Cooling rate of water

Transcription

1 Cooling rate of water Group 5: Xihui Yuan, Wenjing Song, Ming Zhong, Kaiyue Chen, Yue Zhao, Xiangxie Li 目录. Abstract: Introduction: Statement of the problem: objectives: Rationale Literature review Basic Concept Existing experiment on the cooling rate of water Predictive conclusions Methods Experiment variables Design of experiment Instrument Performing the experiment Analysis Data analysis Conclusion In-depth research Future extension Reference Abstract: This thesis aims at exploring factors that can affect the cooling rate of water. Water is the foundation of life; however, although we are very familiar with it, it still has many aspects that we do not quite understand. There are some existing theories and experiments about this topic, on this basis, we will do further study to find cooling rate under different conditions. In order to explore the effectors that have an influence on the cooling condition, we find out different variables, and select proper ones to be considered in our experiment according to the scope of our research. Some 2 level factors, factorial designs will be undertaken, and several analytical methods will be applied to the data got from experiments, until we get to the ultimate goal and get acquaintance of the cooling procedure. / 2

2 2. Introduction: 2.. Statement of the problem: Water is one of the most common things in our daily life. Nearly all the students take a bottle of water or hot drink with them every day. Naturally, how to keep their drink warm in the winter and make hot drink cool as quick as possible becomes a question that many will concern. But at the same time, few of them have put their thoughts into action. As a result, this area of knowledge for most people remains to be relatively blank. 2.2 objectives: In this experiment, we will measure the cooling rate of different drinks in different environment, to draw one or more conclusions about better ways of keeping drink warm in winter and cooling water quickly in summer Rationale Our experiment is focused on cooling rate of different drinks. As cooling rate will be relatively certain and stable under certain conditions, so the data will be meaningful and we can certainly draw some conclusions from it. Different factors of environment conditions, holding bottles and initial temperature will be considered. The cooling rate will be measured by thermometer, guaranteeing the correctness and accuracy of the experiments. This experiment is focused on water or drink s cooling in our daily life, so the experiment subjects will be the most commonly and easily found drink in our life. And the variables of experiment environment will be designed within the limitation of real life as well. 3. Literature review To get some predictive conclusions before the experiment is a very important method to give us some intuition to estimate whether the experiment is performing correctly and whether there are some significant wrong message in the final result. In order to get some intuition and predictive conclusion before the experiment, we find some literature related to our experiment and do some forecasts before we do the experiment. 2 / 2

3 3.. Basic Concept There are some very important formulas related closely to our cooling rate experiment. Heat conduction formular: Q = cm T Q = the quantity of heat c = the specific heat capacity m = the quantity of the liquid T = the difference between the initial temperature and the final temperature Newton's law of cooling Q = q s t T T q = γ d γ = a constant related to the texture of the container T = the initial temperature of the liquid T = the temperature of the environment d = the thickness of the container s = the area of the interface t = time of the cooling experiment After combining all the formulas above, we can have the following formula: T = γ T T S t d cm 3.2 Existing experiment on the cooling rate of water Some experiment show that cooling rate has the relationship with the initial temperature of the water. When the volume is constant, the relationship between freezing time and initial temperature has the following curve as Graph : 3 / 2

4 The time to freeze ( i t ) Initial temperature Graph After finding some materials about this topic, we find that the cooling rate of water may have the relationship with the initial temperature, the volumes, the shape of the container, the material of the container and ambient temperature etc. 3.3.Predictive conclusions. If the other factors are treated as constants, the thicker the container is (d ), the smaller T will be. 2. If the other factors are treated as constants, then the greater the mass is (m ), the smaller T will be. 3. If the other factors are treated as constants, then the larger of the specific heat capacity, the smaller T will be. 4. If the other factors are treated as constants, then the higher the initial temperature is, the larger the T will be. 4. Methods 4.. Experiment variables The key dependent variable of our experiment is the cooling rate of water, there are many aspects can influent it. 4 / 2

5 Graph 2 _ fishbone There are mainly three aspects affecting the cooling rate of water: intrinsic factors of water, environment influences, and anthropic factors. Within each of these topics, there are still many detail factors; these are reflected in the Graph 2 _ fishbone above. As our experiments are restricted to many conditions, some of the variables we can t control, and other of them are not within our scope of study, so we explore further the influence of all variables to our experiment and give them detail analysis. 5 / 2

