Maths / physics / heat transfer question with assumptions allowable.

I have a pond, sunken in the ground, outside in the UK, contains
around 20 cubic metres of water. I was toying with the idea of
preventing it, and the above-ground pumps which supply its' waterfall,
freezing in winter by plumbing a water heater element, of the type
used in a domestic hot water cylinder, into the pumps' pipework,
controlling it thermostatically.

I'm technically very able in terms of being able to physically put the
components together, but would like someone else's opinion as to
wether the physics would add up - would the amount of electricity that
would need to be supplied to prevent freezing be ridiculously high?

I'm looking for a rough calculation for a time period where the
ambient temperature is say, a constant -5 degrees centigrade.

I realise there are a large number of variables involved. In any
calculation feel free to make as many assumptions as you like, but
please state what the major ones are. I doubt if anyone would feel
confident enough to give any specific figures based on the information
given, so rough, napkin-style calculations and brief comments would be
perfectly acceptable as an answer.

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Clarification of Question by gan-ga on 09 Jan 2003 17:54 PST

Thanks for the points you raise, racecar-ga..

Conditions can be taken as calm, and surface area would be roughly 20 square metres.

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Clarification of Question by gan-ga on 09 Jan 2003 22:35 PST

Ballpark figure for cost of electricity: 7 pence per kWh

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Request for Question Clarification by hedgie-ga on 10 Jan 2003 06:54 PST

Hi gan
          Jolly nice of you to allow assumptions.
          
Still want to ask few questions:

1) will you (are you willing to) cover the pond?

 The cost of heating the volume of water 
 from some initial -5 is negligeable. Rather, we have a steady
 heat flow between the pond at +5 to ambient at -5.
 The +5 is a safety margin over the freezing point.


 Your heat flow (which you are supporting by your heater) depends on
this temperature difference and on the INSULATION (which is the quality
of cover, mostly, in your case).

2) are you willing to consider the Heat Pump ? 
In your - fairly mild - climate, it will lower the cost of
operation, but cost a bit more to put in.

I can do some guesstimates on all 4 combinations but
prefer some feedback first.

 3) Did you consider a garden-pump (with fountain)?
It heats the pond, while providing noce sound and benefit to fish
(particularly with biological filter)..

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Clarification of Question by gan-ga on 10 Jan 2003 17:24 PST

Hello Hedgie,

Although I can see the benefits of covering the pond, it's something
I'm trying to avoid from an aesthetic point of view. The pumping /
waterfall system is existing and pretty firmly plumbed in, so again
something I'd pretty much want to keep the same, apart from adding a
chamber to accomodate the immersion heater.

> are you willing to consider the Heat Pump ?

.. I'm intrigued as to what this would entail .. guessing, is it some
kind of peltier effect product or am I wide of the mark in what I'm
imagining there? If it were easy to hide somehow then yes, that would
be interesting.

cipher17-ga, take your point about the extra energy needed to pass
through the solid - to - liquid phase 'barrier', but this is, I guess,
where things become a little blurred .. the ambient temperature might
be taken as say, -5degrees, but in actuality it's only maybe the top
5cm of the pond which freezes to solid ice - so I guess there must
exist an increasing temperature gradient as depth increases. Yesterday
I repaired the air pump, which has an air stone at the bottom of the
pond, the stream of bubbles from which has the effect of causing a
rising column of water under the ice - I noticed this caused the ice
to melt in a circular pattern of diameter approx 2 metres on the
surface where the rising column of water & bubbles hit.

Not as easy to apply a simple mathematical model as I'd hoped, I'm
beginning to suspect, although, still, I'd accept guesstimates in the
answer box based on even drastic assumptions to simplify things, as
long as they're stated.

OK gan,

 This is one of those cases when mathematical formula can
 be misleading, since circumstances and assumptions about
 circumstances are more important then numbers.

 The formula of heat loss is essentially

   J (heat flow in Joules/second) =  Z * ( Te - Tp)

   where  Te is temperature of environment  (e.g. -5 C)
          Tp is temperature of pond (let's say +5C)

	  and Z is the (unknown) Insulation Barrier,

	  basically   length* heat-conductivity / Area

	  That neglects evaporation, which should be small
	  and effect of wind, which may be significant.

	  Strictly it applies to a simple case of

	  Te   | slab |  Tp
	earth  |      |  water

	       where slab would be  some thermal insulation
	       between the water (p) and the earth (e).

