BOOK 02 [ES]

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I Gusti N. DirgantaraIr. Dwi Priyanta,

MSE.01 10/ 3/ 12 Document Format

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DOCUMENT NO. DOC. NO. 02 - 42 09 050 - ES

NUMBER OF PAGES 9 6 3 -

DESIGN-IV: MACHINERY BASIC DESIGN

DISPLACEMENT, LWT AND DWT

ATTACHMENT NO. 01 02 03 - -

Ir. Hari Prastowo,

MSc.

REV. DATE DESCRIPTION PREPARED BY CHECKED BY APPROVED BY


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TABLE OF CONTENTS

PHILOSOPHY

1. INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1 Descript ion 1

1.2 Purpose 1

2. REFERENCES 1

3. ABBREVIATIONS 1

4. DESIGN PARAMETER 2

4.1 Principal Dimensions 2

4.2 Coeff icients and Contants 2

4.3 Proj ect Guide Dat 2

4.4 Another Parameters 3

5. DESIGN REQUIREMENTS 3

5.1 Displacement Calculat ions 3

5.2 Lightweight Tonnage 3

5.3 Deadweight Tonnage 7

5.4 Payload 7

6. SUMMARY 9

LIST OF TABLES

Table 5.2.1 - weight base desig 5LIST OF FIGURES

Figure 5.2.1 - Out f it Weight Graph 6

Figure 5.2.2 - Main Engine Weight 6

Figure 5.2.3 - Weight Remainder 6

ATTACHMENT NO. 01 - CALCULATION

1. Displacement Calculat io 1

2. Ligh Weight Tonnage 1

3. Dead Weight Tonnage 6

4. Payload 6

LIST OF TABLES

Table 1 - weight base design 2

LIST OF FIGURES

Figure 1 - Out f it Weight Graph 2

Figure 2 - Main Engine Weight 4

Figure 3 - Weight Remainder 4

ATTACHMENT NO. 02 - CARGO HOLDS VOLUME

ATTACHMENT NO. 03 - LIQUID CHARACTERISTIC

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: DESIGN IV

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: Table of Contents

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1 INTRODUCTION

1.1 Description

a. Displacement

b. Light Weight Tonnage

1) Structural Weight Approximation (WS)

2) Outfit Weight Calculation (WO)

3) Machinery Weight (WM)

4) Margin Merchant Ship

c. Dead Weight Tonnage

1.2 Objective

2 REFERENCESa. Practical Ship Design : Watson D. G. M.

b. Engine Selection Guide - Two Stroke MC/MC-C Engines, 6th Edition: January 2002, MAN B&W

3 ABBREVIATIONS

Lpp = Length of between perpendicular

Lwl = Length of waterline

B = Breadth of ship

H = Height of ship

T = Draught of ship

Vs = Ships velocity

Cb = Block coefficient

sea water = Sea water density

K = Wet steel weight's constant

The word displacement refers to the weight of the water that the ship displaces while

floating. Another way of thinking about displacement is the amount of water that would spill

out of a completely filled container were the ship to be placed into it. A floating ship always

displaces an amount of water of the same weight as the ship. The weight of water that

would displaced by the volume of the hull measured on the outer surface of the shell plating

below the waterline. Displacement tonnage of a vessel can be obtained directly from

Archimedes principle by multiplying its underwater volume by the density of water. A ship's

displacement is its weight at any given time, generally expressed in metric tons or long tons.

The term is often used to mean the ship's weight when it is loaded to its maximum capacity.

A number of synonymous terms exist for this maximum weight, such as loaded displacement,

full load displacement and designated displacement. Displacement is a measurement ofweight, and should not be confused with similarly named measurements of volume or

capacity such as net tonnage, gross tonnage, or deadweight tonnage.

The components of the lightweight in merchant ship practice consist of the structural

weight, the outfit weight, the machinery weight and the margin.

Is the weight that come from the value of weight displacement minus the light weight

tonnages. That consist of cargo's weight, fuel oil, fresh water, ballast water, provision and

ship's crew weight

The objective of this document is to determine the estimation of displacement, light weight

tonnage, and dead weight tonnage in order to find the relation between among of them.

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SFOC = Specific Fuel Oil Consumption

W res = Reserve weight

= Displacement volume = Ships displacement

Wst = Wet steel weight

E = Steel weights parameter

l1 = Length of forecastle deck

l2 = Length of poop deck

h1 = Height of forecastle deck

h2 = Height of poop deck

Woa = Weight of outfit and accomdation

Wm = Machineriy weight

= Main engines weight

= Maximum continous rating (kW)

