ANALYSIS OF INSULATION OF MATERIAL

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ANALYSIS OF INSULATION OF MATERIAL

Wang Deyu, Li Dejun Zhong Haoyuan

Xu Shanshan Li Yaqiong, Yan Li

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Contents

Executive Summary1

Literature Review2

Data analysis & Conclusion

3 Choice of Experimental Design

4 Performing the Experiment

5 Eliminating Noise

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3

Executive Summary

4

Executive Summary(1/2)

Definition The experiment is aimed to compare the

performance of different kinds of heat insulation materials under normal conditions. The results of the experiment would be quantified into the details including the texture, thickness, exterior color and ventilation.

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Executive Summary(2/3)

Cause and Effects

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Executive Summary(3/3)

Regression Model

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Literature Review

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Literature Review(1/2)

Define how various factors would accelerate or decelerate the cooling rate.

《 Fabric Selection for a Liquid Cooling Garment 》 《水压机泵站工作液体降温问题分析》

Literature Review(2/2)

△T=f(LP, S, M, C, T, HC, CA)

Factors

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properties of the liquid

size of the container

Heat conduction property of container

physical properties of

insulation material

contact of the air

color of the material

thickness of the material

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Choice of Experimental Design

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Preparation & Location

Preparation Material Container: Beaker Kerosene thermometer

Experiment Location C Builiding, Room 306. The room temperature is 26 centigrade.

Flax, Black

Flax, White

Cotton, Black

Cotton, White

Beaker,150ml

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Variables & Blocking

Variable Selection Material, Color, Layer, and Ventilation

Setting Variables

Blocking Two thermometers which have different calibration

Factor Material Color Layer Ventilation+ Heavy Black Multiple Yes- Light White Singular No

4 variables, 2 levels per variable, 2 replications per treatment, 2 blocks, Full factorial

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Performing Experiment

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Performing Experiment

We boil tap water to approximately 100 degrees Celsius, and then quickly pour 200 ml boiling water into the two beakers and read the temperature.

We begin reading when we first see the temperature is steady and begins to drop.

We count 3 minutes before a second reading. Using the two readings with 3-minute interval, the drop of temperature within the 3 minutes could be calculated.

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Eliminating Noise

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Eliminating Noise

Warm up of the beakers and the thermometers To ensure that the heat won’t lose through other

channels.

Wrap the cloth tightly to the beaker Use slim clip to ensure that the least width is

overlapped.

Pad the cup with a paper dish underneath To minimize the heat conducted through the bottom.

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Data analysis & Conclusion

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Initial Model (1/2)

No. Factors No. Factors1 Material 9 Color* ventilation2 Color 10 Layer*Ventilation3 Layer 11 Material*Color*Layer4 Ventilation 12 Material*Color*Ventila

tion5 Material*Color 13 Material*Layer*Ventila

tion6 Material*Layer 14 Color*Layer*Ventilati

on, 7 Material*Ventil

ation15 Material*Color*Layer*

Ventilation8 Color*Layer

Use these 15 factors in a GLM and calculate coefficients in Minitab.

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Initial Model (2/2)

来源 自由度 Seq SS Adj SS Adj MS F P Material 1 6.570 6.570 6.570 22.73 0.000 Color 1 3.445 3.445 3.445 11.92 0.003 Layer 1 2.820 2.820 2.820 9.76 0.007 Ventilation 1 122.853 122.853 122.853 425.00 0.000 Material*Color 1 5.200 5.200 5.200 17.99 0.001 Material*Layer 1 0.263 0.263 0.263 0.91 0.355 Material*Ventilation 1 3.063 3.063 3.063 10.60 0.005 Color*Layer 1 0.015 0.015 0.015 0.05 0.821 Color*Ventilation 1 5.040 5.040 5.040 17.44 0.001 Layer*Ventilation 1 5.200 5.200 5.200 17.99 0.001 Material*Color*Layer 1 0.578 0.578 0.578 2.00 0.177 Material*Color*Ventilation 1 0.000 0.000 0.000 0.00 0.974 Material*Layer*Ventilation 1 0.008 0.008 0.008 0.03 0.871 Color*Layer*Ventilation 1 0.383 0.383 0.383 1.32 0.267 Material*Color*Layer*Ventilation 1 0.015 0.015 0.015 0.05 0.821 误差 16 4.625 4.625 0.289 合计 31 160.080

