How Fast Does Hot Water Cool When Sitting in Pipes? |
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From the School of Hard Knocks. Last updated Nov 18, 2007 |
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Home Page Hot Water Heat--The Big Picture Hot Water Tank Heat Loss Should You Shut Off Your Hot Water Tank When on Vacation? How Quickly Does Water Cool in the Pipes? Newton's Law of Cooling Math Sheets |
When you turn on the hot water at the kitchen faucet the water can take anywhere from a few seconds to a minute to get fully up to temperature. This is because hot water must make its way from the hot water tank through the pipes to the faucet. When you turn off the faucet hot water is left sitting in the pipes and immediately begins to cool. If you come back 10 minutes later then how much will it have cooled? We can answer this question in two different ways. Either we can do an experiment to measure the water temperature directly, or we can use an equation called Newton's Law of Cooling to predict the water temperature. In practice it is useful to take both approaches as we can gain insights from each.
Start with Newton's Law of Cooling: Newton's Law of Cooling states that the hotter an object is, the faster it cools. Please see the web page on this site called Newton's Law of Cooling for a full explanation of the equation and how it is derived. The basic equation for Newton's Law of Cooling applied to water pipes is: T(t) = TA + (TH-TA) e-A/((mwcw+mpcp)R) t where Sample Calculation: As a simple example, suppose 120F water sits in a 1/2 inch diameter CPVC pipe with R-2 insulation and the surrounding air temperature is 60F. What is the water temperature after 10 minutes? Let's consider a 1 foot long section of pipe. As preliminaries we need to compute the surface area of the pipe and the mass of water in the pipe. Surface Area of the Pipe Section: S = circumference x height = 3.14*diameter * height Mass of Water in the Pipe: V = 3.14*radius2 * length Converting this volume in cubic inches to gallons: V = 2.103 in3 * 1 gallon/231 in3 Converting this volume in gallons to pounds: m = 0.009103 gallons * 8.34 lb/gallon Mass of 1 foot section of CPVC Pipe: Summarizing the inputs: So the coefficient A/((mwcw+mpcp)R = 0.124/(0.09135*2) = 0.6787 Substituting into the equation: T(t) = 60 + (120 - 60)e-0.6787 t Setting t=1/6 hour (10 minutes): T(1/6) = 113.6 F Thus the hot water temperature should theoretically drop 6.4 degrees F in 10 minutes. (I think the actual rate of temperature drop is much higher than suggested by this calculation.) With R-4 insulation the temperature after 10 minutes should be 116.7 F, or a 3.3 degree drop. So using higher quality insulation causes the water to stay hot longer and means there is less need to run new hot water from the tank if the last time you used the hot water was a few minutes ago. What if there were no insulation on the pipe? We cannot plug R=0 into the equation or it won't work. In practice there are surface effects that provide an R-value on the order of 1 even with no insulation. We would have to perform an experiment to determine the exact "no insulation" R-value. Even then the R-value may depend on pipe orientation.
Empirical Measurement of Cooling: This will have to wait until January, when our basement temperature is at its coldest. This site is still under construction…to be continued… You can e-mail me at support(@ sign goes here)leaningpinesoftware.com. |