H - h - 0.47(t_{2} - t_{1}) H - h - 0.47(T - t_{1})
= --------------------------- - -----------------------
L L
0.47(T - t_{2})
= --------------- (6)
L
As T and t_{2} are taken with the same thermometer under the same set of
conditions, any error in the reading of the thermometers will be
approximately the same for the temperatures T and t_{2} and the above
method therefore corrects for both the radiation and thermometer errors.
The theoretical readings for dry steam, where there are no losses due to
radiation, are obtainable from formula (5) by letting x = 0 and solving
for t_{2}. The difference between the theoretical reading and the normal
reading for no moisture will be the thermometer and radiation correction
to be applied in order that the correct reading of t_{2} may be
obtained.
For any calorimeter within the range of its ordinary use, such a
thermometer and radiation correction taken from one normal reading is
approximately correct for any conditions with the same or a duplicate
thermometer.
The percentage of moisture in the steam, corrected for thermometer error
and radiation and the correction to be applied to the particular
calorimeter used, would be determined as follows: Assume a gauge
pressure in the trial to be 180 pounds and the thermometer reading to be
295 degrees. A normal reading, taken in the manner described, gives a
value of T = 303 degrees; then, the percentage of moisture corrected for
thermometer error and radiation is,
0.47(303 - 295)
x = ----------------
845.0
= 0.45 per cent.
The theoretical reading for dry steam will be,
1197.7 - 1150.4 - 0.47(t_{2} - 212)
0 = ------------------------------------
845.0
t_{2} = 313 degrees.
The thermometer and radiation correction to be applied to the instrument
used, therefore over the ordinary range of pressure is