(239.)

The pressure of steam used in these machines is such as is never used in European engines, even when worked on railways. A pressure of one hundred pounds per inch is here considered extremely moderate. The captain of one of these boats, plying between Pittsburg and St. Louis, told Mr. Stevenson that "under ordinary circumstances his safety valves were loaded with a pressure equal to one hundred and thirty-eight pounds per square inch, but that the steam was occasionally raised as high as one hundred and fifty pounds to enable the vessel to pass parts of the river in which there is a strong current;" and he added, by way of consolation, that "this pressure was never exceeded except on extraordinary occasions!"

The dimensions and power of the Mississippi steamers may be collected from those of the St. Louis, a boat which was plying on that river in 1837. That vessel measured two hundred and fifty feet on deck, and had twenty-eight feet breadth of beam. Her draught of water was eight feet, and her measured capacity one thousand tons. She was propelled by two engines with thirty-inch cylinders, and ten feet stroke; the safety valve being loaded at one hundred pounds per square inch.

The paddle wheels of these vessels are attached to the paddle shaft, in such a manner as to be thrown into and out of gear, at discretion, by the engineer, so that the paddle shaft may revolve without driving the wheels: by this expedient the power of the engine is used to feed the boilers while the vessel stops at the several stations. The vessel is therefore stopped, not, as is usually the case, by stopping the engines, but by throwing the wheels out of connection with the paddle shaft. The engines continue to work, but their [Pg503] power is expended in forcing water into the boiler. By this expedient the activity of the engines may, within practical limits, be varied with the resistance the vessel has to encounter. In working against a strong current, the feed may be cut off from the boilers, and the production of steam, and consequently the power of the engines, thereby stimulated, while this suspension of the feed may be compensated at the next station.

The stoppages to take in goods and passengers, and for relays of fuel, are frequent. "The liberty which they take with their vessels on these occasions," says Mr. Stevenson, "is somewhat amusing: I had a good example of this on board a large vessel, called the Ontario. She was steered close in shore amongst stones and stumps of trees, where she lay for some hours to take in goods: the additional weight increased her draught of water, and caused her to heel a good deal; and when her engines were put in motion, she actually crawled into the deep water on her paddle wheels: the steam had been got up to an enormous pressure to enable her to get off, and the volume of steam discharged from the escapement pipe at every half stroke of the piston made a sharp sound almost like the discharge of fire-arms, while every timber in the vessel seemed to tremble, and the whole structure actually groaned under the shocks."

Besides the steamers used for the navigation of the Mississippi, innumerable steam tugs are constantly employed in towing vessels between the port of New Orleans and the open sea of the Gulf of Mexico. Before the invention of steam navigation, this southern capital of the United States laboured under the disadvantage of possessing almost the only bad and inconvenient harbour in the vast range of coast by which the country is bounded. New Orleans lies at a distance of about one hundred miles from the Gulf of Mexico. The force of the stream, the frequency of shoals, and the winding course of the channel rendered it scarcely possible for a sailing vessel to pass between the port and the sea with the same wind. The anchorage was every where bad, and great difficulty and risk attended the mooring of large vessels to the banks. The steam engine has, however, overcome all [Pg504] these difficulties, and rendered the most objectionable harbour of the Union a safe and good seaport, perfectly easy of approach and of egress at all times; a small steam tug will take in tow several large ships, and carry them with safety and expedition to the offing, where it will dismiss them on their voyage, and take back vessels which may have arrived.

GREAT WESTERN OFF NEW YORK.

[APPENDIX.]

[Pg505]
[TOC] [INX]

On the Relation between the Temperature, Pressure, and Density of Common Steam.

There is a fixed relation between the temperature and pressure of common steam, which has not yet been ascertained by theory. Various empirical formulæ have been proposed to express it, derived from tables of temperatures and corresponding pressures which have been founded on experiments and completed by interpolation.

The following formula, proposed by M. Biot, represents with great accuracy the relation between the temperature and pressure of common steam, throughout all that part of the thermometric scale to which experiments have been extended.

