Quizsummary
0 of 20 questions completed
Questions:
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 11
 12
 13
 14
 15
 16
 17
 18
 19
 20
Information
STEADY HEAT CONDUCTION TEST 1
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
0 of 20 questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 points, (0)
Average score 

Your score 

Categories
 Not categorized 0%
Pos.  Name  Entered on  Points  Result 

Table is loading  
No data available  
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 11
 12
 13
 14
 15
 16
 17
 18
 19
 20
 Answered
 Review

Question 1 of 20
1. Question
1 pointsConsider a 3mhigh, 5mwide, and 0.3mthick wall whose thermal conductivity is k =0.9 W/m · °C (Fig.). On a certain day, the temperatures of the inner and the outer surfaces of the wall are measured to be 16°C and 2°C, respectively. Determine the rate of heat loss through the wall on that day.
Assumptions 1 Heat transfer through the wall is steady since the surface temperatures remain constant at the specified values. 2 Heat transfer is onedimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors. 3 Thermal conductivity is constant.
Properties The thermal conductivity is given to be k= 0.9 W/m · °C.Correct
SOLUTION :The two surfaces of a wall are maintained at specified temperatures.
The rate of heat loss through the wall is to be determined.Incorrect
SOLUTION :The two surfaces of a wall are maintained at specified temperatures.
The rate of heat loss through the wall is to be determined. 
Question 2 of 20
2. Question
1 pointsThe thermal contact conductance at the interface of two 1cmthick aluminum plates is measured to be 11,000 W/m² · °C. Determine the thickness of the aluminum plate whose thermal resistance is equal to the thermal resistance of the interface between the plates (Fig.)
Properties The thermal conductivity of aluminum at room temperature is k =237 W/m · °C (ANSWER UPTO TWO DECIMAL)
Correct
SOLUTION: The thickness of the aluminum plate whose thermal resistance is equal to the thermal contact resistance is to be determined
Analysis: Noting that thermal contact resistance is the inverse of thermal contact conductance, the thermal contact resistance is
Rc=1/hc=1/11,000 W/m² · °C=0.909 10^4 m²· °C/ W
For a unit surface area, the thermal resistance of a flat plate is defined as R=L/K
where L is the thickness of the plate and k is the thermal conductivity. Setting R= Rc, the equivalent thickness is determined from the relation above to be L=KRc=(237 W/m · °C)(0.909 10^4 m² · °C/ W)=0.0215 m=2.15 cm
Incorrect
SOLUTION: The thickness of the aluminum plate whose thermal resistance is equal to the thermal contact resistance is to be determined
Analysis: Noting that thermal contact resistance is the inverse of thermal contact conductance, the thermal contact resistance is
Rc=1/hc=1/11,000 W/m² · °C=0.909 10^4 m²· °C/ W
For a unit surface area, the thermal resistance of a flat plate is defined as R=L/K
where L is the thickness of the plate and k is the thermal conductivity. Setting R= Rc, the equivalent thickness is determined from the relation above to be L=KRc=(237 W/m · °C)(0.909 10^4 m² · °C/ W)=0.0215 m=2.15 cm