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Modified Model (1/3)

Then we delete the following factors and recalculate one by one. Material*Color*Layer*Ventilation Material*Color*Layer Material* Layer*Ventilation Color*Layer*Ventilation Material*Color*Layer Color*Layer Material*Layer

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Modified Model (2/3) Also we get

S = 0.505930 R-Sq = 96.32% R-Sq （调整） = 95.04%  项 系数 系数标准误 T P 常量 6.65313 0.08944 74.39 0.000 Material Light -0.45313 0.08944 -5.07 0.000 Color White 0.32813 0.08944 3.67 0.001 Layer Singular -0.29688 0.08944 -3.32 0.003 Ventilation No -1.95938 0.08944 -21.91 0.000 Material*Color Light White -0.40312 0.08944 -4.51 0.000 Material*Ventilation Light No 0.30938 0.08944 3.46 0.002 Color*Ventilation White No -0.39687 0.08944 -4.44 0.000 Layer*Ventilation Singular No 0.40313 0.08944 4.51 0.000

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Modified Model (3/3)

C

AD

B

BD

AB

CD

A

D

2520151050

项

标准化效应

2. 07A Mat er i alB Col orC LayerD Vent i l at i on

因子 名称

Pareto 标准化效应的 图 TempDi ff Al pha = . 05)（响应为 ，

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Modified Model (3/3)

2520151050- 5

99

9590

80706050403020

105

1

标准化效应

百分比

A Mater i alB Col orC LayerD Vent i l at i on

因子 名称

不显著显著

效应类型

CD

BD

AD

AB

D

CB

A

标准化效应的正态图 TempDi ff Al pha = . 05)（响应为 ，

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Modified Model (3/3)

10-1

99

90

50

10

1

残差

百分比

1210864

0. 5

0. 0

-0. 5

-1. 0

-1. 5

拟合值

残差

0. 50. 0-0. 5-1. 0-1. 5

8

6

4

2

0

残差

频率

3230282624222018161412108642

0. 5

0. 0

-0. 5

-1. 0

-1. 5

观测值顺序

残差

正态概率图 与拟合值

直方图 与顺序

TempDi ff 残差图

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Modified Model (3/3)

1. 00. 50. 0- 0. 5- 1. 0- 1. 5

99

9590

80706050403020

105

1

1残差

百分比

均值 - 2. 49800E- 16标准差 0. 4358N 32AD 0. 831P 值 0. 029

1 残差 的概率图 - 95% 正态 置信区间

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拟合因子 : TempDiff 与 Material, Color, Layer, Ventilation

TempDiff 的效应和系数的估计（已编码单位） 项 效应 系数 系数标准误 T P 常量 6.7542 0.05487 123.08 0.000 Material 0.9398 0.4699 0.05507 8.53 0.000 Color -0.4542 -0.2271 0.05487 -4.14 0.000 Layer 0.6273 0.3136 0.05507 5.70 0.000 Ventilation 3.8852 1.9426 0.05507 35.28 0.000 Material*Colo r -0.7727 -0.3864 0.05507 -7.02 0.000 Material*Ventilation 0.4167 0.2083 0.05487 3.80 0.001 Color*Ventilation -0.8273 -0.4136 0.05507 -7.51 0.000 Layer*Ventilation 0.6042 0.3021 0.05487 5.50 0.000   S = 0.298239 PRESS = 3.81306 R-Sq = 98.71% R-Sq （预测） = 97.38% R-Sq （调整） = 98.23%

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拟合因子 : TempDiff 与 Material, Color, Layer, Ventilation

对于 TempDiff 方差分析（已编码单位） 来源 自由度 Seq SS Adj SS Adj MS F P 主效应 4 129.498 129.425 32.3564 363.77 0.000 2 因子交互作用 4 13.964 13.964 3.4909 39.25 0.000 残差误差 21 1.868 1.868 0.0889 失拟 7 0.368 0.368 0.0526 0.49 0.826 纯误差 14 1.500 1.500 0.1071 合计 29 145.330

