Page 1 PROBLEM 3.1 KNOWN: One-dimensional, plane wall ...

PROBLEM 3.49KNOWN: Temperature and convection coefficient associated with steam flow through a pipeof prescribed inner and outer diameters. Outer surface emissivity and convection coefficient.Temperature of ambient air and surroundings.FIND: Heat loss per unit length.SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat transfer, (3)Constant properties, (4) Surroundings form a large enclosure about pipe.PROPERTIES: Table A-1, Steel, AISI 1010 (T ≈ 450 K): k = 56.5 W/m⋅K.ANALYSIS: Referring to the thermal circuit, it follows from an energy balance on the outersurface thatT∞,i −Ts,o Ts,o −T∞,o Ts,o −Tsur= +Rconv,i + Rcond Rconv,o Rrador from Eqs. 3.9, 3.28 and 1.7,T∞,i −Ts,o Ts,o −T∞,o 4 4= + επ D( ) ( ) ( )( )i i o i o oo σ T s,o −T1/ π D h + ln D / D / 2πk 1/ π D hsur523K − Ts,oTs,o−293K=2−1 ln ( 75/60)2−1( π× 0.6m× 500 W/m ⋅ K)+ ( π× 0.075m× 25 W/m ⋅K2π56.5 W/m K)× ⋅+0.8π8 2× ( 0.075m)× 5.67× 10−W/m ⋅K4⎡T 4 4 4s,o 293 ⎤K⎢−⎣ ⎥⎦523−Ts,o Ts,o −293 1.07 10−8 T4 4= + × ⎡s,o − 293 ⎤ .0.0106+0.0006 0.170⎢⎣⎥⎦From a trial-and-error solution, T s,o ≈ 502K. Hence the heat loss isq ′ = π D4 4oho Ts,o − T ∞,o + επ DoσTs,o −Tsur( ) ( )2 8 W 4 4 4q ′ −= π( 0.075m) 25 W/m ⋅ K( 502-293) + 0.8 π( 0.075m)5.67 × 10 ⎡502 −243 ⎤K2 4m ⋅ K ⎣ ⎦q ′ =1231 W/m+600 W/m=1831 W/m.