Question 3 of 20
3. Question
1 pointsA 3mhigh and 5mwide wall consists of long 16cm X 22cm cross section horizontal bricks (k =0.72 W/m · °C) separated by 3cmthick plaster layers (k =0.22 W/m · °C). There are also 2cmthick plaster layers on each side of the brick and a 3cmthick rigid foam (k =0.026 W/m · °C) on the inner side of the wall, as shown in Fig. . The indoor and the outdoor temperatures are 20°C and 10°C, and the convection heat transfer coefficients on the inner and the outer sides are h1= 10 W/m² · °C and h2= 25 W/²2 · °C, respectively. Assuming onedimensional heat transfer and disregarding radiation, determine the rate of heat transfer through the wall.
Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer can be approximated as being onedimensional since it is predominantly in the x direction. 3 Thermal conductivities are constant. 4 Heat transfer by radiation is negligible.
Properties The thermal conductivities are given to be k =0.72 W/m · °C for bricks, k =0.22 W/m · °C for plaster layers, and k =0.026 W/m · °C for the rigid foam.
Correct
SOLUTION The composition of a composite wall is given. The rate of heat transfer through the wall is to be determined.
Analysis There is a pattern in the construction of this wall that repeats itself every 25cm distance in the vertical direction. There is no variation in the horizontal direction. Therefore, we consider a 1mdeep and 0.25mhigh portion of the wall, since it is representative of the entire wall. Assuming any cross section of the wall normal to the xdirection to be isothermal, the thermal resistance network for the representative section of the wall becomes as shown . The individual resistances are evaluated as:
Ri=Rconv,1=1/h1A=1/(10 W/m²· °C)(0.25 1 m²)=0.4°C/ W
R1=Rfoam=L/KA=0.03m/(0.026 W/m · °C)(0.25 1 m²)=4.6°C/ W
R2=R6=Rplaster,,SIDE=L/KA=0.02m/(0.22 W/m · °C)(0.25 1 m²)=0.36°C/ W
R3=R5=Rplaster,centre=L/KA=0.16 m/(0.22 W/m · °C)(0.015 1 m²)=48.48°C/ W
R4=Rbrick=L/KA=0.16 m/(0.72 W/m · °C)(0.22 1 m²)=1.01°C/ W
Ro=Rconv,2=1/h2A=1/(25 W/m2 · °C)(0.25 1 m²)=0.16°C/ W
The three resistances R3, R4, and R5 in the middle are parallel, and their equivalent
resistance is determined fromOf course, this result is approximate, since we assumed the temperature within
the wall to vary in one direction only and ignored any temperature change (and
thus heat transfer) in the other two directionsIncorrect
SOLUTION The composition of a composite wall is given. The rate of heat transfer through the wall is to be determined.
Analysis There is a pattern in the construction of this wall that repeats itself every 25cm distance in the vertical direction. There is no variation in the horizontal direction. Therefore, we consider a 1mdeep and 0.25mhigh portion of the wall, since it is representative of the entire wall. Assuming any cross section of the wall normal to the xdirection to be isothermal, the thermal resistance network for the representative section of the wall becomes as shown . The individual resistances are evaluated as:
Ri=Rconv,1=1/h1A=1/(10 W/m²· °C)(0.25 1 m²)=0.4°C/ W
R1=Rfoam=L/KA=0.03m/(0.026 W/m · °C)(0.25 1 m²)=4.6°C/ W
R2=R6=Rplaster,,SIDE=L/KA=0.02m/(0.22 W/m · °C)(0.25 1 m²)=0.36°C/ W
R3=R5=Rplaster,centre=L/KA=0.16 m/(0.22 W/m · °C)(0.015 1 m²)=48.48°C/ W
R4=Rbrick=L/KA=0.16 m/(0.72 W/m · °C)(0.22 1 m²)=1.01°C/ W
Ro=Rconv,2=1/h2A=1/(25 W/m2 · °C)(0.25 1 m²)=0.16°C/ W
The three resistances R3, R4, and R5 in the middle are parallel, and their equivalent
resistance is determined fromOf course, this result is approximate, since we assumed the temperature within
the wall to vary in one direction only and ignored any temperature change (and
thus heat transfer) in the other two directions 
Question 4 of 20
4. Question
1 pointsA 3mmdiameter and 5mlong electric wire is tightly wrapped with a 2mmthick plastic cover whose thermal conductivity is k= 0.15 W/m · °C. Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire. If the insulated wire is exposed to a medium at T∞= 30°C with a heat transfer coefficient of h= 12 W/m² · °C, determine the temperature at the interface of the wire and the plastic cover in steady operation. Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature.
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is onedimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal conductivities are constant. 4 The thermal contact resistance at the interface is negligible. 5 Heat transfer coefficient incorporates the radiation effects, if any.
Properties The thermal conductivity of plastic is given to be k= 0.15 W/m · °C.Correct
SOLUTION An electric wire is tightly wrapped with a plastic cover. The interface temperature and the effect of doubling the thickness of the plastic cover on the interface temperature are to be determined
Analysis Heat is generated in the wire and its temperature rises as a result of resistance heating. We assume heat is generated uniformly throughout the wire and is transferred to the surrounding medium in the radial direction. In steady operation, the rate of heat transfer becomes equal to the heat generated within the wire, which is determined to be
Incorrect
SOLUTION An electric wire is tightly wrapped with a plastic cover. The interface temperature and the effect of doubling the thickness of the plastic cover on the interface temperature are to be determined
Analysis Heat is generated in the wire and its temperature rises as a result of resistance heating. We assume heat is generated uniformly throughout the wire and is transferred to the surrounding medium in the radial direction. In steady operation, the rate of heat transfer becomes equal to the heat generated within the wire, which is determined to be

