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RADIATION HEAT TRANSFER TEST 1
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Question 1 of 20
1. Question
1 pointsDetermine the fraction of the radiation leaving the base of the cylindrical enclosure shown in Figure that escapes through a coaxial ring opening at its top surface. The radius and the length of the enclosure are r1= 10 cm and L= 10 cm, while the inner and outer radii of the ring are r2= 5 cm and r3= 8 cm, respectively
Assumptions The base surface is a diffuse emitter and reflector
Correct
SOLUTION The fraction of radiation leaving the base of a cylindrical enclosure through a coaxial ring opening at its top surface is to be determined.
Analysis We are asked to determine the fraction of the radiation leaving the base of the enclosure that escapes through an opening at the top surface. Actually, what we are asked to determine is simply the view factor F1 → ring from the base of the enclosure to the ringshaped surface at the top. We do not have an analytical expression or chart for view factors between a circular area and a coaxial ring, and so we cannot determine F1 → ring directly. However, we do have a chart for view factors between two coaxial parallel disks, and we can always express a ring in terms of disks. Let the base surface of radius r1= 10 cm be surface 1, the circular area of r2= 5 cm at the top be surface 2, and the circular area of r3= 8 cm be surface 3. Using the superposition rule, the view factor from surface 1 to surface 3 can be expressed as
F1 → 3=F1 → 2+F1 → ring
ince surface 3 is the sum of surface 2 and the ring area. The view factors F1 → 2 and F1 → 3 are determined from the chart in Figure
L/r1=10 cm/10cm=1 AND r2/L=5 cm/10 cm=0.5 → F1 → 2= 0.11 (above Fig. )
L/r1=10 cm/10cm=1 and r3/L=8 cm/10 cm=0.8 → F1 → 3= 0.28
Therefore
F1 → ring= F1 → 3 F1 → 2 =0.28 0.11 =0.17
Incorrect
SOLUTION The fraction of radiation leaving the base of a cylindrical enclosure through a coaxial ring opening at its top surface is to be determined.
Analysis We are asked to determine the fraction of the radiation leaving the base of the enclosure that escapes through an opening at the top surface. Actually, what we are asked to determine is simply the view factor F1 → ring from the base of the enclosure to the ringshaped surface at the top. We do not have an analytical expression or chart for view factors between a circular area and a coaxial ring, and so we cannot determine F1 → ring directly. However, we do have a chart for view factors between two coaxial parallel disks, and we can always express a ring in terms of disks. Let the base surface of radius r1= 10 cm be surface 1, the circular area of r2= 5 cm at the top be surface 2, and the circular area of r3= 8 cm be surface 3. Using the superposition rule, the view factor from surface 1 to surface 3 can be expressed as
F1 → 3=F1 → 2+F1 → ring
ince surface 3 is the sum of surface 2 and the ring area. The view factors F1 → 2 and F1 → 3 are determined from the chart in Figure
L/r1=10 cm/10cm=1 AND r2/L=5 cm/10 cm=0.5 → F1 → 2= 0.11 (above Fig. )
L/r1=10 cm/10cm=1 and r3/L=8 cm/10 cm=0.8 → F1 → 3= 0.28
Therefore
F1 → ring= F1 → 3 F1 → 2 =0.28 0.11 =0.17