= Engine RPM

Wr = Auxiliary engines weight

Wres = Reserve weight

4 DESIGN PARAMETER

4.1 Principal Dimension

1. Lpp = 123 m

2. B = m

3. T = 8.8 m

4. H = m

5. LWL = m

6. Vs = knot = km/hours

7. Distance = Nm = km

8. Time of Voyage = 4 days = 96 hours

4.2 Coefficient and Constants

1. Cb disp =

2. Cb wl =3. Cp disp =

4. Cp wl =

5. Am =

6. Cm =

4.3 Project Guide's Data

1. BHP = kW

2. SFOC = gr/kWh

3. HFO = ton/m3

4. MCR = kW

5. RPM = r/min

6. SLOC = g/BHPh0.95

125.46

RPM

Wd

MCR


: DESIGN IV

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6258.88

127

20.2

14.5

173

0.991

0.694380.717

0.70321

174.916

0.984

6320

127.92

11.5

1200

26.8308

2222.4

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4.4 Another Parameters

l1 = length of full width erectio = m

h1 = height of full width erectio = ml2 = length of houses = m

h2 = height of houses = m

5 DESIGN REQUIREMENTS

5.1 Displacement Calculation

a. Displacement Volume

= Lwl x B x T x Cb (1)

where : = Displacement volume

Lwl = Ships length on the water lineB = Ship width in the middle of ship

T = Draft on fully cargo

Cb = Block coefficients

b. Weight Displacement

= x sea water (2)

where : = ships displacement

= ships displacement volume

sea water = the density of sea water

5.2 Light Weight Tonnage

2.1 Structural Weight Approximations

a.

For those not familiar with the old E number, the formula for this is as follows:

E = L ( B +T ) + 0.85 L ( D -T ) + 0.85 ( l 1 h1 ) + 0.75 ( l2 h2 ) . . . . . . (3)

Where :

L = legth between perpendicular

B = breadth

T = draftD = depth

l1 = length of full width erectio = m

h1 = height of full width erectio = m

l2 = length of houses = m


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2.5

as the result that we found, we can know that value is suitable with our ship or not by the

table 1 below:

31.2

2.5

10.6

The formula to calculate the structural weight of our ship reffer to Practical Ship Design -

Chapter 4 - 4.2 Structural Weight Approximations, 4.2.1 Lloyd's Equipment number method.

10.6

2.5

31.2

2.5

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t able 5.2.1 - weight base design

b. The effect of t he block coefficient on steel-weight

The standard block was set at Cb' = 0.70 measured at 0.8D

WS = WSI (1 + 0.05 ( CB' - 0.7 ) (4)

Where :

WS = Steel weight for actual CB at 0.8

WSI = Steel weight for actual CB = 0.70 as plotted/lifted from graph

WSI = K E

. 6

(5)Where :

K = take from the table 1 - weight base design

The relationship Between CB at moulded and CB at Dept h

CB' = CB + (1 - CB) (0.8D - T) / 3T (6)

2.2 Outfit Weight Calculation

(Pract ical Ship Design - Chapter 4, 4.4 Out f it Weight Calculat ions, page 99)

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: DESIGN IV

The traditional method of estimating the outfit weight for a new merchant ship was by

proportioning the outfit weight of a similar ship on the basis of the relative square

numbers, i.e., L x B, and then making corrections for any known differences in the

specifications of the basis and new ships.

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: Philosophy

0.032

By the same token all steel-weights read from the graph must be corrected from the

standard to the desired block coefficient.

Corrections to the steel-weight for variations in Cb from the standard 0.70 value can be

made using the following approximate relationship :

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Figure 5.2.1 - Outfit Weight Graph

Figure 1, show the value of Wo/(LxB) is approximately 0.3

2.3 Machinery Weight

Divided into two components: propulsion machinery and remainder.

a. Propulsion Machinery Weight

Wd = 12 (MCR/RPM)

0.84

(7)

(Pract ical Ship Design, 4.5.4 Propul sion machinery weight, pages 108)

by ploting to the graph in figure 2 :


: DESIGN IV

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Figure 1, this shows that even for a particular type of ship the ratio outfit weight/square

number is not always constant, although near constant values do seem to apply to general

cargo ships and container ships.

To find the outfit weight value, as figure 1 above, I have drawn my estimation according

to my ships length 123 meters, and then I expand a line throug the tanker line. Cross line

between both of them show us the value of Wo/(LxB).

Approximately values for slow and medium speed diesels can be obtained from figure 2,

the base parameter used in this plot is the maximum torque rating of the engines as

represented by MCR/RPM, by the formula :

Where, MCR (Maximum Continous Rating) can be found in EPM (Engine Propeller

Matching) diagram in Design II. It means the engine power after seamargin and

engine margine added value. Both margins 15% and 10%.

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b. Remainder

Calculate by formula :

Wr = K*MCR0.7

(8)

Where,

Wr = Weight of remainder

K = Constants noted

0.19 for f ri gates and corvett es

0.69 for bulk carr ier and general cargo ship

0.72 for t ankers

0.83 for passenger ship

ploting to graph in figure 3 :

Figure 3 - Weight Remainder


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: Philosophy

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

ploting data on figure 2, shown that the weight of maine engine is approximately 310

kW, that value is close with the formula calculation above.

Figure 5.2.2 - Main Engine Weight

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So the result of Machinery Weight is Wd+Wr,Wm = Wd+Wr

2.4 Margin Merchant Ship

for the result LWT :

LWT = Ws+Wo+Wm (9)

so, we can find the margin = 2%*LWT

LWT total = LWT+Margin (10)

5.3 Dead Weight Tonnage

(Pract ical Ship Design, 4.6.5 Deadweight and Displacement - merchant ships, pages 115)

DWT = - LWT (11)

where,

= Weight Displacement

LWT = Light Weight Tonnage

5.4 Payload

Payload = DWT - Wtotal (12)

where,

DWT = Dead Weight Tonnage

Wtotal = Weight of fuel oil, diesel oil, lubricating oil, crews and provision, fresh water

i. HFO (Heavy Fuel Oil)a. HFO's weight

The formula, as follows :

WHFO = SFOC x BHP x time to voyage x constants addition of fuel . . . (13)

Where,

WHFO = weight of heavy fuel oil

SFOC = specific fuel oil consumption (project guide)

BHP = break horse power of main engine (project guide)

constants addition of fuel = 1.3 - 1.5

b. HFO's tank volume



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If a total deadweight is stipulated the required full displacement is the sum of this and the

lightweight.

