EcoDrain Payback Calculation Challenge

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The crowdsouced calculations are wrong. We had offered our input and explanation of how calculations are done prior to this challenge. The problem with this post is that to someone who quickly glances at it, it has a air of credibility, it is calculated by an engineer who is also an MBA (so am I), on a authoritative site about building, and it uses terms like efficiency, btus and energy transferred. It looks legit even though it is just unchecked calculations by someone who is not a subject matter expert. If we were in university, this type of work would have to submitted to a professor who would review it. Few people get everything right the first time they attempt it and it would be abnormal to expect a 100% score on all assignments. In the real world, it seems you can get published on the internet on the first attempt with no review.

The engineer has correctly realized that energy from the drain water = energy in the makeup water. The basic equation for heating water is Q = m*Cp*dt, where Q is the energy required to heat water, m is the mass of water being heated, Cp is the specific heat of water, and dT is the temperature rise of the water being heated (Tout - Tin).

Qhot = mhot * Cp *(Tdrainhot - Tdrainout)

Qcold =mcold *Cp *(Tpreheatedout - Tcold)

Now he has made some simplifications in combining the equations Qhot =Qcold , he has deleted Cp from both sides and also used the volumes of water rather than masses. These are ok because we are interested in the ratios, so the conversion from volume to mass would cancel out anyways. He makes the following two mistakes:

1. He assumes that Tdrainout = Tpreheatedout, i.e. that the temperature of the drain water leaving the heat exchanger will be the same as the temperature of the preheated water leaving the cold side of the heat exchanger. But there is no reason to assume this. It is like he is pre-solving the problem based on a guess. Without this assumption, and put together correctly, he would end up with one equation and two unknowns, which leaves infinite possible solutions:

Qhot = Qcold

mhot * Cp *(Tdrainhot - Tdrainout) = mcold *Cp * (Tpreheatedout - Tcold)

when we make the cancellations and add the values we know for temperature, we get

75 gallons * (90 - Tdrainout) = 60 * (Tpreheatedout - 55), notice that since Tdrainout does not equal Tpreheatout, we have one equation and two unknowns.

What you end up with is 75/60 * (90 - Tdrainout) = Tpreheatedout -55.

2. He does not understand how efficiency is used in the context of a heat exchanger. He is assuming that 80% efficiency means that 80% of the heat lost on the drain side is transferred to the cold water side. He has taken the drain side energy equation Qhot and multiplied by .80 to account for this:

75/60 * .8 (90 - Tdrainout) = Tpreheatedout - 55.

Since 75/60*.8 = 1 the equation works out to:

90 - Tdrainout = Tpreheatedout - 55 which is equivalent to 90 + 55 = Tdrainout + Tpreheatedout, i.e.

145 = Tdrainout + Tpreheatedout

A casual observer will note that this is still one equation with two unknowns, and an infinite number of solutions. We know that Tpreheatedout must be between 55F and 90F, but with the information as presented, we have no idea what Tpreheatedout actually is.

Efficiency in the context of a heat exchanger takes into account how much heat is lost or gained in the first place, and also how much of it is transferred. The engineer has ignored the first part, how much heat is actually lost, hoping to solve this with Qhot = Qcold, which is not possible.
 

Heat exchanger efficiency is measured through experimentation, and it is referred to as the actual temperature gain (or loss) divided by the total possible temperature gain (or loss).
 

In this case, the cold water is entering the heat exchanger at 55F, and the drain water is entering at 90F. So the maximum possible outlet temperature on the cold water side is 90F, if the heat exchanger is 100% efficient. The minimum outlet temperature on the cold side is 55, if the efficiency is 0. Any efficiency in between will give a temperature rise in between the two limits.

So Efficiency (e) = (Tpreheatedout - Tcold) / (Tdrainhot - Tcold)

Note also that efficiency on the drain side does not necessarily equal efficiency on the cold side since the flow rates (or volumes) may be different.

When the efficiency is given, as it would be in this type of problem, since it is measured experimentally, it can be used to find the preheated outlet temperature.

This can then be used in the equation Qcold = m Cp dT, to solve for Qcold, which is the energy transferred by the heat exchanger to the cold water. The water heater does not need to heat this water so it energy saved.  

This method can be used to solve all of the challenge questions. It is also on our website FAQ section. For natural gas calculations, the energy calculated using this method must be adjusted to account for the lower efficiency of the gas water heater. Electric water heaters have heating elements directly in the cold water. Essentially all of the energy sent to the electric water heater ends up in the water. A gas heater, on the other hand, burns gas, and some of the energy leaves the heater in the form of hot gases. Gas water heaters are rated with an energy factor.

Energy Factor (EF): A measure of water heater overall efficiency, is the ratio of useful energy output from the water heater to the total amount of energy delivered to the water heater. According to Energy Star, new gas water heaters must have an energy factor above 67%. To get btus consumed by a gas water, take btus consumed in an electric water heater and divided by the energy factor.