Let

a = 5·96131330259
log. a1 = 0·82340688193 − 1
log. b1 = −·01309734295
log. a2 = 0·74110951837
log. b2 = −·00212510583

The relation between the temperature t with reference to the centesimal thermometer, and the pressure p in millimètres of mercury at the temperature of melting ice, will then be expressed by the following formula:—

log. p = a − a1b120 + t − a2b220 + t. (1.)

Formulæ have, however, been proposed, which, though not applicable to the whole scale of temperatures, are more manageable in their practical application than the preceding.

For pressures less than an atmosphere, Southern proposed the following formula, where the pressure is intended to be expressed [Pg506] in pounds per square inch, and the temperature in reference to Fahrenheit's thermometer,—

p = 0·04948 + (51·3 + t)5·13.(2.)
155·7256
t = 155·7256 {(p − 0·04948)1⁄5·13 − 51·3}

The following formula was proposed by Tredgold, where p expresses the pressure in inches of mercury:—

p = (100 + t)6.
177

This was afterwards modified by Mellet, and represents with sufficient accuracy experiments from 1 to 4 atmospheres. Let p represent pounds per square inch, and t the temperature by Fahrenheit's thermometer,—

p = (103 + t)6.(3.)
201·18
t = 201·18 p1⁄6 − 103

M. de Pambour has proposed the following formula, also applicable through the same limits of the scale:—

p = (98·806 + t)6. (4.)
198·562
t = 198·562 p1⁄6 − 98·806

MM. Dulong and Arago have proposed the following formula for all pressures between 4 and 50 atmospheres:—

p = (0·26793 + 0·0067585 t)5. (5.)
t = 147·961 p1⁄5 − 39·644

It was about the year 1801, that Dalton, at Manchester, and Gay-Lussac, at Paris, instituted a series of experiments on gaseous bodies, which conducted them to the discovery of the law mentioned in art. ([96].), p. 171. These philosophers found that all gases whatever, and all vapours raised from liquids by heat, as well as all mixtures of gases and vapours, are subject to the same quantity of expansion between the temperatures of melting ice and boiling water; and by experiments subsequently made by Dulong and Petit, this uniformity of expansion has been proved to extend to all temperatures which can come under practical inquiries.

Dalton found that 1000 cubic inches of air at the temperature of melting ice dilated to 1325 cubic inches if raised to the temperature of boiling water. According to Gay-Lussac, the increased volume was 1375 cubic inches. The latter determination has been subsequently found to be the more correct one.[40]

[Pg507]

It appears, therefore, that for an increase of temperature from 32° to 212°, amounting to 180°, the increase of volume is 375 parts in 1000; and since the expansion is uniform, the increase of volume for 1° will be found by dividing this by 180, which will give an increase of 2081⁄3 parts in 100,000 for each degree of the common thermometer.

To reduce the expression of this important and general law to mathematical language, let v be the volume of an elastic fluid at the temperature of melting ice, and let nv be the increase which that volume would receive by being raised one degree of temperature under the same pressure. Let V be its volume at the temperature T. Then we shall have

V = v + nv (T − 32) = v {1 + n (T − 32)}.

If V′ be its volume at any other temperature T′, and under the same pressure, we shall have, in like manner,

V′ = v {1 + n (T′ − 32)}.

Hence we obtain

V = 1 + n (T − 32);(6.)
V′1 + n (T′ − 32)

which expresses the relation between the volumes of the same gas or vapour under the same pressure and at any two temperatures. The co-efficient n, as explained in the text, has the same value for the same gas or vapour throughout the whole thermometric scale. But it is still more remarkable that this constant has the same value for all gases and vapours. It is a number, therefore, which must have some essential relation to the gaseous or elastic state of fluid matter, independent of the peculiar qualities of any particular gas or vapour.

The value of n, according to the experiments of Gay-Lussac, is 0·002083, or 1⁄480.

To reduce the law of Mariotte, explained in ([97].) p. 171., to mathematical language, let V, V′ be the volumes of the same gas or vapour under different pressures P, P′, but at the same temperature. We shall then have

VP = V′P′. (7.)

