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 = W a − 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 + c e + 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 + 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:—

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

For brevity, let

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