Please excuse if this duplicates prior conversations.
From:
http://en.wikipedia.org/wiki/Water_(properties)
http://en.wikipedia.org/wiki/Specific_heat_capacity
http://en.wikipedia.org/wiki/H2o
http://hypertextbook.com/facts/2007/DmitriyGekhman.shtml
The following approximate values for water can be used from the above
refs:
Liquid Density: 1000 kg/m^3 = 1 gm/cm^3
Heat of vaporization: 40.6 kJ/mol = 2260 J/gm
Heat capacity: 4.2 J/(gm K)
Molar mass: 18 gm/mol
Density of steam at 100 C and 760 torr: 0.6 kg/m^3 = 0.0006 gm/cm^3
Now to examine the importance of mass flow vs volume flow
measurements for the steam.
If x is the liquid portion by volume, then x/((x+(1-x)*0.0006)) is
the portion by mass. This gives the following table:
Liquid Liquid Gas
Portion Portion Portion
by Volume by Mass by Mass
--------- ------- -----------
0.000 0.0000 100.00
0.001 0.6252 0.3747
0.002 0.7695 0.2304
0.003 0.8337 0.1662
0.004 0.8700 0.1299
0.005 0.8933 0.1066
0.006 0.9095 0.0904
0.007 0.9215 0.0784
0.008 0.9307 0.0692
0.009 0.9380 0.0619
0.010 0.9439 0.0560
0.011 0.9488 0.0511
0.012 0.9529 0.0470
0.013 0.9564 0.0435
0.014 0.9594 0.0405
We can thus see from this table that if 1 percent by volume of the
steam is entrained water micro-droplets, easily not seen in tubing or
exhaust ports, that only 5.6 percent of the heat of vaporization is
required to produce that mixture.
Rough calculations based on Rossi specifics:
Suppose for the Rossi experiment the mass flow of a system is 292 ml/
min, or 4.9 gm/s, the inlet temperature 13 °C.
The delta T for water heating is 100 °C - 13 °C = 87 °C = 87 K.
If the output gas is 100% gas, we have the heat flow P_liq given by:
P_liq = (4.9 gm/s)*(87 K)*(4.2 J/(gm K))= 1790 J/s = 1.79 kW
and the heat flow H_gas for vaporization given by:
P_gas = (4.9 gm/s)*(2260 J/gm) = 11.1 kW
for a total thermal power P_total of:
P_total = 1.79 kW + 11.1 kW = 12.9 kW
Now, if the steam is 99% gas, we have:
P_liq = 1.79 kW
P_gas = (0.1066)* (11.1 kW) = 1.18 Kw
P_total = 1.79 kW + 1.18 kW = 2.97 kW
or 23% of the originally estimated power out.
The dryness of the steam was measured by measuring the dielectric
constant of the steam.
http://en.wikipedia.org/wiki/Relative_permittivity
At 100 ° C, the dielectric constant of water is roughly 55.3, and
steam is 1.0
The capacitance of two capacitors in series is:
1/C = 1/C1 + 1/C2
so using identical area capacitors with water being 1 unit of
thickness, and vapor being 99 units of thickness, we have the
relative capacitance C:
1/C = 1/(55.3/.01) + 1/(99/1) = 0.01028
C = 97.3
and so there is a 2.7% difference in capacitance for 1% water vapor
vs dry steam. This may be a valid technique, but depends in extreme
on the accuracy of the capacitance measurement. For this reason, it
seems reasonable to do calorimetry on the steam-liquid out. This
could fairly easily be accomplished by running the steam piping
through an ice calorimeter, a very old but accurate and first
principles technique, and thus estimating the heat flow by the ice
melt rate. All it takes is some tubing, a large insulated ice
container and sufficient ice to obtain redundant supporting data and
an improved estimate of error.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