Question 5 of 20
5. Question
1 pointsSteam at T∞1= 320°C flows in a cast iron pipe (k= 80 W/m · °C) whose inner and outer diameters are D1= 5 cm and D2=5.5 cm, respectively. The pipe is covered with 3cmthick glass wool insulation with k= 0.05 W/m · °C. Heat is lost to the surroundings at T ∞2= 5°C by natural convection and radiation, with a combined heat transfer coefficient of h2= 18 W/m² · °C. Taking the heat transfer coefficient inside the pipe to be h1 =60 W/m² · °C, determine the rate of heat loss from the steam per unit length of the pipe. Also determine the temperature drops across the pipe shell and the insulation.
Assumptions: 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is onedimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal conductivities are constant. 4 The thermal contact resistance at the interface is negligible.
Properties: The thermal conductivities are given to be k =80 W/m · °C for cast iron and k= 0.05 W/m · °C for glass wool insulation.Correct
SOLUTION A steam pipe covered with glass wool insulation is subjected to convection on its surfaces. The rate of heat transfer per unit length and the temperature drops across the pipe and the insulation are to be determined
That is, the temperatures between the inner and the outer surfaces of the pipe
differ by 0.02°C, whereas the temperatures between the inner and the outer
surfaces of the insulation differ by 284°C.Incorrect
SOLUTION A steam pipe covered with glass wool insulation is subjected to convection on its surfaces. The rate of heat transfer per unit length and the temperature drops across the pipe and the insulation are to be determined
That is, the temperatures between the inner and the outer surfaces of the pipe
differ by 0.02°C, whereas the temperatures between the inner and the outer
surfaces of the insulation differ by 284°C. 
Question 6 of 20
6. Question
1 pointsA 3mmdiameter and 5mlong electric wire is tightly wrapped with a 2mmthick plastic cover whose thermal conductivity is k =0.15 W/m · °C. Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire. If the insulated wire is exposed to a medium at T∝= 30°C with a heat transfer coefficient of h =12 W/m² · °C, determine the temperature at the interface of the wire and the plastic cover in steady operation. Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature.
Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is onedimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal conductivities are constant. 4 The thermal contact resistance at the interface is negligible. 5 Heat transfer coefficient incorporates the radiation effects, if any.
Properties The thermal conductivity of plastic is given to be k 0.15 W/m · °C.Correct
SOLUTION An electric wire is tightly wrapped with a plastic cover. The interface temperature and the effect of doubling the thickness of the plastic cover on the interface temperature are to be determined.
Incorrect
SOLUTION An electric wire is tightly wrapped with a plastic cover. The interface temperature and the effect of doubling the thickness of the plastic cover on the interface temperature are to be determined.

