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EXAMPLE. – A vessel 260 feet long and finely shaped, having a displacement of 1,000 tons, is to have a speed of 15 knots. What should be the indicated horsepower of the engine?

SOLUTION. – From the table, K = 200. Applying the rule given in Art. 52, we find

153 x $1,000 I. H. P.

3,375 X 100

: 1,687.5. Ans. 200

200

TONNAGE AND DISPLACEMENT

But as

57. By the application of a simple method known as Simpson's rules, the volume of the immersed portion of a ship can be ascertained, which, if considered as water and divided by 35, will give the displacement in tons. vessels vary considerably in form, the mere length, beam, and draft of a ship cannot be multiplied together for the displacement; hence, the coefficient of fineness previously mentioned is used in computing the displacement. Knowing the extreme dimensions of a vessel and the coefficient of fineness, the exact displacement is readily found. For example, take a vessel 100 feet long, 20 feet beam, and floating at 8 feet draft, the coefficient of fineness being .6, the

100 X 20 X 8 X.6 displacement would be

274.2 tons. 35

58. Tonnage refers to the internal capacity, or volume, of a ship. A glance at a tonnage certificate or a register of shipping for any vessel shows two distinct classes of tonnage, namely, gross and net tonnage.

59. Gross tonnage is the entire internal capacity measured according to certain rules as specified in the navigation laws of the United States, and according to size and type of vessel.

*60. Net tonnage is the remainder after having taken from the gross tonnage certain allowances for crew space, engine and boiler rooms, shaft alley, etc. The net tonnage is supposed to represent the earning capacity of the ship, and it is therefore made the basis for port and navigation charges. The detailed rules for computing tonnage are quite complicated, and do not come within the scope of the present text. They will be found, however, in the navigation laws of the United States, or under the Revised Statutes, Chap. I, Title XLVIII, Sec. 4,150 to 4,153, and Chap. 398. If it be required to find, or ascertain, the tonnage of a vessel, the best thing to do is to submit the drawings and plans to the nearest local inspector of the United States Steamboat Inspection Service.

61. Displacement, which is often confused with tonnage, is, as stated before, the weight of the water that the ship displaces, or, what is the same thing, the weight of the ship itself and everything on board. Hence, the displacement of

, a vessel varies from day to day or from one voyage to another, according to the cargo, coal, stores, etc. on board, while tonnage, being determined by the type and internal dimensions of the ship, remains constant.

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62. To Find the Number of Revolutions at Which to Run the Engine in Order to Attain a Certain Speed. Very often the question as to the number of revolutions the engine must be run to drive the vessel at a certain speed comes up before those in charge of a steamer. If the revolutions per minute of the engine for a certain speed of the vessel are known, the question may be readily answered. Assuming the percentage of slip to remain constant, doubling the velocity of the stream projected by the propelling instrument, that is, doubling the revolutions of the engine, and hence of the screw propeller or paddle wheels, doubles the speed of the vessel. In other words, the speed varies directly as the revolutions of the engine.

In actual practice, the percentage of slip varies somewhat at different speeds and under different conditions; hence, the following rule, which is based on the assumption of a constant percentage of slip, does not give the exact number of revolutions per minute required. This can be found only by actual trial. However, it will give a very fair approximation.

Rule. – To find the number of revolutions per minute at which to run the engine in order to give the required speed, divide the product of the revolutions producing any given speed and the required speed by the given speed. Let R = revolutions per minute for a given speed;

S = given speed;
R = required revolutions;

S, = required speed.
Then, the given rule expressed algebraically will be

RS

S.

R.

EXAMPLE 1. - If a vessel is propelled at a rate of 16 knots when the engine is making 32 revolutions per minute, what should be the number of revolutions per minute to give a speed of 14 knots? SOLUTION. -- Applying the foregoing rule, we find

32 x 14 R

28 rev. per min. Ans. 16

EXAMPLE 2.- A steamer leaves Buffalo, N. Y., at 3 P. M., bound for Cleveland, Ohio, the approximate sailing distance between the two places being 173 statute, or 150 nautical, miles. The steamer is capable of a speed of 15 knots (or nautical miles) per hour, at which rate of speed the engine is making 30 revolutions per minute. It was intended that the steamer should call at Erie, Pa., but for some reason this plan is given up and it is decided to run direct for Cleveland. At the ordinary speed of 15 knots, the steamer should arrive at Cleveland at 1 A. M. after a 10 hours' run; this being an inconvenient hour, it is decided to reduce the speed to 10 knots and enter the Cleveland harbor at 6 A. M. In order to do so, at what velocity should the engine be run? SOLUTION. – Applying the given rule, we find

