How Fast Does Hot Water Cool When Sitting in Pipes?

From the School of Hard Knocks.  Last updated November 3, 2008

When you turn on the hot water at the kitchen faucet, it 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 turn on the water 10 minutes later how much will the water have cooled?

We can answer this question in two different ways. Either we can do an experiment to measure the water temperature, 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.

Empirical Measurement of Cooling:

The below graph shows the temperature of a probe attached to the outer surface of a 1/2" CPVC pipe underneath the insulation. The hot water was turned on for the first 5 minutes of the experiment, and the temperature recorded as the pipe surface heated. Then the hot water was turned off, and the temperature was recorded as the pipe cooled over time. The three curves correspond to no pipe insulation, R-2 insulation and R-4 insulation. This experiment wasn't perfect--the basement temperature varied a little between the three different days I did the experiment. However, this variation was small and doesn't change the basic nature of the results.

 
This graph shows that adding pipe insulation greatly slows cooling, and that the higher the R-value the better. A high R-value insulation means there is less need to run new hot water all the way from the tank if the last time you used the hot water was a few minutes ago.

Remember that these curves DO NOT show actual water temperature in the pipe--I couldn't put the probe inside the pipe. Instead, the curves show the surface temperature of the pipe, which is the next best thing. An interesting observation is that the water delivered at the faucet reaches 114 or 115F in all three cases, but the pipe surface temperature is much lower--from 96F to 104F--even after running the hot water for 5 minutes.

Analytic Approach Using 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) = T_A + (T_H-T_A) e^{-A/((m_wc_w+m_pc_p)R) t}

     where
       T(t) = Temperature of water at time t
       T_A = Air temperature surrounding the pipe, in Fahrenheit
       T_H = Initial temperature of the hot water
       A = Surface area in square feet
       m_w = mass of water in pounds
       c_w = specific heat of water in btus/lb/F
       m_p = mass of pipe in pounds
       c_p = specific heat of pipe in btus/lb/F
       t = time in hours
       R = R-value of the insulation in ft²hrF/btu

Simplifying Assumptions:
     The R-value of the CPVC pipe walls is insignificant (assume R-0)
     The specific heat of the foam insulation is insignificant (assume c of 0)
     There is no "surface effect" R-value (due to air in contact with the outside of the insulation)
     We are examining the middle of a very long pipe

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. To use the equation for temperature we must determine all the above parameters.

Surface Area:
The surface area in the formula refers to the area the heat flows through. On a flat surface, like a wall, surface area is easy to compute. Unfortunately, here we are dealing with a curved surface, which presents a dilemma--which surface area should we use--the surface area of the outside of the pipe, or of the outside of the foam insulation? We'll try both and compare. If they are close then it doesn't matter which we use, but if they are far apart we need to research this more.

The outer diameter of a 1/2 inch diameter CPVC pipe is about 1.6 cm, which converts to .6299 inches. The outer diameter of the R-2 foam insulation is 1.5 inches. The formula for the surface area of a cylinder is:

     A = circumference x height = 3.14*diameter * height

So
     A_min = 3.14 * .6299 inches * 12 inches / 144 in/ft²
     A_min = 0.1649 ft²

     A_max = 3.14 * 1.5 inches * 12 inches / 144 in/ft²
     A_max = 0.3927 ft²

Mass of Water in the Pipe:
We must first compute the volume of water in the pipe in order to compute the mass. The inner diameter of the CPVC pipe is 1.2 cm, which converts to .4724 inches. The radius is half of that, or .2362 inches. The formula for the volume of a cylinder is:

     V = 3.14*radius² * length
     V = 3.14*.2362² * 12
     V = 2.103 in³

Converting this volume in cubic inches to gallons:

     V = 2.103 in³ * 1 gallon/231 in³
     V = 0.009103 gallons

Converting this volume in gallons to pounds:

     m = 0.009103 gallons * 8.34 lb/gallon
     m = 0.07592 lb

Mass of 1 foot section of CPVC Pipe:
We cannot ignore the pipe itself, as that weighs about as much as the water inside it. A 1 foot length of pipe weighs 35 grams, which converts to 0.07716 pounds.

Summarizing the inputs:
       T_H = 120 F
       T_A = 60 F
       A_min = 0.1649 ft²
       A_max = 0.3927 ft²
       m_w = 0.07592 lb
       c_w = 1 btu/lb/F for water
       m_p = 0.07716 lb
       c_p = 0.20 btu/lb/F for CPVC pipe
       R = 2

So the coefficient A_min/((m_wc_w+m_pc_p)R) = 0.1649/(0.09135*2) = 0.90255

Substituting into the equation:

     T(t) = 60 + (120 - 60)e^{-0.90255 t}

Setting t=1/6 hour (10 minutes):

     T(1/6) = 111.6 F

Thus the hot water temperature should theoretically drop 8.4 degrees F in 10 minutes if A_min is used.

For A_max the water cools faster:
     A_max/((m_wc_w+m_pc_p)R) = 0.3927/(0.09135*2) = 2.1494

Substituting into the equation:

     T(t) = 60 + (120 - 60)e^-2.1494t

Setting t=1/6 hour (10 minutes):

     T(1/6) = 101.9 F

Thus the hot water temperature should theoretically drop 18.1 degrees F in 10 minutes if A_max is used.

The large difference between the two answers suggests we need to look more carefully at the effect of curvature on insulation R-value. The equation for heat flow used to derive Newton's Law of Cooling in the Newton's Law of Cooling web page assumed a flat surface, with flat insulation above it. But the pipe has a curved surface with heat flowing out from each point through a wedge-shaped piece of insulation. We need calculus to compute the heat flow through curved insulation.