Question 7 of 20
7. Question
1 pointsPower transistors that are commonly used in electronic devices consume large amounts of electric power. The failure rate of electronic components increases almost exponentially with operating temperature. As a rule of thumb, the failure rate of electronic components is halved for each 10°C reduction in the junction operating temperature. Therefore, the operating temperature of electronic components is kept below a safe level to minimize the risk of failure. The sensitive electronic circuitry of a power transistor at the junction is protected by its case, which is a rigid metal enclosure. Heat transfer characteristics of a power transistor are usually specified by the manufacturer in terms of the casetoambient thermal resistance, which accounts for both the natural convection and radiation heat transfers. The casetoambient thermal resistance of a power transistor that has a maximum power rating of 10 W is given to be 20°C/W. If the case temperature of the transistor is not to exceed 85°C, determine the power at which this transistor can be operated safely in an environment at 25°C.
Assumptions 1 Steady operating conditions exist. 2 The transistor case is isothermal at 85°C.
Properties The casetoambient thermal resistance is given to be 20°C/W.Correct
Analysis The power transistor and the thermal resistance network associated with it are shown in( Fig).. We notice from the thermal resistance network that there is a single resistance of 20°C/W between the case at Tc =85°C and the ambient at T∞= 25°C, and thus the rate of heat transfer is
Q=(ΔT/R)case ambient=TcT∞/R case ambient=(85 25)°C/20°C/ W=3W
Therefore, this power transistor should not be operated at power levels above 3 W if its case temperature is not to exceed 85°C.
Incorrect
Analysis The power transistor and the thermal resistance network associated with it are shown in( Fig).. We notice from the thermal resistance network that there is a single resistance of 20°C/W between the case at Tc =85°C and the ambient at T∞= 25°C, and thus the rate of heat transfer is
Q=(ΔT/R)case ambient=TcT∞/R case ambient=(85 25)°C/20°C/ W=3W
Therefore, this power transistor should not be operated at power levels above 3 W if its case temperature is not to exceed 85°C.

Question 8 of 20
8. Question
1 pointsA 60W power transistor is to be cooled by attaching it to one of the commercially available heat sinks shown in Table below. Select a heat sink that will allow the case temperature of the transistor not to exceed 90°C in the ambient air at 30°C.
Assumptions 1 Steady operating conditions exist. 2 The transistor case is isothermal at 90°C. 3 The contact resistance between the transistor and the heat sink is negligible.
Combined natural convection and radiation thermal resistance of various heat sinks used in the cooling of electronic devices between the heat sink and the surroundings. All fins are made of aluminum 6063T5, are black anodized, and are 76 mm (3 in.) long (courtesy of Vemaline Products, Inc.).
Correct
SOLUTION A commercially available heat sink from Table 3–4 is to be selected to keep the case temperature of a transistor below 90°C.
Analysis The rate of heat transfer from a 60W transistor at full power is Q=60 W. The thermal resistance between the transistor attached to the heat sink and the ambient air for the specified temperature difference is determined to be
Q=ΔT/R→R=ΔT/Q=(90 30)°C/60 W=1.0°C/ W
Therefore, the thermal resistance of the heat sink should be below 1.0°C/ W. An examination of Table 3–4 reveals that the HS 5030, whose thermal resistance is 0.9°C/W in the vertical position, is the only heat sink that will meet this requirement.
Incorrect
SOLUTION A commercially available heat sink from Table 3–4 is to be selected to keep the case temperature of a transistor below 90°C.
Analysis The rate of heat transfer from a 60W transistor at full power is Q=60 W. The thermal resistance between the transistor attached to the heat sink and the ambient air for the specified temperature difference is determined to be
Q=ΔT/R→R=ΔT/Q=(90 30)°C/60 W=1.0°C/ W
Therefore, the thermal resistance of the heat sink should be below 1.0°C/ W. An examination of Table 3–4 reveals that the HS 5030, whose thermal resistance is 0.9°C/W in the vertical position, is the only heat sink that will meet this requirement.

Question 9 of 20
9. Question
1 pointsA 30mlong, 10cmdiameter hot water pipe of a district heating system is buried in the soil 50 cm below the ground surface, as shown in (Figure ). The outer surface temperature of the pipe is 80°C. Taking the surface temperature of the earth to be 10°C and the thermal conductivity of the soil at that location to be 0.9 W/m · °C, determine the rate of heat loss from the pipe.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer is twodimensional (no change in the axial direction). 3 Thermal conductivity of the soil is constant.
Properties The thermal conductivity of the soil is given to be k 0.9 W/m · °CCorrect
SOLUTION The hot water pipe of a district heating system is buried in the soil. The rate of heat loss from the pipe is to be determined
Analysis The shape factor for this configuration is given in Table below
S=2πL/iN4Z/D
since z >1.5D, where z is the distance of the pipe from the ground surface, and D is the diameter of the pipe. Substituting, S=2π(30 m)/ln(4 x0.5/0.1)=62.9 m
Then the steady rate of heat transfer from the pipe becomes Q=sk(T1T2)=62.9 m)(0.9 w/m°C)(8010)°c=3963w
Incorrect
SOLUTION The hot water pipe of a district heating system is buried in the soil. The rate of heat loss from the pipe is to be determined
Analysis The shape factor for this configuration is given in Table below
S=2πL/iN4Z/D
since z >1.5D, where z is the distance of the pipe from the ground surface, and D is the diameter of the pipe. Substituting, S=2π(30 m)/ln(4 x0.5/0.1)=62.9 m
Then the steady rate of heat transfer from the pipe becomes Q=sk(T1T2)=62.9 m)(0.9 w/m°C)(8010)°c=3963w

