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HEAT CONDUCTION EQUATION TEST 1
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Question 1 of 20
1. Question
1 pointsConsider an aluminum pan used to cook beef stew on top of an electric range. The bottom section of the pan is L =0.3 cm thick and has a diameter of D= 20 cm. The electric heating unit on the range top consumes 800 W of power during cooking, and 90 percent of the heat generated in the heating element is transferred to the pan. During steady operation, the temperature of the inner surface of the pan is measured to be 110°C. Express the boundary conditions for the bottom section of the pan during this cooking process.(answer upto one decimal)
Correct
The heat transfer through the bottom section of the pan is from the bottom surface toward the top and can reasonably be approximated as being onedimensional. We take the direction normal to the bottom surfaces of the pan as the x axis with the origin at the outer surface, as shown in the figure Then the inner and outer surfaces of the bottom section of the pan can be represented by x =0 and x= L, respectively. During steady operation, the temperature will depend on x only and thus T =T (x). The boundary condition on the outer surface of the bottom of the pan at x= 0 can be approximated as being specified heat flux since it is stated that 90 percent of the 800 W (i.e., 720 W) is transferred to the pan at that surface. Therefore, k=dT(0)/dx=qo qo=Heat transfer rate/Bottom surface area=0.720 kW/π(0.1 m)²=22.9 kW/m²
Incorrect
The heat transfer through the bottom section of the pan is from the bottom surface toward the top and can reasonably be approximated as being onedimensional. We take the direction normal to the bottom surfaces of the pan as the x axis with the origin at the outer surface, as shown in the figure Then the inner and outer surfaces of the bottom section of the pan can be represented by x =0 and x= L, respectively. During steady operation, the temperature will depend on x only and thus T =T (x). The boundary condition on the outer surface of the bottom of the pan at x= 0 can be approximated as being specified heat flux since it is stated that 90 percent of the 800 W (i.e., 720 W) is transferred to the pan at that surface. Therefore, k=dT(0)/dx=qo qo=Heat transfer rate/Bottom surface area=0.720 kW/π(0.1 m)²=22.9 kW/m²

