Met 2014 Foundry Technology I Sample Questions and - most

Met 2014 Foundry Technology I
Sample Questions and Answers for Final Exam, 2008
1. (a) Describe with a neat sketch the cupola operation and explain how would you specify a cupola.
Ans; Page 50,53
(b) Draw a neat sketch of the open hearth furnace indicating and labeling the important parts.
Ans; Page 56
2. Explain briefly the types of electric furnaces generally employed in foundry with neat sketches.
Ans; Page 58
3. Explain briefly the basic operations performed by a molding machine with neat sketches.
Ans; Page 63
4. (a) What are the advantages and disadvantages of molding machines?
Ans; Page 66
(b) How can you ram the sand into a mold by wing different types of machine?
Ans; Page 63
5.(a) What is meant by "mechanization and automation of foundry operation"?
Ans; Page 72
(b) Describe X-ray technique of nondestructive testing.
Ans; Page 75
6.(a) Discuss the merits and demerits of radiographic inspection.
Ans; Page 76
(b) List the types of defects that may be detected by using the following techniques.
(i) Magnetic particle method
(ii) Penetrant tests
(iii) Super sonic testing
Ans; Page 76
7. Classify the casting defects generally found.
Ans; Page 77
8. Describe the causes and remedy for the following casting defects.
(a) Blow holes
(b) Shrinkage
(c) Hot tear
(d)Cold creak
(e) Misrun and cold shut
Ans; Page 77,78
9. Explain briefly the important of solidification of metal in foundry.
Ans; Page 87
10.(a) Classify the types of gate generally employed.
Ans; Page 85
(b) Write short notes on some basic design rules commonly used.
Ans; Page 92
4
11. How do you understand the air furnace?
Ans; Page 57
12. Explain briefly the types of crucible furnace.
Ans; Page 60
13.Explain briefly squeezing and ramming.
Ans; Page 63
14. What do you understand the terms sands lining? Explain briefly.
Ans; Page 64
15.Explain fully how to distribute the prepared sand.
Ans; Page 68
16.Draw the flowsheet the basic steps required in a machined foundry.
Ans; Page 68-70
17.Write a short notes
i)Rough cleaning
ii) Surface cleaning
iii) Trimming
iv) Finishing
Ans; Page 73
18. Explain briefly the design of gating systems.
Ans; Page 86
19.What is the difference between functional design and process design?
Ans; Page 90
20.Explain the fundamental characteristics of cast metal alloys that influence.
Ans; Page 91
21.Briefly explain the fundamental concepts of melting.
Ans; Page 50
22.How do you understand the operational cycle for cupola melting.
Ans; Page 51
23.How do you calculate cupola charges.
Ans; Page 53
24.What are the operating principle of the open hearth furnace.
Ans; Page 56
5
25. How do you understand the converter?
Ans; Page 61
26.Discuss the following defects.
Ans; Page 79-80
27.Draw a neat sketch of complete sand mold labeling the important parts.
Ans; Page 85
28.How do you design the gating system.
Ans; Page 86
29.Briefly explain the mechanics of molten metal flow.
Ans; Page 85
30.Write the short notes Ramming, Blowing.
Ans; Page 65
6
Met 2015 Metallurgical Unit Operations
Sample Questions and Answers for Final Exam, 2008
1.(a) Determine the energy loss if glycerine at 25ЛљC flows 30m through a 150mm diameter pipe with
an average velocity of 4.0m/s
Ans; Chapter 6 Example 6.1 Page 18
(b)Que; Chapter 7 Example 7.1 Page 21
Ans; Chapter 7 Example 7.1 Page 21
2. As a test to determine the wall roughness of an existing pipe installation, water at 10ЛљC is pumped
through it at a rate of 225 L/min. The pipe is standard 1ВЅ" commercial steel tubing with a wall
thickness of 0.109in. Pressure gage located 30m apart in a horizontal run of the pipe read 1032 KPa
and 669KPa. Determine the pipe wall roughness. Di = 32.66mm, A = 8.326 x 10-1 m2
Ans; Chapter 6
Example 6.2 Page 18
3.(a)Que;Chapter 7 Example 7.2 Page 21
Ans; Chapter 7 Example 7.2 Page 21
(b) Air with a specific weight of 127 N/m3 and a kinematic viscosity of 1.3 x 10-5 m2/s is flowing
through a flow nozzle. A manometer using water as the gage fluid reads 18 mm of deflection.
Calculate the volume flow rate if the nozzle dia. is 50mm.Pipe inside dia. is 100mm.
Ans; Chapter 9
Example 9.2 Page 50
4. (a)Que;Chapter 7
Ans; Chapter 7
(b)Que;Chapter 7
Ans; Chapter 7
Example 7.5
Example 7.5
Example 7.6
Example 7.6
Page 24
Page 24
Page 25
Page 25
5. Que;Chapter 7
Example 7.8 Page 26
Ans; Chapter 7 Example 7.8 Page 26
6. It is desired to pump a lubricating oil through a horizontal 6" sch 40 steel pipe with a maximum
pressure drop of 60 KPa per 100m of pipe. The oil has a specific gravity of 0.88 and a dynamic
viscosity of 10 x 10-3 Pa. Calculate the maximum volume flow rate of oil. Is this a class Ii system?
6" sch 40 steel pipe Di = 0.1541m, A = 1.864 x 10-2 m2, С” = 4.6 x 10-5 m ,P1-P2 = 60KPa
Ans; Similar with Chapter 8 Example 8.2 Page 34
7.As a test to determine the wall roughness of an existing pipe installation, water at 10ЛљC is pumped
through it at a rate of 225 L/min. The pipe is standard 1ВЅ" commercial steel tubing with a wall
thickness of 0.109in. Pressure gage located 25m apart in a horizontal run of the pipe read 1030 KPa
and 659KPa. Determine the pipe wall roughness. Di = 32.66mm, A = 8.326 x 10-1 m2
Ans; Similar with Chapter 6 Example 6.2 Page 18
8.As a test to determine the wall roughness of an existing pipe installation, water at 10ЛљC is pumped
through it at a rate of 225 L/min. The pipe is standard 1ВЅ" commercial steel tubing with a wall
thickness of 0.115in. Pressure gage located 25m apart in a horizontal run of the pipe read 1032 KPa
and 669KPa. Determine the pipe wall roughness. Di = 32.66mm, A = 8.326 x 10-1 m2
Ans; Similar with Chapter 6 Example 6.2 Page 18
7
9. Que;Chapter 7
Ans; Chapter 7
Example 7.9 Page 28
Example 7.9 Page 28
10.It is desired to pump a lubricating oil through a horizontal 6" sch 40 steel pipe with a maximum
pressure drop of 60 KPa per 100m of pipe. The oil has a specific gravity of 0.88 and a dynamic
viscosity of 9.5 x 10-3 Pa. Calculate the maximum volume flow rate of oil. Is this a class Ii system?
6" sch 40 steel pipe Di = 0.1541m, A = 1.864 x 10-2 m2, С” = 4.6 x 10-5 m ,P1-P2 = 60KPa
Ans; Chapter 8
Example 8.2 Page 34
11.It is desired to pump a lubricating oil through a horizontal 6" sch 40 steel pipe with a maximum
pressure drop of 55 KPa per 100m of pipe. The oil has a specific gravity of 0.88 and a dynamic
viscosity of 9.5 x 10-3 Pa. Calculate the maximum volume flow rate of oil. Is this a class Ii system?