Question 2 of 20
2. Question
1 pointsDetermine the view factors from the base of the pyramid shown in Figure below to each of its four side surfaces. The base of the pyramid is a square, and its side surfaces are isosceles triangles.
Assumptions The surfaces are diffuse emitters and reflectors.
Correct
SOLUTION The view factors from the base of a pyramid to each of its four side surfaces for the case of a square base are to be determined.
Analysis The base of the pyramid (surface 1) and its four side surfaces (surfaces 2, 3, 4, and 5) form a fivesurface enclosure. The first thing we notice about this enclosure is its symmetry. The four side surfaces are symmetric about the base surface. Then, from the symmetry rule, we have
F12 =F13 =F14= F15
Also, the summation rule applied to surface 1 yields
However, F11= 0, since the base is a flat surface. Then the two relations above yield F12= F13 =F14 =F15= 0.25
Incorrect
SOLUTION The view factors from the base of a pyramid to each of its four side surfaces for the case of a square base are to be determined.
Analysis The base of the pyramid (surface 1) and its four side surfaces (surfaces 2, 3, 4, and 5) form a fivesurface enclosure. The first thing we notice about this enclosure is its symmetry. The four side surfaces are symmetric about the base surface. Then, from the symmetry rule, we have
F12 =F13 =F14= F15
Also, the summation rule applied to surface 1 yields
However, F11= 0, since the base is a flat surface. Then the two relations above yield F12= F13 =F14 =F15= 0.25

Question 3 of 20
3. Question
1 pointsConsider the 5m x 5m x 5m cubical furnace shown in Figure below, whose surfaces closely approximate black surfaces. The base, top, and side surfaces of the furnace are maintained at uniform temperatures of 800 K, 1500 K, and 500 K, respectively. Determine the net rate of radiation heat transfer between the base and the side surfaces,
Assumptions :The surfaces are black and isothermal.
Correct
SOLUTION The surfaces of a cubical furnace are black and are maintained at
uniform temperatures. The net rate of radiation heat transfer between the base
and side surfaces, between the base and the top surface, from the base surface
are to be determined.Analysis : Considering that the geometry involves six surfaces, we may be tempted at first to treat the furnace as a sixsurface enclosure. However, the four side surfaces possess the same properties, and thus we can treat them as a single side surface in radiation analysis. We consider the base surface to be surface 1, the top surface to be surface 2, and the side surfaces to be surface 3. Then the problem reduces to determining Q1 → 3, Q1 → 2, and Q1. The net rate of radiation heat transfer Q 1 → 3 from surface 1 to surface 3 can be determined , since both surfaces involved are black, by replacing the subscript 2 by 3:
Q1 → 3 A1F1 → 3σ(T1^4 – T3^4)
But first we need to evaluate the view factor F1 → 3. After checking the view factor charts and tables, we realize that we cannot determine this view factor directly. However, we can determine the view factor F1 → 2 directly from Figure BELOW to be F1 → 2 0.2, and we know that F1 → 1 0 since surface 1 is a plane. Then applying the summation rule to surface 1 yields
F1 → 1 + F1 → 2 + F1 → 3= 1
OR, F1 → 3 =1 F1 → 1 F1 → 2= 1 0 0.2 =0.8
Substituting,
Q1 → 3=(25 m²)(0.8)(5.67 x 10^ 8 W/m² · K^4)[(800 K)^4 (500 K)^4] = 394 x 10^3 W = 394 kW
Incorrect
SOLUTION The surfaces of a cubical furnace are black and are maintained at
uniform temperatures. The net rate of radiation heat transfer between the base
and side surfaces, between the base and the top surface, from the base surface
are to be determined.Analysis : Considering that the geometry involves six surfaces, we may be tempted at first to treat the furnace as a sixsurface enclosure. However, the four side surfaces possess the same properties, and thus we can treat them as a single side surface in radiation analysis. We consider the base surface to be surface 1, the top surface to be surface 2, and the side surfaces to be surface 3. Then the problem reduces to determining Q1 → 3, Q1 → 2, and Q1. The net rate of radiation heat transfer Q 1 → 3 from surface 1 to surface 3 can be determined , since both surfaces involved are black, by replacing the subscript 2 by 3:
Q1 → 3 A1F1 → 3σ(T1^4 – T3^4)
But first we need to evaluate the view factor F1 → 3. After checking the view factor charts and tables, we realize that we cannot determine this view factor directly. However, we can determine the view factor F1 → 2 directly from Figure BELOW to be F1 → 2 0.2, and we know that F1 → 1 0 since surface 1 is a plane. Then applying the summation rule to surface 1 yields
F1 → 1 + F1 → 2 + F1 → 3= 1
OR, F1 → 3 =1 F1 → 1 F1 → 2= 1 0 0.2 =0.8
Substituting,
Q1 → 3=(25 m²)(0.8)(5.67 x 10^ 8 W/m² · K^4)[(800 K)^4 (500 K)^4] = 394 x 10^3 W = 394 kW