Question 10 of 20
10. Question
1 pointsA 5mlong section of hot and cold water pipes run parallel to each other in a thick concrete layer, as shown in (Figure ) The diameters of both pipes are 5 cm, and the distance between the centerline of the pipes is 30 cm. The surface temperatures of the hot and cold pipes are 70°C and 15°C, respectively. Taking the thermal conductivity of the concrete to be k= 0.75 W/m · °C, determine the rate of heat transfer between the pipes.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer is twodimensional (no change in the axial direction). 3 Thermal conductivity of the concrete is constant.
Properties The thermal conductivity of concrete is given to be k =0.75 W/m · °C.Correct
SOLUTION Hot and cold water pipes run parallel to each other in a thick concrete layer. The rate of heat transfer between the pipes is to be determined.
Analysis The shape factor for this configuration is given in Table below
S=2πL/cosh^1(4z²D1²D2²/2D1D2) where z is the distance between the centerlines of the pipes and L is their length. Substituting,
S=2πx (5m)/ cosh^1(4 x 0.3²0.05²0.05²/2 x 0.05 x 0.05)=6.34 m Then the steady rate of heat transfer between the pipes becomes Q=SK(T1T2)=(6.34 m)(0.75 W/m°C(7015°)C=262W
Incorrect
SOLUTION Hot and cold water pipes run parallel to each other in a thick concrete layer. The rate of heat transfer between the pipes is to be determined.
Analysis The shape factor for this configuration is given in Table below
S=2πL/cosh^1(4z²D1²D2²/2D1D2) where z is the distance between the centerlines of the pipes and L is their length. Substituting,
S=2πx (5m)/ cosh^1(4 x 0.3²0.05²0.05²/2 x 0.05 x 0.05)=6.34 m Then the steady rate of heat transfer between the pipes becomes Q=SK(T1T2)=(6.34 m)(0.75 W/m°C(7015°)C=262W

