2011 CHME 333 Exam 1 Solutions

2011 CHME 333 Exam 1 Solutions - Two boundary conditions...

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Two boundary conditions need to be specified for a 1-dimensional conduction system (steady state or unsteady state) Four boundary conditions need to be specified for a 2-dimensional conduction system (steady state or unsteady state) One initial condition needs to be specified for any type of unsteady state conduction system. Constant Properties w.r.t. thermal conductivity - k is constant with time, with position, and with temperature in the material. w.r.t. convection coefficient - h is the same for the base surface and for the fluid that is flowing around the finned surfaces; h is not changing with time, position, or temperature.
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R insulation 1 Sk insul := R insulation 1.532 K W = Using the thermal resistance concept, we can now calculate the heat transfer rate q T R total = TT 1 T air := T7 3 K = A block w Length 4 := A block 0.4m 2 = R convection 1 h air A block := R convection 0.49 K W = R total R insulation R convection + = R total R insulation 1 h air A block + := R total 2.022 K W = Now calcluate q q T R total := q 36.098W = Answer A 2.5 cm diameter pipe carrying condensing steam at 101.3 kPa passes through the center of a square block of insulating material having k = 0.04 W/m·K. The block is 5 cm x 5cm, and 2 m long. The outside of the block is exposed to room air at 27°C and a convection coefficient of h = 5.1 W/m 2 ·K. Calculate the heat lost by the steam pipe. This is a multidimensional heat transfer problem involving common configurations, and so the Shape Factor can be use Assume 1. Steady State 2. Constant properties 3. Condensing steam - saturated steam at atmospheric pressure - T = 100 C 4. Condensing steam - neglect convection resistance inside pipe, pipe surface temp is 100 C 5. No pipe wall thickness given - neglect conduction resistance D 2.5cm := k insul 0.04 W mK := w 5cm := Length 2m := T air 300K := h air 5.1 W m 2 K := Steam is condensing at atmospheric pressure - saturated steam, so T = 100 C T 1 373K := The Shape factor to use is case 6 in Table 4.1.
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This note was uploaded on 03/11/2011 for the course CHEME 333 taught by Professor Anthony during the Spring '11 term at UNL.

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2011 CHME 333 Exam 1 Solutions - Two boundary conditions...

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