# Heat Loss in a Hot Water Pipe System

1. Mar 25, 2006

### Ouabache

This was a question posed to me by a homeowner. I don't know very much thermodynamics and hope someone here might have a better understanding.

They have a run of 3/8in copper pipe leading from their water heater to various taps. On cold mornings, it takes a long time, for hot water to flow to the faucets and shower heads. Their idea is to do a rough cost analysis to try and justify the cost of replumbing (making shorter run) from water heater to taps.

I thought a reasonable place to start is to calculate the heat lost along their run of pipe (perhaps in BTU/ft-hr). I found some information about this ref a. The Heat Transfer calculation assumes you know the temperature on both the inside and outside surface of the pipe. I could plug in some numbers for inside temperature; such as 130 deg F at the heater and 120 deg F by the time it reaches the tap (with assumption that the hot water temperature deceases along the run of pipe). I don't have any figure to use for the outside pipe surface temperature.

There is also a parameter k , the thermal conductivity, which is puzzling. In Marks' Standard Handbook for Mechanical Engineering, there is a table of Thermal Conductivity of Metals. For copper between 7 and 700 deg F, $k_{to}= 232$ and $a = 0.032$. In caption it states $kt = kt_o - a(t-t_o)$ and k (Btu/hr/ft^2/deg F/ft). The units are different from the k in ref2 which are Btu/hr-ft-deg F. They differ by units of ft^2 in denominator. Now if $kt = kt_o - a(t-t_o)$ , with a little algebra, I find that $k = -a$. But that doesn't seem right either, because this value of $a = 0.032$ is 4 orders of magnitude from the thermal conductivity (108 Btu/hr-ft-deg F) of stainless steel used in example in ref2.

Perhaps there is an easier approach to this part of the question. One online ref b, with some hand waving, state the heat losses from a couple different diameters of copper hot water pipe, with and w/o insulation. They assume a constant inside pipe temperature of 140 deg F and mention the ambient air temperature is 70 deg F. I don't know if there may be standard tables to obtain this sort of information for the heating and plumbing industry.

Your thoughts or suggestions are appreciated.

Last edited: Mar 26, 2006
2. Mar 26, 2006

### Integral

Staff Emeritus
You may be able to do a simpler energy analysis. Since you know the input and output temperatures you know how much energy per gallon (l) of water it lost while flowing through the pipe.

You can compute the approximate volume of water in the pipes to get an idea for how much energy is wasted when the pipes are not flowing. This could be done best by taking some measurements on the water temperature after sitting for a measured amount of time.

3. Mar 26, 2006

### Q_Goest

Strictly speaking, you might include the convective heat transfer coefficient of the water on the inside of the pipe, and then the conduction of heat through the pipe wall in addition to the convective heat transfer from the pipe wall to the air, but this really isn't necessary in your case. The heat transfer from the water to the pipe, and through the pipe wall is very very high compared to the convection from the outer pipe wall to air. Thus you can neglect heat transfer from the water to the outer pipe wall and just assume the outer pipe wall is the same temperature as the water without loosing much accuracy.

Another thing you can do is use simplifed convective heat transfer coefficients for the air on the outer pipe wall if you're only looking for a ballpark estimate. For this situation I'd suggest using h = 1.0 to 1.5 Btu/hr/ft2/F.

Finally, you can assume the water has a large heat capacity and doesn't change in temperature much as it flows through the pipe so that the temperature at the end of the pipe only drops a few degrees at most.

First calculate the outer surface pipe area in square feet (A = pi * OD * L) then use this equation:
Q = A * h * (Tw - Ta)
where Q = heat lost (Btu/hr)
A = outer surface area (ft2)
Tw = water temp (F)
Ta = Air temp (F)

Note this only looks at the steady state heat loss and not the transient of having to warm a cold pipe. I think that can probably be neglected as well, though I'm less sure about that without doing the calculation.

4. Mar 26, 2006

### Averagesupernova

Why don't they just go to Home Depot or someplace and by the foam insulation that wraps around pipes? The stuff works pretty good.

