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ME 375 Heat Transfer
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ME 375 Heat Transfer
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Course Code: ME375
University: California State University, Northridge
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Country: United States
Questions:
(1)Use the finite difference method to determine the temperature for each node.
(2)Plot the temperatures between points A and B for three convection coefficients (h), 1 W/(m2?oC), 10 W/(m2?oC) and 20 W/(m2?oC), and discuss the results.
(3)Plot the temperatures between points A and B for three thermal conductivities (k), 1 W/(m?oC), 20 W/(m?oC) and 50 W/(m?oC), and discuss the results
Answer:
Introduction
Heat transfer can be defined as an amount of energy in transit due to temperature difference, between two local points. Heat usually flows naturally from a high temperature region to a low temperature region. Heat can be transferred through three modes i.e. conduction, convection and radiation. When a boundary is perfectly insulated, then there is no transfer of heat across the wall an adiabatic condition; such a process is mainly found in compressors, turbines and boilers for heat insulation. In the design of energy utilities in which there is to some extent transfer of heat, there is a need to do an analysis of the distribution of heat through the material or thermal changes caused by external temperature variations. This report presents one of the common numerical methods used on the analysis of such distributions. The Finite Difference Method (FDM) is a numerical solution method in which a large body is divided into small discrete shapes which are used to compute the physical properties of the whole body. It uses backward, forward and central differencing schemes to analyze the differential equations.
In this problem temperature distributions are computed at the nodes of each square while considering the boundary conditions. Both Fourier’s law of heat conduction and Newton’s law of cooling are used to determine the nodal temperatures. For a steady state heat flow, the heat equation becomes:
The nodal temperature at node is given by
The boundary conditions on the body may be subjected to a heat flux a convective heat transfer with a coefficient of and ambient temperature or the boundary can be considered insulated. Such are analyzed as shown below
Element subjected to a heat flux along
Where is the length of the squares making up the body while is the thermal conductivity of the solid body.
Element’s wall is perfectly insulated
Element subjected to convective heat transfer at the boundary
Problem statement
Consider steady two-dimensional heat transfer in a solid body whose cross section is given in the figure. The thermal conductivity of the body is and there is no heat generation. The entire top surface is subjected to convection with ambient air at with a convection coefficient of The bottom surface is maintained at. The left side surface of the body is insulated; the right side surface of the body is under constant heat flux The mesh size is
Use the finite difference method to determine the temperature for each node.
Plot the temperatures between points A and B for three convection coefficients and and discuss the results.
Plot the temperatures between points A and B for three thermal conductivities and and discuss the results.
Calculations and Discussions
Temperature at nodes
Temperature at the internal nodes are given by
Considering the left hand boundary (insulated boundary), the temperature at the nodes is given by equation (iv) hence
Considering the heat flux, the temperature at the nodes is obtained through equation (iii)
Considering the convective heat transfer the temperature at the nodes are
When the nodal temperatures are determined using iterative method in an excel workbook then the values are presented in table 1 below.
Table 1: Iterated results using excel spreadsheet
6.131234
8.316527
11.66611
2.210731
1.778982
16.09118
36.11268
36.92069
49.47312
39.08816
42.4907
58.94554
59.88778
63.36645
62.61094
65.69583
79.89595
80.31957
81.49496
82.29416
85.07104
100
100
100
100
100
b) Plotting the temperatures between A and B for various h gives
It is seen that the temperature flows from the right-hand side of the body to the left-hand side in a decreasing manner due to heat loss. There is a momentary temperature rise at the ridge of the block to high heat accumulation. For lower coefficients of convective heat transfer there is minimal temperature reduction gradient compared to higher coefficients of convective heat transfer. Lower coefficients make minimal transfer of heat from the heated body to the moving air.
there is a higher rate of temperature reduction when using blocks made of materials of lower thermal conductivity that those with higher values. At the ridge of the block, more heat is lost for lower k-values than higher values. The above is due to the ability of low thermal conductivity materials to loss heat faster when they encounter a non-uniformity in their structure.
Conclusion
Through the above problem, the solution to complex numerical problem of heat flow was thus approved and it has been confirmed that heat flows from high temperature region to low temperature region, in the process losing its energy. The above process is mainly used in the design of electronic gadgets to simulate the heat distribution throughout the printed circuit board (PCB) so that the designer can determine components that can be adversely affected by heat.
References
Holman, J.P., “Heat Transfer”, Tenth Edition, McGraw-Hill, 2010Chapman, A.J., “Heat Transfer”, Fourth Edition, Macmillan Publishing, 1984Hagen, K.D., “Heat Transfer with Applications”, First Edition, Prentice Hall, 1999
Steven C. Chapra, “Numerical Methods for Engineers”, 7th edition, McGraw-Hill, 2015
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