According to the figure 3 with plot diagram, show that the weight of remainder

approximate 320 ton

The purpose of having a margin is to ensure the attainment of the specified deadweight

even if there has been an underestimate of the lightweight or an overestimate of the load

The figure recommended for the margin for merchant ships was 2% of the lightweight.

Subject to the qualifications made above this still seems as good advice as can be given.

The load capacity that can be transported by ship. In designing, it should be kept to a maximal

capacity of payload to gains the profit. But not out of the minimal requirements of the other

parameters required by the ship. The relation between DWT (Dead Weight Tonnage) with Payload

is shown in formula as follows:

We should consider about the increasing temperature inside the tanks of HFO, so we

add some alocation of expansion margins approximately 2% - 3%.

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VHFO = ((100%+3%)*WHFO)/ HFO (14)

Where,

VHFO = HFO's tanks volumeWHFO = weight of heavy fuel oil

Alocation of expansion = 3%

HFO = 0.991ton/m3

ii. DO (Diesel Oil)

a. DO's weight

The estimation of diesel oil's weight is 10%-20% of heavy fuel oil's weight

for the result :

WDO = 20% x WHFO (15)

b. DO's tanks volume


VDO = ((100%+3%)*WDO)/ DO (16)

Where,

VDO = DO's tanks volume

WDO = weight of heavy fuel oil


DO = 0.85 ton/m

iii. LO (Lubricating Oil)

a. LO's weight


WLO = SLOC x BHP x time to voyage x constant addition of fue(17)

where,

SLOC = Specific Lubricating Oil Consumption = 0.95 g/BHPh

Constants of fuel = 1.3 - 1.5, take 1.4

b. LO's tanks volume


= WLO / LO (18)

where,

LO = 0.9 ton/m3

iv. Fresh Water

a. Consumption for crew

fresh water needs estimation = kg/persons/day

b. Bath and laundry needs

fresh water needs estimation = kg/persons/dayc. Cooking needs


d. Machinery needs


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We should consider about the increasing temperature inside the tanks of LO, so we add

some alocation of expansion margins approximately 2% - 3%.

200

4

20

We should consider about the increasing temperature inside the tanks of DO, so we add


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1. main engine

fresh water needs estimation = 7 gr/kWh

2. auxiliary enginefresh water needs estimation = 0.2 from main engine's fresh water

Total fresh water machinery = fw main engine + fw auxiliary engine

Total Weight of Fresh Water = consumption for crew + bath and laundy + cooking + machinery

Total Volume of Fresh Water = divide the total weight of fresh water by its density.

v. Crew and Provision

a. crew's weight

total crews = 12 persons

average weight of crews = 80 kg

b. provision's weight

average provisions needs = 5 kg/person/day

Weight Total of Ship Supplies

W total = WHFO+WDO+WLO+Wfreshwater+Wcrews+Wprov

PAYLOAD = DWT - W supplies total

determining the type of load () = payload/cargo hold's volume

6 SUMMARY

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0.24Provision Weight

m316.00

0.96

ton

ton

152.73

29.39

16184.29

3935.46

2479.38

3991.04

m3

ton

m3

ton

35.63

0.81

0.92

ton

E Range Number

ton

Weight Displacement

Wr

Wm

margin

E

WSI

CB'

WS

Wd

1.05

Steel weight for actual CB = 0.7

Coefficient Block at Depth

Steel weight for actual CB = 0.8 2523.14

ton

DWT

WHFO

DO tank volume

Weight LO

HFO tank volume

Weight DO

Light Weight Tonnage LWT

VHFO

WDO

VDO

WLO

VLO

Wfw

LO tank volume

15.85

m3

tonWeight of Fresh Water

Volume Tanks Fresh Water

Crew's Weight

Vfw

Displacement Volume

SYMBOL

m3

15789.55

CALCULATION RESULT


: DESIGN IV

Margin Merchant Ship

Dead Weight Tonnage

Weight HFO

78.26

12193.26

146.95

tonton

ton

Main Engine Weight

Remainder Weight

Machinery Weight

316.99

327.25

644.24

ton

ton

ton

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1. Displacement Calculation

a. Displacement Volume

= Lwl x B x T x Cb (1)where : = Displacement volume

Lwl = Ships length on the water line

B = Ship width in the middle of ship

T = Draft on fully cargo

Cb = Block coefficients

for the result :

= Lwl x B x T x Cb

= 127.92 x 20.2 x 8.8 x 0.69438

= m3

b. Weight Displacement

= x sea water (2)

where : = ships displacement

= ships displacement volume

sea water = the density of sea water

for the result :

= x sea water

= 15790 x 1.025

= ton

2. Light Weight Tonnage

2.1 Structural Weight Approximations

a.

