2016年美赛A题美国大学生获奖论文
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54164
Problem Chosen
A
For office use only
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F4________________ 2016
MCM/ICM
Summary Sheet
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Abstract
We present a user-friendly strategy for keeping the water in a bathtub as close as
possible to a certain preferred temperature. This strategy consists in two main
constituents: adjusting the hot water flow from the faucet to keep the bathtub
temperature suitably high, and using hand motions to distribute the heat evenly
throughout the bath water.
As for determining the necessary water flow, we first create a model that determines
the temperature of the bath water at any point in time. We posit that the temperature of
the water in the bath is dependent on the cooling due to the ambient temperature of the
room, which obeys the proportionality given by Newton’s Law of Cooling; and the
heating due to the added hot water from the faucet. Our model combines the
temperature changes over time from these two entities into one consummate
differential equation.
We determine the experimental constant of proportionality for Newton’s Law of Cooling
by proposing that this constant itself is dependent on a combination of volume of the
water and surface area exposed to the air. Taking experimental data from varied
sources, including an experiment we conducted and documented over the time frame of
the MCM competition, we were able to extrapolate values for the constant attributable
to any bathtub given its volume and surface area exposed to air.
We then run a computer simulation to model the temperature changes over time in a
standard bathtub with our differential equation, determining how much hot water
needs to flow in to keep the total deviance over time of the temperature of the bath
from a desired temperature at a minimum. We found the optimal flow rate for a
standard bathtub at standard conditions to be 1.81 gallons/minute, or about 11% of the
maximum flow rate.
Our model was able to predict optimal flow rates for many different combinations of
volume, surface area, tub material, and additives. We found that more effective
insulating materials led to lower optimal flow rates (less hot water needed), as did the
addition of additives. Volume and surface area, surprisingly, showed no clear trend.
As for keeping the temperature roughly even in the bathtub, we suggest that the user
swirl his hands counter-clockwise with respect to the faucet as shown in Figure 1. This
is a simple yet effective way to distribute the heat acceptably evenly throughout the tub
water.
As alternatives for those who do not wish to make their bath into an exercise session,
we invented an original product (Home Jacuzzi) that is low-cost and aids in heating the
water evenly.
The Math Behind the Bath:A”Cool”Investigation on
Cooling
Team#54164
February2016
1Abstract
We present a user-friendly strategy for keeping the water in a bathtub as close as possible to a certain preferred temperature.This strategy consists in two main constituents:adjusting the hot water?ow from the faucet to keep the bathtub temperature suitably high,and using hand motions to distribute the heat evenly throughout the bath water.
As for determining the necessary water?ow,we?rst create a model that determines the tem-perature of the bath water at any point in time.We posit that the temperature of the water in the bath is dependent on the cooling due to the ambient temperature of the room,which obeys the proportionality given by Newton’s Law of Cooling;and the heating due to the added hot water from the faucet.Our model combines the temperature changes over time from these two entities into one consummate di?erential equation.
We determine the experimental constant of proportionality for Newton’s Law of Cooling by proposing that this constant itself is dependent on a combination of volume of the water and surface area exposed to the air.Taking experimental data from varied sources,including an experiment we conducted and documented over the time frame of the MCM competition,we were able to extrapolate values for the constant attributable to any bathtub given its volume and surface area exposed to air.
We then run a computer simulation to model the temperature changes over time in a standard bathtub with our di?erential equation,determining how much hot water needs to?ow in to keep the total deviance over time of the temperature of the bath from a desired temperature at a minimum.We found the optimal?ow rate for a standard bathtub at standard conditions to be1.81 gallons/minute,or about11%of the maximum?ow rate.
Our model was able to predict optimal?ow rates for many di?erent combinations of volume, surface area,tub material,and additives.We found that more e?ective insulating materials led to lower optimal?ow rates(less hot water needed),as did the addition of additives.Volume and surface area,surprisingly,showed no clear trend.
