Dionysius Lardner — The Steam Engine Explained and Illustrated

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. a 1 = 0·82340688193 − 1

log. b 1 = −·01309734295

log. a 2 = 0·74110951837

log. b 2 = −·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 − a 1 b 1 20 + t − a 2 b 2 20 + 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

)

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

)

.(3.)

201·18

t = 201·18 p 1⁄6 − 103

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

p = (

98·806 + t

)

. (4.)

198·562

t = 198·562 p 1⁄6 − 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 p 1⁄5 − 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

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)

1 + n(T − 32)

Eliminating v, we shall obtain

P′

1 + n(T − 32)

;

V′

1 + n(T′ − 32)

or,

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.
1	102·9	20868	1739
2	126·1	10874	1812
3	141·0	7437	1859
4	152·3	5685	1895
5	161·4	4617	1924
6	169·2	3897	1948
7	175·9	3376	1969
8	182·0	2983	1989
9	187·4	2674	2006
10	192·4	2426	2022
11	197·0	2221	2036
12	201·3	2050	2050
13	205·3	1904	2063
14	209·1	1778	2074
15	212·8	1669	2086
16	216·3	1573	2097
17	219·6	1488	2107
18	222·7	1411	2117
19	225·6	1343	2126
20	228·5	1281	2135
21	231·2	1225	2144
22	233·8	1174	2152
23	236·3	1127	2160
24	238·7	1084	2168
25	241·0	1044	2175
26	243·3	1007	2182
27	245·5	973	2189
28	247·6	941	2196
29	249·6	911	2202
30	251·6	883	2209
31	253·6	857	2215
32	255·5	833	2221
33	257·3	810	2226
34	259·1	788	2232
35	260·9	767	2238
36	262·6	748	2243
37	264·3	729	2248
38	265·9	712	2253
39	267·5	695	2259
40	269·1	679	2264
41	270·6	664	2268
42	272·1	649	2273
43	273·6	635	2278
44	275·0	622	2282
45	276·4	610	2287
46	277·8	598	2291
47	279·2	586	2296
48	280·5	575	2300
49	281·9	564	2304
50	283·2	554	2308
51	284·4	544	2312
52	285·7	534	2316
53	286·9	525	2320
54	288·1	516	2324
55	289·3	508	2327
56	290·5	500	2331
57	291·7	492	2335
58	292·9	484	2339
59	294·2	477	2343
60	295·6	470	2347
61	296·9	463	2351
62	298·1	456	2355
63	299·2	449	2359
64	300·3	443	2362
65	301·3	437	2365
66	302·4	431	2369
67	303·4	425	2372
68	304·4	419	2375
69	305·4	414	2378
70	306·4	408	2382
71	307·4	403	2385
72	308·4	398	2388
73	309·3	393	2391
74	310·3	388	2394
75	311·2	383	2397
76	312·2	379	2400
77	313·1	374	2403
78	314·0	370	2405
79	314·9	366	2408
80	315·8	362	2411
81	316·7	358	2414
82	317·6	354	2417
83	318·4	350	2419
84	319·3	346	2422
85	320·1	342	2425
86	321·0	339	2427
87	321·8	335	2430
88	322·6	332	2432
89	323·5	328	2435
90	324·3	325	2438
91	325·1	322	2440
92	325·9	319	2443
93	326·7	316	2445
94	327·5	313	2448
95	328·2	310	2450
96	329·0	307	2453
97	329·8	304	2455
98	330·5	301	2457
99	331·3	298	2460
100	332·0	295	2462
110	339·2	271	2486
120	345·8	251	2507
130	352·1	233	2527
140	357·9	218	2545
150	363·4	205	2561
160	368·7	193	2577
170	373·6	183	2593
180	378·4	174	2608
190	382·9	166	2622
200	387·3	158	2636
210	391·5	151	2650
220	395·5	145	2663
230	399·4	140	2675
240	403·1	134	2687

[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 =

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 =

,(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 =

− b;

∵ de″ =

ads

− bds.

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.

− b(S − S′).(12.)

S′

But by (11.) we have

E′ = a − bS′,

E = a − bS;

∵ E′ − E = b(S − S′);

∵ E″ = a log.

− E′ + E;

S′

∵ E = E″ + E′ = a log.

+ E.(13.)

S′

Or,

E = a (1 + log.

) − 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.)

Since by (10.) we have

P =

− b;

∵ P =

− 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 = W

− 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)

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)

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

P′ =

− 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{

+ log.

1 + c

} − bVA.(19.)

e + c

If the engine work without expansion, e = 1;

∵ E′ =

− bVA,(20.)

1 + c

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

E − E′ = Wa{

−

+ log.

1 + c

};(21.)

e + c

1 + c

e + 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:—

The resistance produced by the load which the engine is required to move. 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. 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{

+ log.

1 + c

} − VAb;

e + c

∵ RVA(1 + m) = Wa{

+ log.