5. Mar 26, 2006

### FredGarvin

Before even running numbers I would like to point out how incredibly difficult it can be to rerough plumbing lines in existing homes. It is very labor intensive and takes quite a long time to do. Granted, there are new products that eliminate the need for copper, but even so, it's a lot of work. Just from gut feel, I would be very surprised if you broke even on the endeavor in 10 years. It does depend on the house of course, but the labor part still has to be considered.

I'd say you're better off spending a day insulating the water heater and the exposed piping first.

6. Mar 26, 2006

### Clausius2

My advice is to take a look at the book "Fundamentals of Heat Transfer" of Incropera and see how they treat the flux of heat through pipelines with T=T(x). In general q''=q''(x) and h=h(x). I don't have the book here but I remind to have made this calculation several times some time ago.

7. Mar 26, 2006

### Ouabache

Q_Goest, excellent input. You reminded me there are transient and steady-state parts to this process. To fully appreciate your comments, I'll need to read up a little on convective heat transfer coeficients and how they're applied. Doing a quick dimensional analysis, your Q appears to be in Btu/hr. I will be sure to include a calculation for the transient condition too.

Averagesupernova, they actually do have new foam insulation around all the exposed pipe, but there is also a significant run of pipe that is not insulated between floors. To more accurately calculate the heat loss, I would need to include the thermal conductivity (k) of the foam, in the calculation for that section of the run.

FredGarvin, good thought, however let me elaborate on their system. They have a relatively new water heater installation. With the old heater, the pipe went straight up to some taps and then across about 50 ft (through cooler ambient temperature room) to 3 more taps. The new water heater was installed in a different location of basement, actually directly below the 3 periferal taps. When they plumbed it, they just ran pipe across the basement to where the old heater was. If you follow me, the run of pipe now goes twice as far to reach the 3 periferal taps. The run across the basement is insulated with foam.

Clausius2, i will followup on your thoughts of Heat Flux. It sounds similar to the one described at the bottom of my reference

Last edited: Mar 27, 2006
8. Mar 26, 2006

### Ouabache

I appreciate your line of thought..The input and output temperatures are still unknown, I plugged in some typical values found on net, for a residential system. They indicated 130 deg F in the tank and 120 deg at the tap.

Back to your discussion, your approach does simplify the analysis. As you point out, the input and output temperatures may be measured. As both Q_Goest and you assert (and something I hadn't considered), this process may be broken into parts.
(i) The *energy loss of the standing water in the pipes between usage (with ambient daytime temperature; nighttime 45 deg, daytime 60 deg). The temperature differential for this water volume, could be as much as 85 deg (130deg cooled to 45 deg F).
(ii) when tap is initially opened, there is energy loss during the transient period to warm up the pipes.
(iii) energy loss during the steady-state, where pipes are warmed up to a maximum temperature and hot water is flowing out the tap.

Parts (i) and (ii) will vary depending on how often hot water is drawn from the pipe. At the same time, there is a variance of ambient air temperature throughout the day. Some reasonable approximations may need to be made (for these variances), to make this calculation managable.

Summarizing my understanding of this approach; by measuring water temperature at the input and output over time, we can make a reasonable estimate for the following losses in an annual period: (a) determine the energy lost between usage of hot water, (b) energy lost in warming up the system (transient), (c) energy lost while the hot water is running (steady-state).

*here using energy and heat loosely. The actual relationship between the two is noted here.

Last edited: Mar 27, 2006
9. Mar 26, 2006

### RVBuckeye

10. Mar 27, 2006

### quark

11. Mar 27, 2006

### FredGarvin

Incropera and Dewitt's book is very good. I have two copies of it myself.

Ouabache, It does indeed sound like you wouldn't have too hard of a time replumbing that. If the new one is directly below the existing vertical run of piping, you could eliminate that new basement run in a half a day. I can now see why you are persuing this.