6" sch 40 steel pipe Di = 0.1541m, A = 1.864 x 10-2 m2, С” = 4.6 x 10-5 m ,P1-P2 = 60KPa
Ans; Similar with Chapter 8 Example 8.2 Page 34
12. Que; From Example 8.4 ,if the supply is from a main line at point A where the pressure is 90 psig,
determine to supply 0.5 cu ft /s of water at 60F.Лљ
Ans; Similar with Chapter 8 Example 8.4 Page 39
13.It is desired to pump a lubricating oil through a horizontal 6" sch 40 steel pipe with a maximum
pressure drop of 65 KPa per 100m of pipe. The oil has a specific gravity of 0.88 and a dynamic
viscosity of 9.5 x 10-3 Pa. Calculate the maximum volume flow rate of oil. Is this a class Ii system?
6" sch 40 steel pipe Di = 0.1541m, A = 1.864 x 10-2 m2, С” = 4.6 x 10-5 m ,P1-P2 = 60KPa
Ans; Similar with Chapter 8 Example 8.2 Page 34
14. Que; From Example 7.9 ,if water at 20ЛљC flows from a large reservoir at the rate of 1.5 x 10-2
m3/s through the system shown, Calculate the pressure at B.
Ans; Similar with Chapter 7 Example 7.9 Page 28
15. (a) What are the three primary reasons for using flow measuring systems.
Ans; Chapter 9
Page 45
(b) Que;Chapter 9 Example 9.1 Page 48
Ans; Chapter 9 Example 9.1 Page 48
16.(a) Define minor losses, flow meter
Ans; Chapter 9 Page 20 and Page 45
(b)Air with a specific weight of 120 N/m3 and a kinematic viscosity of 1.3 x 10-5 m2/s is flowing
through a flow nozzle. A manometer using water as the gage fluid reads 18 mm of deflection.
Calculate the volume flow rate if the nozzle dia. is 50mm.Pipe inside dia. is 100mm.
Ans; Similar with Chapter 9 Example 9.2 Page 50
17.(a) Air with a specific weight of 127 N/m3 and a kinematic viscosity of 2 x 10-5 m2/s is flowing
through a flow nozzle. A manometer using water as the gage fluid reads 18 mm of deflection.
Calculate the volume flow rate if the nozzle dia. is 50mm.Pipe inside dia. is 100mm.
Ans; Similar with Chapter 9 Example 9.2 Page 50
(b)Que;Chapter 7 Example 7.4 Page 23
Ans; Chapter 7 Example 7.4 Page 23
8
18.(a) Air with a specific weight of 127 N/m3 and a kinematic viscosity of 2 x 10-5 m2/s is flowing
through a flow nozzle. A manometer using water as the gage fluid reads 18 mm of deflection.
Calculate the volume flow rate if the nozzle dia. is 50mm.Pipe inside dia. is 110mm.
Ans; Similar with Chapter 9 Example 9.2 Page 50
(b) Determine the energy loss if glycerine at 30ЛљC flows 30m through a 150mm diameter pipe with
an average velocity of 4.0m/s
Ans; Similar with Chapter 6
Example 6.1 Page 18
19.(a)Que;Chapter 10 Example 10.1 Page 54
Ans;Chapter 10 Example 10.1 Page 54
(b)Que; From Chapter 10 Example 10.1,A basic oxygen furnace contains 150 tons of liquid steel.
The steel is tapped into a ladle through a 10cm diameter nozzle. The effective initial head of metal is
1m. Estimate the initial mass flow rate of liquid steel out of the furnace. take CD =0.9,Psteel =
7000kg/m3
Ans; Similar with Chapter 10 Example 10.1 Page 54
20.Que;Chapter 10
Ans;Chapter 10
Example 10.2 Page 55
Example 10.2 Page 55
21. Water at 80ЛљF is being supplied to an irrigation ditch from an elevated storage reservior as shown
in figure. Calculate the volume flow rate of water into the ditch.(Exercise 11)
Given: Вµ/ПЃ = 9.15 x 10-6 ft2/s
4" sch 40 --- Di = 0.3355 ft
A= 0.0884ft2
С” steel = 1.5 x 10-4 ft
Le/D for elbow = 20
Ans;
PA/r + ZA+(VA)2/ 2g - hL = PB/r + ZB +VB2/2g
PA = PB = O ( В·.В· Atm pressure)
VA = negligible
hL = (ZA вЂ“ ZB) вЂ“ VB2/2g = 40 - VB2/2g
hL= hentrance +helbow +hvalve +hfriction
hent: = CL VB2 /2g = VB2 /2g
helbow = f Le VB2 / 2g = 20 f VB2 /2g
halve = f Le VB2 / 2g = 160 f VB2 /2g
hfriction = f L VB2 / 2g = (330) / (0.3355) x f VB2 /2g = 983.6 f VB2 / 2g
hL = VB2 / 2g ( 1 + 20f + 160f + 983.6f )
hL = VB2 /2g (1+ 1163.6f )
40- VB2 / 2g = VB2 / 2g (1 + 1163.6 f )
40 /(VB2/2g) -1 = 1 + 1163.6f
(2g x 40) /VB2 = 2 + 1163.6f
VB2 = 80g / (2 + 1163.6f )
= (80 x 32.2) / (2 + 1163.6f )
2576
VB =
2 + 1163.6 f
9
D VB ПЃ
0.335VB
= 3.66 x 104 VB
в€’6
Вµ
9.15 x10
-4
С”/D =1.5 x 10 / 0.3355 = 0.00044
f = 0.039 for Re = 4000, f = 0.016 for Re = 4.5 x 106
Re =
=
1st trial Assume
f = 0.027
2576
VB =
= 8.7 ft / s
2 + 1163.6 x 0.027
Re = 3.66 x 10 4 x 8. 7 = 3. 18 x 105
f = 0.0175 (from figure )
f = 0.0175
2nd trial Assume
2576
VB =
=10.73 ft / s
2 + 1163.6 x 0.0175
Re = 3.66 x 10 4 x 10.73 = 3.9 x 10 5
f = 0.0175 ( from figure )
Q = ABVB = 0.0884 x 10.73 = 0.946 ft3 / s
22. Figure shows a portion of a hydraulic circuit. The pressure at point B must be 200 psig when the
volume flow rate is 60 gal/min. The hydraulic fluid has a specific gravity of 0.9 and a dynamic
viscosity of 6.0 x 10-5 lb.s/ ft2. The total length of pipe between A & B is 50 ft. The elbows are
standard. Calculate the pressure at the outlet of the pump at A. (Exercise 14)
ВµПЃО±
Given: 2" sch 40 steel pipe -- D1= 0.1738 ft
A= 0.02333ft2
С” = 1.5 x 10-4ft
PA / r + ZA +VA2 / 2g вЂ“ hL = PB /r + ZB + VB2 /2g
PA/r = PB/ r + (ZB вЂ“ ZA) + VB2/ 2g + hL
VA = VB ( same dia)
PA/r = PB/r + (ZB-ZA) + hL
Q = 60 gal/min = (1 ft3/s) / 449 gal/ min = 0.134 ft3 /s
hL= helbows + hvalve + h friction
helbows = (f Le V2) / 2gD = 30 fV2/ 2g
V= Q/A = 0.134 /0.02333 = 5.74 ft /s
