Doc. No. 14 - 42 12 140 - VA

DESAIN IV SISTEM PERMESINAN & KELISTRIKAN KAPALJURUSAN TEKNIK SISTEM PERKAPALAN

FTK - ITS

DOC. NO : 14 - 42 12 140- VA

Rev Tanggal Keterangan

Dikerjakan Oleh Disetujui Oleh

Nama Paraf Dosen Pembimbing Paraf

Atandho Gama M.

TECHNICAL SPECIFICATION OF ENGINE ROOM VENTILATION SYSTEM

Dr. Ir. Aguk Zuhdi M.F, M.Eng

TECHNICAL SPECIFICATION OF ENGINE ROOM

VENTILATION SYSTEM

Project Desain IV Doc. No. 14 - 42 12 140 -VA

Rev. No. 0 By Atandho Gama Magwasyar

1. LIST OF CODE / REFERENCE USED

a. ClassNK

b. Project Guide Wartsila RT FLEX 40

c. ISO 8861 - Engine room ventilation in diesel engined ships

2. CALCULATION ALGORITHM

a. Calculation of Total Air Flow

- Calculation of sum of airflow for combustion

* Calculation of airflow for combustion for main propulsion diesel engine

* Calculation of airflow for combustion for diesel generator engine

* Calculation of airflow for combustion for boiler and thermal fluid heaters

- Calculation of airflow for evaquation of heat emission

* Heat emission from main propulsion diesel engine

* Heat emission from diesel generator engine

* Heat emission from boiler and thermal fluid heater

* Heat emission from steam and condensate pipes

* Heat emission from electrical air-cooled generator

* Heat emission from electrical installations

* Heat emission from exhaust pipes including exhaust gas-fired boilers

* Heat emission from hot tanks

* Heat emission from other components

b. Ducting Design in Engine Room

c. Total Pressure / Pressure Losses

3. INPUT PARAMETER DESIGN

a. Calculation of Total Air Flow (Q)


* Calculation of airflow for combustion for main propulsion diesel engine +

+ Air requirement for combustion for diesel engine(s) (kg/kW.s)+ Density of air (kg/m³)

* Calculation of airflow for combustion for diesel generator engine+

+ Air requirement for combustion for diesel generator engine (kg/kW.s)+ Density of air (kg/m³)

* Calculation of airflow for combustion for boiler and thermal fluid heaters + Total steam capacity (maximum continuous rating) of the boiler (kg/s)+ Fuel consumption (kg.air/kg.steam)+ Air requirement for combustion (kg.air/kg.fuel)+ Density of air (kg/m³)


* Heat emission from main propulsion diesel engine+ Heat loss from main engine (percentage)


Service standard power of the main propulsion diesel engine(s) at maximum continuous power output (kW)

Service standard power of the diesel generator engine(s) at maximum power output (kW)


VENTILATION SYSTEM



+ Heat loss from auxiliary engine (percentage)



* Heat emission from electrical air-cooled generator+ Power of installed air-cooled generator (kW)+ Generator efficiency (percentage)







- Density of air

- Viscosity of air (kg/m.s)

- Roughness factor

- Length of main duct (m)

- Friction factor in pipe

4. OUTPUT PARAMETER DESIGN

a. Calculation of Total Air Flow


* Calculation of airflow for combustion for main propulsion diesel engine

* Calculation of airflow for combustion for diesel generator engine

* Calculation of airflow for combustion for boiler and thermal fluid heaters


* Heat emission from main propulsion diesel engine




* Heat emission from electrical air-cooled generator







(kg/m³)


VENTILATION SYSTEM



5. CALCULATION DETAILS

5.1. Calculation of Total Air Flow (Q)

Q = qc + qh

where :

qc = Sum of airflow for combustion (m³/s)

qh = Airflow for evaquation of heat emission (m³/s)

- Calculation of sum of airflow for combustion (qc)

qc =

where :