As for keeping the temperature roughly even in the bathtub,we suggest that the user swirl his hands counter-clockwise with respect to the faucet as shown in Figure1.This is a simple yet e?ective way to distribute the heat acceptably evenly throughout the tub water.
As alternatives for those who do not wish to make their bath into an exercise session,we invented an original product(Home Jacuzzi)that is low-cost and aids in heating the water evenly.
1.1Non-Technical Explanation
Why is it so di?cult to keep the temperature of the water even throughout when taking a bath?In a standard single-faucet bathtub,hot water comes out of the faucet on only one side of the bath, leaving the other side of the bath colder until enough time passes for the temperature to even out. When left running,this might take more time than a user wants to wait for.
To further complicate things,hot water is also slightly less dense than cold water,and will thus be more concentrated on the surface of the bathtub than the bottom of the tub.
To make the temperature consistent,the easiest solution is to cup your hands and circulate the hot water from the side with the faucet to the other side of the tub,and also circulate the hot water from the water surface to the bottom of the top.
Another option,however,for those of us who don’t want to spend the duration of our bath ?ailing our arms,is to use our invention,”Home Jacuzzi.”Our invention spans the length and
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width of the bathtub,and attaches onto the faucet.The hot water from the faucet is then funneled through the device towards the bottom of the tub,creating a Jacuzzi-like e?ect.The hot water then rises to the surface,mixing with the water above it and creating an evenly distributed,uniform temperature throughout the tub.
Now that we can keep the temperature consistent throughout the tub,we’re still left to?gure out how high we should turn our faucet to keep the water from going cold.
Of course,turning it too high is a waste of water and,on top of that,can increase the temperature to uncomfortably high levels.Too low,and we get to wade around in lukewarm to cold waters that make us not want to take a bath at all.So we want to?nd the optimal?ow rate on our faucet that gets our temperature to where we want it to be.
Not all bathtubs were created equal.The optimal?ow rate isn’t a set rate for all bathtubs.
For the standard60”x30”x14”bathtub,we suggest turning the faucet up to about a ninth of its max.
To conserve even more hot water,you even have the option of purchasing products like the ”Water Blanket”(”Water Blanket”)to keep the temperature of the bath warmer.Interestingly enough,even bubble bath can keep your bath water hotter for longer.And no,it’s not”just for kids.”
Contents
1Abstract1
1.1Non-Technical Explanation (1)
2Introduction3
2.1Background (3)
2.2Restatement of Problem (3)
3The Strategy4
3.1Distribution:A Simple Solution (4)
3.2Global Assumptions (4)
4Simulating Heat Transfer5
4.1Local Assumptions (5)
4.1.1Variable Nomenclature (6)
4.2Creation of the Model (6)
4.2.1Cooling (6)
4.2.2Experimental Data (7)
4.2.3Heating (12)
4.3Simulation Design (13)
4.3.1Selecting Optimal Flow Rate (14)
4.3.2Validation of Model (16)
5Sensitivity Analysis:Varying Bathtub Characteristics16
5.1Varying Surface Area Under Constant Volume (16)
5.2Varying Volume Under Constant Surface Area (24)
5.3Observations (32)
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5.4Material of Bathtub (34)
5.5Additives(bubble bath) (35)
6Conclusion36
6.1Results (36)
6.2Strengths of Model (36)
6.3Limitations of Model (36)
6.4Future Work/Discussion (37)
6.5Home Jacuzzi (37)
Appendices39
A Experimental Design for Real-World Bathtub Data39
B Simulation Code39 2Introduction
2.1Background
Baths,the best way we had to cleanse ourselves–until we became teens and discovered showers,at which point baths became”uncool.”Yet,whenever we want to slow life down and relax a little, we can rest assured that the bath will always be there to ease our nerves.The health bene?ts of taking a bath are plentiful(”5Reasons You Need To Take A Bath”),but no one wants to hop into a steaming hot tub only to?nd him or herself in cold or lukewarm water a half-hour later.Then, how can one keep a bath warm long enough to last a preferred duration?Most people own simple bathtubs,unequipped with secondary heating systems or circulating jets.Several factors can a?ect the temperature change of bath water,including:
?the surface area of water exposed to air
?the volume of water that is being cooled
?the surrounding air temperature
?the rate of hot water?ow into the bathtub
?the material the bathtub is made out of(and how well it is able to provide insulation)?whether any additives are introduced to the water,such as a bubble bath solution
2.2Restatement of Problem
A person has settled into a bathtub for a long,relaxing bath,only to?nd that after some time,the water has become noticeably cooler.In an attempt to salvage his satisfaction,he elects to turn on the faucet and let in some hot water to bring the temperature back up.Excess tub water leaves the tub through an over?ow drain.Of course,he runs into a set of issues:
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?The water in the tub is of inconsistent temperature,being hotter in some areas and cooler in others.