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拟合因子 : TempDiff 与 Material, Color, Layer, Ventilation

TempDiff 的系数估计，使用未编码单位的数据 项 系数 常量 6.75417 Material 0.469886 Color -0.227083 Layer 0.313636 Ventilation 1.94261 Material*Color -0.386364 Material*Ventilation 0.208333 Color*Ventilation -0.413636 Layer*Ventilation 0.302083

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AD

B

CD

C

AB

BD

A

D

403020100

项

标准化效应

2. 08A Mat er i alB Col orC LayerD Vent i l at i on

因子 名称

Pareto 标准化效应的 图 TempDi ff Al pha = . 05)（响应为 ，

拟合因子 : TempDiff 与 Material, Color, Layer, Ventilation

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403020100- 10

99

9590

80706050403020

105

1

标准化效应

百分比

A Mat er i alB Col orC LayerD Vent i l at i on

因子 名称

不显著显著

效应类型

CD

BD

AD

AB

D

C

B

A

标准化效应的正态图 TempDi ff Al pha = . 05)（响应为 ，

拟合因子 : TempDiff 与 Material, Color, Layer, Ventilation

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0. 500. 250. 00-0. 25-0. 50

99

90

50

10

1

残差

百分比

1210864

0. 500. 250. 00

-0. 25-0. 50

拟合值

残差

0. 60. 30. 0-0. 3-0. 6

8

6

4

2

0

残差

频率

30282624222018161412108642

0. 50

0. 25

0. 00-0. 25

-0. 50

观测值顺序

残差

正态概率图 与拟合值

直方图 与顺序

TempDi ff 残差图

拟合因子 : TempDiff 与 Material, Color, Layer, Ventilation

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Regression Model (1/4)

The residuals fits well in a normal distribution, and the

main effects and all the 4 interactions are significant.

Thus,

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Regression Model (2/4)

Bl ackWhi te Mul t i pl eSi ngul ar YesNo10. 0

7. 5

5. 010. 0

7. 5

5. 010. 0

7. 5

5. 0

Mater i al

Col or

Layer

Vent i l at i on

Li ghtHeavy

Materi al

Whi teBl ack

Col or

Si ngul arMul ti pl e

Layer

I nteracti on Pl ot for TempDi ffData Means

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Regression Model (3/4)

HeavyLi ght

98765

Bl ackWhi te

Mul t i pl eSi ngul ar

98765

YesNo

Materi al

平均值

Col or

Layer Venti l at i on

TempDi ff 主效应图数据平均值

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Regression Model (4/4) We transform TempDiff into Exponential form, and get

the residual plot as below:

We see some obvious patterns, we don’t recommend to transform the data in this way.20000100000-10000-20000

99

90

50

10

1

残差

百分比

4500030000150000

30000

20000

10000

0

-10000

拟合值

残差

20000100000-10000

8

6

4

2

0

残差

频率

3230282624222018161412108642

30000

20000

10000

0

-10000

观测值顺序

残差

正态概率图 与拟合值

直方图 与顺序

Exp(Di ff ) 残差图

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Results explanations

No ventilation can remarkably maintain the high level of heat preservation.

Materials have main effect of heat preservation as well. Colors of material have main effect of heat preservation

as well. Thickness of material has less but also main effect of

heat preservation as well. Interaction explanation.

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Possible causes

Ventilation-absence condition has the best ability of maintaining heat may result in that heat is lost mostly from the top of the cup, more than from the wall of cup.

White color surprisingly has better ability of maintaining heat.

Heavy cloth has better heat maintaining ability, which corresponds to our intuition. However, layers have less effect.

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Error sources

Inequity of preliminary heating results the different original conditions of materials such as cloth and the cups.

Two thermometers have different abilities of measuring such as sensitivity to temperature changes and measurement resolution.

System errors from two experimenters reading the thermometers such as view angular.

Water incrustation or impurities in later treatments because of repetitive uses.

Impurities in water may affect the temperature decrease rates. Room temperature may change during the relatively long period

time during the experiment process.

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Q & A

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