=

=

=

*

= (m³/s)

where :=

= 7945 kW

= air requirement for combustion for diesel engine(s) (kg/kW.s)

= 0.0023 kg/kW.s (for 2 stroke engine)= 0.0020 kg/kW.s (for 4 stroke engine)

= 0.0023 kg/kW.s

ρ = Density of air (kg/m³)

= 1.13 kg/m³

=

= 7945(kW) x 0.0023 (kg/kW.s) / 1.13 (kg/m³)

= 16.17 m³/s

*

Calculation of the total air flow using two methods, where the results of the two methods of calculation is taken to the greatest results.

(ISO 8861 - Engine room ventilation in diesel engined ships, 5.1 point (a))

qdp + qdg + qb

qdp Airflow for combustion for main propulsion diesel engine (m3/s)

qdg Airflow for combustion for diesel generator engine (m3/s)

qb Airflow for combustion for boiler and thermal fluid heaters (m3/s)

Calculation of airflow for combustion for main propulsion diesel engine (qdp)

qdp (Pdp x mad) / ρ (ISO 8861 - Engine room ventilation in diesel engined ships, 5.2.2)

Pdp service standard power of the main propulsion diesel engine(s) at maximum continuous power output (kW)

mad

Where specific data for mad are not available, the following values may be used for calculation :

mad

In this case used 2 stroke engine. So the value of mad is :

qdp (Pdp x mad) / ρ

Calculation of airflow for combustion for diesel generator engine (qdg)


VENTILATION SYSTEM



= (m³/s)

where :=

= 200 kW

= air requirement for combustion for diesel generator engine(s) (kg/kW.s)

= 0.0023 kg/kW.s (for 2 stroke engine)= 0.0020 kg/kW.s (for 4 stroke engine)

= 0.0020 kg/kW.s


= 1.13 kg/m³

=

= 200 x 4 genset (kW) x 0.0020 (kg/kW.s) / 1.13 (kg/m³)

= 1.42 m³/s

*

= (m³/s)

(ISO 8861 - Engine room ventilation in diesel engined ships, 5.2.4)

where :

=

= fuel consumption (kg.air/kg.steam)

= air requirement for combustion (kg.air/kg.fuel)


=

= 0 m³/s

qc =

= 16.17 + 1.42 + 0

= 17.59 m³/s

- Calculation of airflow for evaquation of heat emission (qh)

qh =ρ x c x ∆T

qdg (Pdg x mad) / ρ (ISO 8861 - Engine room ventilation in diesel engined ships, 5.2.3)

Pdg service standard power of the diesel generator engine(s) at maximum power output (kW)

mad

Where specific data for mad are not available, the following values may be used for calculation :

mad

In this case used 4 stroke diesel generator engine. So the value of mad is :

qdg (Pdg x mad) / ρ

Calculation of airflow for combustion for boiler and thermal fluid heaters (qb)

qb (ms x mfs x maf) / ρ

ms total steam capacity (maximum continuous rating) of the boiler (kg/s)

mfs

maf

In planning Duct Design on this ship for heating components machining system, such as HFO settling tank is not planned to use as a boiler but using electric heaters.

qb ms x mfs x maf / ρ

So from the calculation of the parameters - the parameters above. It can be seen the amount of air flow for combustion is as follows:

qdp + qdg + qb

Ødp + Ødg + Øb + Øp + Øg + Øel + Øep + Øt + Øo - 0.4 (qdp + qdg) - qb


VENTILATION SYSTEM



(ISO 8861 - Engine room ventilation in diesel engined ships, 5.3)

where :

= Heat emission from main propulsion diesel engine (kW)

= Heat emission from diesel generator engine (kW)

= Heat emission from boiler and thermal fluid heater (kW)

= Heat emission from steam and condensate pipes (kW)

= Heat emission from electrical air-cooled generator (kW)

= Heat emission from electrical installations (kW)

