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Vapor 3000 Steam Vacuum

Steam sparging heating rates?


By Jimmy NY - usenet poster


well lets start with Btu/hour= lbs./hour X Cp. (specific heat) X (T final - T initial)

If its water you want to heat there are 8.33 lbs./gal, Cp. = 1.0 Btu/LB-F, and steam has
970 Btu/LB of latent heat (heat given up at a phase change)

So if you want to heat 100 gallons of water in 30 minutes from 100 to 200 F. We would
need to

833 lbs. X 1.0 X (200 - 100)
all that equals 83,300 Btu/.5 hour or 166,600 Btu/hour

So the amount of steam required is 166,600/970 = 171.5 lbs./hour of steam

This does not include energy from superheat in the steam, or energy to cool the steam to
atmospheric pressure..which if you are deciding how much steam to supply would be small in
comparison.

--
Just my simple input,

John


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Best Solution

posted on Aug 01, 2007
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Duke

Rank:Apprentice Apprentice
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Pete,
try a sparge system designed as an injector. Spirax/Sarco makes them model IN, they are
stainless steel tubes with holes on the side. The steam enters one end of the tube, and
pulls liquid with it. Internally mixes steam and water. Makes for a much quieter
application, and with the added turbulence, you eliminate alot of what you are fearing.

Other sources you may want to try are Penberthy, and Watson-McDaniel suction tees.

--
Just my simple input,

John


Believe it or not, some fine furniture was made
before they made $1600.00 Table saws
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Solution #2

posted on Aug 01, 2007
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Mini Me

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Beware of the 'very simple' approach

It assumes 'lumped heat capacity' and a constant Cp water (or batch fluid).
This is probably good for heating small volumes over a small temperature
range, but not recommended for a 100 m3 tank as posed in the example.
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Solution #3

posted on Aug 01, 2007
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Powe33

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Peter,
start with the first law of thermodynamics for a control volume (the tank)

dE/dt = (mh)in - (mh)out

where dE/dt is the rate of increase of energy (heating rate) in the tank, m
is a mass flow rate, h is an enthalpy, in is at the tank inlet, out is at the
tank outlet, inlet and exit potential and kinetic energy effects are
neglected, and any heat transfer through the tank wall is neglected.

The inlet and outlet enthalpies can be determined by going into a steam table
with the inlet and outlet temperatures and pressures.

dE/dt = dU/dt, with negligible potential and kinetic energies within the tank,
where U is the instantaneous internal energy of the batch.

dU/dt can be approximated as (m2u2 - m1u1)/(t2 - t1), where m2 is the final
mass in the batch (i.e., at time t2), u2 is the internal energy per unit mass
of the batch at t2, m1 is the initial mass in the batch (i.e., at time t1 =
0), u1 is the internal energy per unit mass of the batch at t1. u1 and u2
can be looked up in a steam table , or inversely, temperature data can be
determined from a computed u2

Also, (m2 - m1)/(t2 - t1) can be approximated by min - mout. Note that m1 and
m2 are masses, whereas min and mout are mass flow rates.

If there is no outlet, the problem is not too bad.

Tod


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Solution #4

posted on Aug 01, 2007
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Cornish

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Very simple:

Steam deltaH is about 2000 kJ/kg, cP(water) is 4.18 kJ/kg??C.

Heatup rate (??C/hr)=Steamflow (kg/h)*2000/Water mass (kg)/4.18. Ignore
superheat effect.

A 100 m3 water tank (100000 kg) with 1000 kg/h steam will thus heat up with
4.8??C/hr.

...
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Solution #5

posted on Aug 01, 2007
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man1

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If you can find a mollier diagram you can use it to calculate enthalpy of the added steam per
lbm. After that it becomes a relatively simple math problem.

Pat


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Solution #6

posted on Aug 01, 2007
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Rogers

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I was wondering if anyone knows of any methods for calculating the
heat up rate of tank contents heated by direct steam injection to the
batch. Couldn't find anything in the standard references. I would
imagine the problem can become quite complicated, but there must be
some rules of thumb for design out there.

Thanks in advance,
Peter
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