Heat Transfer with Internal Heat...

Heat Transfer with Internal Heat Generation

Dr. Md. Zahurul Haq

ProfessorDepartment of Mechanical Engineering

Bangladesh University of Engineering & Technology (BUET)Dhaka-1000, Bangladesh

[email protected]://teacher.buet.ac.bd/zahurul/

ME 307: Heat Transfer Equipment Design

http://teacher.buet.ac.bd/zahurul/ME307/

c

Dr. Md. Zahurul Haq (BUET) HT with Internal Heat Generation ME 307 (2017) 1 / 15

Conduction Equation:

∇2T +

_qg

k=

1

α

∂T

∂t

∇2 =

∂2

∂x 2 + ∂2

∂y2 + ∂2

∂z 2 (x , y, z ) coordinate1r

∂∂r

(

r ∂∂r

)

+ 1r2

∂2

∂θ2 + ∂2

∂z 2 (r , θ, z ) coordinate1r2

∂∂r

(

r2 ∂∂r

)

+ 1r2 sin θ

∂∂θ

(

sinθ ∂∂θ

)

+ 1r2 sin2 θ

∂2

∂φ2 (r , θ, φ) coordinate

Conduction Equation: 1D & steady-state:

1

rn

∂

∂r

(

rn ∂T

∂r

)

+

_qg

k= 0

n =

0 (x , y , z ) coordinate (rectangular )

1 (r , θ, z ) coordinate (cylindrical)

2 (r , θ, φ) coordinate (sphecial)

c


Thermal Conductivity, k

T737

Temperature dependence of the thermal conductivity of selected solids.

c


T738

Temperature dependence of k of

selected gases at normal pressures.

T739

Temperature dependence of k of

selected non-metallic liquids under

saturated conditions.

c


Heat Conduction & Convection

T740

Heat transfer through a plane wall. (a) Temperature distribution. (b)

Equivalent thermal circuit.

c


T741

Hollow cylinder with convective surface conditions.

c


System with Internal Heat Generation


Nuclear reactors

Electric heater & conductors

Exothermic chemical reactions

c



Plane Wall with Uniform Heat Generation

T674

BC: T = Tw at x ± L

DE: d2Tdx 2 +

_qg

k = 0

⇒ T (x ) = −

_qg

2k x 2 + C1x + C2

=⇒

T (x ) = Tw +

_qg

2k L2[

1 −(

xL

)2]

if T (x = 0) = T0 :

⇒T0 = Tw +

_qg

2k L2

=⇒

T−TwT0−Tw

= 1 −(

xL

)2

c



T675

Conduction in a plane wall with uniform heat generation. (a)

Asymmetrical boundary conditions. (b) Symmetrical boundary

conditions. (c) Adiabatic surface at midplane.

c



Radial System with Internal Heat Generation

T676

BCs:

T (r0) = Ts

dTdr |r=0 = 0

DE: 1r

ddr

(

r dTdr

)

+

_qg

k = 0

⇒ T (r) = −

_qg

4k r2 + C1 ln r + C2

=⇒

T (r) = Ts +

_qg

4k r20

[

1 −(

rr0

)2]

if T (r = 0) = T0 :

⇒T0 = Ts +

_qg

4k r20

=⇒

T (r )−Ts

T0−Ts= 1 −

(

rr0

)2

c



contd. . . .

T676

Applying an overall energy balance:

_qg(πr20 L) = h(2πr0L)(Ts − T∞)

=⇒Ts = T∞ +

_qg r02h

Tmax = T0 = Ts +

_qg

4k r20 = T∞ +

_qg r04hc

(

2 + hcrok

)

c



Holman Ex. 2-6 ⊲ A current of 200 A is passed through a stainless-steel

wire [k =19 W/moC] 3 mm in diameter. The resistivity of the steel may be

taken as 70 µΩcm, and the length of the wire is 1 m. The wire is submerged

in a liquid at 110oC and experiences a convection heat-transfer coefficient of 4

kW/m2oC. Calculate the center temperature of the wire.

_qg = 560.2 MW/m3

To = 231.6oC

c



Kreith Ex. 2-5 ⊲ Heat is generated at 7.5× 107W/m3 rate in nuclear reactor

uranium rods of 0.05 m diameter. These rods are jacketed by an annulus in

which water at 120oC is circulated. If hav = 55000 W/m2 K, and for uranium,

k = 29.5 W/m K, determine the center temperature of the uranium fuel rods.

T677

Ts = 137.50C

T0 = 534.30C

c



Critical Thickness of Insulation

T700

T701

Rtotal = Rcond + Rconv =ln(r/ri )2πkL + 1

2πrLh∞

For maximum heat loss, dRtotaldr = 1

2πkrL − 12πr2Lh∞

= 0 → r = kh∞

Critical thickness of insulation , rcr = kh∞

c



Dewitt Ex. 3-4 ⊲ Critical thickness of insulation, k = 0.055 W/mK, ri =

5mm, Ti = 100oC, T∞ = 25oC.

T696

0 5 10 15 20 25 30

t [mm]

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

5.5

6.0

6.5

7.0

R[m

K/W

]

Critical Thickness of Insulation

Rcond

Rconv

Rtot

Q

10.0

10.5

11.0

11.5

12.0

12.5

13.0

13.5

14.0

14.5

15.0

Q[W

]

T697

rcr = k/h = 11 mm, t = rcr − ri = 6 mm

Q = 14.5 W

c


Heat Transfer with Internal Heat...

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