THE EXPANSION OF STEAM.
141
THE EXPANSION OF STEAM.
With a given pressure of steam, and the cut-off taking place atany part of the stroke of the piston, to find the mean pressureexerted on the piston during the stroke, by means of the followingtable of hyperbolic logarithms. Rule:—Divide the length of thestroke of the piston by the length of the space when the steam iscut off from the cylinder; find in the table the logarithm of thenumber nearest to the quotient, and add 1 to it: the sum is the ratioof the gain. Then find the terminal pressure by dividing the initialpressure by the proportion of the stroke during which the steam isadmitted into the cylinder, and multiply by the logarithm + 1, asabove; the product will be the mean pressure exerted on the piston.
Example .— Suppose the length of the stroke to be 24 inches,initial pressure 40 Ibs. per square inch, and the steam to be cut offat 6 inches from the commencement of the stroke, we have—
24-=-6 = 4; hyp. log. of 4 is 1-386 + 1 =2-386.
Then 40 = 4 = 10 x 2-386 = 23-86 Ibs. mean pressure.
HYPERBOLIC LOGARITHMS.
Number.
Logarithm.
Number.
Logarithm.
Number.
Logarithm.
Number.
Logarithm.
ro 5
■048
2*05
717
3'°5
rii 5
4-05
1-398
ri
•095
2*1
741
3'i
1-131
4*10
1*410
1-15
•139
2-15
765
3" 1 5
1-147
4-15
1-423
I *2
•182
2'2
•788
3'2
1-163
4-2
1435
1-25
■223
2*25
•810
3' 2 5
1-178
4-25
1-446
i'3
•262
2'3
•832
3'3
C19 3
4'3
1458
i'35
•300
2'35
•854
3'35
I-2o8
435
1470
i'4
•336
2 4
•875
3'4
1-223
4'4
1481
i'45
•371
2'45
•896
3'4S
1-238
445
1-492
r 5
•405
2-5
•916
3'5
I'252
45
1-504
r 55
•438
2-55
•936
3‘55
1-266
455
1-515
r6
•470
2*6
•955
3'6
I'28 o
4-6
1-526
1-65
•50O
2-65
•974
3'6S
1-294
4-65
1-536
17
•530
27
•993
37
1-308
47
1 '547
r 75
•559
275
I*OI I
375
I-32I
475
1-558
r8
•587
2*8
1*029
3-8
i'335
4'8
1-568
1-85
•615
2‘85
T°4 7
3-85
i'348
485
1-578
i'9
•641
2*9
i’o64
3'9
1-360
49
1-589
r 95
•66 7
2 '95
ro8i
3'95
i'373
495
1 '599
2*0
'693
3‘°
i'o98
4'°
1-386
5'°
1-609