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The Absolute Scale.-The establishment of this law indicates the fact that all temperatures should naturally be measured from a point other than that of the freezing temperature of water. Thus it is seen from the above that a point of 459.6° below the ordinary Fahrenheit scale would be known as an absolute zero. Throughout this work then T will represent the absolute scale and t the ordinary scale. Thus

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uct of the pressure and volume is proportional to the change in absolute temperature we shall now write one of the most useful

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FIG. 33.-Saturated steam obeys not at all the simple laws of thermodynamics. (Circulating water pumps operated by saturated steam at the Redondo plant of the Southern California Edison Company.)

formulas in the computation of gas constants, namely that

in which R is a constant.

pv = ᎡᎢ

(5)

In the case of air, let us see if we can compute this constant. Experimentally it is found that the volume of air in a boiler room temperature of 84°F. is 13.71 cu. ft. per lb. when the atmospheric pressure is 14.7 lb. per sq. in. Substituting in the above formula we have 14.7 X 144 X 13.71 R (459.6 +84). There(459.684). fore R = 53.3.

=

A Formula for Gas Density.-If we let y be the density of a gas, it is evident that it has a value equal to the reciprocal of v

in the above equation. In other words the density of a gas is the weight of 1 cu. ft. under standard conditions of pressure and temperature. We may then write without further proof the formula,

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To Compute "R" for Any Gas.-In the measurement of gases there is a standard pressure and temperature to which all gas volumes and densities are reduced in order to have some basis of comparison. These standard conditions are the temperature of freezing water and the pressure of the atmosphere at sea-level. From the equation last writen above, it is now evident that since p and T are constant for all gases under this standardized method of comparison, then the product of R and y must also be constant. This gives us a method or rather formula by which we may obtain the value of R for any gas if we know its density. The molecular weights of all gases may be obtained by reference to any standard book on elementary chemistry.

Let us multiply both sides of the above equation by the molecular weight m and by rearranging the terms, we have

For oxygen Y

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= 0.089222 lb. per cu. ft. at atmospheric pres

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Since this product Rm is always a constant, we have for any perfect gas that

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This formula together with the preceding general formulas for pressures and volumes now enables us to ascertain practically all the constants for perfect gases.

As an example let us assume that the temperature of an escaping chimney gas is 400°F. What would be the density of the nitrogen content of the escaping flue gases? First find the value for R for nitrogen for which m = 28.

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1544

=

54.98

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It is always convenient to express volumes as the number of cu. ft. per lb. Hence when the symbol V is used it will mean the total volume content of the gas under consideration. If M represents the weight of this volume V we have the relationship

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Thus if we have given 18.805 lb. of dry flue gas we can easily compute the volume it would occupy when leaving the chimney at 400°F., if it is known that the value of R for the chimney gas is 51.4. Thus

14.7 X 144V =

....

=

18.805 X 51.4 X (400 + 459.6)
393.5 cu. ft.

Further Illustrative Examples. In order to still further illustrate the wide uses to which the formulas above given may be applied in engineering practice, the following six problems are worked out in full:

1. Find the volume of one pound of air in a compressor at a pressure of 100 lbs. square inch, the temperature being 32°F.

From Boyle's Law:

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at 32°F., v, for 1 lb. of air is 12.39 cu. ft. and po

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= 14.7 X 144

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2. From Charles' Law find the volume of one lb. of air at atmospheric pressure and 72°F.

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3. Find the temperature of two ounces of hydrogen contained in one gallon flask and exerting a pressure of 10,000 lbs. per sq. in.

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4. How large a flask will contain 1 lb. of Nitrogen at 3200 lbs. per sq. in. pressure and 70°F.?

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5. Ten lbs. of air at 200°F. occupy 120 cu. ft. What must be the pressure?

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6. How many lbs. of air does it take to fill 5600 cu. ft. at 15 lbs. per sq. in. pressure and 60°F.?

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CHAPTER VI

WATER AND STEAM

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S we look about us in nature, we find that all inanimate creation presents itself to us in three distinct physical states. Certain bodies, for instance, of themselves readily maintain their shape while others, although non variant in density, nevertheless seem to have no particular physical configuration but seek, due to the force of gravitation, the lowest level attainable and consequently must as a rule be held in a containing vessel. On the other hand, a third class of bodies is found not only possessing no particular physical configuration, but which actually seem inherently desirous of expanding to such an extent that they must as a rule be completely housed, bottom and top, in a containing vessel.

FIG. 34.-Water and steam space

in water tube boiler.

In the class room or in the power plant, it is easy to find illustrations of these three general classifications. Thus, chalk, iron pipe, and coal are instances of the first division and are known as solids. Crude petroleum, water, and kerosene are instances of the second division, and are called liquids. Finally, air, steam, and producer gas illustrate the third division, and are called gases.

These States are Possible to all Bodies.-The most interesting thing about these socalled states of matter, and indeed the item of most importance to the engineer, is that by varying the pressure externally forcing itself against the sides of any one of these bodies and by adding or subtracting the heat that may be held in store within the body itself, any solid may be converted into a liquid and then into a gas, or any liquid may be converted

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