Question 4 of 20
4. Question
1 pointsConsider the 5m x 5m x 5m cubical furnace shown in Figure below, whose surfaces closely approximate black surfaces. The base, top, and side surfaces of the furnace are maintained at uniform temperatures of 800 K, 1500 K, and 500 K, respectively. Determine the net rate of radiation heat transfer between the base and the top surface,
Assumptions The surfaces are black and isothermal.
Correct
SOLUTION The surfaces of a cubical furnace are black and are maintained at uniform temperatures. The net rate of radiation heat transfer between the base and side surfaces, between the base and the top surface, from the base surface are to be determined.
ANALYSIS : The net rate of radiation heat transfer Q1 → 2 from surface 1 to surface 2 is determined in a similar manner
Q1 → 2=A1F1→ 2σ(T1^4 T2^4)
=(25 m²)(0.2)(5.67 x 10 ^8 W/m² K^4)[(800 K)^4 – (1500 K)^4] = 1319 x 10³W=1319 KW
The negative sign indicates that net radiation heat transfer is from surface 2 to surface 1.
Incorrect
SOLUTION The surfaces of a cubical furnace are black and are maintained at uniform temperatures. The net rate of radiation heat transfer between the base and side surfaces, between the base and the top surface, from the base surface are to be determined.
ANALYSIS : The net rate of radiation heat transfer Q1 → 2 from surface 1 to surface 2 is determined in a similar manner
Q1 → 2=A1F1→ 2σ(T1^4 T2^4)
=(25 m²)(0.2)(5.67 x 10 ^8 W/m² K^4)[(800 K)^4 – (1500 K)^4] = 1319 x 10³W=1319 KW
The negative sign indicates that net radiation heat transfer is from surface 2 to surface 1.

Question 5 of 20
5. Question
1 pointsTwo very large parallel plates are maintained at uniform temperatures T1 800 K and T2 500 K and have emissivities ε1 =0.2 andε 2= 0.7, respectively, as shown in Figure below . Determine the net rate of radiation heat transfer between the two surfaces per unit surface area of the plates.
Assumptions Both surfaces are opaque, diffuse, and gray.
Correct
SOLUTION Two large parallel plates are maintained at uniform temperatures. The net rate of radiation heat transfer between the plates is to be determined.
Analysis The net rate of radiation heat transfer between the two plates per unit area is readily determined
Incorrect
SOLUTION Two large parallel plates are maintained at uniform temperatures. The net rate of radiation heat transfer between the plates is to be determined.
Analysis The net rate of radiation heat transfer between the two plates per unit area is readily determined