Question 2 of 20
2. Question
1 pointsConsider a large plane wall of thickness L= 0.2 m, thermal conductivity k= 1.2 W/m · °C, and surface area A= 15 m². The two sides of the wall are maintained at constant temperatures of T1= 120°C and T2 =50°C, respectively, as shown in Figure below Determine the variation of temperature within the wall and the value of temperature at x= 0.1 m
Assumptions: 1 Heat conduction is steady. 2 Heat conduction is onedimensional since the wall is large relative to its thickness and the thermal conditions on both sides are uniform. 3 Thermal conductivity is constant. 4 There is no heat generation.
Properties: The thermal conductivity is given to be k =1.2 W/m · °C.
Correct
SOLUTION :A plane wall with specified surface temperatures is given. The variationof temperature and the rate of heat transfer are to be determined.
Analysis : Taking the direction normal to the surface of the wall to be the xdirection, the differential equation for this problem can be expressed as d²t/dx²=0 with boundary conditions
T(0) =T1 =120°C
T(L)= T2= 50°C
The differential equation is linear and second order, and a quick inspection of it reveals that it has a single term involving derivatives and no terms involving the unknown function T as a factor. Thus, it can be solved by direct integration. Noting that an integration reduces the order of a derivative by one, the general solution of the differential equation above can be obtained by two simple successive integrations, each of which introduces an integration constant. Integrating the differential equation once with respect to x yields
dT/dx=C1
where C1 is an arbitrary constant. Notice that the order of the derivative went down by one as a result of integration. As a check, if we take the derivative of this equation, we will obtain the original differential equation. This equation is not the solution yet since it involves a derivative. Integrating one more time, we obtain
T(x) =C1x + C2
which is the general solution of the differential equation (fig). The general solution in this case resembles the general formula of a straight line whose slope is C1 and whose value at x =0 is C2. This is not surprising since the second derivative represents the change in the slope of a function, and a zero second derivative indicates that the slope of the function remains constant.
Therefore, any straight line is a solution of this differential equation.The general solution contains two unknown constants C1 and C2, and thus we need two equations to determine them uniquely and obtain the specific solution. These equations are obtained by forcing the general solution to satisfy the specified boundary conditions. The application of each condition yields one equation, and thus we need to specify two conditions to determine the constants C1 and C2.
When applying a boundary condition to an equation, all occurrences of the dependent and independent variables and any derivatives are replaced by the specified values. Thus the only unknowns in the resulting equations are the arbitrary constants.
The first boundary condition can be interpreted as in the general solution, replace all the x’s by zero and T (x ) by T1. That is(fig)
T(0)= C1 x 0 +C2 → C2= T1
The second boundary condition can be interpreted as in the general solution, replace all the x’s by L and T (x ) by T2. That is,T(L)= C1L +C2 → T2= C1L +T1 → C1=T2 T1/L Substituting the C1 and C2 expressions into the general solution, we obtain
T(X)=T2 – T1/L x +T1
which is the desired solution since it satisfies not only the differential equation but also the two specified boundary conditions. That is, differentiating with respect to x twice will give d ²T /dx ², which is the given differential equation, and substituting x= 0 and x= L T (0)= T1 and T (L) =T2, respectively, which are the specified conditions at the boundaries. Substituting the given information, the value of the temperature at x= 0.1 m is determined to be
T(0.1 m)=(50120)°C/0.2 m(0.1 m) + 120°C= 85°C
Incorrect
SOLUTION :A plane wall with specified surface temperatures is given. The variationof temperature and the rate of heat transfer are to be determined.
Analysis : Taking the direction normal to the surface of the wall to be the xdirection, the differential equation for this problem can be expressed as d²t/dx²=0 with boundary conditions
T(0) =T1 =120°C
T(L)= T2= 50°C
The differential equation is linear and second order, and a quick inspection of it reveals that it has a single term involving derivatives and no terms involving the unknown function T as a factor. Thus, it can be solved by direct integration. Noting that an integration reduces the order of a derivative by one, the general solution of the differential equation above can be obtained by two simple successive integrations, each of which introduces an integration constant. Integrating the differential equation once with respect to x yields
dT/dx=C1
where C1 is an arbitrary constant. Notice that the order of the derivative went down by one as a result of integration. As a check, if we take the derivative of this equation, we will obtain the original differential equation. This equation is not the solution yet since it involves a derivative. Integrating one more time, we obtain
T(x) =C1x + C2
which is the general solution of the differential equation (fig). The general solution in this case resembles the general formula of a straight line whose slope is C1 and whose value at x =0 is C2. This is not surprising since the second derivative represents the change in the slope of a function, and a zero second derivative indicates that the slope of the function remains constant.
Therefore, any straight line is a solution of this differential equation.The general solution contains two unknown constants C1 and C2, and thus we need two equations to determine them uniquely and obtain the specific solution. These equations are obtained by forcing the general solution to satisfy the specified boundary conditions. The application of each condition yields one equation, and thus we need to specify two conditions to determine the constants C1 and C2.
When applying a boundary condition to an equation, all occurrences of the dependent and independent variables and any derivatives are replaced by the specified values. Thus the only unknowns in the resulting equations are the arbitrary constants.
The first boundary condition can be interpreted as in the general solution, replace all the x’s by zero and T (x ) by T1. That is(fig)
T(0)= C1 x 0 +C2 → C2= T1
The second boundary condition can be interpreted as in the general solution, replace all the x’s by L and T (x ) by T2. That is,T(L)= C1L +C2 → T2= C1L +T1 → C1=T2 T1/L Substituting the C1 and C2 expressions into the general solution, we obtain
T(X)=T2 – T1/L x +T1
which is the desired solution since it satisfies not only the differential equation but also the two specified boundary conditions. That is, differentiating with respect to x twice will give d ²T /dx ², which is the given differential equation, and substituting x= 0 and x= L T (0)= T1 and T (L) =T2, respectively, which are the specified conditions at the boundaries. Substituting the given information, the value of the temperature at x= 0.1 m is determined to be
T(0.1 m)=(50120)°C/0.2 m(0.1 m) + 120°C= 85°C