6 Table response normal meas. precision, relationship of variable operating level accuracy response variable to (units) & range objective Estimate the cooling Cooling rate 5 centigrade in ten minutes Thermometer Precision. centigrade condition of different water under different conditions Table 2 control variable (units) normal level & range meas. precision & setting error proposed settings, based on predicted effects predicted effects Cooling time (in minutes) 5min in ten seconds watch 5 minutes different Range from Initial temperature ( centigrade) 3 centigrade thermometer 3 centigrade according to different different experiment solution Water vs. milk select Same brand, Yili Milk different thickness of -2 layers of paper the containers -2 layers of paper cup count cup according to different different ( in ml) experiments Volume of the liquid -2ml Measuring cylinder 55ml different Table 3 fixed factor (units) desired experimental level & allowable measurement precision How known? how to control in experiment anticipated effects range Cooling 5 minutes.s Cool water of in None to time Within s stopwatch different trails in 5 slight 6 / 2

7 minutes, time controlled by stopwatch Table 4 nuisance factor (units) measurement precision strategy (e.g., randomization, Blocking, etc.) anticipated effects Variation of condition of environment Evaporation amount Difference in paper cups Air velocity Randomization In centigrade experiment By thermometer sequence none In.ml, by Measure measuring volume after slight cylinder heating In o. millimeter Eliminate None to slight Close door and window, stay Too much air flow.-.3 m/s away from the can cause large cups whiling influence cooling 4.3. Design of experiment Factors related to the cooling rate of water are based on comparisons among series of experiments. The comparison in this experiment is basically in two dimensions, the first dimension focus on the cooling rate of different liquid, we would like to find out the cooling speed of some certain kind of liquid, for example, pure water, milk; further more in this dimension, we will compare the differences in cooling rate and use tools to find the related factors to cooling speed. Then we will move forward to design the experiments aiming at finding the relation and the function of these factors. The second dimension aims at finding the factors related to the environment, for example the temperature and the thickness of the cup. The design method we choose is 2 K full factorial design. The response variable is cooling degree in 5 minutes; the factors are initial temperature, volume, thickness of cups, and liquid type. Each factor has two levels, (6 centigrade vs. 8 centigrade), (5ml vs. ml), (layer vs. 2layers), (water vs. milk) Factor Range Actual level Coded level Type Initial temperature Quantitative Quantity of 5ml ml 5 ml Quantitative 7 / 2

8 liquid ml + Thickness of container -2 layers layer (cup) 2layers(cups) + Quantitative Type of liquid Water Milk + Qualitative Use Minitab to generate a 2 4 full factorial design with two replications. Graph Instrument Pure water: solvent and one kind of the experimental liquid. Tap water, milk of the same brand (Yili) Sanitary cup: hold different kinds of liquid, in the same material. Alcohol thermometer: monitor the change and show the temperature. One ml beaker and one 5ml beaker: get the desired amount of the experimental liquid. 8 / 2

9 Air thermometer: get the environment temperature and humidity Performing the experiment In order to improve the efficiency of the experiment, we did some trials to find proper values for the parameters and ranges. Considering the availability, samples of 5ml, 6 water were used to choose the time parameter; the experiment was started at the point when the temperature is 6 and after 3min and 5min separately, the ending temperature was recorded. The difference for 3min was 3, and 8 for 5 min. Considering the error of measurement and the significance of the data, 5min was finally chosen to be the time parameter. The staring temperature was also adjusted. It was originally desired to set the to be 7 and 9, in the trial experiment, 9 was found hard to control, because the procedure of using the beaker to measure the proper volume and then pouring into the cups could always lead to a fall below 9, and with the limitation of the instrument, the original temperature of water in the boiler was not essentially higher than 9. With more trials of monitoring the temperature of water in the boiler and in the cups after the transporting, the range of temperature was set to be 6 and 8. Then, the formal experiment was conducted, first, the temperature of environment was recorded to be 3.2, and the humidity was 23%. The experiment was carried out according to the randomized sequence. A beaker was used to take the desired volume of water out from the boiler, then the water was poured into the numbered cups and the thermometer was used to monitor the temperature until it fell down to the starting point. Then, start the timer and wait for 5 minutes, record the ending temperature and do the next one. As to the milk, it was heated also in the boiler, within its bag. The heated milk was poured into the beaker to get the desired volume, in order to slow down the cooling speed in measuring; the beaker was also heated in the water boiler. After the measurement, the milk was poured into the cup; the thermometer was used to monitor the temperature until it fell down to the starting point. Then, start the timer and wait for 5 minutes, record the ending temperature and do the next one. Perform all the items in the table and record the temperature of the ending point. In the whole process, as the experiment was frequently switched between the water and milk, two beakers were used to measure the volume, and the instrument used should keep clean. Because the temperature could change very quickly around the desired points, great attention was paid on monitoring the temperature. After the experiment, the temperature of the environment was 3. And the humidity was 24%. This tiny change was not taken into consideration as factors in the experiment. 9 / 2