Using Calculus to Compute Heat Flow Through Cylindrical Insulation:

For those familiar with calculus, we will use something called the "method of cylindrical shells." The diagram below shows the basic setup for the problem. The radius of the pipe is a, and the outer radius of the insulation is b. The temperature on the outside of the pipe is T_a, and the temperature on the outside of the insulation is T_b. We consider a thin cylindrical shell within the insulation. As shown, the shell has a radius r on the inside and r+Δr on the outside.

Because the cylindrical shell has a very small thickness, we can apply the formula for heat flow through a flat surface. Imagine cutting the shell down its length and unfolding it to make a flat piece of insulation. The formula for heat flow is then:

     H = A ΔT/(RΔr)

     Where
          Area is computed using the formula for surface area of a cylinder of length L:   A = 2 π r L
          ΔT = temperature difference across the shell
          R = R value per inch (different from previous definition of R-value)
          Δr = thickness of the shell

Substituting for A gives:

     H = 2 π r L ΔT/(RΔr)

Next we need to rearrange this equation. Multiplying both sides by RΔr gives:

     HRΔr = 2 π r L ΔT

Let's group all the r terms on the left. Dividing both sides by r gives:

     HRΔr/r = 2 π L ΔT

For a very small Δr then ΔT will also be very small and we can replace the Δ notation with differential notation as follows:

     HR(dr/r) = 2 π L dT

We are now ready to take integrals, as follows:

Which evaluates to:

     HR(lnb-lna) = 2 π L (T_b-T_a)

     HRln(b/a) = 2 π L (T_b-T_a)

     H = 2 π L ΔT/(Rln(b/a))          (1)

This is the formula for heat flow through the cylindrical insulation around a pipe, where:
          L = Length of the pipe
          ΔT = temperature difference across the insulation (T_b-T_a)
          R = R value per inch
          a = radius of the pipe
          b = radius of the insulation

Contrast this formula to the formula for insulation of thickness Δx on a flat surface:

     H = A ΔT/(RΔx)

Let's define the area A to be the surface area of the pipe, and the thickness Δx to be the difference (b-a). Then this second formula becomes:

     H = 2 π a L ΔT/(R(b-a))          (2)

Now suppose that we wanted to somehow equate these two formulas for H. In particular, let's recompute the R in formula (1) so that formula (1) will give the same H as in formula (2). First, replace R with R' in formula (1), then equate the two formulas:

     2 π a L ΔT/(R(b-a)) = 2 π L ΔT/(R'ln(b/a))

     a/(R(b-a)) = 1/(R'ln(b/a))

     R' = R(b-a)/(aln(b/a))

Now let's subsitute some values and figure out what this means. The foam pipe insulation is made from polyethylene foam. I found one source that listed the R value per inch of polyethylene foam as 2.33. I can't be sure this is the same type of foam as used in the pipe insulation, but let's go with that value for the moment.

          a = outer radius of pipe = .6299 inches/2 = .3150 inches
          b = outer radius of insulation = .75 inches

     R' = R(b-a)/(aln(b/a))

     R' = 2.33(.75-.315)/(.315ln(.75/.315))

     R' = 3.71

This result means that if the R-value per inch of flat insulation is 2.33, then the R-value per inch of cylindrical insulation of the same thickness must be significantly higher in order to perform as well.

Another way of saying this is that if we want to take the formula for Newton's law of cooling for flat insulation and apply it to cylindrical insulation (using the inner surface area), then we must derate the R-value per inch of the insulation according to the formula:

     R_derated = R_original aln(b/a)/(b-a)

The derating factor for the 1/2" pipe example we have been working with is 0.628.

From earlier we computed the coefficient A_min/((m_wc_w+m_pc_p)R) = 0.1649/(0.09135*2) = 0.90255

Multiplying this coefficient by the derating factor 0.628 gives:

     A_min/((m_wc_w+m_pc_p)R*0.628) = 0.1649/(0.09135*2*0.628) = 1.4372

Substituting into the equation for Newton's Law of Cooling gives:

     T(t) = 60 + (120 - 60)e^{-1.4372 t}

Setting t=1/6 hour (10 minutes):

     T(1/6) = 107.2 F

Thus the hot water temperature should theoretically drop 12.8 degrees F in 10 minutes if A_min is used with the correct derating factor.

Picking data off of the graph for R-2 I get an actual temperature drop of 8.2F in 10 minutes. This is very close to the predicted drop of 8.4F for the case where we used surface area A_min. But this is a false comparison because the formula predicts the temperature of the water inside the pipe and we are measuring the temperature on the surface of the pipe. Also we used some simplifying assumptions without knowing how important they were--such as no surface effect R-value.

Thoughts on the Analytic Approach

The analytic approach to computing pipe cooling was a lot of work, and required a number of simplifying assumptions that may throw off the calculation. It is much faster, easier, and more reliable to do the measurements experimentally than to go through all these computations.

Final Aside

How can we use the formula if there were no insulation on the pipe? We cannot plug R=0 into the equation or it won't work because the R is in the denominator of the coefficient. In practice there are surface effects that provide some R-value even with no insulation. We can use our empirical data to compute the "no insulation" R-value. One complexity is that the R-value due to surface effects may depend on pipe orientation.

This site is still under construction…to be continued…

You can e-mail me at support(@ sign goes here)leaningpinesoftware.com.