Question 11 of 20
11. Question
1 pointsDetermine the overall unit thermal resistance (the Rvalue) and the overall heat transfer coefficient (the Ufactor) of a wood frame wall that is built around 38mm x 90mm (2 x4 nominal) wood studs with a centertocenter distance of 400 mm. The 90mmwide cavity between the studs is filled with glass fiber insulation. The inside is finished with 13mm gypsum wallboard and the outside with 13mm wood fiberboard and 13mm x 200mm wood bevel lapped siding. The insulated cavity constitutes 75 percent of the heat transmission area while the studs, plates, and sills constitute 21 percent. The headers constitute 4 percent of the area, and they can be treated as studs. Also, determine the rate of heat loss through the walls of a house whose perimeter is 50 m and wall height is 2.5 m in Las Vegas, Nevada, whose winter design temperature is – 2°C. Take the indoor design temperature to be 22°C and assume 20 percent of the wall area is occupied by glazing.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the wall is onedimensional. 3 Thermal properties of the wall and the heat transfer coefficients are constant.
Properties The Rvalues of different materials are given in Table below
Correct
SOLUTION The Rvalue and the Ufactor of a wood frame wall as well as the rate of heat loss through such a wall in Las Vegas are to be determined.
Analysis The schematic of the wall as well as the different elements used in its construction are shown here. Heat transfer through the insulation and through the studs will meet different resistances, and thus we need to analyze the thermal resistance for each path separately. Once the unit thermal resistances and the Ufactors for the insulation and stud sections are available, the overall average thermal resistance for the entire wall can be determined from
Roverall=1/U overall
where,
U overall=(U x ƒ area)insulation +(U x ƒ area)stud
and the value of the area fraction f area is 0.75 for the insulation section and 0.25 for the stud section since the headers that constitute a small part of the wall are to be treated as studs. Using the available Rvalues from Table 3–6 and calculating others, the total Rvalues for each section can be determined in a systematic manner in the table in this sample. We conclude that the overall unit thermal resistance of the wall is 2.23 m² · °C/W, and this value accounts for the effects of the studs and headers. It corresponds to an Rvalue of 2.23 x 5.68 12.7 (or nearly R13) in English units. Note that if there were no wood studs and headers in the wall, the overall thermal resistance would be 3.05 m² · °C/W, which is 37 percent greater than 2.23 m² · °C/W. Therefore, the wood studs and headers in this case serve as thermal bridges in wood frame walls, and their effect must be considered in the thermal analysis of buildings
Incorrect
SOLUTION The Rvalue and the Ufactor of a wood frame wall as well as the rate of heat loss through such a wall in Las Vegas are to be determined.
Analysis The schematic of the wall as well as the different elements used in its construction are shown here. Heat transfer through the insulation and through the studs will meet different resistances, and thus we need to analyze the thermal resistance for each path separately. Once the unit thermal resistances and the Ufactors for the insulation and stud sections are available, the overall average thermal resistance for the entire wall can be determined from
Roverall=1/U overall
where,
U overall=(U x ƒ area)insulation +(U x ƒ area)stud
and the value of the area fraction f area is 0.75 for the insulation section and 0.25 for the stud section since the headers that constitute a small part of the wall are to be treated as studs. Using the available Rvalues from Table 3–6 and calculating others, the total Rvalues for each section can be determined in a systematic manner in the table in this sample. We conclude that the overall unit thermal resistance of the wall is 2.23 m² · °C/W, and this value accounts for the effects of the studs and headers. It corresponds to an Rvalue of 2.23 x 5.68 12.7 (or nearly R13) in English units. Note that if there were no wood studs and headers in the wall, the overall thermal resistance would be 3.05 m² · °C/W, which is 37 percent greater than 2.23 m² · °C/W. Therefore, the wood studs and headers in this case serve as thermal bridges in wood frame walls, and their effect must be considered in the thermal analysis of buildings

Question 12 of 20
12. Question
1 pointsThe 13mmthick wood fiberboard sheathing of the wood stud wall discussed in the previous example is replaced by a 25mmthick rigid foam insulation. Determine the percent increase in the Rvalue of the wall as a result.
Correct
SOLUTION The overall Rvalue of the existing wall was determined in (table) to be 2.23 m² · °C/W. Noting that the Rvalues of the fiber board and the foam insulation are 0.23 m² · °C/W and 0.98 m² · °C/W, respectively, and the added and removed thermal resistances are in series, the overall Rvalue of the wall after modification becomes
Rnew=Rold Rremoved+Radded = 2.23 0.230.98 = 2.98 m².°C
This represents an increase of (2.98 2.23)/2.23= 0.34 or 34 percent in the Rvalue of the wall. This example demonstrated how to evaluate the new Rvalue of a structure when some structural members are added or removed.
Incorrect
SOLUTION The overall Rvalue of the existing wall was determined in (table) to be 2.23 m² · °C/W. Noting that the Rvalues of the fiber board and the foam insulation are 0.23 m² · °C/W and 0.98 m² · °C/W, respectively, and the added and removed thermal resistances are in series, the overall Rvalue of the wall after modification becomes
Rnew=Rold Rremoved+Radded = 2.23 0.230.98 = 2.98 m².°C
This represents an increase of (2.98 2.23)/2.23= 0.34 or 34 percent in the Rvalue of the wall. This example demonstrated how to evaluate the new Rvalue of a structure when some structural members are added or removed.

