Heat exchanger modeling - Revision history

Ehrennagel at 21:02, 9 January 2007

2007-01-09T21:02:49Z

← Older revision		Revision as of 21:02, 9 January 2007
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	To collect heat transfer data for a heat exchange model we ran three tests using a 12" capped off piece of 3/4" copper pipe filled with cold tap water and a 5 gallon bucket filled with hot water. We held the pipe in the hot water with a pair of tongs and recorded the temperature of the water in the pipe every three seconds using a digital thermometer (carefully inserted into the pipe with plastic spacers so that it didn't touch the sides). Colin has created [http://cs.earlham.edu/~hip/hip/databaseInterface/ graphs] of the change in temperature inside the copper pipe over time for the three test runs.		To collect heat transfer data for a heat exchange model we ran three tests using a 12" capped off piece of 3/4" copper pipe filled with cold tap water and a 5 gallon bucket filled with hot water. We held the pipe in the hot water with a pair of tongs and recorded the temperature of the water in the pipe every three seconds using a digital thermometer (carefully inserted into the pipe with plastic spacers so that it didn't touch the sides). Colin has created [http://cs.earlham.edu/~hip/hip/databaseInterface/ graphs] of the change in temperature inside the copper pipe over time for the three test runs.
	===Flow Rate Through the Copper Heat Exchanger===		===Flow Rate Through the Copper Heat Exchanger===
	Ehren found that the 12" piece of 3/4" diameter copper pipe holds approximately 112 ml of water. During our [http://wiki.cs.earlham.edu/index.php/~~Hot_h2o~~ hot water audit] of Miller Farm we [http://wiki.cs.earlham.edu/index.php/~~Labs~~#Average_GPM_of_shower estimated] that the typical hot to cold water ratio for a shower at Miller Farm is 4:1 and the total flow rate is 2.25 gal/min. We can therefore estimate the hot water flow rate at Miller Farm to be approximately 1.8 gal/min = 6814 ml/min = 114 ml/s. This means that 112 ml will travel through a foot of copper pipe in 112 ml/(114 ml/s) = 0.98 seconds.		Ehren found that the 12" piece of 3/4" diameter copper pipe holds approximately 112 ml of water. During our [http://wiki.cs.earlham.edu/index.php/hotwater-audit hot water audit] of Miller Farm we [http://wiki.cs.earlham.edu/index.php/hotwater-labs#Average_GPM_of_shower estimated] that the typical hot to cold water ratio for a shower at Miller Farm is 4:1 and the total flow rate is 2.25 gal/min. We can therefore estimate the hot water flow rate at Miller Farm to be approximately 1.8 gal/min = 6814 ml/min = 114 ml/s. This means that 112 ml will travel through a foot of copper pipe in 112 ml/(114 ml/s) = 0.98 seconds.

	==Modeling Heat Transfer==		==Modeling Heat Transfer==
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	From our research during the hot water audit we [http://www.eere.energy.gov/buildings/appliance_standards/residential/pdfs/waterheater_lifecycle_1098.pdf found] that the average inlet (i.e. main) water temperature in Indiana ranges from 43-49 degrees Fahrenheit during the year.		From our research during the hot water audit we [http://www.eere.energy.gov/buildings/appliance_standards/residential/pdfs/waterheater_lifecycle_1098.pdf found] that the average inlet (i.e. main) water temperature in Indiana ranges from 43-49 degrees Fahrenheit during the year.

	Our [http://wiki.cs.earlham.edu/index.php/~~Labs~~ findings] at Miller Farm indicate that if we preheated the main water to <math>80^\circ</math>F we could reduce electricity usage on average by 498 kWh per month, thereby providing a savings of $32.37 per month on the Miller Farm electric bill (not to mention the associated reduction in carbon emmisions).		Our [http://wiki.cs.earlham.edu/index.php/hotwater-labs findings] at Miller Farm indicate that if we preheated the main water to <math>80^\circ</math>F we could reduce electricity usage on average by 498 kWh per month, thereby providing a savings of $32.37 per month on the Miller Farm electric bill (not to mention the associated reduction in carbon emmisions).