For those not familiar with the old E number, the formula for this is as follows:

E = L ( B +T ) + 0.85 L ( D -T ) + 0.85 ( l 1 h1 ) + 0.75 ( l2 h2 ) (3)

Where :

L = legth between perpendicular

B = breadth

T = draft

D = depthl1 = length of full width erectio = m

h1 = height of full width erectio = m

l2 = length of houses = m


for the result :

E = L ( B +T ) + 0.85 L ( D -T ) + 0.85 ( l 1 h1 ) + 0.75 ( l2 h2 )

= 123(20.2+8.8)+0.85*123(11.5-8.8)+0.85(31.2*2.5)+0.75(10.6*2.5)

=

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10.6

3935.46


: DESIGN IV

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: Attachment No. 01

16184.3

15789.5

The formula to calculate the structural weight of our ship reffer to Practical Ship Design -

Chapter 4 - 4.2 Structural Weight Approximations, 4.2.1 Lloyd's Equipment number method.

31.2

2.5

2.5

as the result that we found, we can know that value is suitable with our ship or not by the

table 1 below:

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t able 1 - weight base design

b. The effect of t he block coefficient on steel-weight

The standard block was set at Cb' = 0.70 measured at 0.8D

WS = WSI (1 + 0.05 ( CB' - 0.7 ) (4)

Where :

WS = Steel weight for actual CB at 0.8

WSI = Steel weight for actual CB = 0.70 as plotted/lifted from graph

WSI = K E1.36

(5)

Where :

K = take from the table 1 - weight base design

The relationship Between CB at moulded and CB at Depth

CB' = CB + (1 - CB) (0.8D - T) / 3T (6)

then, the calculation

WSI = K E1.36

= 0.032 * 3935.46^1.36

=

CB' = CB + (1 - CB) (0.8D - T) / 3T

= 0.69438+(1-0.69438)*(0.8*(11.5)-8.8)/3*8.8)

=

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2479.38

: Attachment No. 01

1.05297

By the same token all steel-weights read from the graph must be corrected from the standard to

the desired block coefficient.

Corrections to the steel-weight for variations in Cb from the standard 0.70 value can be madeusing the following approximate relationship :

0.032


: DESIGN IV

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WS = WSI (1 + 0.05 ( CB' - 0.7 )

= 2479.38(1+0.05*(1.05297-0.7)

= ton

2.2 Outfit Weight Calculation

(Pract ical Ship Design - Chapter 4, 4.4 Out f i t Weight Calculat ions, page 99)

Figure 1 - Outfit Weight Graph

Figure 1, show the value of Wo/(LxB) is approximately 0.3

and then the calculation :

0.3 = Wo/(LxB) (7)

0.3 = Wo/(123*20.2)

0.3 = Wo/2484.6

Wo = 0.3*2484.6

= ton

2.3 Machinery Weight

Divided into two components: propulsion machinery and remainder.

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

745.4

The traditional method of estimating the outfit weight for a new merchant ship was by

proportioning the outfit weight of a similar ship on the basis of the relative square numbers,

i.e., L x B, and then making corrections for any known differences in the specifications of the

basis and new ships.

Figure 1, this shows that even for a particular type of ship the ratio outfit weight/square

number is not always constant, although near constant values do seem to apply to general cargo

ships and container ships.

To find the outfit weight value, as figure 1 above, I have drawn my estimation according to my

ships length 123 meters, and then I expand a line throug the tanker line. Cross line between both

of them show us the value of Wo/(LxB).

2523.14


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a. Propulsion Machinery Weight

Wd = 12 (MCR/RPM)0.84

(Practi cal Ship Design, 4.5.4 Propul sion machinery weight, pages 108)

for the result :

Wd = 12 (MCR/RPM)0.84

(8)

= 12*((6258.88/127) 0.84)

= ton

by ploting to the graph in figure 2 :

Figure 2 - Main Engine Weight

b. Remainder

Calculate by formula :

Wr = K*MCR0.7

(9)

Where,

Wr = Weight of remainder

K = Constants noted

0.19 for f ri gates and corvet t es0.69 for bulk carrier and general cargo ship

0.72 for t ankers

0.83 for passenger ship

for the result :

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

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

Where, MCR (Maximum Continous Rating) can be found in EPM (Engine Propeller

Matching) diagram in Design II. It means the engine power after seamargin and engine

margine added value. Both margins 15% and 10%.

316.99

Approximately values for slow and medium speed diesels can be obtained from figure 2, the

base parameter used in this plot is the maximum torque rating of the engines as representedby MCR/RPM, by the formula :

ploting data on figure 2, shown that the weight of maine engine is approximately 310 kW,

that value is close with the formula calculation above.


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Wr = K*MCR0.7

= 0.72*(6258.88^0.7)

= ton

ploting to graph :

Figure 3 - Weight Remainder

So the result of Machinery Weight is Wd+Wr,

Wm = Wd+Wr (10)

= 316.99+327.25

= ton

2.4 Margin Merchant Ship

for the result LWT :

LWT = Ws+Wo+Wm (11)

= 2523.14+745.4+644.24

= ton

so, we can find the margin = 2%*LWT

= 2%*3912.78

= ton

LWT total = LWT+Margin (12)

= 3912.78+78.2556

= ton

: 02 - 42 09 050 - ES

: 01

: Attachment No. 01

According to the figure 3 with plot diagram, show that the weight of remainder approximate

320 ton

644.24

3991.04

327.25

The purpose of having a margin is to ensure the attainment of the specified deadweight even if

there has been an underestimate of the lightweight or an overestimate of the load displacement.The figure recommended for the margin for merchant ships was 2% of the lightweight. Subject

to the qualifications made above this still seems as good advice as can be given.

