Calculation of Power

What states does 50% NaOH change to/from at 12°C?
What are the dimensions of the tank?
How much insulation, and of what type, is on the tank?
What temperature is the 50% NaOH when first put into the tank?

You'll never get answers to questions of this nature until you understand what information would be required to solve the problem.

It sounds like homework, although you claim to have a job.

You should be able to find the thermal properties of the materials on the net, and the surface area of the "30 cum" tank. Don't forget to consider if it sits on the floor or on legs that would isolate it from the floor and increase the radiated area. Remember radiation, convection and conduction will all come into play here.

Why don't you just put your tank inside? Or put a power meter on the heater?

Did you ever get anyone to give you the other heater information you requested? Did you try TFE sleeve, as I suggested????

Assuming the insulation is perfect and the NaOH is already at 15degC, the minimum is zero.

Naaahh! The NaOH is at 20 deg C and you can even get energy out from the system!

Waste energy to be used in the process or if in doubt ask around for cooling systems!

There had to be a catch...

I suppose 30 cub.m=3[height]*10[surface] -a swimming pool of 3*2*5 cub.m

50% NaOH solution dens=1.529 kg/l at 15 dgr.C spec.heat=3.234 kJ/kg/dgr.C

I used the formula:

Qheat=Vol*dens*spec.heat*DT Pheat=Qheat/time

time=3600 sec [1 hour]

Pcool=hc*A*DT

hc=Air convection heat transfer=2.5 w/sq.m/dgr.C

A=10 sq.m DT=15-5=10 dgr.C

So, you need 412 kw to heat 30 cub.m NaOH 50% from 5 to 15 dgr.C within 1 hour and 0.25 kw to maintain the temperature.

The phase change doesn't seem to be accounted for, but other than that, a reasonable calculation.

If the walls are not insulated the convection coefficient is a lot higher. During heating there are losses so that the linear assumption is wrong.

If the walls are not insulated the convection heat flow is higher, but not the convection coefficient, that is per °C in #6. But that's not quite right, for natural convection heat flow usually taken as varying as ΔT^1.25.

But the question doesn't ask about warming the contents from 5° - 15°C, only about maintaining 15°C. The nature of the tank contents doesn't come into it, just heat losses.

And where is the 10m² from? A sphere of 30m³ (minimum surface area) has radius 1.9 m and area 47m².

I did some searching on 15deg. C and 50% solution of NaOH and the closest thing I could find is the Specific Gravity is 15.5%. With further reading, maintaining Specific Gravity for this solution makes for easier readings of concentration. The freezing point is also at 12.1 deg. C

Since there may be a desire to maintain the solution close to it's Specific Gravity that this may have something to do with minimizing how much stirring would be needed since the solute becomes neutrally buoyant with the solvent at that temperature.

Upon further reading, I also found that with an open tank, contact with CO2 creates Sodium Carbonate which according to a supplier of 50% NaOH is undesirable. Therefore, it is most likely that the tank is sealed.

Additionally, the solution is typically stored/in contact with carbon steel (with some corrosive action present), stainless steel, polyethylene, PVC, and a number of other polymers.

Sounds like the customer wants to spend the smallest amount of money necessary on an immersion heater for a tank sized roughly 30,000 liters or just under 8,000 gallons. If this customer is wise, the tank will also be insulated already or done after the fact.

Go and do your homework yourself...you are just lazy..........

How rude.

You are right, nick name, the total losses through convection cooling could be up to 1.3 kw if we take all around area. However it is less than 0.4% of heating required power. The radiation is negligible and conduction-if the surround earth is at least 15 dgr.C it is negligible also.

You are furnishing the wrong information to solve the problem. What is important is the shape of the tank (dimensions and thickness), its thermal conductivity, whether the tank is open on the top, etc. I don't believe the contents will have too much effect on the answer because at equilibrium all that is important is the amount of heat escaping.

Thank you, Rixter, for your remarks. You are right: it is a wrong information! It is not only wrong but it is very dangerous!

I put this caustic soda solution -as I already said-in a swimming pool! Do you think someone dare to swim in?

The o.p. -as usually-did not give us a big information-just what he think to be essential.

Since I don't know the NaOH latent heat of melting I neglected it-as drbobwoollery noted.

The Heat required is W x Sp. Heat x Temp. Rise - where W is the Solution weight in Kgs, Sp.Heat in KCal/Deg.C, Temp.Rise in Deg.C This is the Heat required in KCal and this amount of heat is to be added. The Elec. Power demand will depend upon in how much time the Heat is to be added to the solution. However 860 KCal/Hr will be released by 1 KW.

This is with out any Radiation Loss of HEAT ENERGY, and the grade of the INSULATION of the Container, will give how much heat is radiated - this heat is to be compensated.

DHAYANANDHAN.S

But there is no temperature rise the way the question is put. See #14

Really!!

You're more interested in the power requirements to heat a tank of NaOH and really don't care that the NaOH isn't at the proper temperature now? Why does the power matter if your NaOH needs to be at 15 C?

In my opinion, Why does the NaOH have to be at 15 C and what does it "change" to at 12 C.

HMMM!! all the information I've read mentions nothing about "naoh solution changes state at 12 deg c".

We would store our NaOH in a dual walled tank that was usually at approx. 30 C. and had little to no issues with the NaOH. Though we did have to stir it to keep the solution mixed.

OH... And once you get the NaOH up to temperature, your power outlay diminishes greatly, it's easier to keep something at a constant temp than to try to heat up a full tank of cool liquid.

Yep.. Sounds like homework!

BA

By the way, if the NaOH 50% relative enthalpy at 12 dgr.C is 130 btu/lb=300 kJ/kg you need approximately 9-10 hours to melt it , supplying 412 kW.

Sorry for the trouble everyone. It does seem that the technicals were not appropriate. The volume of the tank is 30 cum and it is cylindrical in shape with a dia of 3000mm. The tank is filled with 50% naoh solution and my aim is to maintain the temperature of the tank at 15deg c. During nights, the outside temperature msy drop down to 5 deg c. What should be my kw rating?? This heating or maintaining temp is making me crazy.

It would have been better to provide a link to the specs of the tank and perhaps showed a pic of the tank's level of exposure to it's surroundings. If you can't provide a picture, then describe in enough detail so the reader can imagine with less guess work. Certainly there are many variables which affect the efficiency of maintaining tank temp by at least 40% if not more! (Especially if the tank is single walled stainless and gets rained on)

You still haven't furnished enough information. Please reread post 1.

There are not enough data, as Tornado said.

We need to know:

1)If the tank is underground or in open air.

2)If the tank is underground what is the earth thermal conductivity.

3)The upper side is it in open air?

4)The tank wall is insulated and how.

If the tank is standing all in open air of 5dgr.C and it is not insulated at all then the minimum power required to maintain 15 dgr.C in the tank will be approx. 1.2 kw.

However, you'll need 400 kw in order to melt it in 10 hours.