lows Mariotte's law, pv a constant, and the necessary corrections are made for clearance and compression, the expected mean pressure in practice may be found by multiplying the calculated results by the factor in the following table, according to Seaton.

Particulars of Engine.

Expansive engine, special valve-gear, or with a separate
cut-off valve, cylinder jacketed..

Expansive engine having large ports, etc., and good or-
dinary valves, cylinders jacketed...
Expansive engines with the ordinary valves and gear as
in general practice, and unjacketed.
Compound engines, with expansion valve to h.p. cylin-
der; cylinders jacketed, and with large ports, etc....
Compound engines, with ordinary slide-valves, cylinders
jacketed, and good ports, etc.

Compound engines as in general practice in the merchant
service, with early cut-off in both cylinders, without
jackets and expansion-valves

Fast-running engines of the type and design usually fitted in war-ships..

Factor.

0.94

0.9 to 0.92

0.8 to 0.85

0.9 to 0.92

0.8 to 0.85

0.7 to 0.8

0.6 to 0.8

If no correction be made for clearance and compression, and the engine is in accordance with general inodern practice, the theoretical mean pres sure may be multiplied by 0.96, and the product by the proper factor in the table, to obtain the expected mean pressure.

Given the Initial Pressure and the Average Pressure, to Find the Ratio of Expansion and the Period of Admission.

P

initial absolute pressure in lbs. per sq. in.;

Paverage total pressure during stroke in lbs. per sq. in.;
L= length of stroke in inches;

1= period of admission measured from beginning of stroke;
clearance in inches;

R actual ratio of expansion =

. (1)

To find average pressure p, taking account of clearance,

L+c
R

Given p and P, to find R and I (by trial and error).-There being two unknown quantities R and I, assume one of them, viz., the period of admission 1, substitute it in equation (3) and solve for R. Substitute this value of R in the formula (1), or l = the result is greated than the assumed value of 1, then the assumed value of the period of admission is too long; if less, the assumed value is too short. Assume a new value of 1, substitute it in formula (3) as before, and continue by this method of trial and error till the required values of R and I are obtained.

c, obtained from formula (1), and find l. If

EXAMPLE.-P = 70, p = 42.78, L = 60′′, c = 3′′, to find Assume 7 = 21 in.

hyp log R =

42.78 BL+C

-1=

70

× 60 +3

hyp log R.653, whence R = 1.92,

which is greater than the assumed value, 21 inches.

Now assume l = 15 inches:

Period of Admission Required for a Given Actual Ratio of Expansion:

Pressure at any other Point of the Expansion.-Let L1 = length of stroke up to the given point.

Pl+c)

Pressure at the given point : =

(7)

WORK OF STEAM IN A SINGLE CYLINDER.

To facilitate calculations of steam expanded in cylinders the table on the next page is abridged from Clark on the Steam-engine. The actual ratios of expansion, column 1, range from 1.0 to 8.0, for which the hyperbolic logarithms are given in column 2. The 3d column contains the periods of admission relative to the actual ratios of expansion, as percentages of the stroke, calculated by formula (5) above. The 4th column gives the values of the mean pressures relative to the initial pressures, the latter being taken as 1, calculated by formula (2). In the calculation of columns 3 and 4, clearance is taken into account, and its amount is assumed at 7% of the stroke. The final pressures, in the 5th column, are such as would be arrived at by the continued expansion of the whole of the steam to the end of the stroke, the initial pressure being equal to 1. They are the reciprocals of the ratios of expansion, column 1. The 6th column contains the relative total performances of equal weights of steam worked with the several actual ratios of expansion; the total performance, when steam is admitted for the whole of the stroke, without expansion, being equal to 1. They are obtained by dividing the figures in column 4 by those in column 5.

The pressures have been calculated on the supposition that the pressure of steam, during its admission into the cylinder, is uniform up to the point of cutting off, and that the expansion is continued regularly to the end of the stroke. The relative performances have been calculated without any allow. ance for the effect of compressive action.

