# ch3b - PROBLEM 3.51 KNOWN Pipe wall temperature and...

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PROBLEM 3.51 KNOWN: Pipe wall temperature and convection conditions associated with water flow through the pipe and ice layer formation on the inner surface. FIND: Ice layer thickness δ . SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, steady-state conduction, (2) Negligible pipe wall thermal resistance, (3) negligible ice/wall contact resistance, (4) Constant k. PROPERTIES: Table A.3 , Ice (T = 265 K): k 1.94 W/m K. ANALYSIS: Performing an energy balance for a control surface about the ice/water interface, it follows that, for a unit length of pipe, conv cond qq ′′ = () s,i s,o i 1 ,i s,i 21 TT h2r T T ln r r 2 k π −= Dividing both sides of the equation by r 2 , ( ) s,i s,o 2 2 1 i 2 ,i s,i ln r r k 1.94 W m K 15 C 0.097 rr h r T T 3C 2000W m K 0.05m = × = The equation is satisfied by r 2 /r 1 = 1.114, in which case r 1 = 0.050 m/1.114 = 0.045 m, and the ice layer thickness is r r 0.005m 5mm δ =−= = < COMMENTS: With no flow, h i 0, in which case r 1 0 and complete blockage could occur. The pipe should be insulated.

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PROBLEM 3.52 KNOWN: Inner surface temperature of insulation blanket comprised of two semi-cylindrical shells of different materials. Ambient air conditions. FIND: (a) Equivalent thermal circuit, (b) Total heat loss and material outer surface temperatures. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional, radial conduction, (3) Infinite contact resistance between materials, (4) Constant properties. ANALYSIS: (a) The thermal circuit is, conv,A conv,B 2 RR1 / r h π ′′ == () 21 cond A A ln r / r R k = < 2i cond B ln r / r R k = B The conduction resistances follow from Section 3.3.1 and Eq. 3.28. Each resistance is larger by a factor of 2 than the result of Eq. 3.28 due to the reduced area. (b) Evaluating the thermal resistances and the heat rate AB q=q q , + ( ) 1 2 conv R 0.1m 25 W/m K 0.1273 m K/W × = cond A cond B cond A ln 0.1m/0.05m R 0.1103 m K/W R 8 R 0.8825 m K/W 2 W/m K ⋅= = ×⋅ s,1 s,1 conv conv cond A cond B TT q= RR ∞∞ −− + ++ 500 300 K 500 300 K q = 842 198 W/m=1040 W/m. 0.1103+0.1273 m K/W 0.8825+0.1273 m K/W += + ⋅⋅ < Hence, the temperatures are () () s,1 A s,2 A cond A Wm K T T q R 500K 842 0.1103 407K mW =− = × = < s,1 B s,2 B cond B K T T q R 500K 198 0.8825 325K. = × = < COMMENTS: The total heat loss can also be computed from s,1 equiv q= T T /R , where 1 1 1 equiv conv,A cond(B) conv,B cond A R R R R R 0.1923 m K/W. =++ + =    Hence q = 500 300 K/0.1923 m K/W=1040 W/m. −⋅
PROBLEM 3.53 KNOWN: Surface temperature of a circular rod coated with bakelite and adjoining fluid conditions. FIND: (a) Critical insulation radius, (b) Heat transfer per unit length for bare rod and for insulation at critical radius, (c) Insulation thickness needed for 25% heat rate reduction. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in r, (3) Constant properties, (4) Negligible radiation and contact resistance.

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## This note was uploaded on 04/18/2008 for the course EGGN 471 taught by Professor Kee during the Spring '08 term at Mines.

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ch3b - PROBLEM 3.51 KNOWN Pipe wall temperature and...

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