12. Mar 27, 2006

### Q_Goest

argh... and I only have one copy of that book... How can I keep up with you guys? Gonna have to go buy another copy now... <sigh>

13. Mar 27, 2006

### FredGarvin

I got a second copy by complete luck. I bought a house from a young guy who had just finished school. He left a bunch of his text books in the house after I had closed on it. I called him quite a few times to give them to him, but he never called me back. So....I kinda kept it.

14. Mar 27, 2006

### Clausius2

Somebody told me some time ago that at the end of the game an engineer is a guy who knows how to find how to do something in some book. You know, now I am far away from home and away from my notes, instead of telling this guy how to calculate it, I only diffusely remind that this stuff is made in some book. That's all the reminder left in your brain after studying five years of engineering, guys.:rofl:

15. Mar 28, 2006

### davidddt

As your problem seems to be a real life situation then forget fancy heat loss calculations. The heat loss in real life will be from the total amount of water in the pipe run between the heater and the tap.

Calculate the volume of water in the pipe and how much energy is required to bring it up to a useable heat level from a mean level.

Every time the tap is turned on will be the static heat loss if averaged out over time. So turning on the tap equates to the energy needed to heat that volume of water. Over a period of time, a few weeks, you will find that this can be a lot of energy lost. The solution is to have as short a run as possible. Instant heaters at the point of use can sometimes be a better long term cost effective solution.

DD

16. Mar 28, 2006

### FredGarvin

It only goes downhill from there my friend. It is very tough to be a master of all things in this business. I work at not forgetting everything, but just most everything.

17. Mar 29, 2006

### Ouabache

Okay, i am looking at a library copy of the highly acclaimed Fundamentals of Heat Transfer by Incropera and DeWitt. Hmmmm I notice this was written at a fine institution (and my alma mater Purdue Univ.)
My forté is EE not ME, so I am only grasping rudimentary concepts here.. This book reminds me of an Electromagnetics text (I have two copies of), by the late JD Kraus.

Anyhow, the examples I sought were the ones dealing with a cylindrical tube of some length. I see under steady state conduction of heat, they use the equation I previously posted here ref1 which i called "Heat Transfer Across Length of Cylindrical Tubing" (which is also equation 3.26 in Incropera).

I didn't notice anything applicable, for the transient conduction of heat.

Then under internal flow & convection (chap 8 Incropera), they calculate an average convection heat transfer coeffient $\bar{h}$ which I have just posted here ref2

There's a very nice table towards the end of the book, giving various thermophysical properties of metals. Under copper I found thermal conductivity k = 401 [W/(m-deg K)] and the conversion k=1 [W/(m-deg K) = 0.57782 [Btu/(hr-ft- deg F)]. So I have a useful value of $k_{Cu} \approx$ 231.71 [Btu/(hr-ft- deg F)]

So this reference has me off and running!!

Last edited: Mar 29, 2006
18. Mar 30, 2006

### Cyrus

Ouch,

this problem looks hard because this is not a steady state analysis.

from what I get out of this problem, your concerned with the heat loss from a cold start until steady state value. So you can't use $$Q=\dot{m}C_{p,v}(\Delta T)$$ as a steady state senario.

You see, for your refrences, all the problems are steady state. Your problem is not steady state, your equations wont apply. You would need to have a function that gives you the change in temperature as a function of some properties, because your pipe will start at ambient temperature, and change until it reaches steady state.

Last edited: Mar 30, 2006
19. Apr 1, 2006

### Ouabache

thanks for thinking this through.. Well steady-state energy loss is a part of the whole calculation. In my earlier post I tried to group the 3 parts of this problem, that I've learned from our knowledgable membership.

That's right. I need to bear that in mind after finding the other 2 parts, that for completeness, I should find the transient loss too.

20. Apr 4, 2006

### Artman

If the fixtures are in a floor above the water heater, consider a thermosiphon type hot water return system. The fixtures have to be above the water heater and a pipe is then run from the last fixture back down to the water heater with a check valve in the direction of the water heater to keep water from going back to the fixture when the faucet is opened. The hot water will rise and as it sits in the pipe and cools, it will drop down in the pipe returning to the water heater. Requires no pumping energy and the water should always be hot at the last fixture.