PROBLEM 2.41
KNOWN: An electric cable with an insulating sleeve experiences convection with adjoining air and
radiation exchange with large surroundings.
FIND: (a) Verify that prescribed temperature distributions for the cable and insulating sleeve satisfy
their appropriate heat diffusion equations; sketch temperature distributions labeling key features; (b)
Applying Fourier's law, verify the conduction heat rate expression for the sleeve, q′r , in terms of Ts,1
and Ts,2; apply a surface energy balance to the cable to obtain an alternative expression for q′r in terms
of q and r1; (c) Apply surface energy balance around the outer surface of the sleeve to obtain an
expression for which Ts,2 can be evaluated; (d) Determine Ts,1, Ts,2, and To for the specified geometry
and operating conditions; and (e) Plot Ts,1, Ts,2, and To as a function of the outer radius for the range
15.5 ≤ r2 ≤ 20 mm.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, radial conduction, (2) Uniform volumetric heat generation
in cable, (3) Negligible thermal contact resistance between the cable and sleeve, (4) Constant
properties in cable and sleeve, (5) Surroundings large compared to the sleeve, and (6) Steady-state
conditions.
ANALYSIS: (a) The appropriate forms of the heat diffusion equation (HDE) for the insulation and
cable are identified. The temperature distributions are valid if they satisfy the relevant HDE.
Insulation: The temperature distribution is given as
(
T ( r ) = Ts,2 + Ts,1 − Ts,2
ln r r
) ln ((r r2 ))
(1)
1 2
and the appropriate HDE (radial coordinates, SS, q = 0), Eq. 2.24,
d ⎛ dT ⎞
⎜r
⎟=0
dr ⎝ dr ⎠
d⎛ ⎡
1 r ⎤ ⎞ d ⎛ Ts,1 − Ts,2 ⎞
⎜ r ⎢ 0 + Ts,1 − Ts,2
⎟=0
⎥⎟ = ⎜
dr ⎜⎝ ⎢⎣
ln ( r1 r2 ) ⎥⎦ ⎟⎠ dr ⎝⎜ ln ( r1 r2 ) ⎠⎟
(
)
Hence, the temperature distribution satisfies the HDE.
Cable: The temperature distribution is given as
2 ⎛ r2 ⎞
qr
T ( r ) = Ts,1 + 1 ⎜1 − ⎟
4k c ⎜ r 2 ⎟
1 ⎠
⎝
<
(2)
Continued...
PROBLEM 2.41 (Cont.)
and the appropriate HDE (radial coordinates, SS, q uniform), Eq. 2.24,
1 d ⎛ dT ⎞ q
=0
⎜r
⎟+
r dr ⎝ dr ⎠ k c
2⎛
qr
1 d ⎛ ⎡
2r ⎞ ⎤ ⎞ q
⎜ r ⎢ 0 + 1 ⎜ 0 − ⎟⎥ ⎟ +
=0
r dr ⎜ ⎢
4k c ⎜
r2 ⎟⎥ ⎟ kc
⎝
⎝ ⎣
1
⎠⎦ ⎠
1 d ⎛
2 2r 2 ⎞ q
qr
=0
⎜− 1
⎟+
r dr ⎜ 4k c r 2 ⎟ k c
⎝
1 ⎠
1⎛
2 4r ⎞ q
qr
=0
⎜− 1
⎟+
r ⎜ 4k c r 2 ⎟ k c
⎝
1 ⎠
<
Hence the temperature distribution satisfies the HDE.
The temperature distributions in the cable, 0 ≤ r ≤ r1, and sleeve, r1 ≤ r ≤ r2, and their key features are
as follows:
(1) Zero gradient, symmetry condition,
(2) Increasing gradient with increasing radius,
r, because of q ,
(3) Discontinuous gradient of T(r) across
cable-sleeve interface because of different
thermal conductivities,
(4) Decreasing gradient with increasing radius,
r, since heat rate is constant.