3912.78

78.2556


: DESIGN IV

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

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

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

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3. Dead Weight Tonnage

(Pract ical Ship Design, 4.6.5 Deadweight and Displacement - merchant ships, pages 115)

DWT = - LWT (13)

where,

= Weight Displacement

LWT = Light Weight Tonnage

DWT = - LWT

= 16184.3-3991.04

= ton

4. Payload

Payload = DWT - Wtotal (14)

where,

DWT = Dead Weight Tonnage

Wtotal = Weight of fuel oil, diesel oil, lubricating oil, crews and provision, fresh water

4.1 HFO (Heavy Fuel Oil)a. HFO's weight


WHFO = SFOC x BHP x time to voyage x constants addition of fuel . . . (15)

Where,


SFOC = specific fuel oil consumption (project guide)

BHP = break horse power of main engine (project guide)

constants addition of fuel = 1.3 - 1.5

for the result :WHFO = SFOC x BHP x time to voyage x constants addition of fuel

= 173*6320*96*1.4

= gram

= ton

b. HFO's tank volume


VHFO = ((100%+3%)*WHFO)/ HFO (16)

Where,VHFO = HFO's tanks volume



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

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

146947584

146.95

We should consider about the increasing temperature inside the tanks of HFO, so we add



: DESIGN IV

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: 01

: Attachment No. 01

The load capacity that can be transported by ship. In designing, it should be kept to a maximal

capacity of payload to gains the profit. But not out of the minimal requirements of the other

parameters required by the ship. The relation between DWT (Dead Weight Tonnage) with Payload is

shown in formula as follows:

12193.3

If a total deadweight is stipulated the required full displacement is the sum of this and the

lightweight.

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

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HFO = 0.991ton/m3

for the result :

VHFO = ((100%+3%)*WHFO)/ HFO

= ((100%+3%)*146.95)/0.991

= m3

4.2 DO (Diesel Oil)

a. DO's weight

The estimation of diesel oil's weight is 10%-20% of heavy fuel oil's weight

for the result :

WDO = 20% x WHFO (17)

= 20%*146.95

= tonb. DO's tanks volume


VDO = ((100%+3%)*WDO)/ DO (18)

Where,

VDO = DO's tanks volume

WDO = weight of heavy fuel oil


DO = 0.85 ton/m3

for the result :

VDO = ((100%+3%)*WDO)/ DO

= ((100%+3%)*29.4)/0.85

= m3

4.3 LO (Lubricating Oil)

a. LO's weight


WLO = SLOC x BHP x time to voyage x constant addition of fu (19)

where,SLOC = Specific Lubricating Oil Consumption = 0.95 g/BHPh

Constants of fuel = 1.3 - 1.5, take 1.4

for the result :

WLO = SLOC x BHP x time to voyage x constant addition of fuel

= 0.95*6320*96*1.4

= gram

= 0.8 ton

b. LO's tanks volume


= WLO / LO (20)

where,

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

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

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

We should consider about the increasing temperature inside the tanks of LO, so we add some

alocation of expansion margins approximately 2% - 3%.

VLO

: DESIGN IV

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: Attachment No. 01

806938

35.63

29.4

We should consider about the increasing temperature inside the tanks of DO, so we add


152.73


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LO = 0.9 ton/m3

for the result :

= ((100%+3%)*WLO) / LO

= ((100%+3%)*0.8)/0.9

= m3

4.4 Fresh Water

a. Consumption for crew

fresh water needs estimation = 20 kg/persons/day

fresh water total (1trip) = 20*12*4

= kg

= ton

b. Bath and laundry needsfresh water needs estimation = kg/persons/day


= kg

= ton

c. Cooking needs



= kg

= ton

d. Machinery needs

1. main engine

fresh water needs estimation = 7 gr/kWh

fresh water total (1trip) = 7 gr/kWh*6320 kW*96 hours

= gram

= ton

2. auxiliary engine

fresh water needs estimation = 0.2 from main engine's fresh water

fresh water total (1trip) = 0.2*4.247

= ton

Total fresh water machinery = fw main engine + fw auxiliary engine

= 4.247 + 0.85= ton

Total Weight of Fresh Water = consumption for crew + bath and laundy + cooking + machinery

= 0.96+9.6+0.192+5.097

= ton

Total Volume of Fresh Water = 15849 kg / 1000 kg/m3

= m3

~ m4

4.5 Crew and Provision

a. crew's weight

total crews = 12 persons

average weight of crews = 80 kg

total weight = 12*80

4247040

4.24704

0.85

5.097

15.849

0.92

960

0.96

200

16

15.849

4

192

0.192

9600

9.6

VLO


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= kg

= ton

b. provision's weightaverage provisions needs = 5 kg/person/day

time of 1 trip = 4 days

total weight of provision = 12 persons x 5 kg/person/day x 4 days

= 240 kg

= ton

Weight Total of Ship Supplies

W total = WHFO+WDO+WLO+Wfreshwater+Wcrews+Wprov (21)

= 146.95+29.4+0.8+15.849+0.96+0.24

= ton

PAYLOAD = DWT - W supplies total (22)

= 12193.3 -194.199

= ton

determining the type of load ( = payload/cargo hold's volume

where,

cargo hold's volume = m3

(the calculation reffers to the attachment)

for the result,

determining the type of load ( = payload/cargo hold's volum (23)

= 11999.101/12265.43

=

the closest density of liquid for the ship's load is =Crude Oil, Mexican, with = 0.973

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

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

. . . . . .