If it be required to determine the relation between the volumes of the same gas or vapour, under a change of both temperature and pressure, let V be the volume at the temperature T and under the pressure P, and let V′ be the volume at the temperature T′ and under the pressure P′. Let v be the volume at the temperature T and under the pressure P′.

By formula (7.) we have

VP = vP′;

[Pg508]

and by formula (6.) we have

V′ = 1 + n(T′ − 32)
v1 + n(T − 32)

Eliminating v, we shall obtain

V = P′ · 1 + n(T − 32);
V′P1 + n(T′ − 32)

or,

VP = 1 + n(T − 32);(8.)
V′P′1 + n(T′ − 32)

which is the general relation between the volumes, pressures, and temperatures of the same gas or vapour in two different states.

To apply this general formula to the case of the vapour of water, let T′ = 212°. It is known by experiment that the corresponding value of P′, expressed in pounds per square inch, is 14·706; and that V′, expressed in cubic inches, the water evaporated being taken as a cubic inch, is 1700. If, then, we take 0·002083 as the value of n, we shall have by (8.),

VP = 1700 × 14·706 × {1 + 0·002083 (T − 32)}
1 + 0·002083 × 180
= 18183{1 + 0·002083 (T − 32)}.(9.)

If, by means of this formula (9.), and any of the formulæ (1.), (2.), (3.), (4.), (5.), T were eliminated, we should obtain a formula between V and P, which would enable us to compute the enlargement of volume which water undergoes in passing into steam under any proposed pressure. But such a formula would not be suitable for practical computations. By the formulæ (1.) to (5.), a table of pressures and corresponding temperatures may be computed; and these being known, the formula (9.) will be sufficient for the computation of the corresponding values of V, or the enlargement of volume which water undergoes in passing into steam.

In the following table, the temperatures corresponding to pressures from 1 to 240 lbs. per square inch are given by computation from the formulæ (2.) to (5.), and the volumes of steam produced by an unit of volume of water as computed from the formula (9.).

The mechanical effect is obtained by multiplying the pressure in pounds by the expansion of a cubic inch of water in passing into steam expressed in feet, and is therefore the number of pounds which would be raised one foot by the evaporation of a cubic inch of water under the given pressure. [Pg509]

Total pressure in Pounds per Square Inch.Corresponding Temperature. Volume of the Steam compared to the Volume of the Water that has produced it.Mechanical Effect of a Cubic Inch of Water evaporated in Pounds raised One Foot.
1102·9208681739
2126·1108741812
3141·074371859
4152·356851895
5161·446171924
6169·238971948
7175·933761969
8182·029831989
9187·426742006
10192·424262022
11197·022212036
12201·320502050
13205·319042063
14209·117782074
15212·816692086
16216·315732097
17219·614882107
18222·714112117
19225·613432126
20228·512812135
21231·212252144
22233·811742152
23236·311272160
24238·710842168
25241·010442175
26243·310072182
27245·59732189
28247·69412196
29249·69112202
30251·68832209
31253·68572215
32255·58332221
33257·38102226
34259·17882232
35260·97672238
36262·67482243
37264·37292248
38265·97122253
39267·56952259
40269·16792264
41270·66642268
42272·16492273
43273·66352278
44275·06222282
45276·46102287
46277·85982291
47279·25862296
48280·55752300
49281·95642304
50283·25542308
51284·45442312
52285·75342316
53286·95252320
54288·15162324
55289·35082327
56290·55002331
57291·74922335
58292·94842339
59294·24772343
60295·64702347
61296·94632351
62298·14562355
63299·24492359
64300·34432362
65301·34372365
66302·44312369
67303·44252372
68304·44192375
69305·44142378
70306·44082382
71307·44032385
72308·43982388
73309·33932391
74310·33882394
75311·23832397
76312·23792400
77313·13742403
78314·03702405
79314·93662408
80315·83622411
81316·73582414
82317·63542417
83318·43502419
84319·33462422
85320·13422425
86321·03392427
87321·83352430
88322·63322432
89323·53282435
90324·33252438
91325·13222440
92325·93192443
93326·73162445
94327·53132448
95328·23102450
96329·03072453
97329·83042455
98330·53012457
99331·32982460
100332·02952462
110339·22712486
120345·82512507
130352·12332527
140357·92182545
150363·42052561
160368·71932577
170373·61832593
180378·41742608
190382·91662622
200387·31582636
210391·51512650
220395·51452663
230399·41402675
240403·11342687

