You are on the right track. Always be aware that the listed values are for total heat of solution/dilution/formation, etc., and they NEVER provide for "real" effects of how slow the dilution takes place, nor for effects of heat loss to the environment (they are all generated by chemists and physicists, and there is no way that the system can be explained in universal terms).
Anyway, according to Lange's Handbook of Chemistry, "the value of dHf
(heat of formation) given in its standard state is the apparent molal
enthalpy of formation of the substance in the infinitely dilute real
solution. The experimental value for a heat of dilution is obtained
directly as the difference between the two values of dHf at the
corresponding concentrations.", and the units of dHf are given in
kcal/mol. This is fairly straightforward to a Chem E, but I don't know
how it reads for someone not used to the jargon.
From "Chemical Process Principles" part I, 2nd ed., by Hougan, Watson and Ragatz, pp318 and 319, (I don't know if this is still in print), "the enthalpy change when a substance is dissolved depends on the amounts and natures of the solute and solvent, on the temperature, and on the initial and final concentrations of the solution...". and, "The enthalpy of a solution at temperature T relative to the pure solute and solvent at temperature To is expressed as
Hs = n1H1 + n2H2 + n2(dHs2)
where Hs = enthalpy of n1 + n2 moles of solution of components 1 and 2
at temperature To
H1, H2 = molal enthalpies of pure components 1 and 2 at temperature T
relative to temperature To
dHs2 = integral heat of solution of component 2 at temperature T
(integral heats of solution are the textbook values)
That should get you started. n1 is water, n2 is H2SO4, To is 18 deg C, T is 60 or 70 deg C. Look up molal enthalpies of H2SO4 and water for H1 and H2.
... Bruce D. Bullough ...
Sebesta Blomberg & Associates
2381 Rosegate
Roseville, MN 55113 USA
651-634-7344 direct
651-634-7400 FAX
www.sebesta.com
-----Original Message-----
From: Steve McKenzie [mailto:mechproj@xtra.co.nz]
Sent: Thursday, January 29, 2004 3:46 AM
To: PipingDesign@yahoogroups.com
Subject: [PipingDesign] Sulphuric Acid Heat of Dilution
[bcc][faked-from][bayes]
Gents
I am trying to hotrod a single absorption sulphuric acid plant
running on elemental sulphur. Am chasing a few more tonnes per day of
acid and a few more kW generation.
To evaluate the effect of my proposed changes I need to model the
performance.
The mass balance has come out OK, but I am struggling with a few
areas in the energy balance.
The first is the drier and dilution water.
I need to calculate the heat liberated per unit mass of dilution
water. The drier inlet acid concentration is 98.50w/w and the
dilution/reaction water addition reduces the acid concentration to
97.81w/w The acid inlet temperature is meant to be 60C but sometimes
runs to 70C so I need to consider both cases. Pressure is about 3psig
My schoolboy chemistry calcs (molar fractions and partial molal heats
of SO3 and H20)indicate that at standard conditions (18 C) each pound
of water added will release 887 BTU gross. This ties up with some
ancient data Porter , 1917) which indicates 900BTU from a graph.
However two partial flowsheets (Duecker and some unofficial data)I
have indicate heat releases of 687 and 439 BTU/lb. As they are
flowsheet data, they may be nett heat release. The dilution water
should absorb about 80 BTU and the airflow should remove about
another 120BTU. This brings my 887BTU/lb down to 687 BTU. However
there are some fairly hefty assumptions involved.
My feeling is that the gross heat released by the dilution should
decrease with temperature but I have not alloawe for this.
Does anyone know how I calculate the heat of dilution released at a given temperature?
I expect I will have a similar question for absorbing SO3 into the acid.
References or links would be helpful.
Sorry about the units; I'm working from old books on this one.
Knew I should have gone to chemistry lectures instead of playing snooker.
Cheers
Steve
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