10 5. Analysis 5.. Data analysis Step : calculate the cooling degree. The initial table records the starting temperature and the finishing temperature, so we first calculate the difference of these two columns and get the cooling degree. Step 2: normal verification First of all, we need to check the weather the response variable is normal distributed. So we do the normal test and find that under.5 significant level the response variable 温差 is not normal distributed(graph 4). Then we do the Box-Cox transformation (Graph 5 Box-Cox transformation). We find λ =, so we do the Ln transformation (Graph 6 Normal test of Ln 温差 ). / 2

11 温差的概率图正态 - 95% 置信区间 百分比 均值 3.5 标准差 6.48 N 32 AD.779 P 值.39 5 温差 2 3 Graph 4 温差的 Box-Cox 图 5 CL 下限 CL 上限 Lambda ( 使用 95.% 置信 ) 估计. 4 CL 下限 -.67 CL 上限. 标准差 3 取整值. 2 限 Lambda Graph 5 Box-Cox transformation / 2

12 ln 温差的概率图正态 - 95% 置信区间 百分比 均值 标准差.484 N 32 AD.57 P 值 ln 温差 Graph 6 Normal test of Ln 温差 Step 3: find significant factors and their influences The main effect plot(graph 7 Main effect plot) and the interaction plot(graph 8 Interaction plot of the factors) show that the factors 起温, 水量, 杯厚 and the interaction of 液体 * 水量, 液体 * 杯厚 seem to have significant effect on the response variable. More specifically, when the initial temperature rises, the cooling degree increases, the volume of liquid decreases, the cooling degree increases, and increase the thickness of cup, may result in the decline of cooling degree. What s more, from the main effect plot, we can find that the type of liquid has no significant influence on the cooling rate. 2 / 2

13 ln 温差主效应图数据平均值 液体 起温 平均值 2. 水 水量 牛奶 6 杯厚 Graph 7 Main effect plot ln 温差交互作用图数据平均值 液体 液体水牛奶 起温 起温 6 8 水量 水量 杯厚 杯厚 2 水 牛奶 5 Graph 8 Interaction plot of the factors And we also draw the contour plot (Graph 9) and surface plot(graph ) to show the relationship of the factors and response variable. From the plots, we can see that the factors mainly have liner relationship with the response variable. 3 / 2

14 ln 温差的等值线图 起温 * 液体 水量 * 起温 水量 * 液体 杯厚 * 起温 杯厚 * 液体 杯厚 * 水量 ln 温差 < > 保持值 液体 起温 水量 杯厚... Graph 9 ln 温差的曲面图 l n 温差 起温 液体 2.4 l n 温差 水量 液体 l n 温差 杯厚 液体 保持值 液体 起温 水量 杯厚 l n 温差 水量 起温 l n 温差 杯厚 起温 l n 温差 杯厚 水量 Graph 4 / 2

15 Step 4: explore relationship function The normal probability plot of effect (Figure 8) and the Pareto plot (Figure 9) have showed the same result that we draw from the main effect plot and interaction plot. 标准化效应的正态图 ( 响应为 ln 温差,Alpha =.5) B 效应类型不显著显著 百分比 D AD AC ABD 因子 A B C D 名称液体起温水量杯厚 5 C -5 5 标准化效应 Graph 标准化效应的 Pareto 图 ( 响应为 ln 温差,Alpha =.5) 2.2 B C AC AD D 因子 A B C D 名称液体起温水量杯厚 ABD CD 项 BD ABC AB ABCD BC ACD BCD A 5 5 标准化效应 2 25 Graph 2 5 / 2