Question 3 of 20
3. Question
1 pointsConsider a large plane wall of thickness L= 0.2 m, thermal conductivity k= 1.2 W/m · °C, and surface area A= 15 m². The two sides of the wall are maintained at constant temperatures of T1= 120°C and T2 =50°C, respectively, as shown in Figure below Determine the rate of heat conduction
through the wall under steady conditions.Assumptions: 1 Heat conduction is steady. 2 Heat conduction is onedimensional since the wall is large relative to its thickness and the thermal conditions on both sides are uniform. 3 Thermal conductivity is constant. 4 There is no heat generation.
Properties: The thermal conductivity is given to be k =1.2 W/m · °C.
Correct
SOLUTION: A plane wall with specified surface temperatures is given. The variation of temperature and the rate of heat transfer are to be determined
The rate of heat conduction anywhere in the wall is determined from Fourier’s law to be
Qwall=KA dT/dX= KAC1=KA T2 T1/L=KA T2 T1/L
The numerical value of the rate of heat conduction through the wall is determined by substituting the given values to be
Q=KA T2 T1/L=(1.2 W/m · °C)(15 m²) (12050)°C/0.2 m=6300 W
Incorrect
SOLUTION: A plane wall with specified surface temperatures is given. The variation of temperature and the rate of heat transfer are to be determined
The rate of heat conduction anywhere in the wall is determined from Fourier’s law to be
Qwall=KA dT/dX= KAC1=KA T2 T1/L=KA T2 T1/L
The numerical value of the rate of heat conduction through the wall is determined by substituting the given values to be
Q=KA T2 T1/L=(1.2 W/m · °C)(15 m²) (12050)°C/0.2 m=6300 W

Question 4 of 20
4. Question
1 pointsA 2kW resistance heater wire whose thermal conductivity is k =15 W/m · °C has a diameter of D= 4 mm and a length of L= 0.5 m, and is used to boil water (Fig. ). If the outer surface temperature of the resistance wire is Ts= 105°C, determine the temperature at the center of the wire.
Assumptions 1 Heat transfer is steady since there is no change with time. 2 Heat transfer is onedimensional since there is thermal symmetry about the centerline and no change in the axial direction. 3 Thermal conductivity is constant. 4 Heat generation in the heater is uniform.
Properties The thermal conductivity is given to be k =15 W/m · °CCorrect
SOLUTION The surface temperature of a resistance heater submerged in water is to be determined.
Analysis The 2kW resistance heater converts electric energy into heat at a rate of 2 kW. The heat generation per unit volume of the wire is
g=Qgen/Vwire=Qgen/πr²L=2000 W/π(0.002 m)² (0.5 m)=0.318 109 W/m³
Then the center temperature of the wire is determined from
To=Ts+ gr²/4K=105°C+ (0.318 x 10^9 W/m³)(0.002 m)²/4 x (15 W/m · °C)=126· °C)
Incorrect
SOLUTION The surface temperature of a resistance heater submerged in water is to be determined.
Analysis The 2kW resistance heater converts electric energy into heat at a rate of 2 kW. The heat generation per unit volume of the wire is
g=Qgen/Vwire=Qgen/πr²L=2000 W/π(0.002 m)² (0.5 m)=0.318 109 W/m³
Then the center temperature of the wire is determined from
To=Ts+ gr²/4K=105°C+ (0.318 x 10^9 W/m³)(0.002 m)²/4 x (15 W/m · °C)=126· °C)

Question 5 of 20
5. Question
1 pointsConsider a 2mhigh and 0.7mwide bronze plate whose thickness is 0.1 m. One side of the plate is maintained at a constant temperature of 600 K while the other side is maintained at 400 K, as shown in Figure below The thermal conductivity of the bronze plate can be assumed to vary linearly in that temperature
range as k (T ) =k0(1 + ßT ) where k0 =38 W/m · K and ß= 9.21 x 10^4 K1. Disregarding the edge effects and assuming steady onedimensional heat transfer, determine the rate of heat conduction through the plate.Assumptions: 1 Heat transfer is given to be steady and onedimensional.
2 Thermal conductivity varies linearly. 3 There is no heat generation.
Properties: The thermal conductivity is given to be k (T )= k0(1 ßT )Correct
SOLUTION A plate with variable conductivity is subjected to specified temperatures on both sides. The rate of heat transfer is to be determined.
Analysis The average thermal conductivity of the medium in this case is simply the value at the average temperature and is determined from
Kave=K(Tave)=k0(1+ß T2 T1/2) =38 w/m.k[1 + (9.21 x 10^4 K1)(600+400)k/2]=55.5 w/m k
Then the rate of heat conduction through the plate can be determined from
Q=KaveAT1 – T2/L=(55.5 W/m · K)(2 m x0.7 m) (600400)k/0.1 m=155,400 W
Incorrect
SOLUTION A plate with variable conductivity is subjected to specified temperatures on both sides. The rate of heat transfer is to be determined.
Analysis The average thermal conductivity of the medium in this case is simply the value at the average temperature and is determined from
Kave=K(Tave)=k0(1+ß T2 T1/2) =38 w/m.k[1 + (9.21 x 10^4 K1)(600+400)k/2]=55.5 w/m k
Then the rate of heat conduction through the plate can be determined from
Q=KaveAT1 – T2/L=(55.5 W/m · K)(2 m x0.7 m) (600400)k/0.1 m=155,400 W

