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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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