()

..,,,

,

0""'

,ac -

W,inf8C

--A-" Y.,1nou1

C0'.\1\IE,TS: Since 1hc durJbiht~ l'I 1hc TBC dccri:,~, v.11h m.. rca..,m!_! tcmpr:r.ttur~.. wh1i:h 111..rl.'.b'-"' •,, 1th 111 t~il'.'tllllZ 1h1.d, n.:,-.. hmu<ri. 111 tht! th1d.nc,:1, .Jre ,l\,th,.:liltcd "1th n:hJbllll~ 1..nn,1Ui:u11nrh

PROBLEM 3.24
KNO\VN: Surface area and m3X1mum 1empcn11un, of a chip. Thickness 01 nlummum co,er and chip/cover contact resistance. Fluid convccuon conilition,
FINO: MllX1mum chip power.
SCH!c.\1ATIC:

•9.c

'----------- --

-----~

+-R; '

\SSU\IPTI ONS: (IJ S1c•d,-,1a1e cond1uon,, (2) One-cflmcns1onal heat tr:insfrr. 01 Scgh~tble heat lo~ from >1dcs and bouom, 14) Chip 1> 1sothcnnnJ
PROPERTlES: Table A I , ,\lummum <T = 325 Ki: k ~ 238 W/m K.
ANALYSIS: For a conuol rurfocc about the chap. conservauon of energ) yield• E1 -E,,.,=0
or

p ...... : 5.7 w

<J

COa\l a\lE',TS: The dom111am resi~tnnce 1, 1ha1 due 10con,·ecuon I Re,.,, > R1e > R,_, I

l'ROIIII \I .l..!S
~,o\\ ,: • ,~rJunr conJnmrn tc.•r .J f'u.irJ mnuntc-J .:lur Fl'.'-0: ., t:qu1\'11knt thcmtJI i.;m::utL It'll Chip 1i:mpt<"rJ\\Jr1.: h, I ~1.i,1mum JIIO\\llt,lc h-:.u d1'i.)11'JIO'n I •r
r:: •hd~~ln>- lu.1u1J I h .;;; 1000 \\ 1m 1 J11d air th .:- Ilk.I \\1m· K rffc-..1ur chJn~~ u1 t1ri..uu ho,1rJ
1 ·mrcrJ!Urt'. .anJ com~1.;t rt''\U,tJn;.:~
',Ll{f \1-\1 ll 1

\.~~I'\ IPTICl',S: tht'rrr K" ,t..mi:c

r Str.tJ)' 0 ,u1e ,onJuaons,. 41, 1r1n('-dun.:n..wnJI \.unJu,;hL•n, t 1I ',t:II111I 11..: , 111
~c~1Js1t-ik r,1Jl,Hum. \ 'i\ Cun,t.tnl rr.:,r,entej

PROrl.RT11-.S; rdhtr ~• l ,.\lummun, 0\ldt' lf!IJl~,n·,1.,Um~ '\~8 h.1 ~.: 1~ ~\\Im~

\',\I\ SI~; ,,

- r...

q•,

1lh1

l R·,. .11h. q<

I IJ I .\rrl\ lrl!' ~on-..en·Jt11JR ut cnerJ!} to d u1nuol ~ur1JCt' .lh1,ut lht' '-.htr ( J-

l - T..

T - T...

' 1 ~ I h, • IL,•l, • R'. • I h,

\\ ,,,, 4 :s •• w•\\'1m h !:. 10()1) \\"tm K, ~ - I \V,n1 I.: unJ R~ JIJ"tm· 1'/\V

r :11·c

. - T -::.,re

I I .:,, • r),f111~ I• Ill~ /m i,; \\' 1IIOl"I"' K \\

,.,,~1 ,. 10'\l/m'

-er,J-1'.l(J()T ::.0,11,JO/\V/,11· K

<

~.. ,,7 ihOW/m

<

\\.·1thq :5fl).CX'l!i\\1m anJq: i;~ll-0\\'/m Rc:plJi:in~lh~d1eltcim.·'A11huir1h,cl0!1\\ 111 i,.., ti
(11110" •"F rt,uh ~ .arc l"'hlamr:J for «J1 I11.•,..·m ~l1mt,m,n1on, ,,f ~,. .mJ R"