[Pg511]

In the absence of any direct method of determining the general relation between the pressure and volume of common steam, empirical formulæ expressing it have been proposed by different mathematicians.

The late Professor Navier proposed the following:—Let S express the volume of steam into which an unit of volume of water is converted under the pressure P, this pressure being expressed in kilogrammes per square mètre. Then the relation between S and P will be

S = a,
b + mP

where a = 1000, b = 0·09, and m = 0·0000484.

This formula, however, does not agree with experiment at pressures less than an atmosphere. M. de Pambour, therefore, proposes the following changes in the values of its co-efficients:—Let P express the pressure in pounds per square foot; and let

a = 10000 b = 0·4227 m = 0·00258,

and the formula will be accurate for all pressures. For pressures above two atmospheres the following values give more accuracy to the calculation:—

a = 10000 b = 1·421 m = 0·0023.

In these investigations I shall adopt the following modified formula. The symbols S and P retaining their signification, we shall have

S = a,(10.)
b + P

where

a = 3875969 b = 164.

These values of a and b will be sufficiently accurate for practical purposes for all pressures, and may be used in reference to low-pressure engines of every form, as well as for high-pressure engines which work expansively.

When the pressure is not less than 30 pounds per square inch, the following values of a and b will be more accurate:—

a = 4347826 b = 618.

On the Expansive Action of Steam.

The investigation of the effect of the expansion of steam which has been given in the text, is intended to convey to those who are not conversant with the principles and language of analysis, some notion of the nature of that mechanical effect to which the advantages attending the expansive principle are due. We shall now, however, explain these effects more accurately. [Pg512]

The dynamical effect produced by any mechanical agent is expressed by the product of the resistance overcome and the space through which that resistance is moved.

Let

P = the pressure of steam expressed in pounds per square foot.
S = the number of cubic feet of steam of that pressure produced by the evaporation of a cubic foot of water.
E = the mechanical effect produced by the evaporation of a cubic foot of water expressed in pounds raised one foot.

Then we shall have E = PS; and if W be a volume of water evaporated under the pressure P, the mechanical effect produced by it will be WPS.

By (10.) we have

SP = a − bS.

Hence, for the mechanical effect of a cubic foot of water evaporated under the pressure P we have

E = a − bS.(11.)

Let a cubic foot of water be evaporated under the pressure P′, and let it produce a volume of steam S′ of that pressure. Let this steam afterwards be allowed to expand to the increased volume S and the diminished pressure P; and let it be required to determine the mechanical effect produced during the expansion of the steam from the volume S′ to the volume S.

Let

E′ = the mechanical effect produced by the evaporation of the water under the pressure P′ without expansion.
E″ = the mechanical effect produced during the expansion of the steam.
E = the mechanical effect which would be produced by the evaporation under the pressure P without expansion.
E = the total mechanical effect produced by the evaporation under the pressure P′ and subsequent expansion.

Thus we have

E = E′ + E″.

Let s be any volume of the steam during the process of expansion, p the corresponding pressure, and e″ the mechanical effect produced by the expansion of the steam. We have then by (10.)

p = a − b;
s
∵ de″ = ads − bds.
s

Hence by integrating we obtain

e″ = a log. s − bs + C;

[Pg513]

which, taken between the limits s = S′ and s = S, becomes

E″ = a log.S − b(S − S′).(12.)
S′

But by (11.) we have

E′ = a − bS′,
E = a − bS;
∵ E′ − E = b(S − S′);
∵ E″ = a log. S − E′ + E;
S′
∵ E = E″ + E′ = a log. S + E.(13.)
S′

Or,

E = a (1 + log. S) − bS.(14.)
S′

Hence it appears that the mechanical effect of a cubic foot of water evaporated under the pressure P may be increased by the quantity a log. S/S′, if it be first evaporated under the greater pressure P′, and subsequently expanded to the lesser pressure P.