1 + c

} − VA(b + r).

e + c

For brevity, let

e′ = a{

+ log.

1 + c

};

e + 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 =

.(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′ =

.(27.)

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 =

.(28.)

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 =

{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 =

{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{

+ log.

1·05

e + 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{

+ log.

1·05

e + 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:—

e	Condensing Engines e′.	Non-condensing Engines e′.	e	Condensing Engines e′.	Non-condensing Engines e′.
·10	10126265	11359029	·51	5966367	6692708
·11	9956867	11169008	·52	5903837	6622565
·12	9793136	10985344	·53	5842288	6553525
·13	9634926	10807875	·54	5781693	6485552
·14	9482029	10636364	·55	5722024	6418619
·15	9334219	10470560	·56	5663251	6352693
·16	9191251	10310186	·57	5605353	6287745
·17	9052888	10154978	·58	5548297	6223742
·18	8918896	10004675	·59	5492064	6160662
·19	8789043	9859014	·60	5436628	6098478
·20	8663120	9717760	·61	5381969	6037166
·21	8540918	9580682	·62	5328065	5976699
·22	8422242	9447559	·63	5274896	5917057
·23	8306916	9318193	·64	5222444	5858219
·24	8194770	9192396	·65	5170684	5800159
·25	8085644	9069984	·66	5119605	5742860
·26	7979392	8950796	·67	5069186	5686304
·27	7875870	8834674	·68	5019410	5630469
·28	7774952	8721468	·69	4970263	5575340
·29	7676514	8611048	·70	4921727	5520894
·30	7580447	8503284	·71	4873790	5467121
·31	7486640	8398056	·72	4826434	5414000
·32	7394990	8295250	·73	4779648	5361519
·33	7305407	8194760	·74	4733417	5309659
·34	7217807	8096496	·75	4687728	5258408
·35	7132097	8000352	·76	4642569	5207751
·36	7048206	7906249	·77	4597928	5157676
·37	6966058	7814100	·78	4553794	5108170
·38	6885585	7723832	·79	4510155	5059218
·39	6806720	7635365	·80	4466999	5010808
·40	6729408	7548642	·81	4424317	4962931
·41	6653578	7463580	·82	4382096	4915569
·42	6579187	7380132	·83	4340332	4868720
·43	6506174	7298230	·84	4299010	4822368
·44	6434491	7217822	·85	4258120	4776500
·45	6364099	7138858	·86	4217658	4731113
·46	6294944	7061285	·87	4177613	4686192
·47	6226989	6985058	·88	4137974	4641728
·48	6160190	6910126	·89	4098737	4597713
·49	6094510	6836450	·90	4059893	4554140
·50	6029916	6763992

[Pg520]

In engines which work without expansion we have

e′ =

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.
10	07545	48	127566
11	07660	49	137095
12	07785	50	137635
13	07922	51	147186
14	17069	52	147748
15	17227	53	157321
16	17396	54	157904
17	17576	55	167499
18	17767	56	177104
19	17969	57	177721
20	27182	58	187348
21	27405	59	187986
22	27640	60	197635
23	27885	61	207295
24	37142	62	207966
25	37409	63	217648
26	37687	64	227340
27	37976	65	237044
28	47276	66	237758
29	47587	67	247484
30	47909	68	257220
31	57241	69	257967
32	57585	70	267725
33	57940	71	277494
34	67305	72	287274
35	67681	73	297065
36	77069	74	297867
37	77467	75	307680
38	77876	76	317503
39	87296	77	327338
40	87727	78	337183
41	97168	79	347039
42	97621	80	347907
43	107085	81	357785
44	107559	82	367674
45	117045	83	377574
46	117541	84	387485
47	127048	85	397406

Diameter.	Area.	Diameter.	Area.
Inches.	Sq. Feet.	Inches.	Sq. Feet.
86	407339	124	837863
87	417283	125	857221
88	427237	126	867590
89	437202	127	877970
90	447179	128	897361
91	457166	129	907763
92	467164	130	927175
93	477173	131	937599
94	487193	132	957033
95	497224	133	967479
96	507265	134	977935
97	517318	135	997402
98	527382	136	1007880
99	537456	137	1027369
100	547542	138	1037869
101	557638	139	1057380
102	567745	140	1067901
103	577863	141	1087434
104	587992	142	1097977
105	607132	143	1117532
106	617283	144	1137097
107	627445	145	1147674
108	637617	146	1167261
109	647801	147	1177859
110	657995	148	1197468
111	677201	149	1217088
112	687417	150	1227719
113	697644	151	1247361
114	707882	152	1267013
115	727131	153	1277676
116	737391	154	1297351
117	747662	155	1317036
118	757944	156	1327732
119	777236	157	1347439
120	787540	158	1367157
121	797854	159	1377886
122	817180	160	1397626
123	827516	161	1417377

[Pg521]

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

EXAMPLES.

1. A 36-inch cylinder with 5 1 ⁄ 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.

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′ =

− 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.

FOOTNOTES: