An electric cable of radius r 1 and thermal conductivity k e is enclosed by an insulating sleeve whose outer surface is of radius r 2 and experiences convection heat transfer and radiation exchange with the adjoining air and large surroundings. respectively. When electric current passes through the cable, thermal energy is generated within the cable at a volumetric rate q . . Write the steady-state forms of the heat diffusion equation for the insulation and the cable. Verify that these equations are satisfied by the following temperature distributions: Insulation : T ( r ) = T s , 2 + ( T s , 1 − T s , 2 ) ln ( r / r 2 ) ln ( r 1 / r 2 ) Cable: T ( r ) = T s , 1 + q . r 1 2 4 k c ( 1 − r 2 r 1 2 ) Sketch the temperature distribution, T ( r ) , in the cable and the sleeve, labeling key features. Applying Fourier's law, show that the rate of conduction heat transfer per unit length through the sleeve may be expressed as q . = 2 π k s ( T s , 1 − T s , 2 ) ln ( r 2 / r 1 ) Applying an energy balance to a control surface placed around the cable, obtain an alternative expression for q r ' , expressing your result in terms of q . and r 1 . Applying an energy balance to a control surface placed around the outer surface of the sleeve. obtain an expression from which T s , 2 may be determined as a function of q . , r 1 , h , T ∞ , ε , and T sur . Consider conditions for which 250A are passing through a cable having an electric resistance per unit length of R e ' = 0.005 Ω /m , a radius of r 1 = 15 mm, and a thermal conductivity of k e = 200 W/m ⋅ K . For k s = 15 W/m ⋅ K, r 2 = 15.5 mm , h = 25 W/m 2 ⋅ K, ε = 0.9 , T ∞ = 25 ° C, and T sur = 35 ° C, evaluate the surface temperatures, T s , 1 , and T s , 2 , as well as the temperature To at the centerline of the cable. With all other conditions remaining the same. compute and plot T o , T s , 1 , and T s , 2 as a function of r 2 for 15.5 ≤ r 2 ≤ 20 mm .
An electric cable of radius r 1 and thermal conductivity k e is enclosed by an insulating sleeve whose outer surface is of radius r 2 and experiences convection heat transfer and radiation exchange with the adjoining air and large surroundings. respectively. When electric current passes through the cable, thermal energy is generated within the cable at a volumetric rate q . . Write the steady-state forms of the heat diffusion equation for the insulation and the cable. Verify that these equations are satisfied by the following temperature distributions: Insulation : T ( r ) = T s , 2 + ( T s , 1 − T s , 2 ) ln ( r / r 2 ) ln ( r 1 / r 2 ) Cable: T ( r ) = T s , 1 + q . r 1 2 4 k c ( 1 − r 2 r 1 2 ) Sketch the temperature distribution, T ( r ) , in the cable and the sleeve, labeling key features. Applying Fourier's law, show that the rate of conduction heat transfer per unit length through the sleeve may be expressed as q . = 2 π k s ( T s , 1 − T s , 2 ) ln ( r 2 / r 1 ) Applying an energy balance to a control surface placed around the cable, obtain an alternative expression for q r ' , expressing your result in terms of q . and r 1 . Applying an energy balance to a control surface placed around the outer surface of the sleeve. obtain an expression from which T s , 2 may be determined as a function of q . , r 1 , h , T ∞ , ε , and T sur . Consider conditions for which 250A are passing through a cable having an electric resistance per unit length of R e ' = 0.005 Ω /m , a radius of r 1 = 15 mm, and a thermal conductivity of k e = 200 W/m ⋅ K . For k s = 15 W/m ⋅ K, r 2 = 15.5 mm , h = 25 W/m 2 ⋅ K, ε = 0.9 , T ∞ = 25 ° C, and T sur = 35 ° C, evaluate the surface temperatures, T s , 1 , and T s , 2 , as well as the temperature To at the centerline of the cable. With all other conditions remaining the same. compute and plot T o , T s , 1 , and T s , 2 as a function of r 2 for 15.5 ≤ r 2 ≤ 20 mm .
