# Thermal Fluids Tutorials(1)

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```					Course Assignment for Thermal Fluids Engineering/Viscous Flow Composite Wall The gap between aluminum (kAl = 235 W/mK) and copper (kCu = 400 W/mK) wall is filled with helium (kHe = 0.1513 W/mK), forming a composite wall with three layers. The aluminum wall (left) generates heat at a rate of 200 W/m3 and helium absorbs heat a rate of 175 W/m3. The left face of aluminum wall is kept at a constant temperature 400 K and the right face of the copper wall exchanges heat with the surroundings by convection (h = 20 W/mK, T∞ = 300 K). The walls and the gap have a thickness of 1 m. Find the temperature distribution in the walls and the gap.

T= 400 K

Al

He Cu

h = 20 W/m2K T= 300 K

Single Fin A rectangular fin of thickness 2 mm and length 40.2 mm generates haet at a rate 105 W/m3. The tip and base of the fin are at temperatures 200 and 100oC, respectively. Fin loses heat to the surrounding from its surfaces by convection (h = 50 W/m2K, T∞ = 20oC). Find the temperature distribution in the fin. Compare your results with 1-D analytical solution.

h = 50 W/m2K, T= 20oC T = 100oC T = 200oC

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2-D Heat Conduction within a Rectangular Bar Consider 2-D heat conduction in a steel (k = 20 W/mK) bar with rectangular crosssection, as given in the figure. Boundary conditions are: 1) The top is insulated. 2) The right side has a constant temperature of 100 K. 3) The left side is exposed to a constant heat flux at the rate of 50 W/m2. 4) The bottom is exposed to a convective boundary condition, T∞ = 200 K, h = 50 W/m2K. 5) Heat is uniformly generated in the bar at a rate of 20 W/m3. Determine the temperature distribution in the rectangle.

2-D Heat Conduction within a Furnace Wall Heat loss from a long furnace can be calculated by using a 2-D heat conduction model to find the temperature distribution in the walls. The furnace is made magnesia (KM = 1.41 W/mK), covered with two layers of red (KRB = 0.63 W/mK) and fire (KFB = 0.3 W/mK) brick from inside. The outside and inside surfaces of the walls are at temperatures 300 and 1000 K, respectively. The width and height of the furnace, as measured from outside, are 1.5 and 1.0 m, respectively. Each layer has a thickness of 0.05 m. Find the temperature distribution in each layer of the walls.

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2-D Heat Conduction between Parallel Plates Predicting the thermal behavior of a nano device requires accurate information about the thermal properties such as thermal conductivity and heat capacity. Thermal properties of nano structures, on the other hand, change with size, fabrication method, impurity, etc and are usually smaller than the bulk values. Different techniques are used to measure the thermal properties of the thin films. The steady state technique of uniformly heated suspended bridge is used to measure the lateral thermal conductivity of the thin films. In this method, heat loss from the bridge surface should be found as part of the measurement. A 2-D heat conduction model may be used to predict the heat loss from the lower surface of the bridge (of width w) to the substrate (infinite width) through the air gap. The rate of heat transfer changes with relative width of the bridge and the air gap (w/d). Dimensions are: bridge width, w = 20 x 10-6 m, gap distance, d = 50 x 10-6 m Assume known temperatures for the bridge and substrate. Find the temperature distribution inside the air gap. This can then be used to find a shape factor for heat loss from the bridge to the substrate.

The front view of suspended structure showing the separate distance d

Schematic of the simplified model for ANSYS analysis

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2-D Heat Conduction in a Heat Sink Electronic components generate heat during the course of their operation. To ensure reliable operation of the component, the generated heat needs to be removed and thus the electronic component be cooled. This is done by attaching a heat sink (a fin assembly) to the device to aid in rapid heat transfer to the surroundings. The objective of this problem is to model and analyze the performance of one such heat sink. In this problem heat is generated inside a mock electronic component (a steel block), by an electric heater (a circular rod). Heat is then transferred to the surroundings by convection. The steel block, copper rod, and aluminum fins (including their base plate) have thermal conductivities of 20, 386, and 180 W/mK, respectively. The total amount of heat heat generated in the rod is, Q = 27.2 W. 80% of this total heat generated is lost through the heat sink and the rest is lost from the bottom surface of the steel block. All other surfaces are assumed to be adiabatic. For convection from the fin surfaces, the heat transfer coefficient is h = 50 W/m2K. Outside temperature is assumed o be T∞ = 20oC. The heating rod is 129 mm long. All other dimensions are given on the figure. Use a 2-D model to predict the temperature distribution in the heat sink and mock processor.

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3-D Heat Conduction in a Heat Sink Electronic components generate heat during the course of their operation. To ensure reliable operation of the component, the generated heat needs to be removed and thus the electronic component be cooled. This is done by attaching a heat sink (a fin assembly) to the device to aid in rapid heat transfer to the surroundings. The objective of this problem is to model and analyze the performance of one such heat sink. In this problem heat is generated inside a mock electronic component (a steel block), by an electric heater (a circular rod). Heat is then transferred to the surroundings by convection. The steel block, copper rod, and aluminum fins (including their base plate) have thermal conductivities of 20, 386, and 180 W/mK, respectively. The total amount of heat heat generated in the rod is, Q = 27.2 W. 80% of this total heat generated is lost through the heat sink and the rest is lost from the bottom surface of the steel block. All other surfaces are assumed to be adiabatic. For convection from the fin surfaces, the heat transfer coefficient is h = 50 W/m2K. Outside temperature is assumed o be T∞ = 20oC. All dimensions are given on the figures. Use a 3-D model to predict temperature distribution in the heat sink and mock processor.

Isometric view

Front view

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2-D Flow between Parallel Plates Air at a uniform velocity of 0.1 m/s enter the space between two infinite parallel plate at a separation distance of 0.05 m, see the figure. Solve the viscous flow problem find the velocity distribution and its development in the entrance region. Physical properties of air are given as, density = 1.23 Kg/m3, and viscosity = 1.79e –5 Ns/m2.

Flow over a Flat Plate Air at a uniform velocity (0.15 m/s) flows over a flat plate of 0.8 m length. Solve the viscous flow problem to find velocity distribution in the region near the flat plate. The physical properties of air are given as, Density, ρ =1.23 kg/m3, and Viscosity, µ = 1.79E-5 N-s/m2. (Hint: To solve this problem numerically you need to define a solution domain around the flat plate. You also need to assume some thickness for the flat plate. You may assume that the flat plate is 1 mm thick and use the solution domain as given in the figure. Remember that you have slip conditions for the velocity at the boundaries of the solution domain.)

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 views: 10 posted: 12/23/2009 language: English pages: 6
Jun Wang Dr
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