?Turning the hot water?ow too high is a waste of water(and money).
3The Strategy
Our”strategy”presented to the person in the tub will be a combination of the solutions to queries 1and2—the person will have to take action to evenly distribute the heat in the bathtub water, and the person will have to set the hot water?ow in at a certain?ow rate to keep the bathtub water hot.
3.1Distribution:A Simple Solution
We begin by addressing query number”1”.In a standard single-faucet bathtub,water?ows from above on one side of the bathtub.Since hot water rises,more heat will be concentrated at the top of the bathtub.Because the faucet is only on one side of the tub,more heat will be concentrated on the side of the tub closest to the faucet.Therefore,we advise the person in the tub to cup his hands and make circles in the water as illustrated below:
Figure1:Proposed strategy for equalizing the temperature of the water in any single-faucet bathtub.
3.2Global Assumptions
?The standard bathtub is open and exposed to the air on the top surface,and built with acrylic material that is heavily insulated(”Bathtub Sizes”).
?Heat loss through heavily insulated surfaces is negligible.
?Heat loss through surfaces of poor insulation(thin glass,thin?berglass)behaves as if that surface was directly exposed to air.
?The person in the bathtub follows the suggestion in Figure1,allowing us to assume that water temperature is consistent throughout the tub at any point in time.
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?The temperature sensitivity of the human body is such that we begin noticing appreciable di?erences in temperature at±1.4?C from the original temperature(Davies et al.171)
?The target temperature for the bath water is human body temperature,which is approxi-mately37?C.
?Since the person inside the tub is at thermal equilibrium with the water in the tub,we can treat the person’s body as an extension of the bathtub water.
?Room temperature remains constant throughout the duration of the bath,at21?C,the ac-cepted value for ambient room temperature.
?The faucet from which the hot water?ows has a maximum?ow rate of16gallons/min (”Bathtub Faucets”),and the person is able to toggle the faucet’s?ow rate to any rate between0gallons/min and16gallons/min.
?The hottest water we can release from the faucet has a temperature of48?C to prevent injury from scalding(A.O.Smith Corporation1).This will be the assumed temperature of the additional water from the faucet.
?All bathtubs used in this paper for simulation and data are?lled to the brim with the com-bination of water and person.
4Simulating Heat Transfer
To address our second query,we consider a thermodynamic system of?ve constituents:the water in the tub,the water entering and leaving the tub,the walls of the tub,the human in the tub,and the air outside the tub.Heat transfer with the human and the walls is negligible as justi?ed in our global assumptions.Thus,we need only consider the cooling of the tub water due to the air outside,and the heating of the tub water due to the incoming hot water.
4.1Local Assumptions
?The industry standard bathtub in America is60”×30”×14”,which results in a volume of 60×30×14=25,200in3and surface area exposed to the air of60×30=1,800in2(”Bathtub Sizes”).
