LECTURE IX.

CONTENTS.-Total Heat of Evaporation-Graphic diagram representing the Changes from Ice into Water and Water into Steam-Quantity of Water required for Condensation of Steam, with Examples for a Jet Condenser.

Total Heat of Evaporation.-The total heat of evaporation is the sum of the sensible and the latent heats of evaporation, and is approximately a constant quantity for pressures near the atmospheric pressure.

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The heat required to elevate the temperature of 1 lb. of water from the freezing point, 32° F., to the temperature of evaporation, is called the sensible heat,* and the additional heat required to evaporate it is termed the latent heat (see Lecture VIII.).

The total heat of evaporation for water is, therefore, the quantity of heat in thermal units necessary to raise 1 lb. of water from the freezing point, 32° F., to the particular temperature at which it is being evaporated, and to evaporate it at that temperature.

Let H stand for the Total heat of evaporation in B.T.U.

Now, since we have defined a unit of heat to be the quantity of heat necessary to raise 1 lb. of water by 1° Fah., the amount of heat imparted to 1 lb. of water, in raising its temperature from 32° F. to 212° F., must be (212 32) = 180 such units; or the sensible heat of steam at 212° F., is said to be 180 units per lb., or 180. Again, we saw that the latent heat of steam at 212° F. was in round numbers 966 units per lb., or 966.

Consequently, the total heat of steam at that temperature must H = S+ L

be

= 180 +966

= 1146 Units of Heat.

The following figure will clearly explain this, and should be carefully studied.

* The reason for starting from the freezing point of water, and not from zero Fah., is that we thus avoid the introduction of the latent heat of water.

GRAPHICAL REPRESENTATION OF THE CHANGES FROM ICE INTO WATER, AND WATER INTO STEAM AT ATMOSPHERIC
PRESSURE, DUE TO THE ABSORPTION OF HEAT, WITH THE CORRESPONDING TEMPERATURES AND VOLUMES.

If steam is generated at a higher temperature than 212° F., sensible heat increases, and the latent heat decreases.

the

The following formula, deduced from Regnault's experiments, gives approximately the latent heat of steam produced at other temperatures Fah.:

where L, is the latent heat, and to, the temperature of evaporation on Fahrenheit scale.

Consequently, the total heat of evaporation, at any temperature, t, must be

For example.-Let us find from this equation the total heat of steam at 212° F. Then t = 212° and 3 × 212 = 63.6, which, added to 1082 41146, the number we got before.

The following table gives approximately the sensible, latent, and total heats of evaporation of 1 lb. of steam up to a pressure of 7 atmospheres, i.e., about 88 lbs. on the square inch above the atmosphere:

We see from this table that, notwithstanding the decrease in the latent heat, the total heat of evaporation slowly increases. This point will be found to be of great importance in considering the expansive properties of steam, and one that will explain some curious phenomena connected therewith.

Quantity of Water required for Condensation. Students will be shown further on the importance placed by Watt on the production of a good vacuum behind the piston of his steam engine. He thereby took as full advantage as possible of the natural pressure of the atmosphere in propelling the piston forward, and of getting as much work out of his engine as he could. In order to do so, he attached to his steam cylinder a separate chamber, called

by him a "condenser," whereby the exhausting steam, instead of going straight into the open air (and meeting with a resistance of about 15 lbs. on the square inch), was brought into intimate contact with a spray or douche of cold water, which had instantly the effect of greatly reducing its temperature, and consequently its volume and pressure, as may be observed from an inspection of the graphic diagram on p. 67; from which it will be seen that steam at atmospheric pressure occupies 1642 times the volume of its weight in water. The most economical engines of the present day are "condensing engines," and, consequently, the student will at once see the full importance of thoroughly mastering the following problems relating to the quantity of water required for condensing a known quantity of steam.

First, we shall obtain a formula or rule for determining the minimum weight of condensing water which must be directly mixed with 1 lb. of steam, in order that the mixture may be reduced to water at a certain temperature, or the temperature of the hot-well.

The following is theoretically true:

*

The Loss of Heat from the Steam = the Gain of Heat by the Water. Now, let 1 lb. of steam at a temp. t,° be subjected to an injection of x lbs. of water at a temp. t,, and let the result be water at a temp. t..

Let H

=

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total heat, reckoned from 32° F., in 1 lb. of steam at t; then H = 1082 4+*3t, B.T.U. Hence

and the gain of heat in x lbs. of water = x × (t2-tą).

.. I× {H-(t2-32)} = x × (t3 − t2)

which is the general formula for the weight of water required to

condense 1 lb. of steam at t° to water at t3°.

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An example or two will illustrate the above rule and fix the principle in the memory, as well as clearly show the difference

* In working out examples students may substitute t, for t,; tw for t and th for t, where s stands for steam, w for water and h for hot-well.

between the mixing of two quantities of water, and a quantity of steam and water.

EXAMPLE I.-If I lb. of water at 212° F. be mixed with x lbs. of water at 60° F., what is the value of x when the resulting temperature is 100° F. ? The Loss of Heat from the Water at 212° F. the Gain of Heat by the Water at 60° F.

112

.*. 1 × (212° – 100°) = x × (100° – 60°);.'. 112=40x, or x=— =2.8 lbs.

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Again-If I lb. of steam at 212° F. be mixed with x lbs. of water at 60° F., what is the value of x when the resulting temperature is 100° F. ? The Loss of Heat from the Steam at 212° F. = the Gain of Heat by the Water at 60° F.

The loss of heat from the steam = 1× {1146- (100-32)} = 1078 B T.U.
The gain of heat by the water=xx (100-60)=40x B.T.U.

.. 1078=402, or x= =26.95 lbs.

1078
40

It only

We thus see the great effect of the latent heat of steam. requires 2.8 lbs. of water at 60° to produce the same temperature-result on water at 212° that 26.95 lbs. of water can do on steam at the same temperature.

EXAMPLE II.-In a jet condenser the temperature of the condensing water is 60° F., and that of the exhaust steam is 160° F. The temperature of the hot well is 110° F. Suppose the engines develop 200 I.H.P., and use lb. of steam per I.H. P. per minute, find the weight of condensing water supplied per hour to the condenser.

Here the weight of steam passing through the condenser per hour= 200 × × 60=6,000 lbs. Let x = weight of condensing water required. Then the Loss of Heat from the Steam=the Gain of Heat by the Water. The loss of heat from 6,000 lbs. of steam at 160° F.

=6,000 × {H. - (100 – 32)} B.T.U.

=6,000 × {1082'4+3 × 160 - (110-32)} B.T.U.
=6,000 × 1052*4=6,314,400 B.T.U.

The gain of heat by the water = x(110-60)=50x B.T.U.

.. 6,314,400=50x; or x= 6,314,400=126,288 lbs.=56.47 tons per hour.

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We might have applied the formula already deduced on previous page in answering this question.

Here x=

1114'4+3 × 160 - 110
110-60

=21048 lbs. per lb. of steam.

... since there are 6,000 lbs. of steam passing through the condenser per hour, there will be 6,000 times the above quantity of condensing water required per hour

=21*048 × 6,000=126,288 lbs. =56:47 tons as before.

Although the jet condenser has now been greatly superseded by surface condensers wherever pure fresh condensing water cannot be obtained (owing to the advantage of the condensing water being kept free from contact with the steam), we must refer students to our text-book on Steam and Steam Engines for the problems involved in determining the quantity of condensing water required therewith, owing to their more advanced character.