11999.1

12265.43

0.978

so, we know the density of the load type according to our calculation, and then for the best

load that has the closest value, we can find in the attachment about " the density of liquid".

0.24

194.199

0.96

960


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ATTACHMENT NO. 02 - CARGO HOLDS

VOLUMEDISPLACEMENT, LWT AND DWT

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CARGO HOLD'S VOLUME BY SIMPSON METHOD

CARGO HOLD 1 h= 3.5

no frame y S.Factor y x S. Factor no frame y S.Factor y x S. Factor

45 15.1055 1 15.1055 45 16.9945 1 16.9945

50 16.2111 4 64.8444 50 17.8342 4 71.3368

55 17.1159 2 34.2318 55 18.5112 2 37.0224

60 17.8113 4 71.2452 60 19.0361 4 76.1444

65 18.2491 1 18.2491 65 19.2117 1 19.2117

y x s 203.676 y x s 220.7098

luas = 1/3 h y x s (m2) 237.622 luas = 1/3 h y x s (m2) 257.494767


45 18.1237 1 18.1237 45 18.9873 1 18.9873

50 18.7804 4 75.1216 50 19.1343 4 76.5372

55 19.1312 2 38.2624 55 19.25 2 38.5

60 19.2167 4 76.8668 60 19.25 4 77

65 19.25 1 19.25 65 19.25 1 19.25

y x s 227.6245 y x s 230.2745


no frame y S.Factor y x S. Factor

45 19.2188 1 19.2188

50 19.2431 4 76.9724

55 19.25 2 38.5

60 19.25 4 77

65 19.25 1 19.25

y x s 230.9412

luas = 1/3 h y x s (m2) 269.4314

cargo hold 1 PS & SB h = 2.155

WL Area (m^2 S.Factor Area x S. Factor1.4 237.622 1 237.622

3.6 257.4948 4 1029.979067

5.7 265.5619 2 531.1238333

7.9 268.6536 4 1074.614333

10.0 269.4314 1 269.4314

Area x S. Factor 3142.770633


WL 1.4 m WL 3.575 m

WL 5.73 WL 7.885

WL 10.04 m

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Volume = 1/3 h x (Area x S.Factor)