 But, pond is built, be it gunite or liner,  and we are not adding
 any insulation, right?

 We need to consider other things then exact numbers though.
  Formula is still useful, since it provides some feel for cost
dependence
  on the delta.T = Tp - Te, as function of time and height (we can
take averages, since formula is linear).

  This is the connection with cost:
  
  You will pay  average = J * 1.5  pound / month
      and electric current (in amps) draw will be about
                        J/ 220 (assuming you have 220 voltage in UK).

  To get numbers into perspective, let's use an eaxmple:
 I live in climate with no danger of freezing, but I
  was leaving fountain on continuousl,y during the winter, for the
benefit of the fish.
  The pump was large 5A (at 110 V) and cost was $300 /month. Too high.
 I switched to a very small pump (.5A) and cost dropped to $30 /month.
 Acceptable and fish still survived.
  
  So, even if you see the fountain as separate, it's contribution to
the energy (and cost) budget is essential.
Ergo, First experiment is to determine the power (wattage = Volts *
amps)
  of the pump and then leave the pump on all the time.
 Chances are that pond will not freeze and you will feel the cost.
  It will give you some clue of the Z value of your pond.
  
  Things will be different if you use Heat Pump or floating heaters.
  
Please look at pond FAQ

http://www.absoluteponds.com/faq.htm
particularly  Q3, Q 15 to 17
particularly Q 17 has interesting answer:
We have pond de-icers that are floating heating elements that keep an
area of the pond's surface from freezing.
  This allows toxic gases to escape from the pond as things continue
to decompose during the winter as well
   as provide an "opening" for oxygen to enter the pond.
 (We also have actual "heaters" that will heat the water but they are
quite expensive and expensive to operate).
 

 Now the heat pump issue. First 'what is it?'. 
 It is an 'inverse air conditioner' . 
 The cost of operation is no longer proportional
 on the amount of heat needed. Here is a picture of one:
 http://www.energyright.com/heatpump/
 and here is the rationale of using one:
 http://energyoutlet.com/res/heatpump/how.html

  'Heat pump' is also a search term; There is lot of data on the web.
	    
	    Here is formula for Heat Pump. Note the difference:
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heatpump.html	 



      Now, this kind of problem is likely to need clarification of the
      answer. Feel free to ask for details or explanations.

      
      hedgie

Two additional pieces of information would make a more well-defined
answer possible.  First, the surface area of the pond (more important
than the volume, actually), and second whether the conditions are calm
or windy.  Wind on the pond will increase heat trasfer in the same way
that blowing on a spoonful of soup cools it down.

This is a very interesting problem, and I am very tempted to try it,
but unfortunately I have forgotten a lot of my Heat Transfer...

Here's a good Heat transfer book for reference.
http://web.mit.edu/lienhard/www/ahtt.html

In your case, neglecting heat losses (which we can't, really), the
basic equation to use would be:
Q = m Cp DT

Assuming you want to heat all the contents of the pond, 
taking density of water to be 1000 kg/m3, neglecting the temperature
dependence of density,

Mass, m = 20 m3 * 1000 kg/m3 = 20000 kg
Cp = 4.2 J/kg.C
DT = Tf - T0 = 4 - (-5) = 9°C (assuming you want to bring it to 4
degrees, just to make sure; you can use salt to effect freezing point
depression but that will corrode your pump)

Q = 756000 J = 0.21 kWh.

I would imagine you add the compensation you have to make for heat
losses due to convection (whip out that heat transfer book and
calculate some dimensionless numbers), and there's be your figure.
It's a bit simplistic though... so if anyone can do better, I'd be
interested to know too.

Thanks for your input zhiwenchong-ga, interesting..

I wouldn't be able to add salt, true - probably wouldn't do the pumps
any good as you mention but also, no good for the fish.

Looking at the calculation you suggest so far, maybe the idea isn't so
hare-brained as I'd first suspected it may be, looking purely at the
quantity of energy needed to effect a suitable static raise in temp.
But balancing the rate of heat loss against the energy needed to raise
the temperature through those 9 degrees odd, might still make the cost
prohibitive I guess? I guess there's also the possibility that if the
rate of heat loss were so great, the type of domestic immersion heater
I'm envisaging would not be able to supply enough energy to 'keep up'.

