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= 1.05 m has its inner surface subjected to a uniform heat flux of q1=7 kW/m2. The outer surface of the container has a2-77 A spherical container with an inner radius n = 1 m and an outer radius25°C, and the container wall thermal conductivity is k 1.5 W/m K. Show that the variation of temperature in the container wall can be expressed as T(r)= (4,r^k(1r 1/r.) + T2 and determinetemperature T2the temperature of the inner surface of the container at r = ^FIGURE P2-77SphericalcontainerT2

Question

A spherical container with an inner radius r1 = 1 m and an outer radius r2 = 1.05 m has its inner surface subjected to a uniform heat flux of q1=7kw/m^2. The outer surface of the container has a temperature T2 = 25°C, and the container wall thermal conductivity is k = 1.5 W/m·K. Show that the variation of temperature in the container wall can be expressed as  and determine the temperature of the inner surface of the container at r = r1.

= 1.05 m has its inner surface subjected to a uniform heat flux of q1=7 kW/m2. The outer surface of the container has a
2-77 A spherical container with an inner radius n = 1 m and an outer radius
25°C, and the container wall thermal conductivity is k 1.5 W/m K. Show that the variation of temperature in the container wall can be expressed as T(r)= (4,r^k(1r 1/r.) + T2 and determine
temperature T2
the temperature of the inner surface of the container at r = ^
FIGURE P2-77
Spherical
container
T2
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= 1.05 m has its inner surface subjected to a uniform heat flux of q1=7 kW/m2. The outer surface of the container has a 2-77 A spherical container with an inner radius n = 1 m and an outer radius 25°C, and the container wall thermal conductivity is k 1.5 W/m K. Show that the variation of temperature in the container wall can be expressed as T(r)= (4,r^k(1r 1/r.) + T2 and determine temperature T2 the temperature of the inner surface of the container at r = ^ FIGURE P2-77 Spherical container T2

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check_circleAnswer
Step 1

Using differential equation for heat conduction,

аT
=0
dr
d
dr
r = radius of container
On integration
d
dT
dr
dr
dT
C
dr
--(1)
t 111 1
111
ат с
dr
Again
On integration
dr
T(r)=C 2)
Boundary condition
at rT T(r)
C,
+C..3)
help_outline

Image Transcriptionclose

аT =0 dr d dr r = radius of container On integration d dT dr dr dT C dr --(1) t 111 1 111 ат с dr Again On integration dr T(r)=C 2) Boundary condition at rT T(r) C, +C..3)

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Step 2

Using Fourier’s law in radial r-direction,

dT
k
dr
(4)
here
k - thermal conductivity
dT
temperature gradient
dr
Boundary condition:
atr=T T(r );q,
dT
(5)
dr
dт (t)
-k
dr
dT C
11
dr
dT(rC
(6)
dr
Substituting eqution (6)into (5)
g-k
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Image Transcriptionclose

dT k dr (4) here k - thermal conductivity dT temperature gradient dr Boundary condition: atr=T T(r );q, dT (5) dr dт (t) -k dr dT C 11 dr dT(rC (6) dr Substituting eqution (6)into (5) g-k

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Step 3

Calculation:

&n...

Apply boundary condition equation (3);
atr T T(r, );q = 4
T()C
C T
k
On solving equation 2;
T(r)=C
-4i
T
k
T(r)=
(7)
1
1
T(r)
|+T3
k\r
help_outline

Image Transcriptionclose

Apply boundary condition equation (3); atr T T(r, );q = 4 T()C C T k On solving equation 2; T(r)=C -4i T k T(r)= (7) 1 1 T(r) |+T3 k\r

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Mechanical Engineering

Heat Transfer

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