?Since the majority of bathtubs in America are insulated,we assume that the bathtub that we are simulating is well insulated on all sides except for the top.Therefore,heat can only escape from the bathtub from the top surface.
?Water cooling obeys Newton’s Law of Cooling closely enough to approximate a model for temperature over time.
?The ambient temperature is21?C.
?The temperature of the hot water from the faucet is48?C.
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4.1.1Variable Nomenclature
Below are the variables used throughout our model:
?k=cooling constant
?T=temperature of the water minus ambient room temperature
?T0=initial temperature of the bathtub
?t=time in minutes
?SA E=surface area exposed to air in square meters
?V=volume in liters
?Q=change in heat energy,in joules
?m=mass of the water in the tub,in grams
?C p=speci?c heat of the water,in Joules per gram-degree?C
4.2Creation of the Model
Since the temperature of the water is constantly being altered by the cooling due to the surrounding air temperature and the heating due to the incoming water?ow,we created an equation to model the changes in temperature,based on Newton’s Law of Cooling and the speci?c heat capacity formula.
4.2.1Cooling
The?rst part of the model accounts for the cooling of the water due to the temperature of the environment.We chose to model this cooling with Newton’s Law of Cooling,which states that the di?erence in temperature is directly proportional to the negative of the current temperature,or
dT
dt
=?kT(1) where T is the temperature,and k is the cooling constant that is derived from experimentation. To derive the form we want our regression equation for our experimental data to be in,we solve our di?erential equation by integration with separation of variables:
1 T dT=?kdt(2)
1
T dT=
?kdt(3)
ln(T)=?kt+C(4) where C is a constant.Solving for T,
T=e?kt+C=Ce?kt(5)
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To solve for this constant of integration,C,we substitute in the initial temperature T0of the water at time t=0.
T0=Ce?k×0=C×1(6) giving us the regression equation form
T=T0e?kt(7) Since k is dependent on the sum of physical conditions in each speci?c scenario,it can only be derived from experimental data.Then,to predict the value of k for our industry standard bathtub, we take sets of experimental data and analyze trends between them to extrapolate a value of k for our purposes.
4.2.2Experimental Data
We created a graph to represent the data we found for temperature over time in100mL,300mL, and800mL volumes of water(Gieseking).The ambient room temperature in this setting was23?C. We realized that it is wrong to model these data with exponential regressions that have horizontal asymptotes at0?C.Instead,since logically the temperature of the water should approach—but
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never dip below—the room temperature,we create exponential regressions with horizontal asymp-totes at23?C instead.The easiest way to do this is to subtract23?C from all points for the purpose of regression,and then modifying the regression equation to account for the di?erence.The results are as follows:
The resulting equations are:
y=77e?0.029x+23(8)
y=77e?0.036x+23(9)
y=77e?0.052x+23(10) for the100mL beaker,300mL beaker,and the800mL beaker,respectively.Note that23?C has been added to each equation to account for the shift.
It’s unreasonable to hope that data from small beakers alone can accurately predict the tem-perature changes in a large bathtub.However,we could not?nd any accessible experimental data for large volumes of water.This is likely due to the fact that when studying water cooling,most scientists will choose to use small quantities of water for convenience.Since it was di?cult to obtain large-scale water cooling data,we performed our own experiment with a bathtub to obtain a k-value that pertains to our model.The procedure of our experiment can be found in Appendix A.
The results of our experiment are as follows:
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The regression equation for our bathtub temperature versus time,then,is
which means our experimental k per minute is 0.0044860
=0.000075.However,this k was only relevant to our speci?c bathtub.In order to ?nd k -values for bathtubs of varied volume and surface area,we needed to create a model for k .
We consider that there is a relationship between k and exposed surface area (SA E )and volume (V ).Reasoning through the fact that k is to be a constant,we assumed that k should be unitless.
