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CHE3123 Chap17BW

# CHE3123 Chap17BW - Chapter 17 Steady State Conduction Two...

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Chapter 17 – Steady State Conduction Two approaches to solving problem a) control volume concept b) eliminate terms from general differential equation example – flat wall with no internal heat generation Start with Laplace equation 2 T = 0

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in Cartesian coordinates 2 T 2 T 2 T 2 T 0 x 2 y 2 z 2 integrating T c 1 x T = c 1 x + c 2 boundary conditions @ x = 0 T = T 1 @ x = L T = T 2
T T 1 T 1 – T 2 x L a linear profile as we saw in chapter 15 (see P.233) Example 2 – Page 257 – a “classic” problem steam pipe r 2 r 1 r 3 insulation

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T = T air = constant T 2 = constant = temperature of outer surface of pipe Problem : can the energy loss increase with increase in insulation thickness? from chapter 15 q = U o A o T q A 3 (T 2 – T ) A 3 ln(r 3 /r 2 ) 1 2 π k 2 L h o
A 3 = 2 π r 3 L q 2 π r 3 L (T 2 – T ) 2 π r 3 L ln(r 3 /r 2 ) 1 2 π k 2 L h o r 3 q 2 π L (T 2 – T ) ln(r 3 /r 2 ) 1 k 2 h r 3 increases as r 3 decreases as r 3 increases increases

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maximum suggested, so differentiate dq 2 π L (T 2 – T ) (1/k 2 r 3 – 1/ h r 3 2 ) dr 3 ln(r 3 /r 2 ) 1 2 k 2 h r 3 = 0 solving for r 3 r 3 = k 2 /h the evaluation of the second derivative d 2 q r negative number dr 3 2 r 3 = k/h
q is maximum at r 3 = k/h Typical values given in text k 85% magnesia = 0.0692 W/m.K h = 34 W/m 2 .k (natural convection) r critical = k/h = 0.20cm = 0.0787 inch no practical pipe has a radius this small!

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Conduction with internal energy generation affects temperature profile as expected!
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