	If we take <math>80^\circ</math>F to be our goal and the median main temperature to be <math>46^\circ</math>, then our formula give us		If we take <math>80^\circ</math>F to be our goal and the median main temperature to be <math>46^\circ</math>, then our formula give us

Ehrennagel: /* Scaling our Heat Exchanger */

2006-09-16T18:10:13Z

Scaling our Heat Exchanger

← Older revision		Revision as of 18:10, 16 September 2006
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	If we take <math>80^\circ</math>F to be our goal and the median main temperature to be <math>46^\circ</math>, then our formula give us		If we take <math>80^\circ</math>F to be our goal and the median main temperature to be <math>46^\circ</math>, then our formula give us

	<math>F(t_1) = 80 = 100-\left( 100-46\right)e^(-0.009547t_1)</math>		<math>F(t_1) = 80 = 100-\left( 100-46\right)e^{-0.009547t_1}</math>

	Solving for <math>t_1</math> tells us that incoming water must remain in our copper heat exchanger for <math>t_1 = 79.36</math> seconds, requiring <math>\frac{79.36}{0.98} ~~\thickapprox 81~~</math> feet of copper pipe.		Solving for <math>t_1</math> tells us that incoming water must remain in our copper heat exchanger for <math>t_1 = 79.36</math> seconds, requiring <math>\frac{79.36}{0.98}</math> or about 81 feet of copper pipe.

			Since we already have about 30 feet of copper pipe from our test of the hot water unit, we would need to purchase an additional 50 at a cost of about $150.00. Considering that a solor hot water system with a heat exchanger of this size would save the farm approximately $388.44 per year, this cost seems reasonable (of course this does not take into account the cost of the other materials that have and will go into this project).

Ehrennagel at 18:03, 16 September 2006

2006-09-16T18:03:17Z

← Older revision		Revision as of 18:03, 16 September 2006
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	* <math>k=-0.009547</math>		* <math>k=-0.009547</math>
	With a standard deviation of		With a standard deviation of
	* <math>\sigma=0.~~000557~~</math>		* <math>\sigma=0.001138</math>

