copyright 2006 - michael bush1 industrial application of basic mathematical principles session 7...
Post on 15-Jan-2016
215 views
TRANSCRIPT
Copyright 2006 - Michael Bush
1
Industrial Application of BasicMathematical Principles
Session 7Homework Assignments
Copyright 2006 - Michael Bush
2
Industrial Application of BasicMathematical Principles Session 7 Homework
Unit 27 Problem 1 Page 216Compute the Celsius melting point for
Chromium
a. 5C = F - 32
9
5C = 2,740 - 32
9
5C = 2,708 =
9 1504 C
Copyright 2006 - Michael Bush
3
Industrial Application of BasicMathematical Principles Session 7 Homework
Unit 27 Problem 1 Page 216Compute the Celsius melting point for Cast Iron
a. 5C = F - 32
9
5C = 2,300 - 32
9
5C = 2,268 =
9 1260 C
Copyright 2006 - Michael Bush
4
Industrial Application of BasicMathematical Principles Session 7 Homework
Unit 27 Problem 1 Page 216Compute the Celsius melting point for Copper
a. 5C = F - 32
9
5C = 1,940 - 32
9
5C = 1,908 =
9 1060 C
Copyright 2006 - Michael Bush
5
Industrial Application of BasicMathematical Principles Session 7 Homework
Unit 27 Problem 1 Page 216Compute the Celsius melting point for Aluminum
a. 5C = F - 32
9
5C = 1,200 - 32
9
5C = 1,168 =
9 649 C
Copyright 2006 - Michael Bush
6
Industrial Application of BasicMathematical Principles Session 7 Homework
Unit 27 Problem 1 Page 216Compute the Celsius melting point for Lead
a. 5C = F - 32
9
5C = 620 - 32
9
5C = 588 =
9 327 C
Copyright 2006 - Michael Bush
7
Industrial Application of BasicMathematical Principles Session 7 Homework
Unit 27 Problem 2 Page 217
8 10 12 14 16 18 20 22 24 26 28 30
Degrees Fahrenheit (hundreds)
6
CHROME
CAST
COPPER
ALUMINUM
LEAD
2 4 6 12 14 16 18108
Degrees Celsius (hundreds)
20
Copyright 2006 - Michael Bush
8
Industrial Application of BasicMathematical Principles Session 7 Homework
Unit 27 Problem 3 Page 217Determine difference between Chromium and Lead
a. Chromium = 1,504 C
Lead = 327°C
1,504 C
327 C
1,177 C
Copyright 2006 - Michael Bush
9
Industrial Application of BasicMathematical Principles Session 7 Homework
Unit 27 Problem 3 Page 217Determine difference between Cast Iron and
Aluminum
a. Cast Iron = 1,260 C
Aluminum = 649°C
1, 260 C
649 C
611 C
Copyright 2006 - Michael Bush
10
Industrial Application of BasicMathematical Principles Session 7 Homework
Unit 27 Problem 3 Page 217Determine difference between Copper and Lead
a. Copper = 1,060 C
Lead = 327°C
1,060 C
327 C
733 C
Copyright 2006 - Michael Bush
11
Industrial Application of BasicMathematical Principles Session 7 Homework
Unit 28 Problem 1 Page 217Generate a line graph
10 12 14 16 18 20 22 24 26 28 30
Revolutions Per Minute (hundreds)
TEM
PE
RAT
UR
E (
Fahr
enhe
it)
700
800
900
1000
1100
1200
1300
32 34 36
Copyright 2006 - Michael Bush
12
Industrial Application of BasicMathematical Principles Session 7 Homework
Unit 28 Problem 2 Page 217Determine the average difference in exhaust
temperatures Medium Compression engine
825 920 1,000 1,080 1,150 1,220
Low Compression engine
780 870 950 1,030 1,100 1,160
1032.5 F - 981.7 F= 50.8 F
6195°F 6195°F 6 = 1032.5°F
5890°F 5890°F 6 = 981.7°F
Copyright 2006 - Michael Bush