Cununu.:J

PROULl::M 3.25 I ·ont.1

" I ~. I\\ Im· K I R• till \\

1.f" f \\'/111 I

q 1\~Jm

q" (Win, l

I

10-

115'!

r,~()11

S{),5lJ

I .3:=.~ I

10'
ur-'

~57~
~It,<,

1,5()11
65UU

'107 J
SM,f,

<

3~ ..i

IIJ 1 I 25!i1

1>5(kl

110~ l

t u,1,11-:,T,: I re>r lht ,11nJ11111n" l'tl pan 1t,,. t.h~ IOIJI 1nti:n1JI r~~htani.:1! IS U.IJ.\01 111 1-\.,1\\' \\lllk
h • ,ut ·r ri..: .1~1.1n-e-c-1~ 1111(11 rr Iv\\ H~ni.:1.·
-~ - -fT-- -T.-)-R-~ • l-l-IJ-.1,0-1 - '.*'
•I If r.,)R" <1CXJI
.iru.J onl~ .1ppr,1\lm,1tcl~· Vt 01 thi: hi:Jt 1, d1,,1pJll!d thmu~ lhe" t,l,arJ
:. \\'llh h ":' l' ij} \\ m1 h. Lhl! L,uti:r ri..•~1\l111t1.:i;: 1m,.-rcJ,t:., hl I) OI m • K '\\ m "h1ch ...1,1,; ~ 1• "1.. W' It .. 11 ti~l11.11t 11) c ' I .anJ no~ .ilm,,~1 ~~r;. 111 thl• hL'-il :~ d1,,1p.ui:J through 1ht" i,,,.ud H1..·ncl!' ,lhhvu_!!h
rnl.'a,urt>"' u, rcdUL'C R'" wouh.l h,n·l~ ., nt!p-h~1bt~ clfrLt 11n 1t lor th.: h'-!unJ l..'OHl,tnl. h.'mt unprmL'mc:nl
ma, h,: c.itncd h1r mr~1:o,.1lcd i:unJH1on, /h ~hu\\fl in lht: l.Jhl.: 11f p.trt rt,). U>e nf ~n .Jummum ,•\!Jc
boJrJ1ni.:tc,1..eq, t,) f4~ 1fn11n~l59hJ!57.: \V,m lh~ rct.lucm~ R" trnmUO~OI coU,O~Slm· K.:'\1,
8~.11.Jsc 1hc m11u.1r ~on1.u.:1 rc,1,1.1n,c, R:' 10·~ "1 K f\\' 1'"' alrc.id\ mu1,:h le,, 1han R, . Jfl\ r"-'Llu.:uur: m t1; \ .1Jui.: w,,ulJ h.1\~ "' ncghg1l:,I~ L"itec1 tin q'.• Th'-" 1arti?('S1 l!Jtn "-11Uld bc.> rcJhzcd h~ in..:r~a,an~ h,,
,m,:(' lhc in,1Jc, 111,t'\'ll•'n rtt..,1st.i,11:c mah:, th.: dlm11n.rn1 .:1+ntnb1.1t1on to 1)11! l\il,11 mlcr11;.1J 11.-~1~tu11cc

PROBLEM 3.26
KN O \\-'N: Conducoon m i,1 camcnl \CCoon with pn:scnbed drn.metcr. D. il\ a 1uncuun t1I \ m
the form D = ox 1ll
FIND: l•J Tempcr:uurc dJsu-ibution. T(,1. 1bl Hent uan,kr rate. 4., S CH EMATIC:
T,:bOOK.

\SSUl\fPTIONS: Cl l S1cady,5101e condmons, C:?l Onc-cumcnsion:il c<1nducuon in ,.w.n,cuon, (31 No ,mcrnal hcot gcncrauon. !-1) Consto.nt propcruc,.
PROPERTIES; Tobit,\ ·l, Pure Alummum (500Kl k =!36 \\'/m·I-..

