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KNOWN: Thickness and thermal conductivity of a wall. Heat flux applied to one face and
temperatures of both surfaces.
FIND: Whether steady-state conditions exist.
SCHEMATIC :
ASSUMPTIONS : (1) One-dimensional conduction, (2) Constant properties, (3) No internal energy
generation.
ANALYSIS: Under steady-state conditions an energy balance on the control volume shown is
2 q in ′′ = q out ′′ = q cond ′′ = k T ( 1 (^) − T 2 ) / L = 12 W/m K(50 C⋅ ° − 30 C) / 0.01 m° =24,000 W/m
Since the heat flux in at the left face is only 20 W/m
2
, the conditions are not steady state. <
COMMENTS: If the same heat flux is maintained until steady-state conditions are reached, the
steady-state temperature difference across the wall will be
∆ T =
2 q L k ′′^ / = 20 W/m × 0.01 m /12 W/m K⋅ =0.0167 K
which is much smaller than the specified temperature difference of 20°C.
q ” = 20 W/m^2
L = 10 mm
T 1 = 50°C k = 12 W/m∙K
T 2 = 30°C
q ″cond
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
T T m 40-20 C
x (^2) 1 2
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: Hand experiencing convection heat transfer with moving air and water.
FIND: Determine which condition feels colder. Contrast these results with a heat loss of 30 W/m
2 under
normal room conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Temperature is uniform over the hand’s surface, (2) Convection coefficient is
uniform over the hand, and (3) Negligible radiation exchange between hand and surroundings in the case
of air flow.
ANALYSIS: The hand will feel colder for the condition which results in the larger heat loss. The heat
loss can be determined from Newton’s law of cooling, Eq. 1.3a, written as
q ′′ = h T ( (^) s −T∞)
For the air stream:
( )
qair ′′ = 40 W m ⋅ K 30 − − 8 K =1,520 W m <
For the water stream:
( )
qwater ′′ = 900 W m ⋅ K 30 − 10 K = 18, 000 W m <
COMMENTS: The heat loss for the hand in the water stream is an order of magnitude larger than when
in the air stream for the given temperature and convection coefficient conditions. In contrast, the heat
loss in a normal room environment is only 30 W/m
2 which is a factor of 50 times less than the loss in the
air stream. In the room environment, the hand would feel comfortable; in the air and water streams, as
you probably know from experience, the hand would feel uncomfortably cold since the heat loss is
excessively high.
KNOWN: Width, input power and efficiency of a transmission. Temperature and convection
coefficient associated with air flow over the casing.
FIND: Surface temperature of casing. Thermal convection resistance.
SCHEMATIC:
W = 0.3 m
P = 150 hpi
P (^) o = hP (^) i
q
Too = 30 Co
h (^) i = 200 W/m -K^2
ASSUMPTIONS: (1) Steady state, (2) Uniform convection coefficient and surface temperature, (3)
Negligible radiation.
ANALYSIS: From Newton’s law of cooling,
( ) ( )
q = hAs Ts − T∞ = 6 hW Ts −T∞
where the output power is hPi and the heat rate is
q = Pi − Po = Pi (^) ( 1 − h)= 150 hp × 746 W / hp × 0.07 =7833 W
Hence,
( )
s (^2 ) 2
q 7833 W
T T 30 C 102.5 C
6 hW 6 200 W / m K 0.3m
× ⋅ ×
From Eq. 1.11, the thermal resistance due to convection is
R t,conv = ∆T / q x = T ( s − T∞ ) / q x= ( 102.5 − 30 K / 7833 W) = 0.00926 K/W <
COMMENTS: (1) There will, in fact, be considerable variability of the local convection coefficient
over the transmission case and the prescribed value represents an average over the surface. (2) The
convection thermal resistance could equivalently be calculated from Rt,conv = 1/hA.
KNOWN: Length, diameter, surface temperature and emissivity of steam line. Temperature
and convection coefficient associated with ambient air. Efficiency and fuel cost for gas fired
furnace.
FIND: (a) Rate of heat loss, (b) Annual cost of heat loss.
SCHEMATIC:
ASSUMPTIONS: (1) Steam line operates continuously throughout year, (2) Net radiation
transfer is between small surface (steam line) and large enclosure (plant walls).
ANALYSIS: (a) From Eqs. (1.3a) and (1.7), the heat loss is
( ) (^) ( )
4 4 q q (^) conv q (^) rad A h Ts T∞ Ts Tsur = + = ^ − + −
εs
where (^) A = π DL = π( 0.1m × 25m) =7.85m.^2
Hence,
( ) (^) ( )
2 2 8 2 4 4 4 4 q 7.85m 10 W/m K 150 25 K 0.8 5.67 10 W/m K 423 298 K
= ⋅ − + × × ⋅ −
( ) ( )
2 2
q = 7.85m 1, 250 + 1,095 W/m = 9813 + 8592 W = 18, 405 W <
(b) The annual energy loss is
11 E = qt = 18, 405 W × 3600 s/h × 24h/d × 365 d/y = 5.80 × 10 J
With a furnace energy consumption of
11
E f = E/ ηf = 6.45 × 10 J,the annual cost of the loss is
5
C = C Eg f= 0.02 $/MJ × 6.45 ×10 MJ = $12,900 <
COMMENTS: The heat loss and related costs are unacceptable and should be reduced by
insulating the steam line.
qrad
T s = 150 C,o^ ε
T (^) sur = 25 Co
L = 25 m
D = 100 mm
qconv
W/m -K^2
oo Air
qrad
T s = 150 C,o^ ε
T (^) sur = 25 Co
L = 25 m
D = 100 mm
qconv
W/m -K^2
oo Air
KNOWN: Width, input power, and efficiency of a transmission. Temperature and convection
coefficient for air flow over the casing. Emissivity of casing and temperature of surroundings.