Volume = 2257.56 m^3

CARGO HOLD 2 h= 3.5


65 18.2491 1 18.2491 65 19.2117 1 19.2117

70 18.4876 4 73.9504 70 19.2426 4 76.9704

75 18.5208 2 37.0416 75 19.2473 2 38.4946

80 18.5208 4 74.0832 80 19.25 4 77

85 18.5208 1 18.5208 85 19.25 1 19.25

y x s 221.8451 y x s 230.9267



65 19.25 1 19.25 65 19.25 1 19.25

70 19.25 4 77 70 19.25 4 77

75 19.25 2 38.5 75 19.25 2 38.5

80 19.25 4 77 80 19.25 4 77

85 19.25 1 19.25 85 19.25 1 19.25

y x s 231 y x s 231



65 19.25 1 19.25

70 19.25 4 77

75 19.25 2 38.5

80 19.25 4 77

85 19.25 1 19.25

y x s 231

luas = 1/3 h y x s (m2) 269.5


WL Area (m^2 S.Factor Area x S. Factor

1.4 258.8193 1 258.8192833

3.6 269.4145 4 1077.657933

5.7 269.5 2 539

7.9 269.5 4 1078

10.0 269.5 1 269.5



: DESIGN IV

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: Attachment No. 02

WL 1.4 m WL 3.575 m

WL 5.73 WL 7.885

WL 10.04 m

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Volume = 2315.17 m^3

CARGO HOLD 3 h= 3.325


85 18.5208 1 18.5208 85 19.25 1 19.25

89.75 18.5208 4 74.0832 89.75 19.25 4 77

95 18.5208 2 37.0416 95 19.25 2 38.5

99.25 18.5208 4 74.0832 99.25 19.25 4 77

104 18.5208 1 18.5208 104 19.25 1 19.25

y x s 222.2496 y x s 231



85 19.25 1 19.25 85 19.25 1 19.25

89.75 19.25 4 77 89.75 19.25 4 77

95 19.25 2 38.5 95 19.25 2 38.5

99.25 19.25 4 77 99.25 19.25 4 77

104 19.25 1 19.25 104 19.25 1 19.25

y x s 231 y x s 231



85 19.25 1 19.25

89.75 19.25 4 77

95 19.25 2 38.5

99.25 19.25 4 77

104 19.25 1 19.25

y x s 231

luas = 1/3 h y x s (m2) 256.025



1.4 246.3266 1 246.32664

3.6 256.025 4 1024.1

5.7 256.025 2 512.05

7.9 256.025 4 1024.1

10.0 256.025 1 256.025


WL 1.4 m WL 3.575 m

WL 5.73 WL 7.885

WL 10.04 m


: DESIGN IV

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Volume = 2199.97 m^3



104 18.5208 1 18.5208 104 19.25 1 19.25

108.75 18.5208 4 74.0832 108.75 19.25 4 77

114 18.5233 2 37.0466 114 19.2454 2 38.4908

118.25 18.4635 4 73.854 118.25 19.232 4 76.928

123 18.1327 1 18.1327 123 19.0446 1 19.0446

y x s 221.6373 y x s 230.7134



104 19.25 1 19.25 104 19.25 1 19.25

108.75 19.25 4 77 108.75 19.25 4 77

114 19.25 2 38.5 114 19.25 2 38.5

118.25 19.25 4 77 118.25 19.25 4 77

123 19.1121 1 19.1121 123 19.1352 1 19.1352

y x s 230.8621 y x s 230.8852



104 19.25 1 19.25

108.75 19.25 4 77

114 19.25 2 38.5

118.25 19.25 4 77

123 19.2459 1 19.2459

y x s 230.9959

luas = 1/3 h y x s (m2) 256.0204558



1.4 245.648 1 245.6480075

3.6 255.7074 4 1022.829407

5.7 255.8722 2 511.7443217

7.9 255.8978 4 1023.591053

10.0 256.0205 1 256.0204558


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: Attachment No. 02

WL 5.73 WL 7.885

WL 10.04 m

WL 1.4 m WL 3.575 m


: DESIGN IV

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Volume = 2197.98 m^3

CARGO HOLD 5 h= 3.5


123 18.1327 1 18.1327 123 19.0446 1 19.0446

129 17.3518 4 69.4072 129 18.4656 4 73.8624

134 15.9905 2 31.981 134 17.5231 2 35.0462

139 14.1092 4 56.4368 139 16.1097 4 64.4388

143 11.8546 1 11.8546 143 14.0978 1 14.0978

y x s 187.8123 y x s 206.4898



123 19.1121 1 19.1121 123 19.1352 1 19.1352

129 18.7918 4 75.1672 129 18.9612 4 75.8448

134 17.99 2 35.98 134 18.2331 2 36.4662

139 16.6665 4 66.666 139 17.0688 4 68.2752

143 14.8981 1 14.8981 143 15.248 1 15.248

y x s 211.8234 y x s 214.9694



123 19.2459 1 19.2459

129 18.9405 4 75.762

134 18.4021 2 36.8042

139 17.3309 4 69.3236

143 15.4972 1 15.4972

y x s 216.6329

luas = 1/3 h y x s (m2) 252.7383833



1.4 219.1144 1 219.11435

3.6 240.9048 4 963.6190667

5.7 247.1273 2 494.2546

7.9 250.7976 4 1003.190533

10.0 252.7384 1 252.7383833



: DESIGN IV

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: 01

: Attachment No. 02

WL 10.04 m

WL 1.4 m WL 3.575 m

WL 5.73 WL 7.885

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Volume = 2106.81 m^3


no frme y S.Factor y x S. Factor no frme y S.Factor y x S. Factor

143 11.8546 1 11.8546 143 14.0978 1 14.0978

149.7 8.8536 3 26.5608 149.7 10.9495 3 32.8485

156.3 5.6388 3 16.9164 156.3 7.3898 3 22.1694

163 2.2345 1 2.2345 163 3.9478 1 3.9478

y x s 57.5663 y x s 73.0635


no frme y S.Factor y x S. Factor no frme y S.Factor y x S. Factor

143 14.8981 1 14.8981 143 15.248 1 15.248

149.7 11.9523 3 35.8569 149.7 12.4315 3 37.2945

156.3 8.3358 3 25.0074 156.3 9.2005 3 27.6015

163 4.7537 1 4.7537 163 5.6885 1 5.6885

y x s 80.5161 y x s 85.8325


no frme y S.Factor y x S. Factor

143 15.508 1 15.508149.7 12.8324 3 38.4972

156.3 10.0148 3 30.0444

163 6.7638 1 6.7638

y x s 90.8134

luas = 3/8 h y x s (m2) 158.92345



1.4 100.741 1 100.741025

3.6 127.8611 4 511.4445

5.7 140.9032 2 281.80635

7.9 150.2069 4 600.8275

10.0 158.9235 1 158.92345



Volume = 1187.94 m^3

volume ruang muat t ot al

Volume ruang muat total = V rm 1 + Vrm 2 + Vrm 3 + Vrm 4 + Vrm 5 + Vrm 6

Volume ruang muat total = 12265. 429 m^ 3

WL 10.04 m

WL 1.4 m WL 3.575 m

WL 5.73 WL 7.885


: DESIGN IV

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ATTACHMENT NO. 03 - LIQUID

CHARACTERISTICDISPLACEMENT, LWT AND DWT

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Temperat

ureDensity

t

(o

C) (kg/m3) (

oF) (

oC)

CentiStoke

s(cSt)

Seconds

Saybolt

Universal

(SSU)

AceticAcid 25 1049 61 16.1 0.305

Acetone 25 784.6 68 20 0.295

Acetonitrile 20 782Aceticacid vinegar 10%

CH3COOH59 15 1.35 31.7

Alcohol,ethyl

(ethanol)25 785.1 Aceticacid 50% 59 15 2.27 33

Alcohol,

methyl

(methanol)

