110_-_7_-_Homework_R.. - Skilled Trades Math On-line

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Copyright 2006
-

Michael Bush

1

Industrial Application of Basic

Mathematical Principles

Session 7

Homework Assignments

Copyright 2006
-

Michael Bush

2

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 27 Problem 1 Page 216

Compute the Celsius melting point for Chromium

a.



5
C = F - 32
9



5
C = 2,740 - 32
9
  
5
C = 2,708 =
9
 
1504 C

Copyright 2006
-

Michael Bush

3

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 27 Problem 1 Page 216

Compute the Celsius melting point for Cast Iron

a.



5
C = F - 32
9



5
C = 2,300 - 32
9
  
5
C = 2,268 =
9
 
1260 C

Copyright 2006
-

Michael Bush

4

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 27 Problem 1 Page 216

Compute the Celsius melting point for Copper

a.



5
C = F - 32
9



5
C = 1,940 - 32
9
  
5
C = 1,908 =
9
 
1060 C

Copyright 2006
-

Michael Bush

5

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 27 Problem 1 Page 216

Compute the Celsius melting point for Aluminum

a.



5
C = F - 32
9



5
C = 1,200 - 32
9
  
5
C = 1,168 =
9
 
649 C

Copyright 2006
-

Michael Bush

6

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 27 Problem 1 Page 216

Compute the Celsius melting point for Lead

a.



5
C = F - 32
9



5
C = 620 - 32
9
  
5
C = 588 =
9
 
327 C

Copyright 2006
-

Michael Bush

7

Industrial Application of Basic

Mathematical 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
18
10
8
Degrees Celsius (hundreds)
20
Copyright 2006
-

Michael Bush

8

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 27 Problem 3 Page 217

Determine 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 Basic

Mathematical Principles Session 7 Homework

Unit 27 Problem 3 Page 217

Determine 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 Basic

Mathematical Principles Session 7 Homework

Unit 27 Problem 3 Page 217

Determine 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 Basic

Mathematical Principles Session 7 Homework

Unit 28 Problem 1 Page 217

Generate a line graph

10
12
14
16
18
20
22
24
26
28
30
Revolutions Per Minute (hundreds)
TEMPERATURE (Fahrenheit)
700
800
900
1000
1100
1200
1300
32
34
36
Copyright 2006
-

Michael Bush

12

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 28 Problem 2 Page 217

Determine 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 Basic

Mathematical Principles Session 7 Homework

Unit 28 Problem 2 Page 217

Determine 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 Basic

Mathematical Principles Session 7 Homework

Unit 29 Problem 1 Page 218

Draw a Circle Graph

All
Others
USA
France
Great
Britain
Canada
Copyright 2006
-

Michael Bush

15

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 29 Problem 2 Page 218

Determine 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 Basic

Mathematical Principles Session 7 Homework

Unit 29 Problem 3 Page 218

Determine 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 Basic

Mathematical Principles Session 7 Homework

Unit 29 Problem 4 Page 218

All
Others
USA
and
Great
Britain
Canada
and
France
Draw a Circle Graph

Copyright 2006
-

Michael Bush

18

Industrial Application of Basic

Mathematical 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 Basic

Mathematical 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
CARBON
STEEL TOOLS
320
340
280
260
Construct a Line Graph

Copyright 2006
-

Michael Bush

20

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 30 Problem 2b Page 218

Plain

Carbon 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 Basic

Mathematical Principles Session 7 Homework

Unit 30 Problem 2c Page 218

Plain

Carbon 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