Question 6 of 20
6. Question
1 pointsA furnace is shaped like a long equilateral triangular duct, as shown in Figure below . The width of each side is 1 m. The base surface has an emissivity of 0.7 and is maintained at a uniform temperature of 600 K. The heated leftside surface closely approximates a blackbody at 1000 K. The rightside surface is well insulated. Determine the rate at which heat must be supplied to the heated side externally per unit length of the duct in order to maintain these operating conditions.
Assumptions 1 Steady operating conditions exist. 2 The surfaces are opaque, diffuse, and gray. 3 Convection heat transfer is not considered
Correct
SOLUTION Two of the surfaces of a long equilateral triangular furnace are maintained at uniform temperatures while the third surface is insulated. The external rate of heat transfer to the heated side per unit length of the duct during steady operation is to be determined.
Analysis The furnace can be considered to be a threesurface enclosure with a radiation network as shown in the figure, since the duct is very long and thus the end effects are negligible. We observe that the view factor from any surface to any other surface in the enclosure is 0.5 because of symmetry. Surface 3 is a reradiating surface since the net rate of heat transfer at that surface is zero. Then we must have Q1 = Q2, since the entire heat lost by surface 1 must be gained by surface 2. The radiation network in this case is a simple series– parallel connection, an determine Q1 directly from
Therefore, heat at a rate of 28 kW must be supplied to the heated surface per
unit length of the duct to maintain steady operation in the furnace.Incorrect
SOLUTION Two of the surfaces of a long equilateral triangular furnace are maintained at uniform temperatures while the third surface is insulated. The external rate of heat transfer to the heated side per unit length of the duct during steady operation is to be determined.
Analysis The furnace can be considered to be a threesurface enclosure with a radiation network as shown in the figure, since the duct is very long and thus the end effects are negligible. We observe that the view factor from any surface to any other surface in the enclosure is 0.5 because of symmetry. Surface 3 is a reradiating surface since the net rate of heat transfer at that surface is zero. Then we must have Q1 = Q2, since the entire heat lost by surface 1 must be gained by surface 2. The radiation network in this case is a simple series– parallel connection, an determine Q1 directly from
Therefore, heat at a rate of 28 kW must be supplied to the heated surface per
unit length of the duct to maintain steady operation in the furnace. 
Question 7 of 20
7. Question
1 pointsA thin aluminum sheet with an emissivity of 0.1 on both sides is placed between two very large parallel plates that are maintained at uniform temperatures T1= 800 K and T2 =500 K and have emissivities ε1 =0.2 and ε2 =0.7, respectively, as shown in Fig. below. Determine the net rate of radiation heat transfer between the two plates per unit surface area of the plates and compare
the result to that without the shield.Assumptions The surfaces are opaque, diffuse, and gray
Correct
SOLUTION A thin aluminum sheet is placed between two large parallel plates maintained at uniform temperatures. The net rates of radiation heat transfer between the two plates with and without the radiation shield are to be determined.
Incorrect
SOLUTION A thin aluminum sheet is placed between two large parallel plates maintained at uniform temperatures. The net rates of radiation heat transfer between the two plates with and without the radiation shield are to be determined.

Question 8 of 20
8. Question
1 pointsA thermocouple used to measure the temperature of hot air flowing in a duct whose walls are maintained at Tw= 400 K shows a temperature reading of Tth =650 K (Fig.). Assuming the emissivity of the thermocouple junction to be ε=0.6 and the convection heat transfer coefficient to be h 80 W/m² · °C, determine the actual temperature of the air.
Assumptions The surfaces are opaque, diffuse, and gray
Correct
SOLUTION The temperature of air in a duct is measured. The radiation effect on the temperature measurement is to be quantified, and the actual air temperature is to be determined.
Analysis The walls of the duct are at a considerably lower temperature than the air in it, and thus we expect the thermocouple to show a reading lower than the actual air temperature as a result of the radiation effect.
Incorrect
SOLUTION The temperature of air in a duct is measured. The radiation effect on the temperature measurement is to be quantified, and the actual air temperature is to be determined.
Analysis The walls of the duct are at a considerably lower temperature than the air in it, and thus we expect the thermocouple to show a reading lower than the actual air temperature as a result of the radiation effect.

Question 9 of 20
9. Question
1 pointsConsider an enclosure consisting of 12 surfaces. How many view factors does this geometry involve?
Correct
144
Incorrect
144

Question 10 of 20
10. Question
1 pointsConsider an enclosure consisting of 12 surfaces How many of these view factors can be determined by the application of the reciprocity and the summation rules?
Correct
78
Incorrect
78

Question 11 of 20
11. Question
1 pointsConsider a hemispherical furnace with a flat circular base of diameter D. Determine the view factor from the dome of this furnace to its base.
Correct
0.5
Incorrect
0.5

Question 12 of 20
12. Question
1 pointsConsider a hemispherical furnace of diameter D= 5 m with a flat base. The dome of the furnace is black, and the base has an emissivity of 0.7. The base and the dome of the furnace are maintained at uniform temperatures of 400 and 1000 K, respectively. Determine the net rate of radiation heat transfer from the dome to the base surface during steady operation
Correct
759 kw
Incorrect
759 kw