	~~around with graphs~~ of ~~exponential functions and found that T~~=95-23e^(-s/~~80), where T is temp. and s is time~~ in ~~sec~~., ~~reflects (pretty well--~~we can ~~certainly improve on this)~~ the ~~graph of the experimental data~~ we ~~collected in the first run~~ of ~~the experiment~~. ~~Since the water for this experiment was about 97 degrees Fahrenheit and would most likely reflect~~ the temperature ~~of our storage tank~~ in the ~~mornings when folks are taking showers~~, ~~lets see~~ if we ~~can scale an exchanger~~ to ~~match these findings~~. ~~Our limiting factors are space in~~ the ~~tank, and price~~. ~~The copper pipe costs a little less then $3~~.~~00 per foot~~. ~~Since we'll need to buy some connectors let's just say an average of $3~~.~~00 per foot for~~ the ~~heat exchanger~~. ~~Modifying our formula slightly we should find that Temp~~. ~~Increase(TI~~) ~~= 23~~-~~23e~~^~~(-s~~/~~80). So~~, ~~we can now find~~ a ~~formula that relates Temp. Increase to Length~~ of ~~Pipe, and Temp~~. ~~Increase~~ to ~~Price: (Let L=length of pipe in feet and M=price~~ in ~~dollars~~).		Using the above heat transfer coefficient we have the following formula for determining the temperature in Fahrenheit of water in our copper pipe at time t in seconds.
	**L = (1/~~.98)s = 1.02*s =~~> ~~s = L/1.02~~
	~~**~~and ~~M = 3L =~~> ~~L = M~~/~~3 =~~> ~~s = M/3.06~~		<math>F(t) = A-\left( A-F_0\right) e^{-0.009547t}</math>
	**so, TI = 23-23e^(~~-(L/1.02)/80~~) = ~~23-23*e^(-(M/3.06)/~~80)
	**Also, note that s = ~~80ln(~~-~~23/~~(TI-23)~~), where 1/~~(TI-~~23) <~~ 0. ~~So given a desired temp. increase we can calculate how much it will cost and how much pipe we would need.~~		==Scaling our Heat Exchanger==
	**Thus, L=1.10280ln(-23/(TI-23)~~) and M=3.0680*ln(-23~~/~~(TI-23)).~~		Although we did not gather sufficient data to estimate the average temperature in our 250 gal. thermal bank, from the little data we did collect we can guess that the temperature will normally remain at or above <math>100^\circ</math>F unless we experience a long period of cold overcast weather. Furthermore we can increase the likelihood of such a temperature in the thermal bank at the time of hot water use, if we encourage Miller Farm residents to shower in the afternoon or early evening when the bank will hold the most thermal energy.
	*Now, if we take a look at dTI/~~dM = 0.094(0.996)^M, we see~~ that ~~the slope of the change~~ in ~~Temp~~. ~~over the change in Money is always less then 1. This means that every additional degree of increase comes at a higher price. So~~, ~~we should base our dicision on available space and the long term payback of preheating the water before it goes into the electic hot water heater~~. ~~For example, if we want to increase the temperature of the main water by 10 degrees Fahrenheit we need 1~~.~~0280ln(-23~~/~~(10-23)) = 46.56 ft.~~ of pipe ~~for $139.67. A 15 degree increase would require about 86.17 ft. at a cost of $258.52~~.
			From our research during the hot water audit we [http://www.eere.energy.gov/buildings/appliance_standards/residential/pdfs/waterheater_lifecycle_1098.pdf found] that the average inlet (i.e. main) water temperature in Indiana ranges from 43-49 degrees Fahrenheit during the year.

			Our [http://wiki.cs.earlham.edu/index.php/Labs findings] at Miller Farm indicate that if we preheated the main water to <math>80^\circ</math>F we could reduce electricity usage on average by 498 kWh per month, thereby providing a savings of $32.37 per month on the Miller Farm electric bill (not to mention the associated reduction in carbon emmisions).

			If we take <math>80^\circ</math>F to be our goal and the median main temperature to be <math>46^\circ</math>, then our formula give us

			<math>F(t_1) = 80 = 100-\left( 100-46\right)e^(-0.009547t_1)</math>

			Solving for <math>t_1</math> tells us that incoming water must remain in our copper heat exchanger for <math>t_1 = 79.36</math> seconds, requiring <math>\frac{79.36}{0.98} \thickapprox 81</math> feet of copper pipe.

Ehrennagel at 17:14, 16 September 2006

2006-09-16T17:14:36Z

← Older revision		Revision as of 17:14, 16 September 2006
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	**<math>F(207) = 101.41^\circ</math> F		**<math>F(207) = 101.41^\circ</math> F

	*Now, Ehren ~~played~~ around with graphs of exponential functions and found that T=95-23e^(-s/80), where T is temp. and s is time in sec., reflects (pretty well--we can certainly improve on this) the graph of the experimental data we collected in the first run of the experiment. Since the water for this experiment was about 97 degrees Fahrenheit and would most likely reflect the temperature of our storage tank in the mornings when folks are taking showers, lets see if we can scale an exchanger to match these findings. Our limiting factors are space in the tank, and price. The copper pipe costs a little less then $3.00 per foot. Since we'll need to buy some connectors let's just say an average of $3.00 per foot for the heat exchanger. Modifying our formula slightly we should find that Temp. Increase(TI) = 23-23*e^(-s/80). So, we can now find a formula that relates Temp. Increase to Length of Pipe, and Temp. Increase to Price: (Let L=length of pipe in feet and M=price in dollars).		Now, Ehren substituted these values into our temperature formula and found the heat transfer coefficient for each of the three test runs. Theoretically the heat transfer coefficient should always be the same, so Ehren found the average of these three results and took that to be <math>k</math>.
			* <math>k_1=-0.007995</math>
			* <math>k_2=-0.010693</math>
			* <math>k_1=-0.009954</math>
			This gives us an average
			* <math>k=-0.009547</math>
			With a standard deviation of
			* <math>\sigma=0.000557</math>