Solution Summary: The author explains the heat diffusion equation for insulation and cable and the profile of temperature distribution in Figure 1.
An electric cable of radius
r
1
and thermal conductivity
k
e
is enclosed by an insulating sleeve whose outer surface is of radius
r
2
and experiences convection heat transfer and radiation exchange with the adjoining air and large surroundings. respectively. When electric current passes through the cable, thermal energy is generated within the cable at a volumetric rate
q
.
.
Write the steady-state forms of the heat diffusion equation for the insulation and the cable. Verify that these equations are satisfied by the following temperature distributions:
Insulation
:
T
(
r
)
=
T
s
,
2
+
(
T
s
,
1
−
T
s
,
2
)
ln
(
r
/
r
2
)
ln
(
r
1
/
r
2
)
Cable:
T
(
r
)
=
T
s
,
1
+
q
.
r
1
2
4
k
c
(
1
−
r
2
r
1
2
)
Sketch the temperature distribution,
T
(
r
)
,
in the cable and the sleeve, labeling key features.
Applying Fourier's law, show that the rate of conduction heat transfer per unit length through the sleeve may be expressed as
q
.
=
2
π
k
s
(
T
s
,
1
−
T
s
,
2
)
ln
(
r
2
/
r
1
)
Applying an energy balance to a control surface placed around the cable, obtain an alternative expression for
q
r
'
,
expressing your result in terms of
q
.
and
r
1
.
Applying an energy balance to a control surface placed around the outer surface of the sleeve. obtain an expression from which
T
s
,
2
may be determined as a function of
q
.
,
r
1
,
h
,
T
∞
,
ε
,
and
T
sur
.
Consider conditions for which 250A are passing through a cable having an electric resistance per unit length of
R
e
'
=
0.005
Ω
/m
,
a radius of
r
1
=
15
mm,
and a thermal conductivity of
k
e
=
200
W/m
⋅
K
.
For
k
s
=
15
W/m
⋅
K,
r
2
=
15.5
mm
,
h
=
25
W/m
2
⋅
K,
ε
=
0.9
,
T
∞
=
25
°
C,
and
T
sur
=
35
°
C,
evaluate the surface temperatures,
T
s
,
1
,
and
T
s
,
2
,
as well as the temperature To at the centerline of the cable.
With all other conditions remaining the same. compute and plot
T
o
,
T
s
,
1
,
and
T
s
,
2
as a function of
r
2
for
15.5
≤
r
2
≤
20
mm
.
A hollow aluminum sphere, with an electrical heater in the center, is used in tests to determine the thermal conductivity of insulating materials. The inner and outer radii of the sphere are o.18 and o.21 m, respectively, and testing is done under steady-state conditions with the inner surface of the aluminum maintained at 250°C. In a particular test, a spherical shell of insulation is cast on the outer surface of the sphere to a thickness of o.15 m. The system is in a room for which the air temperature is 20°C and the convection coefficient at the outer surface of the insulation is 30 W/m2. K. If 80 W is dissipated by the heater under steady-state conditions, what is the thermal conductivity of the insulation?
4.Water at a temperature of T∞ = 25°C flows over one of the surfaces of a steel wall (AISI 1010) whose temperature is Ts,1 = 40°C and thermal conductivity of steel is 671 w/m.k. The wall is 0.35 m thick, and its other surface temperature is Ts,2 = 100°C. For steady state conditions what is the convection coefficient associated with the water flow?
19) In fabrication process, steel components are formed hot and then quenched in water. Consider 2 m long, 0.2 m diameter steel cylinder (k = 40 W/m.°C, a = 10-5 m² /s), initially at 400 °C, that is suddenly quenched in water at 50 °C. If heat transfer coefficient is 200 W/m². K, calculate the following 20 min min after immersion: a) center temperature, b) surface temperature, c) The heat transferred to the water during the initial 20 min.
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Principles of Heat Transfer (Activate Learning wi...Mechanical EngineeringISBN:9781305387102Author:Kreith, Frank; Manglik, Raj M.Publisher:Cengage Learning