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NUMERICAL METHODS IN HEAT CONDUCTION TEST 1
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Question 1 of 10
1. Question
1 pointsConsider the base plate of a 800W household iron having a thickness of L =0.6 cm, base area of A= 160 cm² , and thermal conductivity of k =20 W/m · °C. The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. When steady operating conditions are reached, the outer surface temperature of the plate is measured to be 85°C. Disregarding any heat loss through the upper part of the iron and taking the nodal spacing to be 0.2 cm determine the inner surface temperature of the plate by solving those equations.
Correct
100°C
Incorrect
100°C

Question 2 of 10
2. Question
1 pointsUsing EES (or other) software, solve these systems of algebraic equations.
3×1 – x2 +3×3 = 0
x1+ 2×2 +x3= 3
2×1 x2 x3= 2
Correct
x1 =2, x2= 3, x3 = 1
Incorrect
x1 =2, x2= 3, x3 = 1

Question 3 of 10
3. Question
1 pointsUsing EES (or other) software, solve these systems of algebraic equations
4×1 – 2x²2+ 0.5×3 =2
X³1X2+X3=11.964
x1 + x2+ x3 = 3
Correct
x1= 2.33, x2= 2.29 / x3 =1.62
Incorrect
x1= 2.33, x2= 2.29 / x3 =1.62

Question 4 of 10
4. Question
1 pointsConsider steady twodimensional heat transfer in a long solid bar whose cross section is given in the figure. The measured temperatures at selected points of the outer surfaces are as shown. The thermal conductivity of the body is k =20 W/m · °C, and there is no heat generation. Using the finite difference method with a mesh size of Δx =Δy 1.0 cm, determine the temperatures at the indicated pointS in the medium.
Correct
T1 =T4= 143°C, T2 =T3= 136°C
Incorrect
T1 =T4= 143°C, T2 =T3= 136°C

Question 5 of 10
5. Question
1 pointsConsider steady twodimensional heat transfer in a long solid bar of square cross section in which heat is generated uniformly at a rate of g= 0.19 x 10^5 Btu/h · ft³ . The cross section of the bar is 0.4 ft x 0.4 ft in size, and its thermal conductivity is k =16 Btu/h · ft · °F. All four sides of the bar are subjected to convection with the ambient air at T∞= 70°F with a heat transfer coefficient of h= 7.9 Btu/h · ft² · °F. Using the finite difference method with a mesh size of Δx= Δy 0.2 ft, determine the rate of heat loss from the bar through a 1ftlong section.
Correct
3040 Btu/h
Incorrect
3040 Btu/h

Question 6 of 10
6. Question
1 pointsConsider a long concrete dam (k =0.6 W/m · °C, ∝s= 0.7 m² /s) of triangular cross section whose exposed surface is subjected to solar heat flux of qs= 800 W/m² and to convection and radiation to the environment at 25°C with a combined heat transfer coefficient of 30 W/m2 · °C. The 2mhigh vertical section of the dam is subjected to convection by water at 15°C with a heat transfer coefficient of 150 W/m² · °C, and heat transfer through the 2mlong base is considered to be negligible. Using the finite difference method with a mesh size of Δx =Δy= 1 m and assuming steady twodimensional heat transfer, determine the temperature of the top, middle, and bottom of the exposed surface of the dam.(answer upto one decimal)
Correct
21.3°C, 43.2°C, 43.6°C
Incorrect
21.3°C, 43.2°C, 43.6°C

Question 7 of 10
7. Question
1 pointsConsider a long solid bar whose thermal conductivity is k= 12 W/m · °C and whose cross section is given in the figure. The top surface of the bar is maintained at 50°C while the bottom surface is maintained at 120°C. The left surface is insulated and the remaining three surfaces are subjected to convection with ambient air at T∞=25°C with a heat transfer coefficient of h= 30 W/m²· °C. Using the finite difference method with a mesh size of Δx =Δy= 10 cm, determine the unknown nodal temperatures by solving those equations.
Correct
85.7°C, 86.4°C, 87.6°C
Incorrect
85.7°C, 86.4°C, 87.6°C

Question 8 of 10
8. Question
1 pointsConsider a large plane wall of thickness L= 0.3 ft and thermal conductivity k= 1.2 Btu/h · ft · °F in space. The wall is covered with a material having an emissivity of ε=0.80 and a solar absorptivity of ∝s= 0.45. The inner surface of the wall is maintained at 520 R at all times, while the outer surface is exposed to solar radiation that is incident at a rate of qs= 300 Btu/h · ft². The outer surface is also losing heat by radiation to deep space at 0 R. Using a uniform nodal spacing of Δx= 0.1 ft, determine the nodal temperatures by solving those equations.
Correct
522 R, 525 R, 527 R
Incorrect
522 R, 525 R, 527 R

