copyright 2006 - michael bush1 industrial application of basic mathematical principles session 7...

Copyright 2006 - Michael Bush

1

Industrial Application of BasicMathematical Principles

Session 7Homework Assignments


2

Industrial Application of BasicMathematical Principles Session 7 Homework

Unit 27 Problem 1 16Compute the Celsius melting point for

Chromium

a. 5C = F - 32

9

5C = 2,740 - 32

9

5C = 2,708 =

9 1504 C


3


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


4


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


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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


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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


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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


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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


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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


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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


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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


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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


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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


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Unit 29 Problem 1 Page 218Draw a Circle Graph

AllOthers

USA

FranceGreatBritain

Canada


15


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%


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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%


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AllOthers USA

andGreatBritainCanada

andFrance

Draw a Circle Graph


18



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.


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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


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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


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Unit 30 Problem 2c 8

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

copyright 2006 - michael bush1 industrial application of basic mathematical principles session 7...

Documents

celsius melting point

average difference

single sampling

circle graphcopyright

line graphcopyright

manufactured parts

manufactured lot

single sample