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Conduction & Convection Quiz 8 – 2014.01.24 Determine the heat transfer across the 1 inch-thick slab shown at the right, whose thermal conductivity varies linearly with temperature according to the equation 2 ft 3 ft 4 ft k T 20 0 . 5T k 350F B tu , T F o h ft F 2 250F o 1 in TIME IS UP!!! Outline 2. Conduction Heat Transfer 2.1. Series/Parallel Resistances 2.2. Geometric Considerations Outline 3. Convection Heat Transfer 3.1. Heat Transfer Coefficient 3.2. Dimensionless Groups for HTC Estimation Thermal Resistance Circuits Recall: Fourier’s Law of Heat Conduction ONE-DIMENSIONAL ONLY = − Driving force* −∆ = ∆ Thermal Resistanceǂ For a constant cross-section and isotropic thermal conductivity (e.g. a flat slab). *∆ = 2 − 1 where T1> T2 ǂ∆ = − where x > x 2 1 2 1 Series/Parallel Resistances For series resistances (flat slab): Driving Force at A: −∆ = Driving Force at B: −∆ = Overall driving force: ∆ = ∆ = −∆ = + Series/Parallel Resistances For series resistances (flat slab): −∆ = ∆ ∆ + = = = + ∆ = ∆ = −∆ = −∆ + −∆ Series/Parallel Resistances For parallel resistances (flat slab): Driving Force at A: −∆ = Driving Force at B: −∆ = ∆ = ∆ = Overall driving force: −∆ = + Series/Parallel Resistances For parallel resistances (flat slab): Substituting the individual Q: −∆ −∆ −∆ = + = + ∆ = ∆ = 1 1 1 = + −∆ = −∆ = −∆ Series/Parallel Resistances Exercise! (PIChE Quiz Bowl Nationals 2009, Easy Round, 2 min) A composite wall consists of 2-in. corkboard (inner), 6-in. concrete, and 3-in. wood (outer). The thermal conductivities of the materials are 0.025, 0.8, and 0.065 Btu/hr/ft/°F, respectively. The temperature of the inner surface of the wall is 55°F while the outer surface is at 90°F. What are the temperatures in °F: a) Between the cork and concrete? b) Between the concrete and wood? Series/Parallel Resistances Exercise! (Combined series & parallel) Four different materials were joined together as a block of constant width shown below. The cross-sectional area of B is equal to that of C. Materials A, B, C, and D have k = 0.1, 0.5, 0.4, and 0.1 W/mK, respectively. The thickness of blocks A and D is 1” and the thickness of blocks B and C is 6”. If the left of A is exposed to 110°C and the right of D is exposed to 60°C, calculate (a) the temperature right after material A and (b) the temperature right before material D. Geometric Considerations Heat Conduction Through Concentric Cylinders = − = 2 = − 2 Integrating both sides: 2 −2 ln = 2 − 1 1 Geometric Considerations Heat Conduction Through Concentric Cylinders = − = 2 = − 2 Rearranging: 2 = − 2 − 1 ln (2 1 ) Geometric Considerations Heat Conduction Through Concentric Cylinders Define a logarithmic mean area: (2 − 1 ) = 2 ln (2 1 ) 2 = − 2 − 1 ln (2 1 ) …and a logarithmic mean radius: (2 − 1 ) = ln (2 1 ) = 2 = − *Final form 2 − 1 2 − 1 Geometric Considerations Heat Conduction Through Concentric Cylinders When to use logarithmic mean area and when to use arithmetic mean area? r2/r1 (2 − 1 ) = 2 ln (2 1 ) (2 + 1 ) = 2 2 rLM rM 1 #DIV/0! 1 1.05 1.024797 1.025 1.1 1.049206 1.05 1.15 1.073254 1.075 1.2 1.096963 1.1 1.25 1.120355 1.125 1.3 1.143448 1.15 1.35 1.16626 1.175 1.4 1.188805 1.2 Geometric Considerations Heat Conduction Through Concentric Cylinders A tube of 60-mm (2.36-in.) outer diameter is insulated with a 50-mm (1.97-in.) layer of silica foam, for which k = 0.032 Btu/hr-ft-°F, followed by a 40-mm (1.57-in.) layer of cork with k = 0.03 Btu/hr/ft/°F. If the temperature of the outer surface of the pipe is 150°C, and the temperature of the outer surface of the cork is 30°C, calculate the heat loss in W/(m of pipe).