1-D Steady State Heat Transfer With Heat Generation
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It includes derivation of Heat Transfer Through a SPHERE.
![SPHERE WITH UNIFORM HEAT GENERATION
Consider one dimensional radial conduction of heat, under steady
state conduction, through a sphere having uniform heat generation.
Now, general heat conduction equation for sphere is given by:
[
1
𝑟2.
𝜕
𝜕𝑟
𝑟2 𝜕𝑡
𝜕𝑟
+
1
𝑟2
𝑠𝑖𝑛𝜃
𝜕
𝜕𝜃
𝑠𝑖𝑛𝜃
𝜕𝑡
𝜕𝜃
+
1
𝑟2
𝑠𝑖𝑛2
𝜃
𝜕2
𝑡
𝜕∅2] +
𝑞 𝑔
𝑘
=
𝜌𝑐
𝑘
.
𝜕𝑡
𝜕𝜏
According to our consideration equation reduces to
1
𝑟2.
𝜕
𝜕𝑟
𝑟2 𝜕𝑡
𝜕𝑟
+
𝑞 𝑔
𝑘
= 0](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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1-D Steady State Heat Transfer With Heat Generation
- 1. 1-D STEADY STATE HEAT TRANSFER WITH HEAT GENERATION SPHERE WITH UNOFORM HEAT GENERATION PATEL MIHIR H. 130450119107
- 2. SPHERE WITH UNIFORM HEAT GENERATION Consider one dimensional radial conduction of heat, under steady state conduction, through a sphere having uniform heat generation. Now, general heat conduction equation for sphere is given by: [ 1 𝑟2. 𝜕 𝜕𝑟 𝑟2 𝜕𝑡 𝜕𝑟 + 1 𝑟2 𝑠𝑖𝑛𝜃 𝜕 𝜕𝜃 𝑠𝑖𝑛𝜃 𝜕𝑡 𝜕𝜃 + 1 𝑟2 𝑠𝑖𝑛2 𝜃 𝜕2 𝑡 𝜕∅2] + 𝑞 𝑔 𝑘 = 𝜌𝑐 𝑘 . 𝜕𝑡 𝜕𝜏 According to our consideration equation reduces to 1 𝑟2. 𝜕 𝜕𝑟 𝑟2 𝜕𝑡 𝜕𝑟 + 𝑞 𝑔 𝑘 = 0
- 3. • Let, • R = 0utside radius of sphere • K = thermal conductivity(uniform) • qg = uniform heat generation per unit volume, per unit time within the solid • tw =temperature of outside surface (wall) of the sphere , and • ta = ambient temperature. • Consider an element at radius r and thickness dr as shown in figure. • Heat conduction at radius at r, Qr = -kA 𝑑𝑡 𝑑𝑟 = -k*4𝜋𝑟2 ∗ 𝑑𝑡 𝑑𝑟
- 4. Heat generated in the element, Qg = qgAdr = -k*4𝜋𝑟2 ∗ 𝑑𝑡 𝑑𝑟 heat conducted out at radius (r+dr) Q(r+dr) = Qr + 𝑑 𝑑𝑟 Qr.dr Under steady state conduction, we have Qr+Qg= Q(r+dr)= Qr + 𝑑 𝑑𝑟 Qr.dr ∴ Qg= 𝑑 𝑑𝑟 Qr.dr ∴ qg4𝜋𝑟2dr = -4𝜋𝑘 𝑑 𝑑𝑟 [𝑟2 𝑑𝑡 𝑑𝑟 ]dr ∴ 1 𝑟2[𝑟2 𝜕2 𝑡 𝜕𝑟2 + 2r 𝑑𝑡 𝑑𝑟 ] + 𝑞 𝑔 𝑘 = 0 ∴ 𝑑𝑡 𝑑𝑟 r 𝑑𝑡 𝑑𝑟 + 𝑑𝑡 𝑑𝑟 + 𝑞 𝑔 𝑘 = 0 …(multiplying both side by r)
- 5. integrating both side ,we have r 𝑑𝑡 𝑑𝑟 + t + 𝑞 𝑔 𝑘 . (𝑟2/2) = C1 ……(2) 0r 𝑑𝑡 𝑑𝑟 rt + 𝑞 𝑔 𝑘 . (𝑟2/2) = C1 integrating again, we have rt + 𝑞 𝑔 𝑘 . (𝑟3/6) = C1r+ C2 ……(3) At the center of sphere, r=0 so C2=0. Applying boundary condition, at r=R ,t= tw C1= tw + 𝒒 𝒈 𝟔𝒌 . R 𝟐 By substituting values of C1 and C2 in equation(3), we have temperature distribution as,
- 6. t + 𝑞 𝑔 𝑘 . (𝑟3/6) = tw + 𝑞 𝑔 6𝑘 . R2 t = tw + 𝒒 𝒈 𝟔𝒌 . (R 𝟐 − r 𝟐) ……(4) From above equation it is evident that the temperature distribution is parabolic, the maximum temperature occurs at the center(r2=0) is value is given by tmax = tw + 𝑞 𝑔 6𝑘 . R2 ……(5) From above equations (4) and (5) t − tw tmax − tw = R2 − r2 R2 = 1 – 𝑟 𝑅 2 Now, for evaluating heat transfer invoking Fourier’s equation, Q = -kA 𝑑𝑡 𝑑𝑟 r=R
- 7. Q = -k*4𝜋𝑟2 ∗ 𝑑 𝑑𝑟 tw + 𝑞 𝑔 6𝑘 . (R2 − r2) r=R = k*4𝜋𝑅2 ∗ 𝑞 𝑔 3𝑘 .R = 4 3 𝜋𝑅3 ∗ 𝑞 𝑔 Thus heat conducted is equal to heat generated.Under steady state conditions the heat conducted should be equal to heat convected from outer surface of the sphere. 4 3 𝜋𝑅3 ∗ 𝑞 𝑔 = h4𝜋𝑅2(tw − t𝑎) tw = ta + 𝑞 𝑔 3ℎ . R Inserting value of tw in equation (4) t = ta + 𝑞 𝑔 3ℎ . R + 𝑞 𝑔 6𝑘 . (R2 − r2)
- 8. The maximum temperature tmax = ta + 𝑞 𝑔 3ℎ . R + 𝑞 𝑔 6𝑘 .R2 ……(at r=0)