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COOLING OF ELECTRONIC EQUIPMENT TEST 1
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Question 1 of 20
1. Question
1 pointsThe temperature of the case of a power transistor that is dissipating 3 W is measured to be 50°C. If the junctiontocase resistance of this transistor is specified by the manufacturer to be 15°C/ W, determine the temperature at the junction of the transistor
Assumptions Steady operating conditions exist
Correct
SOLUTION The case temperature of a power transistor and the junctiontocase resistance are given. The junction temperature is to be determined.
Analysis The schematic of the transistor is given in Figure . The rate of heat transfer between the junction and the case in steady operation can be expressed as
Q=(ΔT/R) junction case= Tjunction Tcase/Rjunctioncase
Then the junction temperature becomes
Tjunction= Tcase + QRjunctioncase
=50°C + (3 W)(15°C/ W)
=95°C
Therefore, the temperature of the transistor junction will be 95°C when its case
is at 50°CIncorrect
SOLUTION The case temperature of a power transistor and the junctiontocase resistance are given. The junction temperature is to be determined.
Analysis The schematic of the transistor is given in Figure . The rate of heat transfer between the junction and the case in steady operation can be expressed as
Q=(ΔT/R) junction case= Tjunction Tcase/Rjunctioncase
Then the junction temperature becomes
Tjunction= Tcase + QRjunctioncase
=50°C + (3 W)(15°C/ W)
=95°C
Therefore, the temperature of the transistor junction will be 95°C when its case
is at 50°C 
Question 2 of 20
2. Question
1 pointsThis experiment is conducted to determine the junctiontocase thermal resistance of an electronic component. Power is supplied to the component from a 15V source, and the variations in the electric current and in the junction and the case temperatures with time are observed. When things are stabilized, the current is observed to be 0.1 A and the temperatures to be 80°C and 55°C at the junction and the case, respectively. Calculate the junctiontocase resistance of this component.(answer upto one decimal)
Assumptions Steady operating conditions exist
Correct
SOLUTION The power dissipated by an electronic component as well as the junction and case temperatures are measured. The junctiontocase resistance is to be determined.
Analysis The schematic of the component is given in Figure . The electric power consumed by this electronic component is
We=VI=(15 V)(0.1 A)=1.5 W
In steady operation, this is equivalent to the heat dissipated by the component. That is,
Q=(ΔTR/R)junction case =Tjunction Tcase/Rjunctioncase =1.5 W
Then the junctiontocase resistance is determined to be
Rjunctioncase=Tjunction Tcase/Q=(80 55)°C/1.5 W=16.7°C/ W
Incorrect
SOLUTION The power dissipated by an electronic component as well as the junction and case temperatures are measured. The junctiontocase resistance is to be determined.
Analysis The schematic of the component is given in Figure . The electric power consumed by this electronic component is
We=VI=(15 V)(0.1 A)=1.5 W
In steady operation, this is equivalent to the heat dissipated by the component. That is,
Q=(ΔTR/R)junction case =Tjunction Tcase/Rjunctioncase =1.5 W
Then the junctiontocase resistance is determined to be
Rjunctioncase=Tjunction Tcase/Q=(80 55)°C/1.5 W=16.7°C/ W

Question 3 of 20
3. Question
1 pointsA fan blows air at 30°C and a velocity of 200 m/min over a 1.2W plastic DIP with 16 leads mounted on a PCB, as shown in Figure .
Using data from Figure,
determine the junction temperature of the electronic device.
Assumptions Steady operating conditions exist
Correct
SOLUTION A plastic DIP with 16 leads is cooled by forced air. Using data supplied by the manufacturer, the junction temperature is to be determined.
Analysis The junctiontoambient thermal resistance of the device with 16 leads corresponding to an air velocity of 200 m/min is determined from Figure to be
Rjunctionambient=55°C/ W
Tjunction=Tambient + QRjunctionambient=30°C + (1.2 W)(55°C/ W)=96°C
Incorrect
SOLUTION A plastic DIP with 16 leads is cooled by forced air. Using data supplied by the manufacturer, the junction temperature is to be determined.
Analysis The junctiontoambient thermal resistance of the device with 16 leads corresponding to an air velocity of 200 m/min is determined from Figure to be
Rjunctionambient=55°C/ W
Tjunction=Tambient + QRjunctionambient=30°C + (1.2 W)(55°C/ W)=96°C