The calculations have been made for periods of admission ranging from 100%, or the whole of the stroke, to 6.4%, or 1/16 of the stroke. And though, nominally, the expansion is 16 times in the last instance, it is actually only 8 times, as given in the first column. The great difference between the nominal and the actual ratios of expansion is caused by the clearance, which is equal to 7% of the stroke, and causes the nominal volume of steam admitted, namely, 6.4%, to be augmented to 6.4+7= 13.4% of the stroke, or, say, double, for expansion. When the steam is cut off at 1/9, the actual expansion is only 6 times; when cut off at 1/5, the expansion is 4 times; when cut off at 1, the expansion is 2% times; and to effect an actual expansion to twice the initial volume, the steam is cut off at 46% of the stroke, not at half-stroke.

Expansive Working of Steam-Actual Ratios of Expan sion, with the Relative Periods of Admission, Pressures, and Performance.

Steam-pressure 100 lbs. absolute. Clearance atleach end of the cylinder 7%

of the stroke.

ASSUMPTIONS OF THE TABLE.-That the initial pressure is uniform; that the expansion is complete to the end of the stroke; that the pressure in expansion varies inversely as the volume; that there is no back-pressure of exhaust or of compression, and that clearance is 7% of the stroke at each end of the cylinder. No allowance has been made for loss of steam by cyl. inder-condensation or leakage.

Volume of 1 lb. of steam of 100 lbs. pressure per sq. in., or 14,400 lbs. per sq, ft...

......... .....

Product of initial pressure and volume..............................................

..........

4.33 cu. ft. ..62,852 ft.-1

.........

Though a uniform clearance of 7% at each end of the stroke has been assumed as an average proportion for the purpose of compiling the table, the clearance of cylinders with ordinary slides varies considerably-say from 5% to 10%. (With Corliss engines it is sometimes as low as 2%.) With the clearance, 7%, that has been assumed, the table gives approximate results sufficient for most practical purposes, and more trustworthy than results deduced by calculations based on simple tables of hyperbolic logarithms, where clearance is neglected.

Weight of steam of 100 lbs. total initial pressure admitted for one stroke, per cubic foot of net capacity of the cylinder, in decimals of a pound reciprocal of figures in column 9.

=

Total actual work done by steam of 100 lbs. total initial pressure in one stroke per cubic foot of net capacity of cylinder, in foot-pounds = figures in column 7+figures in column 9.

RULE 1: To find the net capacity of cylinder for a given weight of steam admitted for one stroke, and a given actual ratio of expansion. (Column 9 of table.)-Multiply the volume of 1 lb. of steam of the given pressure by the given weight in pounds, and by the actual ratio of expansion. Multiply the product by 100, and divide by 100 plus the percentage of clearance. The quotient is the net capacity of the cylinder.

RULE 2: To find the net capacity of cylinder for the performance of a given amount of total actual work in one stroke, with a given initial pressure and actual ratio of expansion.-Divide the given work by the total actual work done by 1 lb. of steam of the same pressure, and with the same actual ratio of expansion; the quotient is the weight of steam necessary to do the given work, for which the net capacity is found by Rule 1 preceding. NOTE.-1. Conversely, the weight of steam admitted per cubic foot of net capacity for one stroke is the reciprocal of the cylinder-capacity per pound of steam, as obtained by Rule 1.

2. The total actual work done per cubic foot of net capacity for one stroke is the reciprocal of the cylinder-capacity per foot-pound of work done, as obtained by Rule 2.

3. The total actual work done per square inch of piston per foot of the stroke is 1/144th part of the work done per cubic foot.

4. The resistance of back pressure of exhaust and of compression are to be added to the net work required to be done, to find the total actual work. APPENDIX TO ABOVE TABLE-MULTIPLIERS FOR NET CYLINDER-CAPACITY, AND TOTAL ACTUAL WORK DONE.

(For steam of other pressures than 100 lbs. per square inch.)

The figures in the second column of this table are derived by multiplying the total pressure per square foot of any given steam by the volume in cubic feet of 1 lb. of such steam, and dividing the product by 62,352, which is the product in foot-pounds for steam of 100 lbs. pressure. The quotient is the multiplier for the given pressure.

The figures in the third column are the quotients of the figures in the second column divided by the ratio of the pressure of the given steam to 100 lbs.

Measures for Comparing the Duty of Engines.-Capacity is measured in horse powers, expressed by the initials, H.P.: 1 H.P. = 83,000 ft.-lbs. per minute, = 550 ft.-lbs. per second, = 1,980,000 ft.-lbs. per hour

1 ft.-lb. a pressure of 1 lb. exerted through a space of 1 ft. Economy is measured, 1, în pounds of coal per horse-power per hour; 2, in pounds of steam per horse-power per hour. The second of these measures is the more accurate and scientific, since the engine uses steam and not coal, and it is independent of the economy of the boiler.