Re = DVПЃ /Вµ = (0.1733 ft x 5.74 ft/s x 962.4 lb/ft3) / (6 x10-5 lb s/ft3 x 32.2 ft/s2 = 2.9 x104
С” /D = 1.5 x 10-4 /0.1733 = 0.00086
f = 0.025 ( from figure )
h elbow = ( 0.025 x 30 x (5.74)2 ft2/s2 / 2x 32.2 ft/s2 = 0.38 ft
for 2, elbows = 0.38 x 2 = 0.76 ft
hvalve = CL V2/2g = 6.5x (5.74)2 / ( 2 x 32.2) = 3.33 ft
hfriction = fLV2 /2gD = [0.025 x 50 x (5.74 )2 ] / (2 x32.2 x 0.1733) = 3.69ft
hL = helbows + hvalve + hfriction
10
= 0.76 + 3.33 + 3.69 = 7.78 ft
PA / r = PB/ r + (ZB-ZA) + hL
= (200lb/in2 x 144in2/1ft2 ) x 1ft3 /62.4 lb + 25 ft + 7.78 ft
=494 .32 ft
PA = 494.32 ft x 62.4 lb/ft3
=30845.47 lb/ft2
=214.2 psig
23. Figure shows a hydraulic system in which the pressure od B must be 500 psig while the flow rate
is 750 g/min. The fluid is a medium machine tool hydraulic oil. The total length of 4 pipe is 40ft. The
elbows are standard. Neglect the energy loss due to friction in the 6 pipe. Calculate the required
pressure at A if the iol is at 100F. (Exercise 15)
Given: 4" sch 80 -------- Di =.0.3188 ft
A = 0.07986 ft2
6" sch 80 -------- Di = 0.4861
A = 0.181 ft2
Вµ/ПЃ = 7.21 x 10-4 ft2/ s
s.g = 0.895. С” = 1.5 x 10-4 ft
PA / r + ZA + VA2 / 2g вЂ“ hL = PB / r + ZB + VB2 / 2g
PA/r = PB/ r + (ZB вЂ“ ZA ) + VB2 / 2g + hL
Q =750 gpm x 1 ft3/s / 449 gpm = 1..67 ft3/s
VB= 1.67 ft3/s /0.181 = 9.23 ft/s , VA =1.67ft3/s /0.07986 =20.9 ft/s
Re = DV ПЃ /Вµ = 0.3188 x 20.9 / (7.21 x 10-4) = 9241 = 9.2 x 103
С” /D = 1.5 x 10-4 / 0.3188
= 0.00047
f = 0.032 ( from figure)
hL = helbows + hfriction 4вЂі + hsudden enlargement
helbows = fLeV2 / 2gD = [0.032 x 30 x (20.9)2] / (2 x 32.2) = 6.5 ft
for 2 elbows = helbows = 2x6.5 =13 ft
hfriction = fLV2 / 2gD = [0.032 x 40 x ( 20.9 )2 ] / (2 x 32.2 x 0. 3188 ) = 27.23 ft
h sudden enlargement = VA2/ 2g = 0.27 x (20.9)2 / (2 x 32.2) = 1.83 ft
hL = 13 + 27.23 + 1.83 = 42.06 ft
PA / r = PB/r + ( ZB вЂ“ ZA ) + (VB2 вЂ“ VA2) /2g + hL
=(500 lb/in2 x 144 in2 / 1 ft2) x 1 / (62.4 x 0.895) + 4 + (9.232 вЂ“ 20.92) /( 2 x 32.2) + 42.06 ft
= 1329.8ft
PA = 74266.67 lb/ft2
= 515.74 psig
24. Determine the required size of new schedule 80 steel pipe to carry water at 160F with a maximum
pressure drop of 10 psi per 1000ft when the flow rate is 0.5 ft3/s. (Given r = 61 lb/ft, Вµ/ПЃ =4.38 x 10-6
ft2/s,С” = (1.5 x 10-4 ft) (Exercise 24)
Ans; P1/r + Z1 + V12 /2g вЂ“hL = P2/r + Z2 + V22/ 2g
Z1 вЂ“ Z2 = 0 , V1 = V2
11
hl = 1/r ( P1-P2)
hL = 1/ 61.4 lb/ft 3 x 10 lb/in2 x 144in2/1ft = 23.6 ft
hL = fLV2 / 2g D
23.6 = 1000 fV2 /2 x 32.2 x D = 15.52 fV2/D
f = 1.52 D/V2
=0.5 ft3/s
V = Q/A = 4Q/Рџ2 D2 ---------- V2 = 16 Q2/ Рџ2D4
23.6 = 15.52 f/D (16 Q2) / Рџ2 D4 = 15.52 x [16 x (0.5 )2 f]/ (Рџ2 D4 D)
f = 3.75 D5 ---------- D = ( 0.266 f)0.2
Re = DVПЃ/Вµ = D x 4Q / РџD2 x 1 / (4.38 x 10-6 ft 2) = (4 x 0.5) / (РџD x 4.38 x 10-6) = (1.45 x 105) /D
1st trial Assume , f = 0.02
D = ( 0.266 x 0.02) 0.2 = 0.35 ft
Re = (1.45 x 105) / 0.35 = 4.1 x 105
С” /D = (1.5 x 10-4 )/ 0.35 = 0.00043
fnew = 0.0179 ( from graph )
2nd trial Assume , f = 0.0179
D = ( 0.266 x 0.0179 )0.2 = 0.34
Re = 1.45 x 105 / 0.34 = 0.00044
f = 0.0179 ( from graph)
diameter = 0.34 ft
Use 5 in sch 80 steel pipe.
25. A sharp edged orifice meter is placed in a 10 diameter. Pipe carrying ammonia. If the volume
flow rate is 25gpm, calculate the deflection of a water manometer.
(a) if the orifice diameter is 1" and
(b)
if the orifice diameter is 7" . The ammonia has a s.g of 0.83 and a dynamic viscosity of 2.5
x 10-6 lbs/ft2
D = 10" = 0.83 ft, Q = 25gpm x 1ft3/s /449gpm = 0.0557 ft3/s
(a) V = Q/A = 0.0557 ft3/s / [Рџ/4 (0.83)2 ] = 0.103 ft/s
Re = DVПЃ/Вµ
=[10/12 x 0.103 x 51.79] / [2.5 x 10-6 x 32.2] = 5.5 x 104
A2 = Рџ/4 ( 1/12 ft )2 = 5.45 x 10-3 ft2
(A2/A1)2 = [(Рџ /4 x 1) / (Рџ /4 x 10 2) ]2 = 0.01
d/D = 1/10 = 0.1
C = 0.595 (given)
2 g ( P1 в€’ P2 ) / r
Q = CA2
1 в€’ ( A2 / A1 )
0.557 = 0.595 x 5.45 x 10-3
P1-P2 = 237.244 lb /ft2
P1-P2= Rm ( rH2O вЂ“ r ammonia )
-237.244 = Rm (62.4 вЂ“51 .79)
Rm = 22.14 ft
2 x32.2( P1 в€’ P2 )
51.79
12
(b) d = 7 in = 0.583 ft
C = 0.62 ( given )
ПЂ
A2 = /4 (0.583) 2 = 0.267 ft2
ПЂ
ПЂ
(A2/A1) 2 = (( /4 x 72 )/( /4 x 102 )2 = 0.24
2 g ( P1 в€’ P2 ) / r
Q = CA2
1 в€’ ( A2 / A1 )
0.0557 = 0.62 x 0.267
2 x32.2( P1 в€’ P2 )
51.79(1 в€’ 0.24)
P1-P2 =0.069 lb/ft2
P1-P2 = Rm ( rH2 O вЂ“r ammonia)
Rm = 6.5 x 10-3 ft = 0.0065 ft
26. Que;Chapter 8
Ans;Chapter 8
Example 8.1 Page 31
Example 8.1 Page 31
27. Que; From Chapter 8,Example 8.1,calculate the power supplied to the pump shown in the figure,
if it is efficiency is 80%.
Ans; Similar with Chapter 8
Example 8.1 Page 31
28. Que;Chapter 8
Ans;Chapter 8
Example 8.4 Page 39
Example 8.4 Page 39
29.As a test to determine the wall roughness of an existing pipe installation, water at 10ЛљC is pumped
through it at a rate of 225 L/min. The pipe is standard 1ВЅ" commercial steel tubing with a wall
thickness of 0.109in. Pressure gage located 25m apart in a horizontal run of the pipe read 1032 KPa
and 669KPa. Determine the pipe wall roughness. Di = 32.66mm, A = 8.326 x 10-1 m2
Ans; Similar with Chapter 6 Example 6.2 Page 18
30.Que; From Example 7.9 ,if water at 30ЛљC flows from a large reservoir at the rate of 1.5 x 10-2 m3/s
through the system shown, Calculate the pressure at B.