16 We assume that the high order terms are not significant. And we use only the one and two order significant factors to do the regression, the final result is as below: 标准化效应的正态图 ( 响应为 ln 温差,Alpha =.5) 效应类型不显著显著 百分比 D AD AC B 因子 A B C D 名称液体起温水量杯厚 C 标准化效应 Graph 3 标准化效应的 Pareto 图 ( 响应为 ln 温差,Alpha =.5) 2.6 B C 因子 A B C D 名称液体起温水量杯厚 AC 项 AD D A 5 标准化效应 5 2 Graph 4 6 / 2

17 Table 5 一般线性模型 : ln 温差与液体, 起温, 水量, 杯厚 因子类型水平数值 液体固定 2 水, 牛奶 起温固定 2 6, 8 水量固定 2 5, 杯厚固定 2, 2 ln 温差的方差分析, 在检验中使用调整的 SS 来源 自由度 Seq SS Adj SS Adj MS F P 液体 起温 水量 杯厚 液体 * 水量 液体 * 杯厚 误差 合计 S =.8495 R-Sq = 95.89% R-Sq( 调整 ) = 94.9% 项 系数系数标准误 T P 常量 液体 水 起温 水量 杯厚 液体 * 水量 水 液体 * 杯厚水 The fitted model is: ln 温差 = I 起温 = I 水量 = I 杯厚 = So.656I 液体 = 水, 水量 =5.677I 液体 = 水, 杯厚 = 7 / 2

18 温差 = EXP( I 起温 = I 水量 = I 杯厚 =.656I 液体 = 水, 水量 =5.677I 液体 = 水, 杯厚 = ) Step 5: residual analysis: The residual plot (Graph 5 Residual plot) is satisfied. We find that there is an outlier, but we don t know the reason now. Maybe it is the result of reading error. And it is a full factorial design; it may cause unbalance if we delete this point. Then we keep it in the plot. ln 温差残差图 百分比 正态概率图 -.2. 残差.2 残差 与拟合值 2.5 拟合值 直方图 与顺序 频率 残差.. 残差 观测值顺序 Graph 5 Residual plot 5.2. Conclusion. From the equation 温差 = EXP( I 起温 = I 水量 = I 杯厚 =.656I 液体 = 水, 水量 =5.677I 液体 = 水, 杯厚 = ) and analysis of data, we find that high initial temperature, less water yield and thin cup cause large cooling rate. It obeys our common sense. 2. Whether the liquid is milk and water does not affect the cooling rate. We only test these two liquid, so we do not sure whether this conclusion can be extend to other liquid. 8 / 2

19 6. In-depth research In our model, the type of liquid is not a significant factor, however, we want to know whether this is the truth or some other elements influence the result of experiment. After referring to some literatures, we find the heat formulae can calculate the difference of degrees in our experiment conditions According to Q=CM T Where C 水 =4.2 ^3J/(Kg ) C 牛奶 =2.5 ^3J/(Kg ) ρ 水 =.X^3kg/m^3 ρ 牛奶 =.256~.288 X^3kg/m^3 Get the T 2 We compute the difference of cooling centigrade under same experiments for both water and milk, and make a single sample T test, get Table 6 差 异 Table 7 单样本 T: 差异 mu = 与 的检验 平均值 变量 N 平均值标准差标准误 95% 置信区间 T P 差异 (-.9,.4) According the test result, the difference incurred by type of liquid cannot be identified by the measuring system, so in this experiment, the liquid type is not significant. 7. Future extension Enhance the evaluation system: The resolution of the alcohol thermometer is centigrade, this is not precise for our experiment, so in order to enhance the accuracy of the experiment, we 9 / 2

20 need to improve the measuring system. Extend the experiment parameter range: Extend the experiment range: choose more types of liquid and extend the range of initial water, we may get more precise conclusions about cooling rate of different liquid. Also, under this condition, the factors increase, we need to do fractional factorial design or Taguchi design to explore the relationships. 8. Reference f-water 水 盐水降温试验及分析计算 刘永胜, 杨小林, 谢华刚 ( 河南理工大学土木建筑系, 河南焦作 4543) 2 / 2