The logarithms in these formulæ are hyperbolic.

To apply these principles to the actual case of a double acting steam engine,

Let

L = the stroke of the piston in feet.
A = the area of the piston in square feet.
n = the number of strokes of the piston per minute.
∵ 2nAL = the number of cubic feet of space through which the piston moves per minute.

Let

cLA = the clearage, or the space between the steam valve and the piston at each end of the stroke.
∵ The volume of steam admitted through the steam valve at each stroke of the engine will be 2n AL(1 + c).

Let

V = the mean speed of the piston in feet per minute,
∵ 2nL = V.

The volume of steam admitted to the cylinder per minute will therefore be VA (1 + c), the part of it employed in working the piston being VA.

Let

W = the water in cubic feet admitted per minute in the form of steam through the steam valve.
S = the number of cubic feet of steam produced by a cubic foot of water.

[Pg514]

Hence we shall have

WS = VA (1 + c);
∵ S = VA(1 + c).(15.)
W

Since by (10.) we have

P = a − b;
S
∵ P = Wa − b.(16.)
VA(1 + c)

By which the pressure of steam in the cylinder will be known, when the effective evaporation, the diameter of the cylinder, and speed of the piston, are given.

If it be required to express the mechanical effect produced per minute by the action of steam on the piston, it is only necessary to multiply the pressure on the surface of the piston by the space per minute through which the piston moves. This will give

VAP = Wa − VAb;(17.)
1 + c

which expresses the whole mechanical effect per minute in pounds raised one foot.

If the steam be worked expansively, let it be cut off after the piston has moved through a part of the stroke expressed by e.

The volume of steam of the undiminished pressure P′ admitted per minute through the valve would then be

VA (e + c);

and the ratio of this volume to that of the water producing it being expressed by S′, we should have

S′ = VA(e + c).
W

The final volume into which this steam is subsequently expanded being VA(1 + c), its ratio to that of the water will be

S = VA (1 + c).
W

The pressure P′, till the steam is cut off, will be

P′ = Wa − b.(18.)
VA(e + c)

The mechanical effect E′ produced per minute by the steam of full pressure will be

E′ = P′AVe = Wae − AVbe;
e + c

and the effect E″ per minute produced by the expansion of the steam will by (12.) be [Pg515]

E″ = Wa log.1 + c − bVA(1 − e).
e + c

Hence the total effect per minute will be

E = Wa{e + log. 1 + c} − bVA.(19.)
e + ce + c

If the engine work without expansion, e = 1;

∵ E′ = Wa − bVA,(20.)
1 + c

as before; and the effect per minute gained by expansion will therefore be

E − E′ = Wa{e − 1 + log.1 + c};(21.)
e + c1 + ce + c

which therefore represents the quantity of power gained by the expansive action, with a given evaporating power.

In these formulæ the total effect of the steam is considered without reference to the nature of the resistances which it has to overcome.

These resistances may be enumerated as follows:—

  1. The resistance produced by the load which the engine is required to move.
  2. The resistance produced by the vapour which remains uncondensed if the engine be a condensing engine, or of the atmospheric pressure if the engine do not condense the steam.
  3. The resistance of the engine and its machinery, consisting of the friction of the various moving parts, the resistances of the feed pump, the cold water pump, &c. A part of these resistances are of the same amount, whether the engine be loaded or not, and part are increased, in some proportion depending on the load.

When the engine is maintained in a state of uniform motion, the sum of all these resistances must always be equal to the whole effect produced by the steam on the piston. The power expended on the first alone is the useful effect.

Let

R = the pressure per square foot of the piston surface, which balances the resistances produced by the load.
mR = the pressure per square foot, which balances that part of the friction of the engine which is proportional to the load.
r = the pressure per square foot, which balances the sum of all those resistances that are not proportional to the load.