Question 6 of 20
6. Question
1 pointsIn a nuclear reactor, heat is generated uniformly in the 5cmdiameter cylindrical uranium rods at a rate of 7 x 10^7 W/m³ . If the length of the rods is 1 m, determine the rate of heat generation in each rod.
Correct
137.4 kW
Incorrect
137.4 kW

Question 7 of 20
7. Question
1 pointsIn a nuclear reactor, heat is generated uniformly in the 5cmdiameter cylindrical uranium rods at a rate of 7 x 10^7 W/m³ . If the length of the rods is 1 m, determine the rate of heat generation in each rod.
Correct
137.4 kW
Incorrect
137.4 kW

Question 8 of 20
8. Question
1 pointsConsider a large 3cmthick stainless steel plate in which heat is generated uniformly at a rate of 5 x 10^6 W/m³ Assuming the plate is losing heat from both sides, determine the heat flux on the surface of the plate during steady operation.
Correct
75,000 W/m²
Incorrect
75,000 W/m²

Question 9 of 20
9. Question
1 pointsConsider a large plane wall of thickness L 0.4 m, thermal conductivity k= 2.3 W/m · °C, and surface area A= 20 m² . The left side of the wall is maintained at a constant temperature of T1= 80°C while the right side loses heat by convection to the surrounding air at T ∞=15°C with a heat transfer coefficient of h= 24 W/m² · °C. Assuming constant thermal conductivity and no heat generation in the wall, evaluate the rate of heat transfer through the wall.
Correct
6030 W
Incorrect
6030 W

Question 10 of 20
10. Question
1 pointsConsider the base plate of a 800W household iron with 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, evaluate the inner surface temperature.
Correct
100°C
Incorrect
100°C

Question 11 of 20
11. Question
1 pointsConsider a large plane wall of thickness L =0.3 m, thermal conductivity k =2.5 W/m · °C, and surface area A =12 m². The left side of the wall at x= 0 is subjected to a net heat flux of q0 =700 W/m² while the temperature at that surface is measured to be T1= 80°C. Assuming constant thermal conductivity and no heat generation in the wall, evaluate the temperature of the right surface of the wall at x= L.
Correct
4°c
Incorrect
4°c

Question 12 of 20
12. Question
1 pointsA long homogeneous resistance wire of radius r0= 0.25 in. and thermal conductivity k =8.6 Btu/h · ft · °F is being used to boil water at atmospheric pressure by the passage of electric current. Heat is generated in the wire uniformly as a result of resistance heating at a rate of g= 1800 Btu/h · in³ The heat generated is transferred to water at 212°F by convection with an average heat transfer coefficient of h= 820 Btu/h · ft² · °F. Assuming steady onedimensional heat transfer, determine the temperature at the centerline of the wire.(answer upto one decimal)
Correct
290.8°F
Incorrect
290.8°F

Question 13 of 20
13. Question
1 pointsA 6mlong 2kW electrical resistance wire is made of 0.2cmdiameter stainless steel (k =15.1 W/m · °C). The resistance wire operates in an environment at 30°C with a heat transfer coefficient of 140 W/m² · °C at the outer surface. Determine the surface temperature of the wire by using the applicable relation and
Correct
409°C
Incorrect
409°C

Question 14 of 20
14. Question
1 pointsA 6mlong 2kW electrical resistance wire is made of 0.2cmdiameter stainless steel (k= 15.1 W/m · °C). The resistance wire operates in an environment at 30°C with a heat transfer coefficient of 140 W/m² · °C at the outer surface. Determine the surface temperature of the wire by setting up the proper differential equation and solving it.
Correct
409°C
Incorrect
409°C

Question 15 of 20
15. Question
1 pointsA long homogeneous resistance wire of radius r0= 5 mm is being used to heat the air in a room by the passage of electric current. Heat is generated in the wire uniformly at a rate of g=5 x 10^7 W/m³ as a result of resistance heating. If the temperature of the outer surface of the wire remains at 180°C, determine the temperature at r =2 mm after steady operation conditions are reached. Take the thermal conductivity of the wire to be k= 8 W/m · °C.(answer upto one decimal)
Correct
212.8°C
Incorrect
212.8°C

Question 16 of 20
16. Question
1 pointsConsider a large plane wall of thickness L= 0.05 m. The wall surface at x= 0 is insulated, while the surface at x= L is maintained at a temperature of 30°C. The thermal conductivity of the wall is k =30 W/m · °C, and heat is generated in the wall at a rate of g= g0e ^0.5x/L W/m³ where g0= 8 x 10^6 W/m³. Assuming steady onedimensional heat transfer, determine the temperature of the insulated surface of the wall.
Correct
314°C
Incorrect
314°C

Question 17 of 20
17. Question
1 pointsConsider a 1.5mhigh and 0.6mwide plate whose thickness is 0.15 m. One side of the plate is maintained at a constant temperature of 500 K while the other side is maintained at 350 K. The thermal conductivity of the plate can be assumed to vary linearly in that temperature range as k(T)= k0(1 ßT) where k0 =25 W/m · K and ß= 8.7 x 10^4 K1 . Disregarding the edge effects and assuming steady onedimensional heat transfer, determine the rate of heat conduction through the plate.
Correct
30,800 W
Incorrect
30,800 W

Question 18 of 20
18. Question
1 pointsConsider a large plane wall of thickness L= 0.5 ft and thermal conductivity k= 1.2 Btu/h · ft · °F. The wall is covered with a material that has an emissivity of ε=0.80 and a solar absorptivity of α=0.45. The inner surface of the wall is maintained at T1= 520 R at all times, while the outer surface is exposed to solar radiation that is incident at a rate of qsolar =300 Btu/h · ft² .The outer surface is also losing heat by radiation to deep space at 0 K. Determine the temperature of the outer surface of the wall and the rate of heat transfer through the wall when steady operating conditions are reached.(answer upto one decimal)
Correct
530.9 R, 26.2 Btu/h · ft²
Incorrect
530.9 R, 26.2 Btu/h · ft²

Question 19 of 20
19. Question
1 pointsThe boiling temperature of nitrogen at atmospheric pressure at sea level (1 atm pressure) is 196°C. Therefore, nitrogen is commonly used in low temperature scientific studies since the temperature of liquid nitrogen in a tank open to the atmosphere will remain constant at 196°C until the liquid nitrogen in the tank is depleted. Any heat transfer to the tank will result in the evaporation of some liquid nitrogen, which has a heat of vaporization of 198 kJ/kg and a density of 810 kg/m³ at 1 atm. Consider a thickwalled spherical tank of inner radius r1= 2 m, outer radius r2= 2.1 m , and constant thermal conductivity k =18 W/m · °C. The tank is initially filled with liquid nitrogen at 1 atm and 196°C, and is exposed to ambient air at T ∞=20°C with a heat transfer coefficient of h= 25 W/m² · °C. The inner surface temperature of the spherical tank is observed to be almost the same as the temperature of the nitrogen inside. Assuming steady onedimensional heat transfer determine the rate of evaporation of the liquid nitrogen in the tank as a result of the heat transfer from the ambient air.(answer upto two decimal)
Correct
1.32 kg/s
Incorrect
1.32 kg/s

Question 20 of 20
20. Question
1 pointsConsider a large plane wall of thickness L =0.4 m and thermal conductivity k =8.4 W/m · °C. There is no access to the inner side of the wall at x= 0 and thus the thermal conditions on that surface are not known. However, the outer surface of the wall at x =L, whose emissivity is ε= 0.7, is known to exchange heat by convection with ambient air at T∞=25°C with an average heat transfer coefficient of h =14 W/m² · °C as well as by radiation with the surrounding surfaces at an average temperature of Tsurr =290 K. Further, the temperature of the outer surface is measured to be T2= 45°C. Assuming steady onedimensional heat transfer, evaluate the inner surface temperature of the wall at x= 0
Correct
64.3°C
Incorrect
64.3°C