			around with graphs of exponential functions and found that T=95-23e^(-s/80), where T is temp. and s is time in sec., reflects (pretty well--we can certainly improve on this) the graph of the experimental data we collected in the first run of the experiment. Since the water for this experiment was about 97 degrees Fahrenheit and would most likely reflect the temperature of our storage tank in the mornings when folks are taking showers, lets see if we can scale an exchanger to match these findings. Our limiting factors are space in the tank, and price. The copper pipe costs a little less then $3.00 per foot. Since we'll need to buy some connectors let's just say an average of $3.00 per foot for the heat exchanger. Modifying our formula slightly we should find that Temp. Increase(TI) = 23-23e^(-s/80). So, we can now find a formula that relates Temp. Increase to Length of Pipe, and Temp. Increase to Price: (Let L=length of pipe in feet and M=price in dollars).
	**L = (1/.98)s = 1.02*s => s = L/1.02		**L = (1/.98)s = 1.02*s => s = L/1.02
	**and M = 3L => L = M/3 => s = M/3.06		**and M = 3L => L = M/3 => s = M/3.06

Ehrennagel at 17:01, 16 September 2006

2006-09-16T17:01:44Z

← Older revision		Revision as of 17:01, 16 September 2006
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	===Adapting Newton's Law of Cooling===		===Adapting Newton's Law of Cooling===
	*Let <math>F(t) =</math> temperature in degrees Fahrenheit of water in the copper pipe at time <math>t</math>.		*Let <math>F(t) =</math> temperature in degrees Fahrenheit of water in the copper pipe at time <math>t</math> in seconds.
	*Let <math>F(0) = F_0 = </math>starting temperature in degrees Fahrenheit of the water in the copper pipe.		*Let <math>F(0) = F_0 = </math>starting temperature in degrees Fahrenheit of the water in the copper pipe.
	*Let <math>A =</math> the ambient temperature in degrees Fahrenheit of the thermal bank, in this case the water in the 5 gallon bucket.		*Let <math>A =</math> the ambient temperature in degrees Fahrenheit of the thermal bank, in this case the water in the 5 gallon bucket.
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	<math>F(t) = A-\left( A-F_0\right) e^{kt}</math>		<math>F(t) = A-\left( A-F_0\right) e^{kt}</math>

			===Determining the Heat Transfer Coefficient of our Copper Pipe===
			With the formula for temperature that we found in the preceding section, we can determine the heat transfer coefficient or our copper pipe. We begin by noting the results of our three test runs.
			*Test 1
			**<math>A=97.5^\circ</math> F
			**<math>F_0 = 72.72^\circ</math> F
			**<math>F(207) = 92.74^\circ</math> F
			*Test 2
			**<math>A=107^\circ</math> F
			**<math>F_0 = 72.28^\circ</math> F
			**<math>F(207) = 103.44^\circ</math> F
			*Test 3
			**<math>A=105^\circ</math> F
			**<math>F_0 = 72.28^\circ</math> F
			**<math>F(207) = 101.41^\circ</math> F

