Amount of direct steam to heat 2000kg Water? 2

[Jetjager]

Dear all

I have been put in a position where I have to estimate the amount of 3bar overheated steam, which is injected directly in to 2000kg of water, to heat the water from 10-85 degrees Celsius.
I doesn't take loss in to consideration - instead I will add a factor to the final result.

As it is some while ago that I have read in my Thermodynamics, could I please use a little help on this one. I have tried on my own below, but is it correct?


Q = m * c * dt = 2000kg * 4,19 kJ/(kg*C) * 75C = 628500kJ

m_steam = Q / h = 628500kJ / 2724kJ/kg = 231kg steam at 3bar


Br Morten

 

[zdas04]

I think your last value should be m_steam = Q/Δh since there still is considerable entropy in the exhaust steam (even if you are dumping the steam directly into the water, in that case the h_final would be for water at your final temp). You just don't get to take credit for all of the energy in the steam unless your heat sink is at absolute zero.

David Simpson, PE
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist

 

[racookpe1978]

1. You are mixing completely and inimately the steam and water, right?

Therefore, your final mass = 2000 kg + m_steam_used

2. Your thermal balance is "odd" ... 2000 kg water @ 10C x cp water_10_deg_C + M_steam @ WHAT Temperature "superheated" steam at 3 bar? Your (homework ?) problem is incomplete.

3. You are adding the superheated steam of unknown mass which will lose its heat three different ways: cooling from superheated steam at 3 bar down to saturated steam at 3 bar (actually a little less, but ignore that for now), then the latent heat of vaporization as that steam condenses to water, then that mass of water cooling from Tsat @ 3 bar down to the final mass of (2000 kg water + unknown mass of steam) at the final approximate temperature of the mix (85 deg C).

4. What "final factor" do you anticipate, and why are you adding it? Don't guess! Figure it out. A safety factor? Or a margin for future corrosion and contamination somehow? You're mixing the two fluids intimately, there are no other heat losses at this stage of the process.

So.

What is T_sat for 3 bar steam?
What is the degree of superheat for your initial steam?

 

[zdas04]

If you believe his enthalpy number then at 3 bar(a) he has about 18C of superheat. If you get a rigorous enthalpy of water at 85C and 3 bar(a) then it includes the cooling to saturation, the latent heat of vaporization, and the cooling to exit temp. I get a number that is about 10% of the superheated steam value so the mass required is more than 10% higher than the OP.

David Simpson, PE
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist

 

[chicopee]

3 bars of steam at what temperature? anotherword is it saturated steam or superheated steam?

 

[LittleInch]

2000kg of water is only 2 cubic metres. Unless you really dribbled the steam in I can't see it all being absorbed before it escapes out the top of the water surface, hence it will have some heat energy which you need to subtract from your last line.

I assume when you say losses, what you mean is the amount of heat lost during the temperature rise time. What is the intended rise time by the way? - this will help everyone understand a little more what is being planned here.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[Jetjager]

At first - thank you all for your time.

The intent of the heating is to heat up a Peber Sauce in a Batch Mixer vessel - from the customer is it also a demand to heat up a batch of water from 10 to 85 C Celsius by means of a steam valve (or more) in the bottom of the vessel.
I need to estimate the amount of steam to choose the correct amount of valves in the vessel, as well as the dimensions of the inlet pipes.

The heatlosses from the vessel and the energy to heat up the stainless steel does of course have influence in this matter. But in this case i do not need to calculate the amount of steam to perfection - I only need to know whether I need to use around 230kg or 330kg or...-this is why I will add a safety factor in the very end...

At this point I can assume that:

-the steam at 3bar(a) is 133 C Celsius
-the degree of superheated steam is x = 0
-the steam is super heated
-the desired heating time from 10 - 85 C is around 10 minutes, but is not that important


Does my little estimate in the OP look totally wrong, when it is only an estimate?


Br Morten

 

[racookpe1978]

Is this a "soup" or thick "paste" type of mixture? Even if water, you'll need a spray-type dispersion nozzle to spread the steam out through the water, and even the very low viscosity water will tend to splash and "vigorously" be agitated around. You'll need a lid or firmly attached cover over the container to prevent massive spillage.

And that problem gets worse if the mix is a thick paste or stew.

Yes, find the difference in enthaly from your superheated steam down to 85 deg C water. Then iterate: Find the mass you need to heat the pure water from 10 C to 85 C, then add that mass to the original 2000 kg at 85 deg C, then find the new mass of steam to heat the new total amount of water from 10 C to 85 C.

Is the final pressure of the water at 85 C at 3 bar?

 

[Jetjager]

The vessel is complete closed, and with both an top-agitator and a high shear mixer in the bottom, but equipped with an pressure relief valve to equal the pressure in the vessel, with the atmospheric pressure - the final pressure is therefore at atmospheric pressure.

 

[gruntguru]

racoopie. The mass of steam required to heat the water does not need to be considered as part of the water to be heated up if the final enthalpy of the steam used in the calculation is for liquid at 85*C.

je suis charlie

 

[IRstuff]

From the steam tables, 2724 kJ/kg vapor enthalpy plus about 84 kJ/kg liquid enthalpy down to 85C, which works out to about 225.2509726 kg, assuming perfect transfer

ΔHmix (10C to 85C) = 355.946081 - 42.02108865 = 313.9249924 kJ/kg

ΔHsteam (133C to 100C to 85C) = 419.099155 - 355.946081 + 2724.182215 = 2787.335289 kJ/kg

(ΔHmix / ΔHsteam) * 2000kg = 225.2509726 kg

So, that's the minimum amount of steam required.


TTFN
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[IRstuff]

oops, I didn't read the output correctly

ΔHsteam (133C to 85C) = 2724.2 - 355.946081 = 2368.2 kJ/kg

(ΔHmix / ΔHsteam) * 2000kg = 265.112915 kg

TTFN
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Need help writing a question or understanding a reply? forum1529

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