Part 15 (1/2)

When steaher to a lower pressure, as is the case with the throttling calori a resistance Hence, if there is no loss from radiation, the quantity of heat in the stea the orifice as before passing If the higher steae and the lower pressure that of the atmosphere, the total heat in a pound of dry steam at the former pressure is 11959 B t u and at the latter pressure 11504 B t u, a difference of 454 B t u As this heat will still exist in the steam at the lower pressure, since there is no external work done, its effectthe specific heat of superheated steah will be superheated 454/047 = 966 degrees If, however, the steam had contained one per cent of moisture, it would have contained less heat units per pound than if it were dry Since the latent heat of steae pressure is 8528 B t u, it follows that the one per cent of moisture would have required 85 B t u to evaporate it, leaving only 454 - 85 = 369 B t u available for superheating; hence, the superheat would be 369/047 = 785 degrees, as against 966 degrees for dry stearee of superheat for other percentages ofcalori facts, as shown below

Let H = total heat of one pound of steam at boiler pressure, L = latent heat of steam at boiler pressure, h = total heat of stea orifice, t_{1} = temperature of saturated steam at the reduced pressure, t_{2} = teh the orifice in the disc, 047 = the specific heat of saturated steaht of moisture in steam

The difference in B t u in a pound of stea the orifice is the heat available for evaporating thethe steam Therefore,

H - h = xL + 047(t_{2} - t_{1})

H - h - 047(t_{2} - t_{1}) or x = --------------------------- (4) L

Almost invariably the lower pressure is taken as that of the atmosphere

Under such conditions, h = 11504 and t_{1} = 212 degrees The formula thus becomes:

H - 11504 - 047(t_{2} - 212) x = ------------------------------ (5) L

For practical work it is more convenient to dispense with the upper thermometer in the calorimeter and to measure the pressure in the steae

A chartthe value of x for approximate ithout the necessity for co 15 and its use is as follows: assu of 295 degrees The intersection of the vertical line from the scale of temperatures as shown by the caloriauge pressures will indicate directly the per cent of onal scale In the present instance, this per cent is 10

Sources of Error in the Apparatus--A slight error may arise from the value, 047, used as the specific heat of superheated steam at atmospheric pressure This value, however is very nearly correct and any error resulting froer source of error due to the fact that the steth, to an initial error in the thermometer and to radiation losses

With an ordinary therrees rees would be about 3 degrees and the true terees[19]

The steah radiation fro nipple The heat available for evaporating h the orifice into the lower pressure will be dih radiation and the value of t_{2}, as shown by the calorimeter thermometer, will, therefore, be lower than if there were no such loss

Thefor the thermometer and radiation error recommended by the Power Test Coineers is by referring the readings as found on the boiler trial to a ”nor is the reading of the lower calorimeter thermometer for dry saturated stea the instrument to a horizontal stea nozzle projects upward to near the top of the pipe, there being no perforations in the nozzle and the steah its open upper end The test should be made with the steam in a quiescent state and with the steam pressure maintained as nearly as possible at the pressure observed in the main trial, the calorimeter thermometer to be the same as was used on the trial or one exactly si thus obtained for a pressure approxie of moisture in the steam, that is, with the proper correction made for radiation, may be calculated as follows:

Let T denote the nor in the trial The effect of radiation from the instrument as pointed out will be to lower the temperature of the steam at the lower pressure Let x_{1} represent the proportion of water in the steam which will lower its temperature an amount equal to the loss by radiation Then,

H - h - 047(T - t_{1}) x_{1} = ----------------------- L

This ainally but is the result of condensation in the instruh radiation Hence, the true amount of moisture in the steam represented by X is the difference between the a from condensation, or,

X = x - x_{1}

H - h - 047(t_{2} - t_{1}) H - h - 047(T - t_{1}) = --------------------------- - ----------------------- L L