	*Now, Ehren played around with graphs of exponential functions and found that T=95-23e^(-s/80), where T is temp. and s is time in sec., reflects (pretty well--we can certainly improve on this) the graph of the experimental data we collected in the first run of the experiment. Since the water for this experiment was about 97 degrees Fahrenheit and would most likely reflect the temperature of our storage tank in the mornings when folks are taking showers, lets see if we can scale an exchanger to match these findings. Our limiting factors are space in the tank, and price. The copper pipe costs a little less then $3.00 per foot. Since we'll need to buy some connectors let's just say an average of $3.00 per foot for the heat exchanger. Modifying our formula slightly we should find that Temp. Increase(TI) = 23-23*e^(-s/80). So, we can now find a formula that relates Temp. Increase to Length of Pipe, and Temp. Increase to Price: (Let L=length of pipe in feet and M=price in dollars).		*Now, Ehren played around with graphs of exponential functions and found that T=95-23e^(-s/80), where T is temp. and s is time in sec., reflects (pretty well--we can certainly improve on this) the graph of the experimental data we collected in the first run of the experiment. Since the water for this experiment was about 97 degrees Fahrenheit and would most likely reflect the temperature of our storage tank in the mornings when folks are taking showers, lets see if we can scale an exchanger to match these findings. Our limiting factors are space in the tank, and price. The copper pipe costs a little less then $3.00 per foot. Since we'll need to buy some connectors let's just say an average of $3.00 per foot for the heat exchanger. Modifying our formula slightly we should find that Temp. Increase(TI) = 23-23*e^(-s/80). So, we can now find a formula that relates Temp. Increase to Length of Pipe, and Temp. Increase to Price: (Let L=length of pipe in feet and M=price in dollars).

Ehrennagel: /* Adapting Newton's Law of Cooling */

2006-09-16T16:43:25Z

Adapting Newton's Law of Cooling

← Older revision		Revision as of 16:43, 16 September 2006
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	Thus, our equation becomes		Thus, our equation becomes

	<math>F(t) = A-(A-F_0)e^{kt}</math>		<math>F(t) = A-\left( A-F_0\right) e^{kt}</math>

	*Now, Ehren played around with graphs of exponential functions and found that T=95-23e^(-s/80), where T is temp. and s is time in sec., reflects (pretty well--we can certainly improve on this) the graph of the experimental data we collected in the first run of the experiment. Since the water for this experiment was about 97 degrees Fahrenheit and would most likely reflect the temperature of our storage tank in the mornings when folks are taking showers, lets see if we can scale an exchanger to match these findings. Our limiting factors are space in the tank, and price. The copper pipe costs a little less then $3.00 per foot. Since we'll need to buy some connectors let's just say an average of $3.00 per foot for the heat exchanger. Modifying our formula slightly we should find that Temp. Increase(TI) = 23-23*e^(-s/80). So, we can now find a formula that relates Temp. Increase to Length of Pipe, and Temp. Increase to Price: (Let L=length of pipe in feet and M=price in dollars).		*Now, Ehren played around with graphs of exponential functions and found that T=95-23e^(-s/80), where T is temp. and s is time in sec., reflects (pretty well--we can certainly improve on this) the graph of the experimental data we collected in the first run of the experiment. Since the water for this experiment was about 97 degrees Fahrenheit and would most likely reflect the temperature of our storage tank in the mornings when folks are taking showers, lets see if we can scale an exchanger to match these findings. Our limiting factors are space in the tank, and price. The copper pipe costs a little less then $3.00 per foot. Since we'll need to buy some connectors let's just say an average of $3.00 per foot for the heat exchanger. Modifying our formula slightly we should find that Temp. Increase(TI) = 23-23*e^(-s/80). So, we can now find a formula that relates Temp. Increase to Length of Pipe, and Temp. Increase to Price: (Let L=length of pipe in feet and M=price in dollars).