Question 9 of 10
9. Question
1 pointsHot combustion gases of a furnace are flowing through a square chimney made of concrete (k= 1.4 W/m · °C). The flow section of the chimney is 20 cm x 20 cm, and the thickness of the wall is 20 cm. The average temperature of the hot gases in the chimney is Ti= 300°C, and the average convection heat transfer coefficient inside the chimney is hi =70 W/m² · °C. The chimney is losing heat from its outer surface to the ambient air at To =20°C by convection with a heat transfer coefficient of ho =21 W/m² · °C and to the sky by radiation. The emissivity of the outer surface of the wall is ε=0.9, and the effective sky temperature is estimated to be 260 K. Using the finite difference method with Δx =Δy= 10 cm and taking full advantage of symmetry, determine the temperatures at the nodal points of a cross section and the rate of heat loss for a 1mlong section of the chimney
Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer through the chimney is twodimensional since the height of the chimney is large relative to its cross section, and thus heat conduction through the chimney in the axial direction is negligible. It is tempting to simplify the problem further by considering heat transfer in each wall to be onedimensional, which would be the case if the walls were thin and thus the corner effects were negligible. This assumption cannot be justified in this case since the walls are very thick and the corner sections constitute a considerable portion of the chimney structure. 3 Thermal conductivity is constant.
Properties The properties of chimney are given to be k= 1.4 W/m · °C and ε= 0.9.
Correct
Analysis The cross section of the chimney is given in (Figure ) The most striking aspect of this problem is the apparent symmetry about the horizontal and vertical lines passing through the midpoint of the chimney as well as the diagonal axes, as indicated on the figure. Therefore, we need to consider only oneeighth of the geometry in the solution whose nodal network consists of nine equally spaced nodes. No heat can cross a symmetry line, and thus symmetry lines can be treated as insulated surfaces and thus “mirrors” in the finite difference formulation.
Then the nodes in the middle of the symmetry lines can be treated as interior nodes by using mirror images. Six of the nodes are boundary nodes, so we will have to write energy balances to obtain their finite difference formulations. First we partition the region among the nodes equitably by drawing dashed lines between the nodes through the middle. Then the region around a node surrounded by the boundary or the dashed lines represents the volume element of the node. Considering a unit depth and using the energy balance approach for the boundary nodes (again assuming all heat transfer into the volume element for convenience) and the formula for the interior nodes, the finite difference equations for the nine nodes are determined as follows:Node 1. On the inner boundary, subjected to convection,(fig)
This problem involves radiation, which requires the use of absolute temperature, and thus all temperatures should be expressed in Kelvin. Alternately, we could use °C for all temperatures provided that the four temperatures in the radiation terms are expressed in the form (T +273)^4. Substituting the given quantities, the system of nine equations for the determination of nine unknown nodal temperatures in a form suitable for use with the GaussSeidel iteration method becomes
which is a system of nonlinear equations. Using an equation solver, its solution
is determined to beThe variation of temperature in the chimney is shown in Figure . Note that the temperatures are highest at the inner wall (but less than 300°C) and lowest at the outer wall (but more that 260 K), as expected. The average temperature at the outer surface of the chimney weighed by the surface area is
Then the rate of heat loss through the 1mlong section of the chimney can be determined approximately from
The difference between the two results is due to the approximate nature of the numerical analysis.
Incorrect
Analysis The cross section of the chimney is given in (Figure ) The most striking aspect of this problem is the apparent symmetry about the horizontal and vertical lines passing through the midpoint of the chimney as well as the diagonal axes, as indicated on the figure. Therefore, we need to consider only oneeighth of the geometry in the solution whose nodal network consists of nine equally spaced nodes. No heat can cross a symmetry line, and thus symmetry lines can be treated as insulated surfaces and thus “mirrors” in the finite difference formulation.
Then the nodes in the middle of the symmetry lines can be treated as interior nodes by using mirror images. Six of the nodes are boundary nodes, so we will have to write energy balances to obtain their finite difference formulations. First we partition the region among the nodes equitably by drawing dashed lines between the nodes through the middle. Then the region around a node surrounded by the boundary or the dashed lines represents the volume element of the node. Considering a unit depth and using the energy balance approach for the boundary nodes (again assuming all heat transfer into the volume element for convenience) and the formula for the interior nodes, the finite difference equations for the nine nodes are determined as follows:Node 1. On the inner boundary, subjected to convection,(fig)
This problem involves radiation, which requires the use of absolute temperature, and thus all temperatures should be expressed in Kelvin. Alternately, we could use °C for all temperatures provided that the four temperatures in the radiation terms are expressed in the form (T +273)^4. Substituting the given quantities, the system of nine equations for the determination of nine unknown nodal temperatures in a form suitable for use with the GaussSeidel iteration method becomes
which is a system of nonlinear equations. Using an equation solver, its solution
is determined to beThe variation of temperature in the chimney is shown in Figure . Note that the temperatures are highest at the inner wall (but less than 300°C) and lowest at the outer wall (but more that 260 K), as expected. The average temperature at the outer surface of the chimney weighed by the surface area is
Then the rate of heat loss through the 1mlong section of the chimney can be determined approximately from
The difference between the two results is due to the approximate nature of the numerical analysis.