Ans; Similar with Chapter 7 Example 7.9 Page 28
13
Met 2016 Extractive Metallurgy
Sample Questions and Answers for Final Exam, 2008
1. Define the following.
a. Electrolysis (Electrometallurgy Page 23)
b. Faraday's first law (Electrometallurgy Page 24)
c. Current efficiency (Electrometallurgy Page 25)
d. Basic considerations of electrolysis (Electrometallurgy Page 23)
e. Energy Efficiency (Electrometallurgy Page 25)
f. Heap leaching (Hydrometallurgy Page 13)
g. Sand leaching (Hydrometallurgy Page 14)
2. Write short notes about them.
a. Slime leaching
Slime leaching is done by agitation. The slime and the leach solution are agitated in one or
more agitators until the ore minerals have been dissolved.
b. Cyanidation
c. Electrofining
d. The purpose of roasting
3. How do you understand solvent qualification and leaching- flotation.
Ans; (Hydrometallurgy Page 12)
4. (a) What are the advantages of pressure leaching at atmospheric pressure.
(b) Construct a chart showing how the following ore minerals are treated by hydrometallurgical
means. Give both the preliminary and final treatment method.
a. Bauxite
b. Quartz
c. Carnotite
Ans;(Hydrometallurgy Page 16)
5.(a) Fill in the blanks
Sr. Item
No
1.
Lead Electrorefining
Anode
Cathode
Impure lead
Starting sheet
2.
Nickel Electrorefining
Crude nickel
3.
Gold Electrorefining
Cast of impure
gold
Electrolyte
Lead Fluosilicate (PbS)
Free Fluosilicic acid
Starting sheet
Acid nickel sulfate solution
containing some boric acid
Pure
rolled Hot solution of gold chloride
gold
and free hydrochloric acid
(b) Define i. Electrowinning
ii. Faraday's second law (Electrometallurgy Page 24)
iii. Pressure leaching iv. Cementation
v. Current density (Electrometallurgy Page 25)
6.Compare the electrorefining and Electrodeposition of copper.
Ans;(Electrometallurgy Page 28)
14
7. Briefly explain the required step (or) processes of hydrometallurgical operation.
8.You have Cu ore that contain both oxide and sulphide copper. Which method can you choose to get
Cu metal from that ore? Briefly explain.
9. Explain the gold cyanidation process including with chemical reactions.
Ans;(Hydrometallurgy Page 20)
10.(a) How much copper can be deposited from a CuSO4 solution by a current of 30 amperes flowing
for 8 hours?(Cu = 63.75)
Ans;Electro-metallurgy Example 1 Page 24
(b)Determine the current efficiency of a chromium plating cell in which a current of 10A flowing
for 90 minutes deposits 0.9 g of chromium from chromium acid electrolyte(CrCO3) comprising Cr6+
ions. The relative atomic mass of chromium is 52.
Ans;Electro-metallurgy Example 2 Page 25
11.A copper electroplating process utilizes 15 A of current by chemically dissolving a copper anode
and electroplating a copper cathode. If it is assumed that there are no side reactions how long with it
take to corrode 8.5g of copper from the anode?(Cu = 63.75)
Ans; I = 15A, m = 8.50g , n = 2, F = 96500, W = 63.75, t = ?
m = WIT / (nf)
8.5 = (63.75 x 15 x t)/(2 x 965 )
956.25 t = 1.64 x 106
t = 1715.5s
12.An electroplating process uses 15A of current by chemically corroding a Cu anode. What is the
corrosion rate of the anode in grams per hours?(Cu = 63.75)
Ans; Similar with no.11
13.A silver electroplating process uses 12A of current and chemically corrodes silver anode. How
long will it take to corrode 5.8g of silver from the anode?(Ag =108)
Ans; Similar with no.11
14. (a) Define hydrometallurgy and state the two essential steps required in every hydro metallurgical
operation.
Ans;(Hydrometallurgy Page 11)
(b) Which materials are suitable for leaching?. Divide into two classes .
Ans;(Hydrometallurgy Page 11)
15.(a) Which qualifications must passes the solvent for leaching.?
Ans;(Hydrometallurgy Page 12)
(b) Give the reasons why the materials to be leached is roasted.
Ans;(Hydrometallurgy Page 12)
16. Describe the different solvent have been used for leaching process.
Ans;(Hydrometallurgy Page 13)
17. How do you understand about the methods of leaching.
15
Ans;(Hydrometallurgy Page 13)
18.(a) How can we separate the solution and residue .
Ans;(Hydrometallurgy Page 15)
(b) What are the advantages of pressure leaching over leaching at atmospheric pressure.
Ans;(Hydrometallurgy Page15)
19. Construct a chart showing how the following ore minerals are treated by hydrometallurgical
means. Give both the preliminary and final treatment method.
Oxidized, sea water, sphalerite, oxidized (sulphied ).
Ans;(Hydrometallurgy Page 16)
20. Draw the schematic diagram of process used in acid leaching of oxidized copper.
Ans;(Hydrometallurgy Page 17)
21. Briefly describe the leaching of oxidized copper with sulphuric Acid.
Ans;(Hydrometallurgy Page 16.17)
22. Draw a schematic diagram of process used in acid leaching of roasted zinc concentrate
(calcine)(the double-leach method)
Ans;(Hydrometallurgy Page 19)
23. Describe the recovery of gold by amalgamation. Ans;(Hydrometallurgy Page 20)
24.(a) How much copper can be deposited from a CuSO4 solution by a current of 35 amperes flowing
for 7 hours?(Cu = 63.75)
Ans;Electro-metallurgy Example 1 Page 24
(b)Determine the current efficiency of a chromium plating cell in which a current of 12A flowing
for 80 minutes deposits 0.9 g of chromium from chromium acid electrolyte(CrCO3) comprising Cr6+
ions. The relative atomic mass of chromium is 52.
Ans;Electro-metallurgy Example 2 Page 25
25. Briefly explain the differences between Arc and induction furnace.
Ans;(Electrometallurgy Page 26)
26 Compare the electro refining and electrodepositing methods of Cu.
Ans;(Electrometallurgy Page 26)
27.Explain tow gold and silver are recovery from ore by capitation process.
Ans;(Hydrometallurgy Page 20)
28.Draw a diagram of a simple electrolytic cell.
Ans;(Electrometallurgy Page 28)
29.Explain why the amalgamation process has a low recovery ? What are the limitations and
restrictions of this process?
Ans;(Hydrometallurgy Page 20)
16
30. Explain the reasons why higher power requirements ore needed when electron fining and electro
winning with insoluble anodes as compared with that when using soluble anodes.
17
Met вЂ“ 2023 Concepts of Materials Science
Sample Questions and Answers for Final Exam, 2008
1. Define the following.
i. Frenkel defect
ii.Interstitial defect
iv. Surface defect
v. Schottky defect.
vii. Burgers vector
viii. Grain boundary
x. Vacancy
Ans; Chapter 1
Page 19,20
2.(a)
(b)
17
iii.Point defect
vi. Schmid's Law
ix.Slip system
3.(a)
(b)
4.(a)
18
(b)Que;
Ans;
5.(a)Que;
Ans;
Chapter 1
Chapter 1
Example(1-6) Page 13
Example(1-6) Page 13
Chapter 1 Sample Que; 1-8
n = 4 atom/u.c/(3.6151 x 10-8cm)3 = 8.47 x 1022 atom/cm3
nv = 8.47 x 1022 exp[-20000/(1.987)(1358)]
= 8.47 x 1022 exp(-7.41195) = 5.1 x 1019 vacancies/cm3
(b) Determine the interlunar spacing and the length of Burgers vector for slip on the expected slip
system in BCC tantalum. (lattice parameter is 3.3026AЛљ). Repeat, assuming that the slip system is a
(111)/[110]. What is the ration between the shear stress required for slip for the two systems?