25 786.5 Aceticacid 80% 59 15 2.85 35

Alcohol,

propyl

25 800Aceticacid concentrated

glacial

59 15 1.34 31.7

Ammonia

(aqua)25 823.5

Aceticacidanhydride

(CH3COO)2O59 15 0.88

Aniline 25 1019 AcetoneCH3COCH3 68 20 0.41

Automobile

oils15 880940 68 20 1.6

Beer(varies) 10 1010 104 40 0.90cp

Benzene 25 873.8 Alcohol butyln 68 20 3.64 38

Benzil 15 1230 68 20 1.52 31.7

Brine 15 1230 100 37.8 1.2 31.5

Bromine 25 3120 59 15 0.74

ButyricAcid 20 959 32 0 1.04

Butane 25 599 68 20 2.8 35

nButyl

Acetate20 880 122 50 1.4 31.7

nButyl

Alcohol20 810

Aluminumsulfate 36%

solution68 20 1.41 31.7

n

Butylhloride20 886 Ammonia 0 17.8 0.3

Caproicacid 25 921 68 20 4.37 40

Carbolicacid 15 956 50 10 6.4 46.4

Carbondisulfide

25 1261 77 25 159324 737

1.5M

Carbon

tetrachloride25 1584 100 37.8 60108 280500

Carene 25 857Automaticcrankcaseoil

SAE10W0 17.8 1295max 6Mmax

Castoroil 25 956.1Automaticcrankcaseoil

SAE10W0 17.8 12952590 6M12M

Chloride 25 1560Automaticcrankcaseoil

SAE20W0 17.8

2590

1035012M48M

Chlorobenzen

e20 1106

Automaticcrankcaseoil

SAE20

210 98.9 5.79.6 4558

Densityofliquid Kinematicviscosityofliquid

Aniline

AsphaltRC0,MC0,SC0

Alcohol allyl 31.8

Alcohol ethyl(grain)

C2H5OH

Alcohol methyl(wood)

CH3OH

Alcohol propyl

Liquid Liquid

Temperature KinematicViscosity

AcetaldehydeCH3CHO 36

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Temperat

ureDensity

t

(o

C) (kg/m3) (

oF) (

oC)

CentiStoke

s(cSt)

Seconds

Saybolt

Universal

(SSU)

Chloroform 20 1489 AutomaticcrankcaseoilSAE30

210 98.9 9.612.9 5870

Chloroform 25 1465Automaticcrankcaseoil

SAE40210 98.9 12.916.8 7085

Citricacid 25 1660Automaticcrankcaseoil

SAE50210 98.9 16.822.7 85110

Coconutoil 15 924AutomotivegearoilSAE

75W210 98.9 4.2min 40min

Cottonseed

oil15 926

AutomotivegearoilSAE

80W210 98.9 7.0min 49min

Cresol 25 1024AutomotivegearoilSAE

85W210 98.9 11.0min 63min

Creosote 15 1067 AutomotivegearoilSAE90W

210 98.9 1425 74120

Crudeoil,48o

API60

oF 790

AutomotivegearoilSAE

140210 98.9 2543 120200

Crudeoil,40o

API60

oF 825

Automotivegearoil

SAE150210 98.9 43 min 200min

Crudeoil,

35.6oAPI

60o

F 847 Beer 68 20 1.8 32

Crudeoil,

32.6oAPI

60o

F 862 32 0 1

Crude

oil,alifornia60

oF 915 68 20 0.74

Crudeoil,Mexican

60o

F 973 130 54.4 47.5 220

Crudeoil,

Texas60

oF 873 212 100 11.6 65

Cumene 25 860 Bromine 68 20 0.34

Cyclohexane 20 779 50 0.52

Cyclopentane 20 745 30 0.35

Decane 25 726.3 68 20 1.61

Dieselfueloil

20

to

60

15 820950 32 0 2.3cp

Diethylether 20 714 Calciumchloride5% 65 18.3 1.156

o

Dichlorobenz

ene

20 1306 Calciumchloride25% 60 15.6 4 39

Dichlorometh

ane20 1326 65 18.3 11.83

Diethylene

glycol15 1120 194 90 1.26cp

Dichlorometh

ane20 1326 68 20 0.612

Dimethyl

Acetamide20 942 100 37.8 0.53


Liquid Liquid


Benzene(Benzol)C6H6 31

Boneoil

Butanen 1.1

Butyricacidn 31.6

Carbolicacid(phenol) 65

CarbontetrachlorideCCl4

Page2of3

7/31/2019 BOOK 02 [ES]

33/33

Temperat

ureDensity

t

(o

C) (kg/m3)

N,N

Dimethylform

amide

20 949 32 0 0.33

Dimethyl

Sulfoxide20 1100 68 20 0.298

Dodecane 25 754.6 100 37.8 259325 12001500

Ethane 89 570 130 54.4 98130 450600

Ether 25 713.5 69 20.6 308.5 1425

Ethylamine 16 681 100 37.8 125.5 580


Liquid Liquid


CarbondisulfideCS2

Castoroil

Chinawoodoil

Seconds

Saybolt

Universal

(SSU)

CentiStoke

s(cSt)(

oC)(

oF)

BOOK 02 [ES]

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Transcript of BOOK 02 [ES]