Ehrennagel: /* Adapting Newton's Law of Cooling */

2006-09-16T16:41:35Z

Adapting Newton's Law of Cooling

← Older revision		Revision as of 16:41, 16 September 2006
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	Thus, our equation becomes		Thus, our equation becomes
	~~<br>~~<math>F(t) = A-(A-F_0)e^{kt}.</math>
			<math>F(t) = A-(A-F_0)e^{kt}</math>

	*Now, Ehren played around with graphs of exponential functions and found that T=95-23e^(-s/80), where T is temp. and s is time in sec., reflects (pretty well--we can certainly improve on this) the graph of the experimental data we collected in the first run of the experiment. Since the water for this experiment was about 97 degrees Fahrenheit and would most likely reflect the temperature of our storage tank in the mornings when folks are taking showers, lets see if we can scale an exchanger to match these findings. Our limiting factors are space in the tank, and price. The copper pipe costs a little less then $3.00 per foot. Since we'll need to buy some connectors let's just say an average of $3.00 per foot for the heat exchanger. Modifying our formula slightly we should find that Temp. Increase(TI) = 23-23*e^(-s/80). So, we can now find a formula that relates Temp. Increase to Length of Pipe, and Temp. Increase to Price: (Let L=length of pipe in feet and M=price in dollars).		*Now, Ehren played around with graphs of exponential functions and found that T=95-23e^(-s/80), where T is temp. and s is time in sec., reflects (pretty well--we can certainly improve on this) the graph of the experimental data we collected in the first run of the experiment. Since the water for this experiment was about 97 degrees Fahrenheit and would most likely reflect the temperature of our storage tank in the mornings when folks are taking showers, lets see if we can scale an exchanger to match these findings. Our limiting factors are space in the tank, and price. The copper pipe costs a little less then $3.00 per foot. Since we'll need to buy some connectors let's just say an average of $3.00 per foot for the heat exchanger. Modifying our formula slightly we should find that Temp. Increase(TI) = 23-23*e^(-s/80). So, we can now find a formula that relates Temp. Increase to Length of Pipe, and Temp. Increase to Price: (Let L=length of pipe in feet and M=price in dollars).
	**L = (1/.98)s = 1.02*s => s = L/1.02		**L = (1/.98)s = 1.02*s => s = L/1.02

Ehrennagel at 16:36, 16 September 2006

2006-09-16T16:36:42Z

@@ Line 6: / Line 6: @@
 Ehren found that the 12" piece of 3/4" diameter copper pipe holds approximately 112 ml of water.  During our [http://wiki.cs.earlham.edu/index.php/Hot_h2o hot water audit] of Miller Farm we [http://wiki.cs.earlham.edu/index.php/Labs#Average_GPM_of_shower estimated] that the typical hot to cold water ratio for a shower at Miller Farm is 4:1 and the total flow rate is 2.25 gal/min.  We can therefore estimate the hot water flow rate at Miller Farm to be approximately 1.8 gal/min = 6814 ml/min = 114 ml/s. This means that 112 ml will travel through a foot of copper pipe in 112 ml/(114 ml/s) = 0.98 seconds.
 **Now, Ehren played around with graphs of exponential functions and found that T=95-23*e^(-s/80), where T is temp. and s is time in sec., reflects (pretty well--we can certainly improve on this) the graph of the experimental data we collected in the first run of the experiment.  Since the water for this experiment was about 97 degrees Fahrenheit and would most likely reflect the temperature of our storage tank in the mornings when folks are taking showers, lets see if we can scale an exchanger to match these findings.  Our limiting factors are space in the tank, and price.  The copper pipe costs a little less then $3.00 per foot.  Since we'll need to buy some connectors let's just say an average of $3.00 per foot for the heat exchanger.  Modifying our formula slightly we should find that Temp. Increase(TI) = 23-23*e^(-s/80).  So, we can now find a formula that relates Temp. Increase to Length of Pipe, and Temp. Increase to Price: (Let L=length of pipe in feet and M=price in dollars).
 ***L = (1/.98)*s = 1.02*s => s = L/1.02