Thus,when relating k to V and SA E ,k should be dependent on a power of V 2
3E to cancel the units of meters in the numerator and denominator.
The calculations for SA E must take into account the fact that bathtubs are heavily insulated on ?ve of six faces,but beakers are not.The e?ective SA E of the bathtub,then,is its top surface’s area,or 55×23=1,265in 2.The e?ective SA E of the beakers are their total surface area,because of our assumption that thin glass provides virtually no insulation.Then,their SA E is given by:
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SA cylinder=2πr2+πrh(11) where r is the radius of the beaker and h is the height of the water in the beaker.Because the beakers were not?lled in the experiments,we recalculate the e?ective heights with the following relationship:
h=h beaker×V water
V beaker
(12)
giving us a surface area equation for the beakers of
SA cylinder=2πr2+rh beaker×V water
V beaker
(13)
Below is a table of V,SA E,V2
SA3
E ,and k values for our four data sets.
SA E(m)V(L)k
0.0521504380.80.026259385
0.0307876080.30.030316642
0.0157079630.10.040508908
0.8161274269.485270.000075
Proceeding,we test an exponential regression on the four sets of data relating k to V2
SA3
E
.First
we tested V2
3
E
,
which gave us an R2value of0.99764.We then test(V2
SA3
E
)2
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which gave us an R 2
value of 0.996529.Seeing as this is worse,we test
V 2
SA 3E
,
which gave us an R 2
value of 0.999294.Noticing improvement,we test 3
V
SA 3E
,noting a possible connection between the three dimensions our heat transfer acts in,and the third-degree root we
take on our V 2
SA 3E
,
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which gives us the regression equation:
k=0.385387190197681e ?0.16708193785762233
V2
SA3
E(14)
This model?t the data with an accuracy of R2=0.99929854436231100000,which was even bet-ter than the previous https://www.sodocs.net/doc/0f1314949.html,ing this model for k,we predict a k-value of0.385387190197681 for our standard60”×30”×14”bathtub.Thus,our?nal equation for the change in temperature due to the environment is
dT
dt
=?0.385387190197681T(15) 4.2.3Heating
The second part of the model calculates the amount of heat added by the person as he turned on the water faucet to a certain?ow rate.To achieve this,we used the speci?c heat capacity formula, which is
?Q=mC p T(16) where?Q is the change in heat,m is mass,C p is the speci?c heat capacity of water,and T is the change in temperature.There are two instances of heat change resulting from the addition of hot water:heat is being added through the addition of hot water from the faucet,and heat is being removed through the drainage of excess water.The heat contributed by the water?owing in is given by the following equation:
Q1=dm
dt
C p T hot(17)
where Q1is the heat from the water?owing in,dm
dt is the mass?ow rate of the water?owing
in,and T is the absolute temperature of the water?owing in(Kelvins).
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The heat lost from excess water drainage is given by the following equation:
Q2=dm lost
dt
C p T lost(18)
where Q2is the heat from the water draining out,dm lost
dt is the mass?ow rate of the water
draining out,and T lost is the absolute temperature of the water draining out(Kelvins).By our global assumption that temperature is consistent throughout the tub,we have that T lost is simply the temperature of the tub,T.
Because the volume of water?owing in at any point in time is equivalent to the volume of water?owing out due to over?ow,the mass?ow rate out is equal to the mass?ow rate in,or
dm lost
dt =dm
dt
.Then,the heat change of the water in the tub can be obtained by subtracting the
above two equations:
?Q=dm
dt
C p(T hot?T)(19)
We also have that
?Q=m tub C p T tub(20)
Since?Q is common to both equations,we can substitute the expression dm
dt C p(T hot?T)for
?Q in the second equation and solve for T tub,which becomes:
4.3Simulation Design
All that is left now is to combine the cooling and heating of the water to simulate the overall temperature change.
