Water Heating Process, what is more productive?

Question

Suppose we have 5kWa (5000J) or around 1190cal to heat water.

a) An electric resistor (thin thermal wire) of 10cm long, installed in the exact center of a plastic pipe, 5cm diameter, 30cm long, flowing 200ml of water per second, it needs to increase water temp from 20 to 40°C at the exit of the pipe.

b) A copper pipe, 3 meter long, 5mm internal diameter, heated by a strong electric current, low voltage, same final power consumed by the item (a) above, 200ml of water entering the pipe per second, approximated flow speed is 1 meter per second, so water takes 3 seconds to heat +20°C and exit the heating pipe.

The point of discussion here is that in item (a) above, the small contact area of super heating element will boil the little water that contacts it and air bubbles will be created around the short and thin heating element. Those air bubbles will create problems for better contact of next little water to contact the heating element, creating an isolation and not 100% of the water flowing the plastic pipe will be contacting this hot filament. Water boils at 100°C, the other water that crosses the plastic pipe mm away from the filament will heat less, the final temperature will be a warm water at the exit.

In the item (b) above, the copper pipe will have lots of contact with the water inside, it doesn't need to boil the water, different from item (a) where super heated water needs to mix with cold water, water is kept in contact with the copper warm pipe for much longer time during the flow, so the overall temperature of the copper pipe doesn't need to be super hot.

What do you think? A short heating element in the middle of lot of water with super power consumed works better than a long serpentine with much lower temperature all along?

I am not expecting a conclusive answer, but some points of view, please.

Answer 1

Hi there, that’s a nice question you ask in my opinion direct water heater load control-estimating program effectiveness using an engineering model

Gustafson, M.W.; Baylor, J.S.; Epstein, G.

Power Systems, IEEE Transactions on

Volume 8, Issue 1, Feb 1993 Page(s):137 - 143

Digital Object Identifier   10.1109/59.221259

A method is described for evaluating the potential effectiveness of a direct water heater load control program. Developed through a comprehensive load research program coupled with engineering insights into the energy use of residential hot water systems, this method results in an algorithm that has been utilized to evaluate the potential for load control on three electric utility systems in the western US. Variables to be monitored through the load research are presented and a procedure that will allow the system planner to determine if such a program will be cost-effective and successful compared to a program developed through a more traditional pilot or demonstration approach is developed. It is shown how the same methodology can be used for determining a procedure for the dispatcher to properly initiate and terminate a load control program without hot water recovery problems. If you want more information visit the website which is given below it helps you a lot I search it on a net.

Answer 2

Something is not right with your numbers. The required heat to change the temp of 200 ml/sec by 20 C is more like 16.7 kW. Secondly in (b), that pipe diameter and flow rate does not result in a 1 m/s flow velocity. More like 10 m/s.

Anyway, one other consideration is natural convection. The bubbles will tend to move upward and enhances the mixing of the liquid. Heat transfer rates are expected to be much higher.

But there are other practical issues you need to consider. For one, the water may affect the electrical conduction in the wire. The high temperatures in the copper wire may also lead to faster material degradation.

I think the best option would be to use the heated pipes, but slow down the rate of flow.

Answer 3

First, Joules and calories are measures of energy, not power.  I'll assume you have 5,000 Watts to play around with.

(a) 200mL of water through a 5cm diameter pipe means the water is flowing at ~10cm/sec.  That means the water is only in contact with the resistor for 1 second.  It takes 4.184J per mL of water to heat it up one degree, so you need to put ~17kW of power through that resistor.  That power requirement is three times what you say you have.

Also, using the 120V coming out of a normal electrical socket, that's a current of 140A, way beyond most home circuit breakers.  I don't think many thin wires can handle the power or the current.  Why not use something thick, like the heating coil of an electric stove?

Heat doesn't diffuse through water that quickly.  You'll end up with a little water near the center of the pipe being very hot and the rest at 20°C.  I like your hunch about the bubbles forming near the wire preventing the heating of the rest of the water.

(b) Your calculation of the flow speed is correct, but you will only need one-third the power since the water will be in contact with the pipe for three times as long as the thermal wire from (a).  So, your power requirement is only 5.6kW.  Still beyond what you specified, but that's a much more manageable difference.  The problem is that you'll need more power than that because the pipe will also heat the air outside of it in addition to the water inside of it.

Also, the pipe will be too hot to touch, so any plumber should be warned.

The easiest method is to pool the water in a tank (or, if you want to keep the water moving, a long pipe with a large diameter) and heat it there.  That way you have a longer heating time and thus much lower power requirements.

Answer 4

I hate the idea of trying to heat water using the electrical resistance of a copper pipe.  You are going to have to draw a huge amount of current.   You could of course modify your design by adding having heat tracing wires around the pipe, all enclosed by insulation.  This would be effective.

But power is power.  Put in the same amount of watts, and the same flow, and barring some escape route for the heat, you will end up with the same temperature.

I assume that in both cases the length to pipe diameter is such that you will get sufficient mixing to ensure that any "bubble" will cavitate pretty quickly.  Given the high flow rate and the presumptive mixing,  I am also not sure if you will get any bubbles.

Your long serpentine copper tubing sounds just like a tried and true heat exchanger.  Proven technology.  Has worked for the past 150 years.  Tends to get a little gunky on the surface.  The super hot filament might avoid this problem and be self-cleaning.