= Heat emission from exhaust pipes including exhaust gas-fired boilers (kW)

= Heat emission from hot tanks (kW)

= Heat emission from other components (kW)

= Air flow for main propulsion diesel engine combustion(m³/s)

= Air flow for diesel generator engine combustion (m³/s)

= Air flow for boiler combustion (m³/s)

ρ = 1.13 kg/m3

c = 1.01 kJ/(kg.K)

∆T = 12.5 K

*

= (kW)

where :

=

= 7945 kW

= heat loss from the diesel engine (percentage)

Ødp

Ødg

Øb

Øp

Øg

Øel

Øep

Øt

Øo

qdp

qdg

qb

Calculation of heat emission from main propulsion diesel engine (Ødp)

Ødp Pdp / (∆hd / 1000) (ISO 8861 - Engine room ventilation in diesel engined ships, 6.1)

Pdp service standard power of the main propulsion diesel engine(s) at maximum continuous rating (kW)

∆hd


VENTILATION SYSTEM



From the graph at attachment, the heat emission is obtained from the main engine:

= 135 kW

*

= (kW)

where :

=

= 200 kWTotal of auxiliary engines are 3, so :

= 200 x 4

= 800 kW

= heat loss from the diesel engine (percentage)

From the graph above, the heat emission is obtained from the auxiliary engine:

=

If specific date fo Ødp not aviable,so the value of Ødp can be obtained by use graphic by ISO 8861 - Engine room ventilation in diesel engined ships, 7.1

Ødp

Calculation of heat emission from diesel generator engine (Ødg)

Ødg Pdp / (∆hd / 1000) (ISO 8861 - Engine room ventilation in diesel engined ships, 6.2)

Pdp service standard power of the diesel generator engine(s) at maximum continuous rating (kW)

∆hd

If specific date of Ødp not aviable,so the value of Ødp can be obtained by use graphic by ISO 8861 - Engine room ventilation in diesel engined ships, 7.1

Ødg0,0396 x P0,70


VENTILATION SYSTEM



=

= 1.62 kW x 4

= 6.46 kW

= 6.46 kW

*

=

or;

=

where :

= total steam capacity (maximum continuous rating) of the boiler (kg/s)

= fuel consumption (kg.air/kg.steam)

= heat loss from the boiler (percentage)

=

Q = maximum continuous rating of the thermal fluid heater or boiler (kW)

=

= 0 kW

*

= (kW)

where :

= total steam consumption (kW)

= heat loss from steam and condensate pipes (percentage)

=

= 0 kW

*

= Pg (1 - (η / 100)) (kW)

where :

Pg = power of installed air-cooled generator (kW)

= 200 kW

Total of auxiliary engines are 4

η = generator efficiency (percentage)

= 94 %

= Pg (1 - (η / 100))

= (4 x 200) x (1 - (94/100))

= 48.00 kW

0,0396 x (200)0,70

Ødg

Calculation of heat emission from boiler and thermal fluid heater (Øb)

Øb ms x mfs x h x (∆hb / 1000) x B1

(ISO 8861 - Engine room ventilation in diesel engined ships, 6.3)Øb Q x B1 x (∆hb / 1000)

ms

mfs

∆hb

B1 constant that applies to the location of the boiler(s) and other heat exchangers in the engine room

Øb ms x mfs x h x (∆hb / 1000) x B1

Calculation of heat emission from steam and condensate pipes (Øp)

Øp msc x (∆hp / 1000) (ISO 8861 - Engine room ventilation in diesel engined ships, 6.4)

msc

∆hp

Øp msc x (∆hp / 1000)

Calculation of heat emission from electrical air-cooled generator (Øg)

Øg (ISO 8861 - Engine room ventilation in diesel engined ships, 6.5)

Øg


VENTILATION SYSTEM



*

Calculation of heat emission is taken 20% of the electric power equipment and lighting.