In order to simulate the cooling of the water when the person has not turned on the faucet,we use the Newton’s law di?erential obtained earlier:
dT
dt
=?0.385387190197681T(21) And to simulate the change in temperature due to the addition of hot water,we add the equation obtained in the previous section to the above:
dT dt =?0.385387190197681T+
dm
dt
(T hot?T)m tub(22)
We used a computer program written in Python(code included in Appendix B)to simulate the change in temperature of the bathtub according to the di?erential equation over a period of40 minutes.All units are converted to standard SI units in the simulation.The simulation accounts for the loss of heat to the environment,as well as the heating of the water due to the person operating the faucet.In order to achieve this,we set up a loop to run the simulation between time t=0minutes to t=40minutes,what we assumed to be the end of the bath.During each one-minute”frame”in this interval,we calculated the new temperature of the water using the previous temperature with the two equations for change in temperature per minute calculated in the previous section.
The simulation will begin by applying the?rst equation to the temperature of the water every frame,since the hot water faucet is initially o?.When the temperature of the water reaches a
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noticeably cooler temperature(of1.4?C below its initial temperature),the person will turn on the hot water,and the simulation will no longer apply the?rst equation,but instead the second equation,in order to account for the temperature gained from the hot water.
4.3.1Selecting Optimal Flow Rate
In order to evaluate the results for the various?ow rates,we used the deviance value,obtained by ?nding the sum of the squares of the di?erences between the”current”temperature and the initial temperature every frame:
?total=
N
n
(x n?ˉx)2(23)
Comparing the deviance of every?ow rate produces the following:
Now,selecting the optimal?ow rate is trivial.Since the optimal?ow rate is characterized by the least di?erence from the original temperature,it is represented by the point indicated at the bottom of the graph,at(1.81,10.179381314125882).This means that our optimal?ow rate to be recommended to the person is1.81gallons/minute,which results in a total deviance of10.179 from the original temperature of his bath.The following is a temperature vs.time graph of the simulation using the optimal?ow rate of1.81gallons per minute.Note that the curve initially decreases,as the temperature of the water is only a?ected by the cooler temperature of the room. However,once the temperature of the water has dropped below1.4?C from the initial temperature of37?C,the person turns on the hot water faucet which begins to increase the temperature of the water.
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4.3.2Validation of Model
To con?rm that our cooling simulation accurately models what occurs in the real world,we ran the simulation on the circumstances of our real-world experiment in Appendix B.Fitting the temperatures predicted by the simulation to the actual experimentally obtained data,we?nd that the simulation is able to very accurately predict the temperatures at given points in time retroactively.
As a matter of fact,this di?erential simulation is even more accurate than the previous regression model we had obtained for the data:
y=16e?0.00448x+21(24) as that regression led to an R-squared value of0.97326,and the simulation gave an R-squared value of0.999836.
While the relationship between temperature and time in this graph appears to be linear,it is still exponential–the very small k-value is what makes it appear linear.Indeed,there is a very slight,hardly noticeable curvature to the above graph.
5Sensitivity Analysis:Varying Bathtub Characteristics
We now address query number”3”by creating various bathtub designs,varying the surface areas exposed and volumes,to observe the e?ect of the surface area to volume ratio on the rate of cooling. The changes in surface area and volume change the k value,so we recalculate k for each case and test each bathtub design with our simulation.
5.1Varying Surface Area Under Constant Volume
The standard bathtub model has a volume of25,200in3(”Bathtub Sizes”),so we keep that volume constant while creating three other common bathtub https://www.sodocs.net/doc/0f1314949.html,ing our model,we predict the
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optimal?ow rate for following designs created using Autodesk Inventor,choosing?ow rates with the lowest total deviance from initial temperature.
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Standard bathtub model
Dimensions(length x width x height):60”×30”×14”
Surface area:1,800in2
Volume:25,200in3
The standard bathtub model has a predicted optimal?ow rate of1.81gallons/minute and a deviance of10.18?C2.
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