Question 4 of 20
4. Question
1 pointsA fan blows air at 30°C and a velocity of 200 m/min over a 1.2W plastic DIP with 16 leads mounted on a PCB, as shown in Figure
Using data from Figure ,
.
What would the junction temperature be if the fan were to fail?
Assumptions Steady operating conditions exist.
Correct
SOLUTION A plastic DIP with 16 leads is cooled by forced air. Using data supplied by the manufacturer, the junction temperature is to be determined.
Analysis The junctiontoambient thermal resistance of the device with 16 leads corresponding to an air velocity of 200 m/min is determined from Figure to be
Rjunctionambient=55°C/ W
When the fan fails, the airflow velocity over the device will be zero. The total
thermal resistance in this case is determined from the same chart by reading
the value at the intersection of the curve and the vertical axis to beRjunctionambient=70°C/ W
which gives
Tjunction=Tambient+QRjunctionambient=30°C + (1.2 W)(70°C/ W)=114°C
Incorrect
SOLUTION A plastic DIP with 16 leads is cooled by forced air. Using data supplied by the manufacturer, the junction temperature is to be determined.
Analysis The junctiontoambient thermal resistance of the device with 16 leads corresponding to an air velocity of 200 m/min is determined from Figure to be
Rjunctionambient=55°C/ W
When the fan fails, the airflow velocity over the device will be zero. The total
thermal resistance in this case is determined from the same chart by reading
the value at the intersection of the curve and the vertical axis to beRjunctionambient=70°C/ W
which gives
Tjunction=Tambient+QRjunctionambient=30°C + (1.2 W)(70°C/ W)=114°C

Question 5 of 20
5. Question
1 pointsConsider a 10cm x 15cm glass–epoxy laminate (k = 0.26 W/m · °C) whose thickness is 0.8 mm, as shown in Figure . Determine the thermal resistance of this epoxy layer for heat flow along the 15cmlong side
Assumptions 1 Heat conduction in the laminate is onedimensional in either case. 2 Thermal properties of the laminate are constant.
Correct
SOLUTION The dimensions of an epoxy–glass laminate are given. The thermal resistances for heat flow along the layers and across the thickness are to be determined.
Analysis The thermal resistance of a plane parallel medium in the direction of heat conduction is given by
R=L/KA
where L is the length in the direction of heat flow, k is the thermal conductivity, and A is the area normal to the direction of heat conduction. Substituting the given values, the thermal resistances of the board for both cases are determined to be
Ralong length=(L/KA)along length
=0.15 m/(0.26 W/m · °C)(0.1 m)(0.8x 10^3m)=7212°C/ W
Incorrect
SOLUTION The dimensions of an epoxy–glass laminate are given. The thermal resistances for heat flow along the layers and across the thickness are to be determined.
Analysis The thermal resistance of a plane parallel medium in the direction of heat conduction is given by
R=L/KA
where L is the length in the direction of heat flow, k is the thermal conductivity, and A is the area normal to the direction of heat conduction. Substituting the given values, the thermal resistances of the board for both cases are determined to be
Ralong length=(L/KA)along length
=0.15 m/(0.26 W/m · °C)(0.1 m)(0.8x 10^3m)=7212°C/ W

Question 6 of 20
6. Question
1 pointsonsider a 10cm x 15cm glass–epoxy laminate (k = 0.26 W/m · °C) whose thickness is 0.8 mm, as shown in Figure . Determine the thermal resistance of this epoxy layer for heat flow across its thickness
Assumptions 1 Heat conduction in the laminate is onedimensional in either case. 2 Thermal properties of the laminate are
Correct
SOLUTION The dimensions of an epoxy–glass laminate are given. The thermal resistances for heat flow along the layers and across the thickness are to be determined.
Analysis The thermal resistance of a plane parallel medium in the direction of heat conduction is given by
R=L/KA
where L is the length in the direction of heat flow, k is the thermal conductivity, and A is the area normal to the direction of heat conduction. Substituting the given values, the thermal resistances of the board for both cases are determined to be
Racross thickness=(L/KA)across thickness
=0.8 cx 10^3m/(0.26 W/m · °C)(0.1 m)(0.15 m)=0.21°C/ W
Incorrect
SOLUTION The dimensions of an epoxy–glass laminate are given. The thermal resistances for heat flow along the layers and across the thickness are to be determined.
Analysis The thermal resistance of a plane parallel medium in the direction of heat conduction is given by
R=L/KA
where L is the length in the direction of heat flow, k is the thermal conductivity, and A is the area normal to the direction of heat conduction. Substituting the given values, the thermal resistances of the board for both cases are determined to be
Racross thickness=(L/KA)across thickness
=0.8 cx 10^3m/(0.26 W/m · °C)(0.1 m)(0.15 m)=0.21°C/ W

Question 7 of 20
7. Question
1 pointsReconsider the 10cm x 15cm glass–epoxy laminate (k = 0.26 W/m · °C) of thickness 0.8 mm discussed in Example . In order to reduce the thermal resistance across its thickness from the current value of 0.21°C/W, cylindrical copper fillings (k = 386 W/m · °C) of 1mm diameter are to be planted throughout the board with a centertocenter distance of 2.5 mm, as shown in Figure . Determine the new value of the thermal resistance of the epoxy board for heat conduction across its thickness as a result of this modification.
Assumptions 1 Heat conduction along the board is onedimensional. 2 Thermal properties of the board are constant.
Correct
SOLUTION Cylindrical copper fillings are planted throughout an epoxy glass board. The thermal resistance of the board across its thickness is to be determined
Analysis Heat flow through the thickness of the board in this case will take place partly through the copper fillings and partly through the epoxy in parallel paths. The thickness of both materials is the same and is given to be 0.8 mm. But we also need to know the surface area of each material before we can determine the thermal resistances. It is stated that the distance between the centers of the copper fillings is 2.5 mm. That is, there is only one 1mmdiameter copper filling in every 2.5mm 2.5mm square section of the board. The number of such squares and thus the number of copper fillings on the board are
n=Area of the board/Area of one square =(100 mm)(150 mm)/(2.5 mm)(2.5 mm)=2400
Then the surface areas of the copper fillings and the remaining epoxy layer become
Incorrect
SOLUTION Cylindrical copper fillings are planted throughout an epoxy glass board. The thermal resistance of the board across its thickness is to be determined
Analysis Heat flow through the thickness of the board in this case will take place partly through the copper fillings and partly through the epoxy in parallel paths. The thickness of both materials is the same and is given to be 0.8 mm. But we also need to know the surface area of each material before we can determine the thermal resistances. It is stated that the distance between the centers of the copper fillings is 2.5 mm. That is, there is only one 1mmdiameter copper filling in every 2.5mm 2.5mm square section of the board. The number of such squares and thus the number of copper fillings on the board are
n=Area of the board/Area of one square =(100 mm)(150 mm)/(2.5 mm)(2.5 mm)=2400
Then the surface areas of the copper fillings and the remaining epoxy layer become

Question 8 of 20
8. Question
1 pointsConsider a thermal conduction module with 100 chips, each dissipating 3 W of power. The module is cooled by water at 25°C flowing through the cold plate on top of the module. The thermal resistances in the path of heat flow are Rchip = 1°C/W between the junction and the surface of the chip, Rint = 8°C/W between the surface of the chip and the outer surface of the thermal conduction module, and Rext = 6°C/W between the outer surface of the module and the cooling water. Determine the junction temperature of the chip.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through various components is onedimensional.
Correct
SOLUTION A thermal conduction module TCM with 100 chips is cooled by water. The junction temperature of the chip is to be determined
Analysis Because of symmetry, we will consider only one of the chips in our analysis. The thermal resistance network for heat flow is given in Figure . Noting that all resistances are in series, the total thermal resistance between the junction and the cooling water is
Rtotal=Rjunctionwater=Rchip +Rint + Rext=(1 + 8 + 6)°C/ W = 15°C/ W
Noting that the total power dissipated by the chip is 3 W and the water temperature is 25°C, the junction temperature of the chip in steady operation can be determined from
Q=ΔT/R=junctionwater=Tjunction Twater/Rjunctionwater
Solving for Tjunction and substituting the specified values gives
Tjunction=Twater+QRjunctionwater=25°C +(3 W)(15°C/ W)=70°C
Incorrect
SOLUTION A thermal conduction module TCM with 100 chips is cooled by water. The junction temperature of the chip is to be determined
Analysis Because of symmetry, we will consider only one of the chips in our analysis. The thermal resistance network for heat flow is given in Figure . Noting that all resistances are in series, the total thermal resistance between the junction and the cooling water is
Rtotal=Rjunctionwater=Rchip +Rint + Rext=(1 + 8 + 6)°C/ W = 15°C/ W
Noting that the total power dissipated by the chip is 3 W and the water temperature is 25°C, the junction temperature of the chip in steady operation can be determined from
Q=ΔT/R=junctionwater=Tjunction Twater/Rjunctionwater
Solving for Tjunction and substituting the specified values gives
Tjunction=Twater+QRjunctionwater=25°C +(3 W)(15°C/ W)=70°C

Question 9 of 20
9. Question
1 pointsA 0.2W small cylindrical resistor mounted on a PCB is 1 cm long and has a diameter of 0.3 cm, as shown in Figure . The view of the resistor is largely blocked by the PCB facing it, and the heat transfer from the connecting wires is negligible. The air is free to flow through the parallel flow passages between the PCBs. If the air temperature at the vicinity of the resistor is 50°C, determine the surface temperature of the resistor.
Assumptions 1 Steady operating conditions exist. 2 The device is located at sea level so that the local atmospheric pressure is 1 atm. 3 Radiation is negligible in this case since the resistor is surrounded by surfaces that are at about the same temperature, and the net radiation heat transfer between two surfaces at the same temperature is zero. This leaves natural convection as the only mechanism of heat transfer from the resistor
Correct
SOLUTION A small cylindrical resistor mounted on a PCB is being cooled by natural convection and radiation. The surface temperature of the resistor is to be determined.
Incorrect
SOLUTION A small cylindrical resistor mounted on a PCB is being cooled by natural convection and radiation. The surface temperature of the resistor is to be determined.

Question 10 of 20
10. Question
1 pointsThe desktop computer shown in Figure is to be cooled by a fan. The electronics of the computer consume 75 W of power under fullload conditions. The computer is to operate in environments at temperatures up to 40°C and at elevations up to 2000 m, where the atmospheric pressure is 79.50 kPa. The exit temperature of air is not to exceed 70°C to meet reliability requirements. Also, the average velocity of air is not to exceed 75 m/min at the exit of the computer case, where the fan is installed, to keep the noise level down. Determine the flow rate of the fan that needs to be installed (answer upto three decimal)
Assumptions 1 Steady operation under worst conditions is considered. 2 Air is an ideal gas.
Correct
SOLUTION A desktop computer is to be cooled by a fan safely in hot environments and high elevations. The airflow rate of the fan and the diameter of the casing are to be determined
Analysis We need to determine the flow rate of air for the worstcase scenario. Therefore, we assume the inlet temperature of air to be 40°C and the atmospheric pressure to be 79.50 kPa and disregard any heat transfer from the outer surfaces of the computer case. Note that any direct heat loss from the computer case will provide a safety margin in the design. Noting that all the heat dissipated by the electronic components is absorbed by air, the required mass flow rate of air to absorb heat at a rate of 75 W can be determined from
Incorrect
SOLUTION A desktop computer is to be cooled by a fan safely in hot environments and high elevations. The airflow rate of the fan and the diameter of the casing are to be determined
Analysis We need to determine the flow rate of air for the worstcase scenario. Therefore, we assume the inlet temperature of air to be 40°C and the atmospheric pressure to be 79.50 kPa and disregard any heat transfer from the outer surfaces of the computer case. Note that any direct heat loss from the computer case will provide a safety margin in the design. Noting that all the heat dissipated by the electronic components is absorbed by air, the required mass flow rate of air to absorb heat at a rate of 75 W can be determined from

Question 11 of 20
11. Question
1 pointsA logic chip used in an IBM 3081 computer dissipates 4 W of power and has a heat transfer surface area of 0.279 cm2, as shown in Figure . If the surface of the chip is to be maintained at 80°C while being cooled by immersion in a dielectric fluid at 20°C, determine the necessary heat transfer coefficient and the type of cooling mechanism that needs to be used to achieve that heat transfer coefficient.
Assumptions Steady operating conditions exist.
Correct
SOLUTION A logic chip is to be cooled by immersion in a dielectric fluid. The minimum heat transfer coefficient and the type of cooling mechanism are to be determined.
Analysis The average heat transfer coefficient over the surface of the chip can be determined from Newton’s law of cooling,
Q=hAs(Tchip Tfluid)
Solving for h and substituting the given values, the convection heat transfer coefficient is determined to be
h=Q/As(Tchip Tfluid)=4W/(0.279 x 10^4m²)(80 20)°C=2390 W/m² · °C
Incorrect
SOLUTION A logic chip is to be cooled by immersion in a dielectric fluid. The minimum heat transfer coefficient and the type of cooling mechanism are to be determined.
Analysis The average heat transfer coefficient over the surface of the chip can be determined from Newton’s law of cooling,
Q=hAs(Tchip Tfluid)
Solving for h and substituting the given values, the convection heat transfer coefficient is determined to be
h=Q/As(Tchip Tfluid)=4W/(0.279 x 10^4m²)(80 20)°C=2390 W/m² · °C