Question 13 of 20
13. Question
1 pointsTwo parallel disks of diameter D =0.6 m separated by L =0.4 m are located directly on top of each other. Both disks are black and are maintained at a temperature of 700 K. The back sides of the disks are insulated, and the environment that the disks are in can be considered to be a blackbody at T∞=300 K. Determine the net rate of radiation heat transfer from the disks to the environment.
Correct
5505 W
Incorrect
5505 W

Question 14 of 20
14. Question
1 pointsA spherical tank of diameter D= 2 m that is filled with liquid nitrogen at 100 K is kept in an evacuated cubic enclosure whose sides are 3 m long. The emissivities of the spherical tank and the enclosure are ε1 =0.1 and ε2= 0.8, respectively. If the temperature of the cubic enclosure is measured to be 240 K, determine the net rate of radiation heat transfer to the liquid nitrogen.
Correct
228 W
Incorrect
228 W

Question 15 of 20
15. Question
1 pointsConsider a circular grill whose diameter is 0.3 m. The bottom of the grill is covered with hot coal bricks at 1100 K, while the wire mesh on top of the grill is covered with steaks initially at 5°C. The distance between the coal bricks and the steaks is 0.20 m. Treating both the steaks and the coal bricks as blackbodies, determine the initial rate of radiation heat transfer from the coal bricks to the steaks
Correct
1674 W
Incorrect
1674 W

Question 16 of 20
16. Question
1 pointsConsider a circular grill whose diameter is 0.3 m. The bottom of the grill is covered with hot coal bricks at 1100 K, while the wire mesh on top of the grill is covered with steaks initially at 5°C. The distance between the coal bricks and the steaks is 0.20 m. Treating both the steaks and the coal bricks as blackbodies determine the initial rate of radiation heat transfer to the steaks if the side opening of the grill is covered by aluminum foil, which can be approximated as a reradiating surface.
Correct
3757 W
Incorrect
3757 W

Question 17 of 20
17. Question
1 pointsTwo parallel disks of diameter D= 3 ft separated by L= 2 ft are located directly on top of each other. The disks are separated by a radiation shield whose emissivity is 0.15. Both disks are black and are maintained at temperatures of 1200 R and 700 R, respectively. The environment that the disks are in can be considered to be a blackbody at 540 R. Determine the net rate of radiation heat transfer through the shield under steady conditions
Correct
866 Btu/h
Incorrect
866 Btu/h

Question 18 of 20
18. Question
1 pointsA clothed or unclothed person feels comfortable when the skin temperature is about 33°C. Consider an average man wearing summer clothes whose thermal resistance is 0.7 clo. The man feels very comfortable while standing in a room maintained at 20°C. If this man were to stand in that room unclothed, determine the temperature at which the room must be maintained for him to feel thermally comfortable. Assume the latent heat loss from the person to remain the same.
Correct
26.4°C
Incorrect
26.4°C

Question 19 of 20
19. Question
1 pointsConsider a large classroom with 150 students on a hot summer day. All the lights with 4.0 kW of rated power are kept on. The room has no external walls, and thus heat gain through the walls and the roof is negligible. Chilled air is available at 15°C, and the temperature of the return air is not to exceed 25°C. Determine the required flow rate of air, in kg/s, that needs to be supplied to the room(answer upto two decimal)
Correct
1.45 kg/s
Incorrect
1.45 kg/s

Question 20 of 20
20. Question
1 pointsConsider a sealed 8in.high electronic box whose base dimensions are 12 in. x 12 in. placed in a vacuum chamber. The emissivity of the outer surface of the box is 0.95. If the electronic components in the box dissipate a total of 100 W of power and the outer surface temperature of the box is not to exceed 130°F, determine the highest temperature at which the surrounding surfaces must be kept if this box is to be cooled by radiation alone. Assume the heat transfer from the bottom surface of the box to the stand to be negligible.
Correct
43°F
Incorrect
43°F