In gas-engine tests the common measure is the number of cubic feet of gas (measured at atmospheric pressure) per horse-power, but as all gas is not of the same quality, it is necessary for comparison of tests to give the analysis of the gas. When the gas for one engine is made in one gas-producer, then the number of pounds of coal used in the producer per hour per horse-power of the engine is the proper measure of economy.

Economy, or duty of an engine, is also measured in the number of footpounds of work done per pound of fuel. As 1 horse-power is equal to 1,980,000 ft.-lbs. of work in an hour, a duty of 1 lb. of coal per H.P. per hour would be equal to 1,980,000 ft.-lbs. per lb. of fuel; 2 lbs. per H.P. per hour equals 990,000 ft.-lbs. per lb. of fuel, etc.

The duty of pumping-engines is commonly expressed by the number of foot-pounds of work done per 100 lbs. of coal.

When the duty of a pumping-engine is thus given, the equivalent number of pounds of fuel consumed per horse-power per hour is found by dividing 198 by the number of millions of foot-pounds of duty. Thus a pumping engine giving a duty of 99 millions is equivalent to 198/99 = 2 lbs. of fuel per horse-power per hour.

Efficiency Measured in Thermal Units per Minute.Some writers express the efficiency of an engine in terms of the number of thermal units used by the engine per minute for each indicated horse-power, instead of by the number of pounds of steam used per hour.

The heat chargeable to an engine per pound of steam is the difference between the total heat in a pound of steam at the boiler-pressure and that in a pound of the feed-water entering the boiler. In the case of condensing engines, suppose we have a temperature in the hot-well of 101° F., corresponding to a vacuum of 28 in. of mercury, or an absolute pressure of 1 lb. per sq. in. above a perfect vacuum: we may feed the water into the boiler at that temperature. In the case of a non-condensing-engine, by using a portion of the exhaust steam in a good feed-water heater, at a pressure a trifle above the atmosphere (due to the resistance of the exhaust passages through the heater), we may obtain feed-water at 212°. One pound of steam used by the engine then would be equivalent to thermal units as follows: Pressure of steam by gauge: 50

Total heat in steam above 32° :

75

100

125

150

175

200

1172.8 1179.6 1185.0 1189.5 1193.5 1197.0 1200.2 Subtracting 69.1 and 180.9 heat-units, respectively, the heat above 32o in feed-water of 101° and 212° F., we have

Heat given by boiler:

Feed at 101°

Feed at 212°

1103.7 1110.5

1115.9 1120.4 1124.4 1127.9 1131.1 991.9 998.7 1004.1 1008.6 1012.6 1016.1 1019.3

Thermal units per minute used by an engine for each pound of steam used per indicated horse-power per hour: Feed at 101°...... 18.40 18.51 18.60 18.67 18.74 18.80 18.85

Feed at 212°.

16.53 16.65 16.74 16.81

16.88

16.94 16.99 EXAMPLES.-A triple-expansion engine, condensing, with steam at 175 lbs., gauge and vacuum 28 in., uses 13 lbs. of water per I.H.P. per hour, and a high-speed non-condensing engine, with steam at 100 lbs. gauge, uses 30 lbs. How many thermal units per minute does each consume?

Ans. 13X18.80 244.4, and 30 × 16.74 502.2 thermal units per minute. A perfect engine converting all the heat-energy of the steam into work would require 33,000 ft.-lbs. 778 42.4164 thermal units per minute per indicated horse-power. This figure, 42.4164, therefore, divided by the number of thermal units per minute per I.H.P. consumed by an engine, gives its efficiency as compared with an ideally perfect engine. In the examples above, 42.4164 divided by 244.4 and by 502.2 gives 17.35% and 8.45% efficiency, respectively.

Total Work Done by One Pound of Steam Expanded in a Single Cylinder. (Column 7 of table.)-If 1 pound of water be converted into steam of atmospheric pressure =2116.8 lbs. per sq. ft., it occu pies a volume equal to 26.36 cu. ft. The work done is equal to 2116.8 lbs.