Assume the K=2
Ans; (a)For (110)/[111]:
b = (1/2)(в€љ3)(3.3026 AЛљ)= 2.860AЛљ
d110= 3.3026/в€љ12+12+02 AЛљ= 2.335AЛљ
(b) (111)/[110]:
b = в€љ2(3.3026 AЛљ)= 4.671AЛљ
d111= 3.3026/в€љ12+12+12 AЛљ= 1.907AЛљ
6.(a) A single crystal of a BCC metal is oriented so that the [001] direction is parallel to the applied
stress. If the critical resolved shear stress required for slip is 12000 psi, calculate the magnitude of
the applied stress required to cause slip to begin in the [111] direction on the (011), (101), (101) .
Ans; CRSS = 12000psi = О± cosГ� cosО»
О» = 54.76 Лљ 12000psi/ cosГ� cosО» = О±
Г�110 =90Лљ
О± = infinity
Г�011 =90Лљ
О± = 29412 psi
Г�101 =90Лљ
О± = 29412 psi
19
(b)
7.(a) A 0.05 cm layer of MgO is deposited between layers of nickel and tantalum to provide a
diffusion barrier that prevents reactions between the two metals. (Figure 2-8). At 1400ЛљC, nickel
ions are created and diffuse through the MgO ceramic to the tantalum. Determine the number of
nickel ions that pass through the MgO per second. The diffusion coefficient of nickel in MgO is 9
x10-12 cm2/s, and the lattice parameter of nickel at 1400ЛљC is 3.6 x 10-8 cm.
20
(b)Define Dislocation, Critical resolved shear stress, Slip plane, Slip Direction.
Ans;
Chapter 1
Page 19,20
8.Define the following;
a. Diffusion b.Diffusion coefficient c.Interstitial diffusion d. Self вЂ“diffusion e.Surface
diffusion f. Volume diffusion g.Fick's first law
h.Fick's first law
i.Powder metallurgy
j.Sintering
Ans; Chapter 1 Page 44
9.(a)
21
(b)
22
10.
23
11.
12.(a) We find that 10h are required to successfully carburize a batch of 500 steel gears at 900ЛљC,
Where the iron has the FCC structure. We find that it costs $ 1000 per hour to operate the
carburizing furnace at 900ЛљC and $ 1500 per hour to operate the furnace at 1000ЛљC. Is it economical
to increase the carburizing temperature to 1000ЛљC?
24
( b) Define activation energy, Diffusion bonding, Grain growth.
Ans;
Page 44
13.Define the following;
Creep test, % Elongation, Endurance Limit, Endurance Ratio, Fatigue life, Elastic deformation,
Engineering Stress and Strain, Fatigue test, Hooke's law.
Ans;
Page 44
14.(a)Atoms are found to move from one lattice position to another at the rate of 5 x 105 jumps per
second at 400ЛљC when the activation energy for their movement is 30,000 cal/mol. Calculate the
jump rate at 750ЛљC.
Ans; Rate = (5 x 105) /x = C0 exp[-30000/(1.987)( 673)] / C0 exp[-30000/(1.987)( 1023)
= exp(-22.434 + 14.759) = 4.64 x 10-4
5
(5 x 10 ) /x = 4.64 x 10-4
x = 1.08 x 10-4 jumps/s
(b)A BCC steel containing 0.001% N is nitrided at 550ЛљC for 5h. If the nitrogen content at the
steel surface is 0.08%, determine the nitrogen content at 0.25 mm from the surface.
Ans; (0.08-cs)/ (0.08-0.001) = erf[0.025/2в€љDt]
t = (5h)(3600s/h) = 1.8 x 104 s
D = 0.0047 exp[-18300/(1.987)(823) = 6.488 x 10-8 cm2/s
в€љDt = 0.0342
erf[0.025/(2) (0.0342)]= erf(0.3655) = 0.394
(0.08-cs)/ (0.079) = 0.394 or cs= 0.049%N
15.(a)
(b)
25
16.(a)
(b)
26
17.
18.(a)
27
(b) Que;
Ans;
Chapter 3
Chapter 3
Example(3-7) Page 73
Example(3-7) Page 73
19.How do you understand the impact test? Briefly explain with figure
Ans; Chapter 3
Page 59
.
20.(a)The following data were collected from a standard 0.505-in, diameter test specimen of a
copper alloy:
Ans; О±= F/(Рџ /4)(0.505)2 = F/0.2
С” = (l-2)/2
Load (lb)
0
3,000
6,000
7,500
9.000
10,500
12,000
12,400
11,400
Gage Length (in.)
2.00000
2.00167
2.00333
2.00417
2.0090
2.040
2.26
2.50(maximum load)
3.02(fracture)
40
stress(ksi)
l
Stress(psi)
0
15000
30000
37500
45000
52500
60000
62000
57000
0.2 offset
30
20
10
0.001
0.01
strain (in/in)
28
Strain (in/in)
0.0
0.000835
0.001665
0.002085
0.0045
0.02
0.13
0.25
0.51
After fracture, the gage length is 3.014 in, and the diameter is 0.374 in .Plot the data and calculate
(a) the 0.2% offset yield strength = 45000psi
(b) the tensile strength = 62000psi
(c) the modulus of elasticity E = (30000-0) / (0.001665-0) = 18 x 106 psi
(d) the % elongation = (3.014-2)/2 x 100 = 50.7%
(e) the % reduction in area = [(Рџ /4)(0.505)2-(Рџ /4)(0.374)2 /(Рџ /4)(0.505)2 x 100 = 45.2%
(f) the engineering stress at fracture = 57000psi
(g) the true stress at fracture = 11400lb/(Рџ /4)(0.374)2 = 103770 psi
(h) from the graph, yielding begins at about 37500psi.Thus: ВЅ (yield strength)(strain at
yield) = ВЅ(37500)(0.002085) = 39.1 psi
(b)The following data were collected from a 12 -mm diameter test specimen of magnesium.
Load(N)
0
5,000
10,000
15,000
20,000
25,000
26,500
27,000
26,500
25,000
Gage Length (mm)
30.0000
30.0296
30.0592
30.0888
30.15
30.51
30.90
31.50 (maximum load)
32.10
32.79(fracture)
After fracture, the gage length is 32.61 mm and the diameter is 11.74mm. Plot the data and calculate
(i) the 0.2% offset yield strength.
(j) the tensile strength,
(k) the modulus of elasticity,
(l) the % elongation,
(m) the % reduction in area,
(n) the engineering stress at fracture,
(o) the true stress at fracture, and
(p) the modulus of resilience.
Ans; Similar with No.20(a)
21.(a)An aluminum alloy that has a plane strain fracture toughness of 25000 psi in fails when a
stress of 42,000 psi is applied. Observation of the fracture surface indicates that fracture began at
the surface of the part. Estimate the size of the flaw that initiated fracture. Assume that f = 1.1.
Ans; Klc = fПѓ ПЂв€‚ or в€‚ = (1 / ПЂ )[ K lc / fПѓ ] 2
в€‚ = (1 / ПЂ )[25000 / 1.1x 42000] 2 = 0.093in
(b)A ceramic part for a jet engine has a yield strength of 75.000 psi and a plane strain fracture
toughness of 5.000 psi in . To be sure that the part does not fail, we plan to assure that the
maximum applied stress is only one-third the yield strength. We use a nondestructive test that will
detect any internal flaws greater than 0.05 in long. Assuming that f = 1.4, does our nondestructive
test have the required sensitivity? Explain.