Question 13 of 20
13. Question
1 pointsConsider a 1.2mhigh and 2mwide doublepane window consisting of two 3mmthick layers of glass (k= 0.78 W/m · °C) separated by a 12mmwide stagnant air space (k= 0.026 W/m · °C). Determine the steady rate of heat transfer through this doublepane window and the temperature of its inner surface for a day during which the room is maintained at 24°C while the temperature of the outdoors is – 5°C. Take the convection heat transfer coefficients on the inner and outer surfaces of the window to be h1 =10 W/m² · °C and h2= 25 W/m² · °C, and disregard any heat transfer by radiation
Correct
114 W, 19.2°C
Incorrect
114 W, 19.2°C

Question 14 of 20
14. Question
1 pointsConsider a person standing in a room at 20°C with an exposed surface area of 1.7 m². The deep body temperature of the human body is 37°C, and the thermal conductivity of the human tissue near the skin is about 0.3 W/m · °C. The body is losing heat at a rate of 150 W by natural convection and radiation to the surroundings. Taking the body temperature 0.5 cm beneath the skin to be 37°C, determine the skin temperature of the person.
Correct
35.5° C
Incorrect
35.5° C

Question 15 of 20
15. Question
1 pointsHeat is to be conducted along a circuit board that has a copper layer on one side. The circuit board is 15 cm long and 15 cm wide, and the thicknesses of the copper and epoxy layers are 0.1 mm and 1.2 mm, respectively. Disregarding heat transfer from side surfaces, determine the percentages of heat conduction along the copper (k =386 W/m · °C)
Correct
0.8 percent
Incorrect
0.8 percent

Question 16 of 20
16. Question
1 pointsHeat is to be conducted along a circuit board that has a copper layer on one side. The circuit board is 15 cm long and 15 cm wide, and the thicknesses of the copper and epoxy layers are 0.1 mm and 1.2 mm, respectively. Disregarding heat transfer from side surfaces, determine the percentages of heat conduction along the epoxy (k= 0.26 W/m · °C) layers.(answer upto one decimal)
Correct
99.2 percent,
Incorrect
99.2 percent,

Question 17 of 20
17. Question
1 pointsHeat is to be conducted along a circuit board that has a copper layer on one side. The circuit board is 15 cm long and 15 cm wide, and the thicknesses of the copper and epoxy layers are 0.1 mm and 1.2 mm, respectively. Disregarding heat transfer from side surfaces, determine the effective thermal conductivity of the board.(answer upto one decimal)
Correct
29.9 W/m · °C
Incorrect
29.9 W/m · °C

Question 18 of 20
18. Question
1 pointsTwo 5cmdiameter, 15–cmlong aluminum bars (k= 176 W/m · °C) with ground surfaces are pressed against each other with a pressure of 20 atm. The bars are enclosed in an insulation sleeve and, thus, heat transfer from the lateral surfaces is negligible. If the top and bottom surfaces of the twobar system are maintained at temperatures of 150°C and 20°C, respectively, determine the rate of heat transfer along the cylinders under steady conditions and
Correct
142.4 W
Incorrect
142.4 W

Question 19 of 20
19. Question
1 pointsTwo 5cmdiameter, 15–cmlong aluminum bars (k= 176 W/m · °C) with ground surfaces are pressed against each other with a pressure of 20 atm. The bars are enclosed in an insulation sleeve and, thus, heat transfer from the lateral surfaces is negligible. If the top and bottom surfaces of the twobar system are maintained at temperatures of 150°C and 20°C, respectively, determine the temperature drop at the interface.( answer upto one decimal)
Correct
6.4°C
Incorrect
6.4°C

Question 20 of 20
20. Question
1 pointsConsider a 6in. 8in. epoxy glass laminate (k= 0.10 Btu/h · ft · °F) whose thickness is 0.05 in. In order to reduce the thermal resistance across its thickness, cylindrical copper fillings (k= 223 Btu/h · ft · °F) of 0.02 in. diameter are to be planted throughout the board, with a centertocenter distance of 0.06 in. Determine the new value of the thermal resistance of the epoxy board for heat conduction across its thickness as a result of this modification.(answer upto five decimal)
Correct
0.00064 h · °F/Btu
Incorrect
0.00064 h · °F/Btu