EXERCICIOS DE MECANICA

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 Fourier’s law,

q q q A = -k dT dx A = kA TT Lcond xx 12== ′′ ⋅⋅ − .

Solving for T2 gives

TT qL kA21 cond=− .

Substituting numerical values, find

T C- 3000W 0.025m 0.2W/ m K 10m2 2 = × ⋅× 415$

T C-37.5 C 2 = 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 38°C.

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 Fourier’s law, it is evident that the gradient, x dT dx q k ′′=− , is a constant, and hence the temperature distribution is linear, if x q′′ 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 = -15°C are ( ) 212 x 25 C 15 C dT T T q k k 1W m K 133.3W m dx L 0.30m −−− ′′ = − = = ⋅ = $$ . (1)

22

xx q q A 133.3W m 20m 2667W ′′ = × = × = . (2) <

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

-20 -10 0 10 20 30 40

Ambient air temperature, T2 (C)

-1500

-500

500

1500

2500

3500

Heat loss, qx (W)

Wall thermal conductivity, k = 1.25 W/m.K k = 1 W/m.K, concrete wall k = 0.75 W/m.K For the concrete wall, k = 1 W/m⋅K, 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

() () 12 T T 7 C q k LW 1.4W/ m K 11m 8m 4312 W t 0.20m −° = = ⋅ × = <

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

() () g d 6 f qC 4312W $0.01/MJ C t 24h /d 3600s/h $4.14/d 0.9 10 J /MJ η × = ∆ = × = × <

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 prescribed thickness.

FIND: Thermal conductivity, k, of the wood.

SCHEMATIC:

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

ANALYSIS: Subject to the foregoing assumptions, the thermal conductivity may be determined from Fourier’s law, Eq. 1.2. Rearranging,

()

L W 0.05m k=q 40 TT m 40-20 C x 2 12 ′′ = −

k = 0.10 W / m K. ⋅ <

COMMENTS: Note that the °C or K temperature units may be used interchangeably when evaluating a temperature difference.

PROBLEM 1.5

KNOWN: Inner and outer surface temperatures of a glass window of prescribed dimensions.

FIND: Heat loss through window.

SCHEMATIC:

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

ANALYSIS: Subject to the foregoing conditions the heat flux may be computed from Fourier’s law, Eq. 1.2.

()

TT q k L 15-5 CW q 1.4 m K 0.005m q 2800 W/m . 12 x x 2 x − ′′ = ′′ = ⋅ ′′ =

Since the heat flux is uniform over the surface, the heat loss (rate) is

q = qx A q = 2800 W/ m2 3m2 ′′ × × q = 8400 W. <

COMMENTS: A linear temperature distribution exists in the glass for the prescribed conditions.

PROBLEM 1.6

KNOWN: Width, height, thickness and thermal conductivity of a single pane window and the air space of a double pane window. Representative