29
Ans;
Пѓ = (1 / 3)(75000) = 25000 psi
в€‚ = (1 / ПЂ )[ K lc / fПѓ ] 2 = (1 / ПЂ )[5000 / 1.4 x 25000] 2 = 0.0065in
22.(a) A 2-cm diameter, 20-cm long bar of an acetal polymer is loaded on one end and is expected
to survive one million cycles of loading, with equal maximum tensile and compressive stresses,
during its lifetime. What is the maximum permissible load that can be applied?
Figure 3-21 The S-N fatigue curve for an acetal polymer
Ans;From the figure,we find that the fatigue strength must be 22MPa in order for the polymer to
survive one million cycles. Thus, the maximum load is
F = (fatigue strength) d3/10.18L
F = (22MPa)(20mm)3 = (10.18)(200mm) = 86.4 N
(b)Write short notes for Rupture time, Poisson's ratio, Fracture toughness, Facture mechanics
Ans; Chapter 3 Page 76
23.What factors are depended on the ability of a material to resist the growth of a crack.
Ans; Chapter 3 Page 71
24.(a)
Ans; 11.98 g/cc = (x)(106.4 g/mol)/ (3.8902 x 01-8 cm)3 (6.02 x 1023 atoms/mol)
x = 3.9905
fraction = (4.0-3.9905) /4 = 0.002375
(b) number = (0.0095 vacancies/u.c) / (3.8902 x 10-8cm)3 = 1.61 x 1020 vacancies/cc
(b)
Ans;
(a) 1 vacancy/(200)(3.5089 x 10-8 cm)3 = 1.157 x 1020 vacancies/cc
(b) In 200 unit cells, there are 399 Li atoms.The atoms/cell are 399/200:
ПЃ = (399/200)(6.94 g/mol)/ (3.5089 x 10-8 cm)3 (6.02 x 1023 atoms/mol) = 0.532 g/cc
25.(a)
Ans; (a) 6 x 10-15 /(1 x 10-9) = D0 exp [-Q/(1.987) (1000)] / D0 exp [-Q/(1.987) (1673)]
6 x 10-6 = exp [-Q/(0.000503-0.00030)] = exp[-0.000203 Q]
30
(b)
Q = 59230 cal/mol
(b) 1 x10-9 = D0 exp [-59230/(1.987) (1673)]
D0 = 0.055 cm2/s
Ans; Similar with 25(a)
26.(a)
Ans;
1.1 в€’ 0.5
= 0.667 = erf [0.05 / 2 Dt ]
1.1 в€’ 0.2
0.05 / 2 Dt = 0.685 or Dt = 0.00133
t = (2h)(3600 s / h) = 7200 s
D= 0.00133/7200 = 1.85 x 10-7 = 0.23 exp[-32900/1.987T]
exp(-16558/T) = 8.043 x 10-7
T = 1180K = 907ЛљC
(b)
Ans; Similar with No.26(a)
27.(a)
Ans; Similar with No.26(a)
(b)
Ans; Similar with No.26(a)
31
28.(a)
Ans; Similar with No.26(a)
(b)
Ans; Similar with No.26(a)
29.(a)
Ans; (a) First determine the stress acting on the wire:
Пѓ = ( F / A) = 850lb /(ПЂ / 4)(0.15in) 2 = 48100 psi
Because О± is greater than the yield strength of 45000psi, the wire will plastically
deform.
(b) Because О± is less than the tensile strength of 55000psi, no necking will occur.
(b)
Ans; Similar with No.29(a)
30.(a)
Ans; The strain is С” = О±/E = 40000/17 x 106 psi = 0.00235 in/in
С” = (2in -d0)/do = 0.00235in./in.
2-do = 0.00235 do
do = 1.995in
The opening in the die must be smaller than the final diameter.
(b) Chapter 3 Sample Que;3-4
Ans; Similar with No.30(a)
32
Met-2024 Metal Process Engineering I
Sample Questions and Answers for Final Exam, 2008
1.(a) Short notes on the liquid state welding.
Ans; Chapter 6 Joining Page 35
(b) Short notes on the liquid-solid state bonding .
Ans; Chapter 6 Joining Page 38,39
2.(a) Describe and distinguish features between liquid state welding and liquid solid state welding.
Ans; Chapter 6 Joining Page 35,38,39
(b) How do brazing and soldering differ from the liquid-solid state welding processes?
Ans; Chapter 6 Joining Page 41,42
3.(a) What is the technical difference between brazing and soldering?
Ans; Chapter 6 Joining Page 39,42
(b) Discuss briefly the difference between braze welding and brazing.
Ans; Chapter 6 Joining Page 39,41
4.(a) What are the three basic steps in the conventional powder metallurgy shaping process? Explain
each briefly.
Ans; Powder Metallurgy Page 46,47,48
(b) What is the technical difference between mixing and blending in PM?
Ans; Powder Metallurgy Page 46
5.(a) What are three steps in the sintering cycle in PM?. Explain each briefly.
Ans; Powder Metallurgy Page 48,49
(b) Describe briefly the any three techniques of power production with neat sketches.
Ans; Powder Metallurgy Page 49,50
6.Describe briefly the applications of PM products.
Ans; Powder Metallurgy Page 51,52
7. Draw and label the various form of welding cracks and imperfect shapes of weld beads. .
Ans; Chapter 8 Welding Process Page 69,70
8. Define the following with neat sketches.
i. SMAW ii. GMAW iii. FCAW
iv. GTAW
Ans; Chapter 8 Welding Process Page 55 to 58
v. FCAW
9. (a) Chapter 8 Arc Welding Problem no.1
Ans; (a) HRw = f1f2 E I = (0.85)(0.75)(30)(225)=4303.1W
1055W =1 Btu
HRw = 4303.1 /1055 =4.079 But/sec
(b) WVR = (4.079 Btu/sec) / (153 Btu/in3) =0.0266 in3/sec
33
(b) Chapter 8 Arc Welding Problem no.2
Ans; (a) HRW = f1f2 EI=UmAwV
0.65 (0.70)(25)(125) = 9.3 (Aw) (5)
1421.9 = 46.5 Aw
Aw =1421 .9 /46.5 = 30.6mm2
(a) AwV = 30.6 (5) =153mm3/s
filler wire A = ПЂ D2/4 = ПЂ (3)2/4 =7.07 mm2
At 50% filler metal, feed rate of filler wire = 153 (0.50)/ 7.07
=10.82 mm/s
10.(a) Chapter 8 Arc Welding Problem no.3
Ans; (a) f1f2 EI=UmAw V
V= f1f2 EI/UmAw
1 Btu/sec = 1055 J/s = 1055W, so 150 Btu/sec = 158,250W
V= 0.95 (0.7) (25) (300)/(158,250 x 0.120) = 0.263 in /sec = 15
Circumference C = ПЂ D =12 x 6 ПЂ =226.2 in/rev
Rotational speed N = (15.76 in/min ) (226.2 in/rev ) = 0.696
(b) Chapter 8 Resistance Welding Problem no.4
Ans; (a) PD = I2R/A
A = ПЂD2/4 = ПЂ (0.19)2/4 = 0.02835 in2
I2R = (9500)2 (100 x 10 вЂ“6) = 9025 W
1 Btu/sec = 1055 W, so 9025 W = 8. 554 Btu/sec
PD = 8.554 / 0.02835 = 302 Btu/sec-in2
(b)H = I Rt = (9500)6 (100 x10-6) (0.17) = 1534 W.sec
Weld nugget volume V= ПЂ D2d/4 = ПЂ (0.19)2 (0.060)/4
Heat required for melting = UmV = (150 Btu/in3)(0.0017)
Proportion of heat for welding = 0.255 / 1.454 = 0.175
11.(a) Chapter 8 Resistance Welding Problem no.5
Ans; H = I2Rt = (5000)2(75 x 10-6) (0.15) = 405 W . sec =405J
Weld nugget volume V = ПЂ D2d/4 = ПЂ (5)2 (2.5)/4 = 49.1 min3
Heat required for melting = UmV = (2.9J/mm3) ( 49.1 mm3)=142.4J
Proportion of heat for welting = 142.4/405 = 0.351
(b) Draw the flow sheets of classification of welding processes and fusion welding.
Ans; Welding Page 53
12. (a) Chapter 8 Electron Beam Welding Problem no.6
Ans; PD = f1 EI/A
A= ПЂ D2/4 = ПЂ (0.020)2/4 = 0.000314 in3
Power P = 0.9 (25 x 103) (100 x 10-3)/1055 =2.133 Btu/sec
PD = 2.1333/0.000314 = 6792 Btu/sec in3
34
(b) Chapter 8 Electron Beam Welding Problem no.7
Ans; PD = f1EI/A
A= ПЂ D2/4 = ПЂ (0.30)2/4 = 0.0707mm3
Power P = 0.85 (50 x 103) (65 x 10-3) = 2762.5W
PD =2762.5 / 0.0707 = 39,074W/mm2
13.(a) Define Braze welding
Ans; Chapter 6 Joining Page 41
(b) What is consolidation? Briefly explain each.
Ans; Page 6, 35
14.(a)Que; Chapter 7 Problem no.2 Page 51
Ans; .(a) Um = 3.33 x 10-6 Tm2
for austenitic stainless steel = 1670K
Um = 3.33 x 10-6 ( 1670)2 =9.29 J/mm3
Volume of metal melted V =20 (200)=4000mm3
Hm =9.29 (4000) =37.148 J at weld
(b) Given f1 = 0.8 and f2 =0.6
H =37.148/ (0.8 x 0.6) =77.392 at source
(b) Que; Chapter 7 Problem no.1 Page 51
Ans; (a) Um = 1.467 x 10-5 T2m
Tm for medium carbon steel = 3060 R
Um = 1.467 x 10-5 (3060)2 = 137.4 Btu/in3
Volume of metal melted V = 0.045 (10) =0.45 in3
Hm = 137.4 (0.45) = 61.8 Btu at weld
(b)Given f1 = 0.9 and f2 = 0.7
H = 61.8/(0.9 x 0.7 ) = 98.1 Btu at source
15.(a) What are the three basic steps in the conventional power metallurgy shaping process? Briefly
explain each.
Ans; Chapter 9 Page
(b) Define welding and describe the different types of welding processes.
Ans; Chapter 7 Page
16.(a)What is the technical difference between brazing and soldering?
Ans; Chapter 6 Page
(b) What are the three steps in sintering cycle in PM? Briefly explain.
Ans; Chapter 9 Page
17.(a)Que; Chapter 8 Resistance Welding Problem no.5
Ans; H = I2Rt = (5000)2(75 x 10-6) (0.15) = 405 W . sec =405J
Weld nugget volume V = ПЂ D2d/4 = ПЂ (5)2 (2.5)/4 = 49.1 min3
Heat required for melting = UmV = (2.9J/mm3) ( 49.1 mm3)=142.4J
Proportion of heat for welting = 142.4/405 = 0.351
(b) Que; Chapter 7 Fundamental of Welding Problem no.4 Page 52
35
In a certain welding operation to make a fillet weld, Aw = 0.025 in2. and v = 15 in/min. If f1=
0.95, f2= 0.5, and УЁm = 2000F for the metal to be welded, determine the rate of heat generation
required at the welding source to accomplish this weld.
Ans; Um = 1.467 x 10-5 (2000+460)2 = 88.80 Btu/in3
V = 15= f1f2 HR/UmAw = 0.95(0.5)HR /(88.8 x 0.025) =0.214 in3
HR =15/0.214 + 70.1 Btu/min
18.(a)Que; Chapter 7 Fundamental of Welding Problem no.5 Page 51
Ans; . Tm =1350K for copper
Um =3.33 x 10-6 (1350)2 = 6.07 J/mm3
V= f1f2 HR/UmAw =0.9(0.25)(3000)/(6.07 x 15 ) = 7.4 mm
(b) Welding Processes Example 4.3 Page 64
19.(a) Chapter 7 Fundamental of Welding Example 3.1 Page 50
(b) Chapter 7 Fundamental of Welding Example 3.2 Page 50
20.What is the difference between fusion welding and solid state welding? Explain.
Ans; Welding Processes Page 55 to 58
21.(a) In a certain welding operation to make a fillet weld, Aw = 0.035 in2. and v = 15 in/min. If f1=
0.95, f2= 0.5, and УЁm = 2000F for the metal to be welded, determine the rate of heat generation
required at the welding source to accomplish this weld.
Ans; Similar with no.17 (b)
(b) A resistance spot welding operation is performed on two pieces of 0.030 in thick sheet steel
(low carbon). The unit melting energy for steel = 150 Btu/in3. Process parameters result in a weld
nugget of diameter = 0.2 in and thickness = 0.040 in. Assume the resistance = 100Вµв„¦. Determine
a)the average power density in the interface area defined by the weld nugget and (b) the proportion
of energy generated that went into formation of the weld nugget.
Ans; Similar with no.10 (b)
22.(a)The voltage in an EBW operation = 55 kV and the beam current = 60 mA. The electron beam
is focused on the circular area that is 0.3 mm in diameter. The heat transfer efficiency f1 = 0.85 .
Calculate the average power density in the area in W/mm2.
Ans; Similar with no.12 (b)
(b) A resistance spot welding operation is performed on two pieces of 0.035 in thick sheet steel
(low carbon). The unit melting energy for steel = 150 Btu/in3. Process parameters result in a weld
nugget of diameter = 0.2 in and thickness = 0.060 in. Assume the resistance = 100Вµв„¦. Determine
a)the average power density in the interface area defined by the weld nugget and (b) the proportion
of energy generated that went into formation of the weld nugget.
Ans; Similar with no.10 (b)
23.Defie the following with neat sketch .
i) Resistance sport welding
ii) Shielded metal arc welding (SMAW)
iii) Submerged arc welding (SAW)
iv) Gas metal arc welding (GTAW)
36
Ans; Joining Processes Page 36 to 38
24.Describe the different types of brazing methods.
Ans; Joining Processes Page 41
25. Welding processes Example 4.1 Page 55
26. A gas tungsten arc welding operation is performed at a current 400A and voltage of 15V. The
melting efficiency f2 = 0.5, and the unit melting energy for the metal Um = 150 Btu/in3. Determine
(a)power in the operation, (b) rate of heat generation at the weld, and (c) volume rate of metal
welded.
Ans; Similar with Welding processes Example 4.1 Page 55
27.(a)In a certain welding operation to make a fillet weld, Aw = 0.030 in2. and v = 25 in/min. If f1=
0.95, f2= 0.5, and УЁm = 2000F for the metal to be welded, determine the rate of heat generation
required at the welding source to accomplish this weld.
Ans; Similar with no.17 (b)
(b)The voltage in an EBW operation = 50 kV and the beam current = 60 mA. The electron beam is
focused on the circular area that is 0.32 mm in diameter. The heat transfer efficiency f1 = 0.85.
Calculate the average power density in the area in W/mm2.