Question 12 of 20
12. Question
1 pointsAn 8W chip having a surface area of 0.6 cm² is cooled by immersing it into FC86 liquid that is maintained at a temperature of 15°C, as shown in Figure . Using the boiling curve in 2nd Figure , estimate the temperature of the chip surface
Assumptions The boiling curve in Figure is prepared for a chip having a surface area of 0.457 cm2 being cooled in FC86 maintained at 5°C. The chart can be used for similar cases with reasonable accuracy.
Correct
SOLUTION A chip is cooled by boiling in a dielectric fluid. The surface temperature of the chip is to be determined
Analysis The heat flux is
q=Q/As=8W/0.6 cm²=13.3 W/cm²
Corresponding to this value on the chart is Tchip Tfluid =60°C. Therefore
Tchip=Tfluid + 60 = 15 +60 =75°C
Incorrect
SOLUTION A chip is cooled by boiling in a dielectric fluid. The surface temperature of the chip is to be determined
Analysis The heat flux is
q=Q/As=8W/0.6 cm²=13.3 W/cm²
Corresponding to this value on the chart is Tchip Tfluid =60°C. Therefore
Tchip=Tfluid + 60 = 15 +60 =75°C

Question 13 of 20
13. Question
1 pointsThe temperature of the case of a power transistor that is dissipating 12 W is measured to be 60°C. If the junctiontocase thermal resistance of this transistor is specified by the manufacturer to be 5°C/W, determine the junction temperature of the transistor
Correct
120°C
Incorrect
120°C

Question 14 of 20
14. Question
1 pointsConsider a 0.1kΩ resistor whose surfacetoambient thermal resistance is 300°C/W. If the voltage drop across the resistor is 7.5 V and its surface temperature is not to exceed 150°C, determine the power at which it can operate safely in an ambient at 30°C
Correct
0.4 W
Incorrect
0.4 W

Question 15 of 20
15. Question
1 pointsThe temperature of air in high winds is measured by a thermometer to be 12°C. Determine the true temperature of air if the wind velocity is 90 km/h.(answer upto one decimal)
Correct
11.7°C
Incorrect
11.7°C

Question 16 of 20
16. Question
1 pointsHeat is to be conducted along a PCB with copper cladding on one side. The PCB is 12 cm long and 12 cm wide,and the thicknesses of the copper and epoxy layers are 0.06 mm and 0.5 mm, respectively. Disregarding heat transfer from the side surfaces, determine the percentages of heat conduction along the copper (k = 386 W/m · °C)(answer upto one decimal)
Correct
0.6 percent
Incorrect
0.6 percent

Question 17 of 20
17. Question
1 pointsHeat is to be conducted along a PCB with copper cladding on one side. The PCB is 12 cm long and 12 cm wide,and the thicknesses of the copper and epoxy layers are 0.06 mm and 0.5 mm, respectively. Disregarding heat transfer from the side surfaces, determine the percentages of heat conduction epoxy (k = 0.26 W/m · °C) layers.(answer upto one decimal)
Correct
99.4 percent,
Incorrect
99.4 percent,

Question 18 of 20
18. Question
1 pointsHeat is to be conducted along a PCB with copper cladding on one side. The PCB is 12 cm long and 12 cm wide,and the thicknesses of the copper and epoxy layers are 0.06 mm and 0.5 mm, respectively. Disregarding heat transfer from the side surfaces, determine the effective thermal conductivity of the PCB.(answer upto one decimal)
Correct
41.6 W/m · °C
Incorrect
41.6 W/m · °C

Question 19 of 20
19. Question
1 pointsConsider a 15–cm x 20cm doublesided circuit board dissipating a total of 30 W of heat. The board consists of a 3mmthick epoxy layer (k = 0.26 W/m · °C) with 1mmdiameter aluminum wires (k=237 W/m · °C) inserted along the 20cmlong direction, as shown in Figure . The distance between the centers of the aluminum wires is 2 mm. The circuit board is attached to a heat sink from both ends, and the temperature of the board at both ends is 30°C. Heat is considered to be uniformly generated on both sides of the epoxy layer of the PCB. Considering only a portion of the PCB because of symmetry, determine the magnitude and location of the maximum temperature that occurs in the PCB. (answer upto one decimal)
Correct
88.7°C
Incorrect
88.7°C

Question 20 of 20
20. Question
1 pointsA 6in.x 8in. 0.06in. copper heat frame is used to conduct 20 W of heat generated in a PCB along the 8in.long side toward the ends. Determine the temperature difference between the midsection and either end of the heat frame(answer upto one decimal)
Correct
12.8°F
Incorrect
12.8°F