1
$\begingroup$

The dimensions of the container for the water is 6.06m by 2.44m the depth of water is at 1m equating to 14.786m^3 or 14,786 ℓ of water.

The sytems is a closed system with a pump circulating the water 150 ℓ/min @ 1100 bar. How long would it take for the water to heat up from 20*C to 30*C?

How would the time change if the depth of water increase to 2m (ie. 29.572 ℓ).

Q: What formula(s) do I need to use to find the time required for the water heat up from 20*C to 30*C. Also what other variable is required for me to calculate this time?

Any pointers will be highly appreciated.

$\endgroup$
9
  • $\begingroup$ 1100 bar or 1.100 bar? $\endgroup$ – Nicolas Raoul Jun 9 '15 at 8:59
  • $\begingroup$ one thousand one hundred bar. $\endgroup$ – 3kstc Jun 9 '15 at 9:20
  • $\begingroup$ 1100 atmospheres? Way above Scuba tank filling type pressures? 1 bar = 14.5 psi So 1100 bar = ~ 16,000 psi. ie utterly immense and dangerous stuff. What are you working with? Or is that a mistake. ie what is the application. $\endgroup$ – Russell McMahon Jun 9 '15 at 9:34
  • $\begingroup$ Pressure ? psi? ...? $\endgroup$ – Russell McMahon Jun 9 '15 at 14:29
  • $\begingroup$ I can confirm that 1100 bar is correct, this is the water blast pressure. $\endgroup$ – 3kstc Jun 10 '15 at 0:24
3
$\begingroup$

These two very simple to use formulas will probably allow easy order of magnitude scoping.

Time to heat with given heating power

T = V x 1000 cc/l x 4.17 x K / W seconds or T = 4170 V.K/W

Power required to heat in time T

Power = W = = V x 1000 cc/l x 4.17 x K / T Watts or W = 4170 V.K/T

Where

T = seconds
V = litres
K = degrees K (or C) rise
W = Watts heating power
4.17 = Specific heat of water over range 20 C to 30 C adjusted for mean water density so units are J/cc/K

This is increased by any losses
eg electric kettle uninsulated is 85% - 95% efficient Pumping energy adds partially to heating.


The key factor is the amount of heat stored in water per cc per degree K (or degree C) rise in temperature. Once this is known all else can be calculated.
A more useful than many table of water thermal properties may be found here on the well-worth-bookmarking "Engineering Toolbox" site. Rather than giving just a few figures at a few selected temperatures this gives properties at a wide range of temperatures, so it can be established if temperature has a significant effect on the relevant properties in the context of the current problem.

The specific heat of water is an average of about 4.18 Joule per gram* per degree K across the 20 C to 30C temperature range, or 4.17 Joule per cc per degree K. The 0.01 J/unit/K is not going to make much difference to your result.

  • The per cc and per gram figures differ slightly because the density of water is very slightly less than 1 g/cc across this range. The specific heat will be about 4.18 Joule per gram and 4.17 J/cc per degree K. The usual value given is 4.182 J/g/K, often without reference to temperature range.

Other useful (and necessary) facts:

  • One Joule is provided by 1 Watt of heating per second or
    1 J = 1W/s

    1 litre = 1000 cc

    1 metre^3 = 1000 litre And 1 litre = 10 x 10 x 10 cm or 100 x 100 x 100 mm

    Total_energy = SH x cc x delta-C = SH x litres x 1000 x delta-C

    Power = energy /second = Total_energy / seconds.

    Pressure matters little here except as it may affect water density.
    Look up compressibility of water to see how much this matters.

Doubling water head affects volume (of course) but has an insignificant effect on pressure.


Power input

Energy in Volume of V litres = V x 1000 cc/l x SG g/cc x SH x delta_T

V = 14,786 (user figure)
delta_T = 10 K (user figure)
SG = 1 (actual figiure rolled into SH - see text)
SH = 4.17 J/cc/K (SG is included in this - see text)

Energy = 14,786 x 1000 x 1 x 4.17 x 10 = 616,576,000 J or W.s

So you can heat it in 1 second using a 617,000! kW heater.
This may be hard.

or in one hour with 616,576/3600 seconds = 171,000 Watts of heat (+ losses)

Or using a 3kW element as used in domestic hot water heaters.
(Max size available is usually 3 kW - larger for special uses). 616576000/3000 =~ 57 hours!

To heat it in and '8 hour day' = 616.6 MJ/(8 x 3600) = 21.4 kW.


Heat capacity of tank may be significant. Insulation of tank will be significant.
Other losses may occur.
But that gives you an idea.

$\endgroup$
4
  • $\begingroup$ This is not course work, at work we are in process of designing a test rig. And I am doing the theoretical values - so, any help is much appreciated. $\endgroup$ – 3kstc Jun 9 '15 at 9:22
  • $\begingroup$ Thanks Russell, I can confirm that the pressure is indeed 1100 bar and not 1100 milli bar. $\endgroup$ – 3kstc Jun 10 '15 at 0:26
  • $\begingroup$ Many thanks for the clear and concise explanation. Sincerely appreciate it. But we didn't use the rate of flow of 150L/min also the pressure? Would these values affect anything? $\endgroup$ – 3kstc Jun 10 '15 at 2:32
  • $\begingroup$ @3kstc It would have been good to have know the application up front. "water blast" gives some clue. Water is minimally compressible - I haven't looked it up but I doubt it would make much difference compared to losses from cooling etc. Flow rate mainly only affects the number of passes through the heater. As long as water has many passes AND/OR is well mixed post heating it makes minimal difference. BUT a percentage of pump energy will heat water - how much tbd but say work on 50% to start. $\endgroup$ – Russell McMahon Jun 10 '15 at 3:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.