Ans; Similar with no.12 (b)
28.Defie the following with neat sketch .
i) Submerged arc welding (SAW)
ii) Gas metal arc welding (GTAW)
iii) Gas tungsten arc welding(GTAW)
iv) Plasma arc welding (PAW)
Ans; Joining Processes Page 36 to 38
29.A gas tungsten arc welding operation is performed at a current 350A and voltage of 15V. The
melting efficiency f2 = 0.5, and the unit melting energy for the metal Um = 150 Btu/in3. Determine
(a)power in the operation, (b) rate of heat generation at the weld, and (c) volume rate of metal
welded.
Ans; Similar with Welding processes Example 4.1 Page 55
30. Discuss about the soldering as possible as you can.
Ans; Joining Processes Page 42
37
Met-2033 Physical Metallurgy
Sample Questions and Answers for Final Exam, 2008
1.(a) List the four classifications of steels, (b) For each, briefly describe the properties and typical
applications.
Ans; Chapter 4 Page 57
(b)Compare white and nodular cast irons with respect to (a) Composition and heat treatment, (b)
microstructure, and (c) mechanical characteristics.
Ans; Chapter 4 Page 64,68
2.Calculate the amounts of ferrite, cementite, primary
following steels. (a) 1015, (b) 1035 (c)1095 (d) 10130
Ans; Chapter 4 Page 52 with figure 3.1
microconstituents and pearlite in the
3..(a) Briefly explain why ferritic and austenitic stainless steels are not heat treatable.
Ans; Chapter 4 Page 61
(b) Compute the volume percent of graphite VGr in a 3.5 wt%C cast iron, assuming that all the
carbon exists as graphite phase. Assume densities of 7.9 and 2.3 g/cm3 for ferrite and graphite,
respectively.
Ans; Chapter 4 Page 52
4. Estimate the AISI-SAE number for steels having the following microstructures:
(a) 38% pearlite-62 % primary ferrite
(b) 97% ferrite- 3% cementite.
(c) 85% ferrite -10% cementite
(d) 90% ferrite -25% cementite (20-marks)
Ans; Chapter 4 Page 52
5.(a) Why must rivets of a 2017 aluminum alloy be refrigerated before they are used?
Ans; Chapter 5 Page 72
(b) What is the chief difference between heat-treatable and non head-treatable alloys?
Ans; Chapter 5 Page 70
6.(a) Give the distinctive features, limitations, and applications of the following alloy groups:
titanium alloys, refractory metals, super-alloys, and noble metals.
Ans; Chapter 5 Page 76
(b) Draw a schematic diagram for the various ferrous alloys.
Ans; Chapter 5 Page 52
7. Write short notes on the following;
(a) Copper and its alloy
(b) Aluminium and its alloy
(c) Magnesium and its alloy
(d) Titanium and its alloy
Ans; Chapter 5 Page 74 to 78
38
8. (a) Compute the volume percent of graphite VGr in a 3.5 wt%C cast iron, assuming that all the
carbon exists as graphite phase. Assume densities of 7.9 and 2.3 g/cm3 for ferrite and graphite,
respectively.
Ans; Similar with Chapter 4 Page 52
(b) Compare structure, properties and application of bronze and brass.
Ans; Page 71
9. Draw the schematic diagram of iron-carbon diagram. Predict the phase 1085 steel at 650ЛљC.
Ans; Page 52 Fig 3.1
10.Calculate the amounts of ferrite, cementite, primary
following steels. (a) 1025, (b) 1045 (c)1085 (d) 10130
Ans; Similar with Chapter 4 Page 52 with figure 3.1
microconstituents and pearlite in the
11. Estimate the AISI-SAE number for steels having the following microstructures.
(a) 25% pearlite
75% primary ferrite
(b) 35% pearlite
15% primary cementate
(c) 83% ferrite
17% cementite
(d) 77% ferrite
23% cementite (20-marks)
Ans; Similar with Chapter 4 Page 52
12. Compare gray and malleable cast irons with respect to (a) composition and heat treatment (b)
microstructure (c) mechanical properties.
Ans; Page 64,68
13. How many kinds of steels? Explain their mechanical, physical properties and applications.
Ans; Page 57
14.Calculate the amounts of ferrite, cementite, primary micro constituent and partite in the
following steel. (a) 1025 (b)1010 (d)1085 (e)10150 .
Ans; Similar with Page 52 Fig 3.1
15. Estimate the AISI-SAE number for steels having the following microstructures.
(a) 30% pearlite
70% primary ferrite
(b) 90% pearlite
7% primary cementate
(c) 95% ferrite
5% cementite
(d) 80% ferrite
20% cementite (20-marks)
Ans; Similar with Page 52 Fig 3.1
16 Draw the iron-carbon phase diagram. Explain the phase transformation according to
composition and temperature. (20-marks)
Ans; Similar with Page 52 Fig 3.1
17. (a) Compute the volume percent of graphite VGr in a 4.0 wt%C cast iron, assuming that all the
carbon exists as graphite phase. Assume densities of 7.9 and 2.3 g/cm3 for ferrite and graphite,
respectively.
Ans; Similar with Chapter 4 Page 52
39
(b) What is the function of alloying elements in tool steels.
Ans; Example 4.3
18. List the four classifications of cast iron and explain the properties, applications of any two cast
iron (20-marks)
Ans; Page 64-68
19. Explain the physical and mechanical properties, structure and applications of the following
(a) Copper and its alloy
(b) aluminum and its alloy
(c) magnesium and its alloy
(d) titanium and its alloy
Ans; Page 74-79
20. Short not the following materials
(a) titanium alloys
(b) refectory metal
(c) Super alloys
(d) retile metals
Ans; Page 78
21. (a) Explain the reaction, temperature and composition at eutectoid and eutectic of iron-carbon
and iron-graphite phase diagram . (10-marks)
Ans; Page 52,63
(b) Why does the tool steel hard by alloys elements? (10-marks)
Ans; Page 60
23.(a) Calculate the amount of phase and composition of 0.7% C steel at 600ЛљC (10-marks)
Ans; Page 52,Figure 3.1
(b) Calculate the amount of phase and composition of 10130 steel at 600ЛљC
Ans; Page 52,Figure 3.1
24.(a) What is the main difference between a brass and bronze?
Ans; Page 71
(b) Describe the disadvantages of steel and other ferrous alloys.]
Ans; Page 70
25.How many classified the cast iron family? Short note any two cast iron.
Ans; Page 64-68
26.Draw the schematic diagram of iron-carbon and iron-graphite phase diagram. Describe the
different facts of above the two phase diagram
Ans; Page 52,Figure 3.1
27. (a) List the four classifications of steels, (b) For each, briefly describe the properties and typical
applications.
Ans; Chapter 4 Page 57
40
(b)Compare white and nodular cast irons with respect to (a) Composition and heat treatment, (b)
microstructure, and (c) mechanical characteristics.
Ans; Chapter 4 Page 64,68
28.(a) How many types of cast iron? Briefly explain. (10-marks)
Ans; Page 64-68
(b) Compute the volume percent of graphite VGr in a 5.5 wt%C cast iron, assuming that all the
carbon exists as graphite phase. Assume densities of 7.9 and 2.3 g/cm3 for ferrite and graphite,
respectively.
Ans; Similar with Chapter 4 Page 52
29.(a) Calculate the amount of phase and composition of 0.7% C steel at 600ЛљC (10-marks)
Ans; Page 52,Figure 3.1
(b) Calculate the amount of phase and composition of 10130 steel at 600ЛљC
Ans; Page 52,Figure 3.1
30. (a) Compute the volume percent of graphite VGr in a 4.0 wt%C cast iron, assuming that all the
carbon exists as graphite phase. Assume densities of 7.9 and 2.3 g/cm3 for ferrite and graphite,
respectively.
Ans; Similar with Chapter 4 Page 52
(b) What is the function of alloying elements in tool steels.
Ans; Example 4.3
41

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