PROBLEM 1.1

KNOWN: Heat rate, q, through one-dimensional wall of area A, thickness L, thermal conductivity k and inner temperature, T1.

FIND: The outer temperature of the wall, T2.

SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant properties.

ANALYSIS: The rate equation for conduction through the wall is given by Fouriers law,

q q q A = -k dTdx

A = kA T TLcond x x

1 2= =

.

Solving for T2 gives

T Tq L

kA2 1cond

= .

Substituting numerical values, find

T C - 3000W 0.025m0.2W / m K 10m2 2

=

415$

T C - 37.5 C2 = 415$ $

T C.2 = 378$

<

COMMENTS: Note direction of heat flow and fact that T2 must be less than T1.

PROBLEM 1.2

KNOWN: Inner surface temperature and thermal conductivity of a concrete wall.

FIND: Heat loss by conduction through the wall as a function of ambient air temperatures ranging from -15 to 38C.

SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant properties, (4) Outside wall temperature is that of the ambient air.

ANALYSIS: From Fouriers law, it is evident that the gradient, xdT dx q k= , is a constant, and hence the temperature distribution is linear, if xq and k are each constant. The heat flux must be constant under one-dimensional, steady-state conditions; and k is approximately constant if it depends only weakly on temperature. The heat flux and heat rate when the outside wall temperature is T2 = -15C are

( ) 21 2x

25 C 15 CdT T Tq k k 1W m K 133.3W mdx L 0.30m

= = = =

$ $

. (1)

2 2x xq q A 133.3W m 20m 2667 W= = = . (2) <

Combining Eqs. (1) and (2), the heat rate qx can be determined for the range of ambient temperature, -15 T2 38C, with different wall thermal conductivities, k.

-20 -10 0 10 20 30 40

Ambient air temperature, T2 (C)

-1500

-500

500

1500

2500

3500

Hea

t los

s, q

x (W

)

Wall thermal conductivity, k = 1.25 W/m.Kk = 1 W/m.K, concrete wallk = 0.75 W/m.K

For the concrete wall, k = 1 W/mK, the heat loss varies linearily from +2667 W to -867 W and is zero when the inside and ambient temperatures are the same. The magnitude of the heat rate increases with increasing thermal conductivity.

COMMENTS: Without steady-state conditions and constant k, the temperature distribution in a plane wall would not be linear.

PROBLEM 1.3

KNOWN: Dimensions, thermal conductivity and surface temperatures of a concrete slab. Efficiency of gas furnace and cost of natural gas.

FIND: Daily cost of heat loss.

SCHEMATIC:

ASSUMPTIONS: (1) Steady state, (2) One-dimensional conduction, (3) Constant properties.

ANALYSIS: The rate of heat loss by conduction through the slab is

( ) ( )1 2T T 7 Cq k LW 1.4 W / m K 11m 8m 4312 Wt 0.20 m

= = = <

The daily cost of natural gas that must be combusted to compensate for the heat loss is

( ) ( )gd 6fq C 4312 W $0.01/ MJC t 24h / d 3600s / h $4.14 / d

0.9 10 J / MJ

= = =

<

COMMENTS: The loss could be reduced by installing a floor covering with a layer of insulation between it and the concrete.

PROBLEM 1.4

KNOWN: Heat flux and surface temperatures associated with a wood slab of prescribedthickness.

FIND: Thermal conductivity, k, of the wood.

SCHEMATIC:

ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-stateconditions, (3) Constant properties.

ANALYSIS: Subject to the foregoing assumptions, the thermal conductivity may bedetermined from Fouriers law, Eq. 1.2. Rearranging,

( )L W 0.05mk=q 40

T T m 40-20 Cx 21 2 =

k = 0.10 W / m K. Tm) is appropriate,

( ) ( )4 / 5 0.44/5 0.4 2i Dk 0.634 W/m K

h 0.023Re Pr 0.023 12,611 4.16 1230 W/m K.D 0.040 m

= = =

For the external flow,-6 2 5

DRe VD/ 100 m/s 0.040 m/38.79 10 m /s 1.031 10n= = =

and from Table 7.4, C = 0.26 and m = 0.6; Pr 10, n = 0.37, and Pr Prs,

( ) ( )1 / 4m no D sh k/D CRe Pr Pr/Pr=

( ) ( ) ( )3 0.6 0.37 1 / 45 2

o40.7 10 W/m K

h 0.26 1.031 10 0.684 1 234 W/m K.0.040 m

- = =

Substituting numerical values into Eq. 8.46 with P = pD and U = 197 W/m2K,

( )m,o pm,i

T Texp PLU/mc

T T

-= -

-& (1)

( ) 2m,o0.040 m 4 m

T 225 C 225 30 Cexp 197 W/m K 47.6 C0.25 kg/s 4179 J/kg K

p = - - - =

oo o (2)

To estimate the convection coefficient, use Eq. 10.9,

( )D

1/43v fgconv

v v v e

g h Dh DNu C

k k T

r rn

- = = D

l(3)

where C = 0.62 for the horizontal cylinder and ( )fg fg p,v s sath h 0.8 c T T . = + - Find

( ) ( )

( )

1 / 42 3 3 3

conv 6 2

9 .8m/s 957.9 31.55 k g / m 2257 10 0.8 4640 355 J/kg 0.020m0.0583W/m Kh 0.62

0.020 m 18.6 10 /31.55 m / s 0.0583W/m K 355K-

- + =

2convh 690 W / m K.=

To estimate the radiation coefficient, use Eq. 10.11,

( ) ( )4 4 8 2 4 4 4 4s sat 2rad

s sat

T T 0.9 5.67 10 W / m K 728 373 Kh 37.6W/m K.

T T 355K

e s -- -= = =

-Substituting numerical values into the simpler form of Eq. (2), find

( )( ) 2 2h 690 3 / 4 37.6 W / m K 718W/m K.= + = Using Eq. (1), the heat rate, with As = p D L, is

( )2sq 718 W / m K 0.020m 0.200m 355K 3.20kW.p= = hrad , the simpler form of Eq. 10.10b is appropriate. Find,( )( ) 2 2h 2108 3 4 28 W m K 2129 W m K= + =

Continued...

PROBLEM 10.28 (Cont.)

The heat rate is

( )2q 2129 W m K 0.002m 455K 6.09 kW m. = = 10). The Dittus-Boelter correlation with n = 0.4 is appropriate,

( ) ( )0.8 0.40.8 0.4D i i f D fNu h D k 0.023Re Pr 0.023 21, 673 5.2 130.9= = = =Continued...

PROBLEM 11.35 (Cont.)

2fi D 3i

k 0.620W m Kh Nu 130.9 6057 W m K

D 13.4 10 m

= = =

.

Substituting numerical values into Eq. (1), the overall heat transfer coefficient is

( ) 13o 2 2

15.9 10 m 21 15.9 15.9 1U n

115 W m K 13.4 13.413,500 W m K 6057 W m K

= + +

"

15 5 5 2 2oU 7.407 10 1.183 10 19.590 10 W m K 3549 W m K

= + + = .