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 Fourier’s law,
dT T −T
q cond = q x = q ′′x ⋅ A = -k ⋅ A = kA 1 2 .
dx L
Solving for T2 gives
q cond L
T2 = T1 − .
kA
Substituting numerical values, find
3000W × 0.025m
T2 = 415$ C -
0.2W / m ⋅ K × 10m2
T2 = 415$ C - 37.5$ C
T2 = 378$ C. <
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, dT dx = − q′′x k , is a constant, and
hence the temperature distribution is linear, if q′′x 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
q′′x = − k
dT
=k
T1 − T2
= 1W m ⋅ K
25$ C − −15$ C
= 133.3W m 2 .
( ) (1)
dx L 0.30 m
q x = q′′x × A = 133.3 W m 2 × 20 m 2 = 2667 W . (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.
3500
2500
Heat loss, qx (W)
1500
500
-500
-1500
-20 -10 0 10 20 30 40
Ambient air temperature, T2 (C)
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
T −T 7°C
q = k ( LW ) 1 2 = 1.4 W / m ⋅ K (11m × 8 m ) = 4312 W <
t 0.20 m
The daily cost of natural gas that must be combusted to compensate for the heat loss is
q Cg 4312 W × $0.01/ MJ
Cd = ( ∆t ) = ( 24 h / d × 3600s / h ) = $4.14 / d <
ηf 0.9 ×106 J / MJ
COMMENTS: The loss could be reduced by installing a floor covering with a layer of insulation
between it and the concrete.
, PROBLEM 1.4
KNOWN: Heat flux and surface temperatures associated with a wood slab of prescribed
thickness.
FIND: Thermal conductivity, k, of the wood.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state
conditions, (3) Constant properties.
ANALYSIS: Subject to the foregoing assumptions, the thermal conductivity may be
determined from Fourier’s law, Eq. 1.2. Rearranging,
L W 0.05m
k=q′′x = 40
T1 − T2 m2 ( 40-20 ) C
k = 0.10 W / m ⋅ K. <
COMMENTS: Note that the °C or K temperature units may be used interchangeably when
evaluating a temperature difference.
KNOWN: Heat rate, q, through one-dimensional wall of area A, thickness L, thermal
conductivity k and inner temperature, T1.
FIND: The outer temperature of the wall, T2.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions,
(3) Constant properties.
ANALYSIS: The rate equation for conduction through the wall is given by Fourier’s law,
dT T −T
q cond = q x = q ′′x ⋅ A = -k ⋅ A = kA 1 2 .
dx L
Solving for T2 gives
q cond L
T2 = T1 − .
kA
Substituting numerical values, find
3000W × 0.025m
T2 = 415$ C -
0.2W / m ⋅ K × 10m2
T2 = 415$ C - 37.5$ C
T2 = 378$ C. <
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, dT dx = − q′′x k , is a constant, and
hence the temperature distribution is linear, if q′′x 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
q′′x = − k
dT
=k
T1 − T2
= 1W m ⋅ K
25$ C − −15$ C
= 133.3W m 2 .
( ) (1)
dx L 0.30 m
q x = q′′x × A = 133.3 W m 2 × 20 m 2 = 2667 W . (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.
3500
2500
Heat loss, qx (W)
1500
500
-500
-1500
-20 -10 0 10 20 30 40
Ambient air temperature, T2 (C)
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
T −T 7°C
q = k ( LW ) 1 2 = 1.4 W / m ⋅ K (11m × 8 m ) = 4312 W <
t 0.20 m
The daily cost of natural gas that must be combusted to compensate for the heat loss is
q Cg 4312 W × $0.01/ MJ
Cd = ( ∆t ) = ( 24 h / d × 3600s / h ) = $4.14 / d <
ηf 0.9 ×106 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 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′′x = 40
T1 − T2 m2 ( 40-20 ) C
k = 0.10 W / m ⋅ K. <
COMMENTS: Note that the °C or K temperature units may be used interchangeably when
evaluating a temperature difference.