The total resistance, therefore, being R + mR + r, which, when the mean motion of the piston is uniform, must be equal to the mean pressure on the piston. The total mechanical effect [Pg516] must therefore be equal to the total resistance multiplied by the space through which that resistance is driven. Hence we shall have

{R(1 + m) + r}VA = Wa{e + log. 1 + c} − VAb;
e + ce + c
∵ RVA(1 + m) = Wa{e + log. 1 + c} − VA(b + r).
e + ce + c

For brevity, let

e′ = a{e + log. 1 + c};
e + ce + c
∵ RVA(1 + m) = We′ − VA(b + r).(22.)

By solving this for VA, we obtain

VA = We′;
R(1 + m) + b + r
∵ RVA = We′R.(23.)
R(1 + m) + b + r

This quantity RVA, being the product of the resistance RA, of the load reduced to the surface of the piston, multiplied by the space through which the piston is moved, will be equal to the load itself multiplied by the space through which it is moved. This being, in fact, the useful effect of the engine, let it be expressed by U, and we shall have

U = We′R.(24.)
R(1 + m) + b + r

Or by (22.),

U(1 + m) = We′ − VA(b + r).(25.)

The value of the useful effect obtained from these formulæ will be expressed in pounds, raised one foot per minute, W being the effective evaporation in cubic feet per minute, A the area of the piston in square feet, and V the space per minute through which it is moved, in feet.

Since a resistance amounting to 33,000 pounds moved through one foot per minute is called one-horse power, it is evident that the horse power H of the engine is nothing more than the useful effect per minute referred to a larger unit of weight or resistance; that is to 33,000 pounds instead of one pound. Hence we shall have

H = U.(26.)
33000

Since the useful effect expressed in (24.) and (25.) is that due to a number of cubic feet of water, expressed by W, we shall obtain the effect due to one cubic foot of water, by dividing U by W. If, therefore, U′ be the effect produced by the effective evaporation of a cubic foot of water, we shall have [Pg517]

U′ = U.(27.)
W

If the quantity of fuel consumed per minute be expressed by F, the effect produced by the unit of fuel, called the DUTY of the engine, will, for like reason, be

D = U.(28.)
F

If the fuel be expressed in hundredweights of coal, then D will express the number of pounds' weight raised one foot by a hundredweight of coal.

By solving (24.) and (25.) for W, we obtain

W = U{R(1 + m) + b + r},(29.)
Re′
W = 1{U(1 + m) + VA(b + r)}.(30.)
e′

By eliminating U, by (26.), we shall have

W = 33000 H{R(1 + m) + b + r},(31.)
Re′
W = 1{33000 H(1 + m) + VA(b + r)}.(32.)
e′

The evaporation necessary per horse power per minute will be found by putting H = 1 in these formulæ.[41]

It will be observed that the quantities A and V, the area of the cylinder and the speed of the piston, enter all these formulæ as factors of the same product. Other things, therefore, being the same, the speed of the piston will be always inversely as the area of the cylinder. In fact, VA is the volume of steam per minute employed in working the piston, and if the piston be increased or diminished in magnitude, its speed must be inversely [Pg518] varied by the necessity of being still moved through the same number of cubic feet by the same volume of steam.

It has been already stated in the text, that no satisfactory experiments have yet been made, by which the numerical value of the quantity r can be exactly known. In engines of different magnitudes and powers, this resistance bears very different proportions to the whole power of the machine. In general, however, the larger and more powerful the engine, the less that proportion will be.

That part of this resistance which arises from the reaction of the uncondensed vapour on the piston is very variable, owing to the more or less perfect action of the condensing apparatus, the velocity of the piston, and the magnitude and form of the steam passages. M. de Pambour states, that, by experiments made with indicators, the mean amount of this resistance in the cylinder is 21⁄2 lbs. per square inch more than in the condenser, and that the pressure in the latter being usually 11⁄2 lb. per square inch, the mean amount of the pressure of the condensed vapour in the cylinder is about 4 lbs. per square inch. Engineers, however, generally consider this estimate to be above the truth in well-constructed engines, when in good working order.