Thanks for the link to the heat transfer text, It might be a little
'over my head' but I'll certainly go & have a check.

Meanwhile, I'd still be very interested if there are any researchers
comfortable in this area, who might be able to fill the answer box.

as per zhiwenchong's calculations, it will only come to 4 degrees of
ice.. you would need to supply more heat (called latent heat) to bring
it from 4 degrees of ice to 4 degrees of water.
PS:Latent heat of liquefaction of ice is 80 kcal/kg at 0 degree and 1
atmosphere.

Thanks, cipher-ga!
That was a really really egregious error... not including latent heat.
I should be ashamed of myself!

Heat due to latent heat of fusion for water would be:
Q = m L
  = 20000 kg * 336000 J/kg
  = 6720000000 J = 1,866.6 kWh

Using the approximate relation

Q = 0.0008*(rho)*(C_p)*u*(T_s - T_a)

where 
Q = surface heat flux (W/m^2)
rho = density of air (~1 kg/m^3)
C_p = heat capacity of air at constant pressure (~1000 J/kg K)
u = wind speed (~1 m/s)
T_s = surface temperature (5C)
T_a = air temperature (-5C)

and the dimensionless constant .0008 is valid for stable conditions,

I get 8 W/m^2, which may as well be rounded to 10 since the wind speed
1 m/s was pulled straight from my nether regions.

This means about 200 W is required, or about 150 kWh per month, or
about 10 pounds per month.

I got the equation from Atmosphere-Ocean Dynamics by Adrian E. Gill. 
It's meant to be used with respect to the ocean, and there may be an
additional effect due to evaporation which hasn't been considered.  So
this answer is very ballpark.  More of a guess I suppose.

Hi gan! I have a pond in the Chicago USA area which I judge to be
about one third the volume of yours, which I calculated at about 4000
U.S.gallons (not imperial gallons). The Chicago climate is likely much
colder in winter than yours. Ice forms 3 to 5 inches thick each
winter. I have fish in my pond which have survived five winters so
far. To get to the point: By my estimation (I cannot provide the
mathematical calculations others have) it would be prohibitively
expensive to keep my pond ice-free. I have a pond heater which merely
produces a small opening in the ice to allow oxygen exchange, and it
adds $50 US to my monthly electric bill in winter. I estimate it would
cost me thousands of US dollars per month to keep my pond above
freezing with no apparent benefit. Guess what? I won't. Even if your
climate is milder, you have a much larger pond. Hope this helps.
knowitall22-ga

zhiwenchong, I don't think you should beat yourself up for ignoring
latent heat - the requirement isn't to raise the temperature through
the freezing point, it's to stop it getting that low in the first
place. Latent heat's working in our favour to help prevent freezing.

Trying to calculate the rate of heat loss is a fascinating problem,
especialy if you factor in the radiative losses, but I wonder if a
different approach could be taken? Part of the requirement is to stop
the pump freezing - I wonder what sort of numbers we'd get if we took
the pump's flow rate and worked out the cost of heating an hour's flow
by 5 degrees (from 0-5 degC)? If the waterfall could be kept going,
the agitation would probably be enough to prevent the pond freezing
over completely.

Ian G.

iang-ga, yes that could well be a way forward.. as long as the pumps
and pipes don't freeze solid, & there is partial prevention of ice on
the pond itself, things should be fine. Heaters on the inlet pipes
where they leave the pond?

Pumps are:

2 x Stuart Turner 904's

http://www.stuart-turner.co.uk/products/industrial/cent_range.htm

Each one has it's own 2.5cm diameter feed pipe; guess I could use 2
small heaters as opposed to 1 large one.

Head would be about 2.5 metres.

When I first saw this, it scared me even without calculations. I'd
hate to pay for it.  Have you given any thought to putting a small,
clear plastic bubble over a portion of the pond?  I know you don't
want to cover it all, but the greenhouse effect under it might add a
reasonable amount of heat, which your pump would distribute.

I'd realy appreciate someone checking my math (I have a habit of
putting decimal points in the wrong place!) but I work out the cost of
running the heater(s) as about £1.10/hour/degree C raise in
temperature. That's assuming no heat losses and 100% efficiency in
transfering the heat from the heater element to the water. To get
anywhere near that, I think you'd need some sort of header tank
arrangement - heating the water "on the fly" would need a more
sophisticated heat exchanger.

Ian G.