Ehrennagel at 15:09, 16 September 2006

2006-09-16T15:09:09Z

New page

==Copper Pipe Heat Exchanger Test==
In order to preheat the main water before it enters Miller Farm's electric hot water heater, we will route the supply through a copper pipe heat exchanger inside of a 250 gallon thermal bank, which is heated by our solor panel. This design was chosen because it is non-pressurized and simple to install; however, this design does create one significant challenge: we have a relatively short period of time to transfer heat to the main water traveling through the heat exchanger. Therefore, we must properly scale the heat exchanger (i.e. determine how many feet of copper pipe we need) if we hope to produce a significant change in the temperture of the main supply.
===Data Collection===
To collect heat transfer data for a heat exchange model we ran three tests using a 12" capped off piece of 3/4" copper pipe filled with cold tap water and a 5 gallon bucket filled with hot water. We held the pipe in the hot water with a pair of tongs and recorded the temperature of the water in the pipe every three seconds using a digital thermometer (carefully inserted into the pipe with plastic spacers so that it didn't touch the sides). Colin has created [http://cs.earlham.edu/~hip/hip/databaseInterface/ graphs] of the change in temperature inside the copper pipe over time for the three test runs.
===Flow Rate Through the Copper Heat Exchanger===
Ehren found that the 12" piece of 3/4" diameter copper pipe holds approximately 112 ml of water. During our [http://wiki.cs.earlham.edu/index.php/Hot_h2o hot water audit] of Miller Farm we [http://wiki.cs.earlham.edu/index.php/Labs#Average_GPM_of_shower estimated] that the typical hot to cold water ratio for a shower at Miller Farm is 4:1 and the total flow rate is 2.25 gal/min. We can therefore estimate the hot water flow rate at Miller Farm to be approximately 1.8 gal/min = 6814 ml/min = 114 ml/s. This means that 112 ml will travel through a foot of copper pipe in 112 ml/(114 ml/s) = 0.98 seconds.

**Now, Ehren played around with graphs of exponential functions and found that T=95-23*e^(-s/80), where T is temp. and s is time in sec., reflects (pretty well--we can certainly improve on this) the graph of the experimental data we collected in the first run of the experiment. Since the water for this experiment was about 97 degrees Fahrenheit and would most likely reflect the temperature of our storage tank in the mornings when folks are taking showers, lets see if we can scale an exchanger to match these findings. Our limiting factors are space in the tank, and price. The copper pipe costs a little less then $3.00 per foot. Since we'll need to buy some connectors let's just say an average of $3.00 per foot for the heat exchanger. Modifying our formula slightly we should find that Temp. Increase(TI) = 23-23*e^(-s/80). So, we can now find a formula that relates Temp. Increase to Length of Pipe, and Temp. Increase to Price: (Let L=length of pipe in feet and M=price in dollars).
***L = (1/.98)*s = 1.02*s => s = L/1.02
***and M = 3*L => L = M/3 => s = M/3.06
***so, TI = 23-23*e^(-(L/1.02)/80) = 23-23*e^(-(M/3.06)/80)
***Also, note that s = 80*ln(-23/(TI-23)), where 1/(TI-23) < 0. So given a desired temp. increase we can calculate how much it will cost and how much pipe we would need.
***Thus, L=1.102*80*ln(-23/(TI-23)) and M=3.06*80*ln(-23/(TI-23)).
**Now, if we take a look at dTI/dM = 0.094*(0.996)^M, we see that the slope of the change in Temp. over the change in Money is always less then 1. This means that every additional degree of increase comes at a higher price. So, we should base our dicision on available space and the long term payback of preheating the water before it goes into the electic hot water heater. For example, if we want to increase the temperature of the main water by 10 degrees Fahrenheit we need 1.02*80*ln(-23/(10-23)) = 46.56 ft. of pipe for $139.67. A 15 degree increase would require about 86.17 ft. at a cost of $258.52.