FIND: Surface temperature of casing. Resistances due to convection and radiation.
SCHEMATIC:
ASSUMPTIONS: (1) Steady state, (2) Uniform convection coefficient and surface temperature, (3)
Radiation exchange with large surroundings.
ANALYSIS: Heat transfer from the case must balance heat dissipation in the transmission, which
may be expressed as q = Pi – Po = Pi (1 - η) = 150 hp × 746 W/hp × 0.07 = 7833 W. Heat transfer
from the case is by convection and radiation, in which case
q As h Ts T∞ εs Ts Tsur
where As = 6 W
2
. Hence,
(^2 2 2 4 4 4 ) s s
7833 W 6 0.30 m 200 W / m K T 303K 0.8 5.67 10 W / m K T 303 K
= ^ ⋅ − + × × ⋅ −
A trial-and-error solution yields
T s ≈ 373K = 100 C° <
The thermal resistances can be found from Eq. 1.11, that is, Rt = ∆T/q. For convection, the relevant
temperature difference is Ts − T∞.
R t,conv = ( Ts − T∞ ) / qconv = ( Ts − T∞ ) / ^ hAs ( Ts − T∞ )=1/ hAs
R t,conv= 1/ 6 0.30 m^ 200 W / m ⋅ K =0.00926 K / W
For radiation, the relevant temperature difference is Ts − Tsurand
Continued …
KNOWN: Radial distribution of heat dissipation in a cylindrical container of radioactive
wastes. Surface convection conditions.
FIND: Total energy generation rate and surface temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible temperature drop across thin
container wall.
ANALYSIS: The rate of energy generation is
( )
( )
r (^2) g o 0 o
2 2 g o o o
E qdV=q 1- r/r 2 rLdr
E 2 Lq r / 2 r / 4
o
or per unit length,
E′ =.
q r
g
o o
2 π
Performing an energy balance for a control surface about the container yields, at an instant,
E^ ^ ′ - E′ =
g out 0
and substituting for the convection heat rate per unit length,
( )( )
2 o o o s
q r h 2 r T T 2
T T
q r
4h
s
o o
COMMENTS: The temperature within the radioactive wastes increases with decreasing r
from Ts at ro to a maximum value at the centerline.
KNOWN: One-dimensional system with prescribed thermal conductivity and thickness.
FIND: Unknowns for various temperature conditions and sketch distribution.
SCHEMATIC:
qx “
T 1
T 2
dT
dx
, Temperature gradient
k = 50 W/m∙K
L = 0.35 m
x
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) No internal heat
generation, (4) Constant properties.
ANALYSIS: The rate equation and temperature gradient for this system are
2 1 x
dT dT T T q k and. dx dx L
Using Eqs. (1) and (2), the unknown quantities for each case can be determined.
(a)
dT (^) ( 20 50 K) 200 K/m dx 0.35m
2 x
W K
q 50 200 10.0 kW/m. m K m
′′ = − × − =
(b)
dT (^10 (^30 ))K 57 K/m dx 0.35m
2 x
W K
q 50 57 2.86 kW/m. m K m
′′ = − × = −
(c)
2 x
W K
q 50 160 8.0 kW/m m K m
′′ = − × = −
2 1
dT K T L T 0.35m 160 70 C. dx m
= ⋅ + = × +
T 2 = 126 C.
Continued …
q (^) x “
50°C
x
-20°C
-10°C
x
-30°C
qx “
x
70°C
qx “
dT dx
= 160 K/m
KNOWN: One-dimensional system with prescribed thermal conductivity and thickness.
FIND: Unknowns for various temperature conditions and sketch distribution.
SCHEMATIC:
qx “
T 1
T 2
dT
dx
, Temperature gradient
k = 50 W/m∙K
L = 0.35 m
x
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) No internal heat
generation, (4) Constant properties.
ANALYSIS: The rate equation and temperature gradient for this system are
2 1 x
dT dT T T q k and. dx dx L
Using Eqs. (1) and (2), the unknown quantities for each case can be determined.
(a)
dT (^) ( 20 50 K) 200 K/m dx 0.35m
2 x
W K
q 50 200 10.0 kW/m. m K m
′′ = − × − =
(b)
dT (^10 (^30 ))K 57 K/m dx 0.35m
2 x
W K
q 50 57 2.86 kW/m. m K m
′′ = − × = −
(c)
2 x
W K
q 50 160 8.0 kW/m m K m
′′ = − × = −
2 1
dT K T L T 0.35m 160 70 C. dx m
= ⋅ + = × +
T 2 = 126 C.
Continued …
q (^) x “
50°C
x
-20°C
-10°C
x
-30°C
qx “
x
70°C
qx “
dT dx
= 160 K/m
PROBLEM 2.6 (Cont.)
(d)
2 x
W K
q 50 80 4.0 kW/m m K m
′′ = − × − =
1 2
dT K T T L 40 C 0.35m 80 dx m
T 1 = 68 C.
(e)
2 x
W K
q 50 200 10.0 kW/m m K m
′′ = − × = −
1 2
dT K T T L 30 C 0.35m 200 40 C. dx m
x
40°C
qx “
dT dx
= -80 K/m
x
30°C qx “^ dT dx
= 200 K/m
