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Water heating theory question

Water heating theory question

Water heating theory question

(OP)
Hi all,

I have a water heating theory question that is hurting my head a bit.

Scenario:
1. Say I have two water storage tanks - one of 10L one of 5L.
2. The tanks have a secondary sealed surround - filled with PUF insulation so the only method of cooling is via conduction (rough assumption but go with it for the question).
3. Both tanks are to be maintained using a heating element cutting in at 93degC and out at 96degC.

I know by Q = m*Cp*deltaT that the energy into the 10L tank is double that into the 5L tank. The 10L and 5L tanks roughly transfer HEAT at the same rate by their design - so Q'1 = Q'2.

So the 5L tank cools twice as fast as the 10L one because it has half the stored energy Q' = Q/deltaT.

Essentially it seems that the more efficient option is to heat a 10L tank as it takes the same amount of energy to keep 5L at 94.5degC as it does a 10L tank?
(I know that the intial heating of the tanks to 94.5 will take more energy in the 10L but thereafter the energy in to the tanks is common)

Any assistance to my Friday brain would be appreciated.
Thanks
James

RE: Water heating theory question

It sort of depends on what you consider by "efficient".

Assume PERFECT insulation.  You put energy in once to heat of the water in the tank, then the only future energy needed is for heating the make-up water that replaces water used.

So in the perfect insulation case, efficiency is directly m*Cp*Dt/Input.

If you consider losses from the tank:

efficiency = ((m*Cp*Dt)-losses)/Input

If you go with your assumption the losses from both tanks are the same, then there is no difference in efficiency (at steady state).

If you consider the initial heat-up, then the larger tank is more efficient because m is larger.

RE: Water heating theory question

(OP)
Excellent quick reply MintJulep - brain pain easing already.

Thanks!

RE: Water heating theory question

How about this.  The heat loss from a tank is decribed by the heat transfer equation Q = U * A * deltaT.  which is lost energy = heat transfer coeficient or insulating value * the tank AREA * the difference in temperature between inside and outside.  If both tanks have the same insulation and differential temperatures, the the heat loss is related to the surface area of the tanks.

If the 10 liter is a sphere container and the 5 liter a veseel with a 7.5 lenght/diameter tank, the 5 liter will lose heat faster!

RE: Water heating theory question

5 liter will have higher ratio of surface/volume than the 10 L tank and will loose heat quicker.

RE: Water heating theory question

Assuming both are the same shape- the temperature in the 5L will drop faster with no supplemental heating. But the surface area of the 10L is higher- so it will lose more heat than the 5L. It will take more energy to keep the 10L hot than the 5L (due to the higher surface area)- but the efficiency of energy spent per L or water kept hot will be higher for the 10L due to the lower surface area:volume ratio.

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