13
Industrial Application of BasicMathematical Principles Session 7 Homework
Unit 28 Problem 2 Page 217Determine the difference in exhaust
temperatures at 3,250 RPM
High Compression engine
3,250 RPM 1,240 F Low Compression engine
3,250 RPM 1,130 F
1240 C - 1130 C= 110 F
Copyright 2006 - Michael Bush
14
Industrial Application of BasicMathematical Principles Session 7 Homework
Unit 29 Problem 1 Page 218Draw a Circle Graph
AllOthers
USA
FranceGreatBritain
Canada
Copyright 2006 - Michael Bush
15
Industrial Application of BasicMathematical Principles Session 7 Homework
Unit 29 Problem 2 Page 218Determine the Percentages
175,000
90,000
125,000
100,000
+ 60,000 550,000
175,000 550,000= .3182= 31.82%
125,000 550,000= .2273= 22.73%
90,000 550,000= .1636= 16.36%
100,000 550,000= 60,000 550,000=
.1818=
.1091=
18.18%
10.91%
Copyright 2006 - Michael Bush
16
Industrial Application of BasicMathematical Principles Session 7 Homework
Unit 29 Problem 3 Page 218Determine the Percentages
175,000
90,000
125,000
100,000
+ 60,000 550,000
175,000 550,000= .3182= 31.82%
125,000 550,000= .2273= 22.73%
31.82%
+ 22.73% 54.55%
Copyright 2006 - Michael Bush
17
Industrial Application of BasicMathematical Principles Session 7 Homework
Unit 29 Problem 4 Page 218
AllOthers USA
andGreatBritainCanada
andFrance
Draw a Circle Graph
Copyright 2006 - Michael Bush
18
Industrial Application of BasicMathematical Principles Session 7 Homework
Unit 30 Problem 1 Page 218
a.
Sampling provides an economical method of establishing the acceptability of a manufactured lot by a limited sampling verses complete inspection of all manufactured parts.
b.
In single sampling, a random, single sample of a specified number of parts is inspected. A larger number of parts is required verses the first samples in sequential sampling. Sequential requires greater numbers of samples but fewer parts with each sample.
Copyright 2006 - Michael Bush
19
Industrial Application of BasicMathematical Principles Session 7 Homework
Unit 30 Problem 2a Page 218
Tempering Temperature Fahrenheit
220 240 300
Tempering Temperature Celsius
430 450 470 490 510 530 550 570 590 610 630 650
A
B
C
D
E
F
G
H
C
AR
BO
NS
TEE
L TO
OLS
320 340280260
Construct a Line Graph
Copyright 2006 - Michael Bush
20
Industrial Application of BasicMathematical Principles Session 7 Homework
Unit 30 Problem 2b Page 218
PlainCarbon Steel Tools
Tempering Temperatures
°F °C
A Roughing Mills 430 221
B Counterbores 460 238
C Knurls 485 251
D Tube Cutters 485 251
E Taps 500 260
F Threading Dies 530 277
G Pneumatic Tools
580 302
H Non-Cutting Tools
640 338
4110°F
2138°C
4110°F ÷ 8 =513.75°F
2138°C ÷ 8 =
267.25°C
Copyright 2006 - Michael Bush
21
Industrial Application of BasicMathematical Principles Session 7 Homework
Unit 30 Problem 2c Page 218
PlainCarbon Steel Tools
Tempering Temperatures
°F °C
A Roughing Mills 430 221
B Counterbores 460 238
C Knurls 485 251
D Tube Cutters 485 251
E Taps 500 260
F Threading Dies 530 277
G Pneumatic Tools
580 302
H Non-Cutting Tools
640 338
485°F + 500 F =985°F
251°C + 260 C =511 C
Median
985°F ÷ 2 =492.5°F
511°C ÷ 2 =255.5°C