10 x 30 R,

20 rev. per min. Ans. 15

63. To Find the Number of Revolutions the Propeller Should Make in Order to Advance at a Required Rate of Speed, the Pitch of the Propeller Being Known. – In order to solve this problem, the slip of the

propeller for the required speed must be known, and if not
known from trial-trip records, must be assumed at a con-
servative figure. The slip of well-designed propellers varies
between 5 and 15 per cent., the average being about 10 per
cent. Owing to the slip of the propeller, it must be run at
a higher number of revolutions than would be the case
otherwise.
Let P= pitch of propeller in feet;

K = required speed in knots;
R= revolutions per minute at required speed;
N = number of feet in a knot (6,080).

S = per cent. of slip, expressed as a decimal. The number of revolutions for the required speed is then found by the following proportion:

60 XP:N = K:RX(1 - S); whence, 6,080 X K

60 X PXRX (1 - S) R= and K

. 60 x P x (1 – S)'

6,080

=

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EXAMPLE 1. – The pitch of a propeller is 16 feet; how many revolutions per minute must it make to drive the ship at the rate of 10 knots per hour, the slip being estimated at 10 per cent. ?

SOLUTION. – Applying the first formula given, and substituting values, we get

6,080 x 10 R =

704 rev. per min., nearly. Ans. 60 X 16 X(1 — .1)

EXAMPLE 2. – A propeller having a pitch of 20 feet makes 70 revolutions per minute. From a trial-trip record the slip is known to be 12 per cent. at that number of revolutions. What is the speed of the ship? SOLUTION. – Applying the second formula given, we get 60 X 20 X 70 X (1 – .12)

= 12.15 knots per hr. Ans. 6,080

K =

NOTES ON FUEL CONSUMPTION

64. The fuel consumption may be said to vary directly as the horsepower developed (this is not exactly true, but only approximately). The horsepower varies directly as the cube of the speed, whence it follows that the fuel consumption will also vary as the cube of the speed (approximately).

=

Ic S2

3

Then,

Let S = certain speed of vessel;

C = coal consumption at speed S;
s = new speed;
(= coal consumption at speed s.

s C

and s= S3

VC: EXAMPLE 1. -A steamer consumes 100 tons of coal per day at a speed of 10 knots. What should be the speed in order to cut the coal consumption down to 50 tons per day? SOLUTION. – Using the second formula, we find 50 X 102

V500 100

7.9, or 8 knots, nearly. Ans.

S =

EXAMPLE 2. - A steamer consumes 80 tons of coal per day at a speed of 12 knots per hour. Suppose the speed is to be reduced to 10 knots per hour; what would be the fuel consumption per day at that rate of speed? SOLUTION. – Using the first formula, we find 10% X 80 1,250

- 44.8 tons per da. Ans. 123

27

C =

EXAMPLE 3.- If a steamer consumes 15 tons of coal per day to produce a speed of 9 knots per hour, how many knots would she steam if the coal consumption were reduced to 12 tons per day?

SOLUTION.- In this case, c = 12, S = 9, and C = 15. Inserting these values in the second formula, we find the new speed, or 12 x 93 3/4 X 729

1583.2 = 8.3 knots per hr., nearly. Ans. 15

5

S =

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EXAMPLE 4.-A steamer consumes 20 tons of coal per day at a normal speed of 10 knots per hour. The distance to the nearest port where coal can be had is 600 miles, and the estimated quantity of coal in the bunkers is but 35 tons. Find what speed should be maintained in order to reach the coaling station with the coal supply on hand.

SOLUTION. - The best way to proceed in a case of this kind is to assume a lower speed, say 8 knots, and calculate the new coal con

89 x 20 256 sumption for that speed; thus, s =

10.24 tons per day,

25 or .43 ton per hour. The time required to cover a distance of 600 miles at a speed of 8 knots per hour is go 75 hours, and at a coal consumption of .43 ton per hour the total quantity of coal required at that speed is 75 X .43 = 324 tons. Hence, if a speed of 8 knots per hour is maintained, the supply of coal on hand (35 tons) will suffice to reach the coaling station under ordinary weather conditions. Ans.

103

600

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