In condensing low pressure engines of forty horse power and upwards, working with an average load, it is generally considered that the resistance produced by the friction of the machine and the force necessary to work the pumps may be taken at about 2 lbs. per square inch of piston surface.

Thus the whole resistance represented by r in the preceding formulæ, as applied to the larger class of low pressure engines, may be considered as being under 6 lbs. per square inch, or 864 lbs. per square foot, of the piston. It is necessary, however, to repeat, that this estimate must be regarded as a very rough approximation; and as representing the mean value of a quantity subject to great variation, not only in one engine compared with another, but even in the same engine compared with itself at different times and in different states.

In the same class of engines, the magnitude of the clearage is generally about a twentieth part of the capacity of the cylinder, so that c = 0·05.

That part of the resistance which is proportional to the load, and on which the value of m in the preceding formulæ depends, is still more variable, and depends so much on the form, magnitude, and the arrangement of its parts, that no general rule can be given for its value. It must, in fact, be determined in every particular case.

In the practical application of the preceding formulæ in condensing engines we shall have [Pg519]

a = 3875969 b = 164 c = 0·05;
e′ = 3875969{e + log.1·05}.
e + 0·05e + 0·05

In engines which work without condensation, and therefore with high pressure steam, we shall have

a = 4347826 b = 618 c = 0·05;
e′ = 4347826{e + log.1·05}.
e + 0·05e + 0·05

To facilitate computation, the values of e′ corresponding to all values of e, from e = ·10 to e = ·90, are given in the following table:—

eCondensing Engines e′.Non-condensing Engines e′.eCondensing Engines e′.Non-condensing Engines e′.
·101012626511359029·5159663676692708
·11995686711169008·5259038376622565
·12979313610985344·5358422886553525
·13963492610807875·5457816936485552
·14948202910636364·5557220246418619
·15933421910470560·5656632516352693
·16919125110310186·5756053536287745
·17905288810154978·5855482976223742
·18891889610004675·5954920646160662
·1987890439859014·6054366286098478
·2086631209717760·6153819696037166
·2185409189580682·6253280655976699
·2284222429447559·6352748965917057
·2383069169318193·6452224445858219
·2481947709192396·6551706845800159
·2580856449069984·6651196055742860
·2679793928950796·6750691865686304
·2778758708834674·6850194105630469
·2877749528721468·6949702635575340
·2976765148611048·7049217275520894
·3075804478503284·7148737905467121
·3174866408398056·7248264345414000
·3273949908295250·7347796485361519
·3373054078194760·7447334175309659
·3472178078096496·7546877285258408
·3571320978000352·7646425695207751
·3670482067906249·7745979285157676
·3769660587814100·7845537945108170
·3868855857723832·7945101555059218
·3968067207635365·8044669995010808
·4067294087548642·8144243174962931
·4166535787463580·8243820964915569
·4265791877380132·8343403324868720
·4365061747298230·8442990104822368
·4464344917217822·8542581204776500
·4563640997138858·8642176584731113
·4662949447061285·8741776134686192
·4762269896985058·8841379744641728
·4861601906910126·8940987374597713
·4960945106836450·9040598934554140
·5060299166763992

[Pg520]

In engines which work without expansion we have

e′ = a.
1 + c

For condensing engines without expansion, we shall then have

e′ = 3875969 = 3691399;(33.)
1·05

and for non-condensing engines,

e′ = 4347826 = 4140787.(34.)
1·05

As the diameters of the cylinders of engines are generally expressed in inches, the corresponding areas of the pistons expressed in square feet are given in the following table, so that the values of A may be readily found:—

