Heat exchanger modeling

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 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 hot water audit of Miller Farm we 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

Our model of heat transfer is based on the assumption that the temperature of the water in our thermal bank (or in this case our 5 gallon bucket) remains fixed over time. This allows us to use Newton's Law of Cooling to describe the temperature change of the water in our heat exchanger:

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> in seconds.
• 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.

Newton's Law of Cooling tells us that

<math>\frac{dF}{dt} = k(A-F(t)).</math>

where <math>k</math> is a constant heat transfer coeficient.

We can quickly solve this differential equation:
<math>\frac{dF}{dt} = k(A-F(t))</math>
<math>\Rightarrow \frac{dF}{A-F(t)} = k</math> <math>dt</math>
<math>\Rightarrow \int \frac{dF}{A-F(t)} = \int k</math> <math>dt</math>
<math>\Rightarrow ln \mid A-F(t)\mid = kt+C_1</math>
<math>\Rightarrow A-F(t) = C_2e^{kt}</math>
<math>\Rightarrow F(t) = A-C_2e^{kt}.</math>

Now, we find the constant <math>C_2</math>:
<math>F(0) = A-C_2e^{k\cdot 0}</math>
<math>\Rightarrow F_0 = A-C_2</math>
<math>\Rightarrow C_2 = A - F_0.</math>

Thus, our equation becomes

<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 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=0.001138</math>

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.

<math>F(t) = A-\left( A-F_0\right) e^{-0.009547t}</math>

Scaling our Heat Exchanger

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.

From our research during the hot water audit we found that the average inlet (i.e. main) water temperature in Indiana ranges from 43-49 degrees Fahrenheit during the year.

Our 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.