Question 10 of 10
10. Question
1 pointsConsider a large uranium plate of thickness L =4 cm and thermal conductivity k= 28 W/m · °C in which heat is generated uniformly at a constant rate of g=5 x 10^6 W/m³. One side of the plate is maintained at 0°C by iced water while the other side is subjected to convection to an environment at T 30°C with a heat transfer coefficient of h =45 W/m² · °C, as shown in (Figure) Considering a total of three equally spaced nodes in the medium, two at the boundaries and one at the middle, estimate the exposed surface temperature of the plate under steady conditions using the finite difference approach.
Assumptions 1 Heat transfer through the wall is steady since there is no indication of any change with time. 2 Heat transfer is onedimensional since the plate is large relative to its thickness. 3 Thermal conductivity is constant. 4 Radiation heat transfer is negligible.
Properties The thermal conductivity is given to be k= 28 W/m · °CCorrect
SOLUTION A uranium plate is subjected to specified temperature on one side and convection on the other. The unknown surface temperature of the plate is to be determined numerically using three equally spaced nodes
Analysis The number of nodes is specified to be M= 3, and they are chosen to be at the two surfaces of the plate and the midpoint, as shown . Then the nodal spacing Δx becomes
Δx=L/M1=0.04 m/31=0.02 m
We number the nodes 0, 1, and 2. The temperature at node 0 is given to be T0 =0°C, and the temperatures at nodes 1 and 2 are to be determined. This problem involves only two unknown nodal temperatures, and thus we need to have only two equations to determine them uniquely. These equations are obtained by applying the finite difference method to nodes 1 and 2.
Node 1 is an interior node, and the finite difference formulation at that node is obtained directly
T02T1+T2/Δx²+ g1/K=0 →02T1+T2/ΔX²+g1/K=0→2T12T2=g1ΔX²/K (1)
Node 2 is a boundary node subjected to convection, and the finite difference formulation at that node is obtained by writing an energy balance on the volume element of thickness Δx/2 at that boundary by assuming heat transfer to be into the medium at all sides:
hA(T∞T2)+KA T1T2/ΔX+g2(AΔX/2)=0
Canceling the heat transfer area A and rearranging give
T1(1+hΔx/K)T2=hΔx/K T∞ g2Δx²/2K (2)
Equations (1) and (2) form a system of two equations in two unknowns T1 and T2. Substituting the given quantities and simplifying gives
2T1T2=71.43 (in °C)
T11.032T=36.88 (in °C)
This is a system of two algebraic equations in two unknowns and can be solved easily by the elimination method. Solving the first equation for T1 and substituting into the second equation result in an equation in T2 whose solution is
T2=136.1 °C
This is the temperature of the surface exposed to convection, which is the desired result. Substitution of this result into the first equation gives T1= 103.8°C, which is the temperature at the middle of the plate
Incorrect
SOLUTION A uranium plate is subjected to specified temperature on one side and convection on the other. The unknown surface temperature of the plate is to be determined numerically using three equally spaced nodes
Analysis The number of nodes is specified to be M= 3, and they are chosen to be at the two surfaces of the plate and the midpoint, as shown . Then the nodal spacing Δx becomes
Δx=L/M1=0.04 m/31=0.02 m
We number the nodes 0, 1, and 2. The temperature at node 0 is given to be T0 =0°C, and the temperatures at nodes 1 and 2 are to be determined. This problem involves only two unknown nodal temperatures, and thus we need to have only two equations to determine them uniquely. These equations are obtained by applying the finite difference method to nodes 1 and 2.
Node 1 is an interior node, and the finite difference formulation at that node is obtained directly
T02T1+T2/Δx²+ g1/K=0 →02T1+T2/ΔX²+g1/K=0→2T12T2=g1ΔX²/K (1)
Node 2 is a boundary node subjected to convection, and the finite difference formulation at that node is obtained by writing an energy balance on the volume element of thickness Δx/2 at that boundary by assuming heat transfer to be into the medium at all sides:
hA(T∞T2)+KA T1T2/ΔX+g2(AΔX/2)=0
Canceling the heat transfer area A and rearranging give
T1(1+hΔx/K)T2=hΔx/K T∞ g2Δx²/2K (2)
Equations (1) and (2) form a system of two equations in two unknowns T1 and T2. Substituting the given quantities and simplifying gives
2T1T2=71.43 (in °C)
T11.032T=36.88 (in °C)
This is a system of two algebraic equations in two unknowns and can be solved easily by the elimination method. Solving the first equation for T1 and substituting into the second equation result in an equation in T2 whose solution is
T2=136.1 °C
This is the temperature of the surface exposed to convection, which is the desired result. Substitution of this result into the first equation gives T1= 103.8°C, which is the temperature at the middle of the plate