ANAi.\'SIS: fol Bnscd upon the :usumpoon,. and followmg th< ,amc mclhodolu~ nl

or,. Exnmplc 3.~. q, Is• con,1nn1 mdcpcndcm

Accordingl),

q, = -kJ\ ddxT = -kl~Cnx 11:: ll /4] ~ dT

1l l

usirli, A= 1t01/4 where D = .,.,r. Sep:ir.aung van•blcs and 1dc1111fy1ng hm1«,

lme/!l,lt1n¥ ond solving forT(x1 wid then for T::.

Sohing E.q. 1,1 l for tt, ond then <11b,urnung into Eq. I3l ~ives the result,.

'!. =-47t ,·'k rT1-T1 1/ln 1,1/,1 1
Tl•l=T1 + 1T,-T,) -In-r-x't,-1,• In (x 1/•::J
From £4, Ill note lhar !dT/dx1·x = Con\lan1
J.bovc.

<I
It follnw< that T(x) ha~ the d.Jsmbuuun ;hawn

(hi 11ie hc31 mtc follow, Imm Eq (51

<I

PROBLEM J.27
KSOWN: Geome~· md ~umcc:: condinon\ of a D'\lncted sohd 1.:une
Fl'ID: ta) Tempcnature dtstnbuuon. lbl R,ce of he>1 u-.w,fer ocro<.< the tone
SCHEMATIC:
"!.,,.-..-t.,.-.x.-1=• 0-. 0=7.S-m, - x
Alum,n u m - --s'}'-y ,D,.,xi -- -i'I'--=-"
~. zo·c-c~::...: "'- - .L
\SSLIMPTIOI\S: I 11 Steady-,tall: condmons. (21 One-duncnswnal conducnon 1n x. r3i
Cc:m~1am pn:,pcmcs.
r ROPERT IF.S: T<1b/e .•I -/ , ,\lummum t3'.13 K> k = 23~ W/m K
\ro.A L YS IS· la I From Founer·s luw, EQ (~ .I). with A = itD2/4 = (:ta=/JI< l. It folio.- sthar
-lq.dx , , =-kdT
1trx·
Hcni:<', ~mcc 4, i, rndcpcndcn1 or :i,._
~ f.' d.: =-kfTdT
>r Jt:l... '-1 ~ -
or

<J

1bl From the forcJ:;ot.ng C'(prc-.-.ion, u ill~ follow-, th,u

ita1~ T1-T1

q, i=.--:;--

..

•

- III•! - 1/•il

q, ., n11m· 11238 W/m K

1:t0-11)(11°C

<J
COMME',iS: Tho lm~o,ng r.-ult, an: approx1m,1e due 10 use of a one•dtmcnsionaJ model
1n trcaan~ J two--d1mcns1onttl problem

PROBLEM 3.28
KNO~V"I: Temperature dependence oi the thcnnnJ conducavny, k. FIJl,D: He1u Aux and form of tcmpcrmurc d1stnbunon for• plane wall. SCHEt">1AT I C:

-k• k0 ,-aT

T

i1>0

Ta

'< To> Ti

7;,

; :a=O

(erbtfrary

i1<0

s election)

9': '---T1.

Ti

l+x

I
L

0

)(
L

ASSUl\1PTIONS: 111 One•dimen.iono.J conducuon through • plane wull, 121 S1cnd)•si.1tc condmons. (3) Np m1emal hent gcnernaon.
A NALYSIS: For 1hc ussurrn:.d cond1t1on, q, nnd ACX) arc constant ond Eq. J.21 ,;ive,
q.1Ldx = -~:• t1,0 +aTldT
q•, = LI fk,,fr,-T,l- IafT"~-·fT"JI
From Founer', 1011>,
q•, = -1k0 +an dT/dx Hence. since the product of lk,,raTJ nnd 1dT/dx1 ,. consiam. dccrca~mg T ,-uh mcrc;isin~ ,
tmphes.
• > 0 . dccrensmg ck,,+aTl and mcrca,m!? JdT/cb I with 1ncrca,1ng,
a = 0: k=kn => con~t.101 CdT/d,,
• < 0. mcrcn.<ing (k,,+nn and decrca<ing fdT/dx I wnh increasing, The temperature distnbuuon~ appear a~ shown m the obovc sketch