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=0.000557</math>

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.