Diameter.Area.Diameter.Area.
Inches.Sq. Feet.Inches.Sq. Feet.
100754548127566
110766049137095
120778550137635
130792251147186
141706952147748
151722753157321
161739654157904
171757655167499
181776756177104
191796957177721
202718258187348
212740559187986
222764060197635
232788561207295
243714262207966
253740963217648
263768764227340
273797665237044
284727666237758
294758767247484
304790968257220
315724169257967
325758570267725
335794071277494
346730572287274
356768173297065
367706974297867
377746775307680
387787676317503
398729677327338
408772778337183
419716879347039
429762180347907
4310708581357785
4410755982367674
4511704583377574
4611754184387485
4712704885397406
Diameter.Area.Diameter.Area.
Inches.Sq. Feet.Inches.Sq. Feet.
86407339124837863
87417283125857221
88427237126867590
89437202127877970
90447179128897361
91457166129907763
92467164130927175
93477173131937599
94487193132957033
95497224133967479
96507265134977935
97517318135997402
985273821361007880
995374561371027369
1005475421381037869
1015576381391057380
1025677451401067901
1035778631411087434
1045879921421097977
1056071321431117532
1066172831441137097
1076274451451147674
1086376171461167261
1096478011471177859
1106579951481197468
1116772011491217088
1126874171501227719
1136976441511247361
1147078821521267013
1157271311531277676
1167373911541297351
1177476621551317036
1187579441561327732
1197772361571347439
1207875401581367157
1217978541591377886
1228171801601397626
1238275161611417377

[Pg521]

The practical application of the preceding formulæ will be shown by the following examples.

EXAMPLES.

1. A 36-inch cylinder with 51⁄2 feet stroke is supplied by a boiler evaporating effectively 60 cubic feet of water per hour, and the piston makes 20 strokes per minute without expansion;—what is the power of the engine and the pressure of steam in the cylinder?

Let it be assumed that r = 6 × 144 = 864 and m = 0·1. Since the engine is a condensing engine, we have b = 164 and e′ = 3691399. By the formulæ (25.) and (26.) we have

H = We′ − VA(b + r);
33000(1 + m)

and since by the data we have

W = 1 A = 7·069 V = 2nL = 40 × 5·5 = 220,

the formula, by these substitutions, becomes

H = 3691399 − 220 × 1028 × 7·069;
33000 × 1·1
∵ H = 57·6.

Since e = 1, the pressure P of steam in the cylinder, by (18.), is

P = We′ − b.
VA

Therefore

P = 3691399 − 164 = 2210;
1555·18

which being the pressure in pounds per square foot, the pressure per square inch will be 151⁄3 lbs.

2. To find the effective evaporation necessary to produce a power of 80 horses with the same engine. Also, find the pressure of steam in the cylinder, the speed of the piston being the same.

By the formula (32.), with the above substitutions, we have

W = 33000 × 80 × 1·1 + 220 × 7069 × 1028 = 1·22.
3691399

The evaporating power would therefore be only increased 22 per cent., while the working power of the engine would be increased nearly 40 per cent.

The pressure P in the cylinder will be given, by (18.), as before.

P = 1·22 × 3691399 − 164 = 2732;
1555·18

which is equivalent to 19 lbs. per square inch. [Pg522]

3. What must be the diameter of a cylinder to work with a power of a hundred horses, supplied by a boiler evaporating effectively 70 cubic feet of water per hour, the mean speed of the piston being 240 feet per minute, and the steam being cut off at half stroke? Also, what will be the full pressure of steam on the piston?

Taking, as in the former examples, m = 0·1, b = 164, and r = 864, we shall have

H = 100 W = 7⁄6 V = 240,

and by the column for condensing engines, in table, p. 519, we have e′ = 6029916, where e = 0·50. Making these substitutions in

We′ = 33000 H(1 + m) + VA(b + r),

we shall have

(7⁄6) × 6029916 = 3300000 × 1·1 + 240 × 1028 × A.

Whence we find

A = 13·8;

and by the table, p. 520, the corresponding diameter of the cylinder will be 501⁄3 inches.

If P′ be the full pressure of the steam, we shall have, by (18.),

P′ = Wa − b.
VA(e + c)

Making in this the proper substitutions, we have

P′ = 7⁄6 × 3875969 − 164 = 2318;
240 × 13·8 × 0·55

which being in pounds per square foot, the pressure per square inch will be 161⁄10 lbs.