(b) Using Fourier’s law for the radial-cylindrical coordinate, the heat rate through the insulation
(sleeve) per unit length is
q′r = − kA′r
dT
dr
= −k2π r
dT
<
dr
and substituting for the temperature distribution, Eq. (1),
⎡
(
q′r = − k s 2π r ⎢ 0 + Ts,1 − Ts,2
⎣
⎤
) ln (1r r r ) ⎥ = 2π ks
1 2 ⎦
( Ts,1 − Ts,2 )
ln ( r2 r1 )
(3)
<
Applying an energy balance to a control surface placed around the cable,
E in − E out = 0
q ∀c − q′r = 0
Continued...
PROBLEM 2.41 (Cont.)
c represents the dissipated electrical power in the cable
where q∀
( )
q π r12 − q′r = 0
or
12
q′r = π qr
(4)
<
(5)
<
(c) Applying an energy balance to a control surface placed around the outer surface of the sleeve,
′
qconv
E in − E out = 0
q ′r − q′conv − q′rad = 0
(
)
4
4
12 − h ( 2π r2 ) ( Ts,2 − T∞ ) − ε ( 2π r2 ) σ Ts,2
π qr
− Tsur
=0
This relation can be used to determine Ts,2 in terms of the variables q , r1, r2, h, T∞, ε and Tsur.
(d) Consider a cable-sleeve system with the following prescribed conditions:
r1 = 15 mm
r2 = 15.5 mm
kc = 200 W/m⋅K
ks = 0.15 W/m⋅K
h = 25 W/m2⋅K
T∞ = 25°C
ε = 0.9
Tsur = 35°C
For 250 A with R ′e = 0.005 Ω/m, the volumetric heat generation rate is
(π r12 )
2
q = ( 250 A ) × 0.005 Ω / m (π × 0.0152 m 2 ) = 4.42 × 105 W m3
q = I 2 R ′e ∀c = I 2 R ′e
Substituting numerical values in appropriate equations, we can evaluate Ts,1, Ts,2 and To.
Sleeve outer surface temperature, Ts,2: Using Eq. (5),
π × 4.42 × 105 W m3 × ( 0.015m )2 − 25 W m 2 ⋅ K × ( 2π × 0.0155m ) ( Ts,2 − 298K )
(
)
4
−0.9 × ( 2π × 0.0155m ) × 5.67 × 10−8 W m 2 ⋅ K 4 Ts,2
− 3084 K 4 = 0
Ts,2 = 395 K = 122D C
<
Sleeve-cable interface temperature, Ts,1: Using Eqs. (3) and (4), with Ts,2 = 395 K,
12 = 2π k s
π qr
( Ts,1 − Ts,2 )
ln ( r2 r1 )
π × 4.42 × 105 W m3 × ( 0.015 m )2 = 2π × 0.15 W m ⋅ K
Ts,1 = 406 K = 133D C
( Ts,1 − 395 K )
ln (15.5 15.0 )
<
Continued...
PROBLEM 2.41 (Cont.)
Cable centerline temperature, To: Using Eq. (2) with Ts,1 = 133°C,
To = T(0) = Ts,1 +
12
qr
4k c
To = 133D C + 4.42 × 105 W m3 × ( 0.015 m )
2
( 4 × 200 W
m ⋅ K ) = 133.1D C
<
(e) With all other conditions remaining the same, the relations of part (d) can be used to calculate To,
Ts,1 and Ts,2 as a function of the sleeve outer radius r2 for the range 15.5 ≤ r2 ≤ 20 mm.
Temperature, Ts1 or Ts2 (C)
200
180
160
140
120
100
15
16
17
18
19
20
Sleeve outer radius, r2 (mm)
Inner sleeve, r1
Outer sleeve, r2
On the plot above To would show the same behavior as Ts,1 since the temperature rise between cable
center and its surface is 0.12°C. With increasing r2, we expect Ts,2 to decrease since the heat flux
decreases with increasing r2. We expect Ts,1 to increase with increasing r2 since the thermal resistance
of the sleeve increases.
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