= 20% x Ptotal (kW)

Known electric equipment:

1. Electric equipment in the engine room :

Ptotal = 385.70 kW2. Lighting equipment in the engine room :

Ptotal = 3.86 kW

So the total power and electric lighting is :

Ptotal = 389.56 kW

= 20% x Ptotal

= 20% x 389.56

= 77.91 kW

*

Known :

Main engine exhaust pipe diameter = 900 mm

Auxiliary engine exhaust pipe diameter = 300 mm

= 320 K for 4 stroke

= 250 K for 2 stroke

Calculation of heat emission from electrical installations (Øel)

Øel (ISO 8861 - Engine room ventilation in diesel engined ships, 6.6)

Øel

Calculation of heat emission from exhaust pipes including exhaust gas-fired boilers (Øep)

This value can be determined by looking at the parameters of the exhaust pipe diameter of main engine and auxiliary engines. Once it is known, then searched by using the exhaust pipe emissions by ISO 8861 - Engine room ventilation in diesel engined ships, 7.3

∆t

∆t

(ISO 8861 - Engine room ventilation in diesel engined ships, 6.7)


VENTILATION SYSTEM



From the graph at attachment, can be obtained :

= 0.51 kW

= 0.25 kW

= 0.51 + 0.25

= 0.76 kW

*

In planning the design duct in the engine room tanks known to have a hot tank them;

1. Settling Tank

Temperature of the tank = 70 ˚C

= 0.234 uninsulated

2 pieces of settling tanks = 0.468 kW

2. Service Tank

Temperature of the tank = 115 ˚C= 0.515 uninsulated

2 pieces of day tanks = 1.03 kW

So can be obtained heat emission from hot tanks

= 0,468 + 1,03

= 1.498 kW

*

1. Compressor

Øep (main engine)

Øep (auxiliary engine)That can be known of the heat emission from exhaust pipes including exhaust gas-fired boilers :

Øep

Calculation of heat emission from hot tanks (Øt)Calculation of heat emission from hot tanks (Øt) according to ISO 8861 - Engine room ventilation in diesel engined ships, 6.7; tabel 1 - Heat emission from hot tanks, so can be obtained :

Øt

Calculation of heat emission from other components (Øo)


VENTILATION SYSTEM



2. Steam

3. Turbine

4. Pump

5. Separator

6. Heat Exchanger

7. Piping and hidraulic sistem

= 11.00 kW

ρ =

= 1.13

c = specific heat capacity of the air

= 1.01 kJ/(kg.K)

=

= 12.5 K

qh =ρ x c x ∆T

=135 + 6,46 + 0 + 0 + 48 + 77.91 + 0.76 + 1.498 + 11 - 0.4 (16,17 + 1,42) - 0

1.13 x 1.01 x 12.5

= 12.64

It can be seen the value of the air flow requirements in the engine room is

Q = qc + qh

= 17.59 + 12.64

= 30.22

If Q> 50% qc, then the Q calculation using the following formula:

Q = 1.5 x qc

where :

qc = Sum of airflow for combustion (m³/s)

= 17.59 m³/s

it can be seen the value of air flow needs in engine room needs is

Q = 1.5 x qc

= 1.5 x 17,59

(not applicable)

(not applicable)

Of these components is not known how much of the heat generated emissions. So the calculation using assumsion by:

Øo

density of air, at +30˚C, 70% RH and 101,3 kPa

kg/m³

∆T (the increase of the air temperature in the engine room i.e. the difference between inlet and outlet temperature measured at design conditions. The outlet temperature shall be measured at the outlet from engine room to casing or funnel without heat-sensitive installations).

So from the calculation of the above, it can be seen the value of the amount of air flow for heat discharge of emissions are:

Ødp + Ødg + Øb + Øp + Øg + Øel + Øep + Øt + Øo - 0.4 (qdp + qdg) - qb

m³/s

m³/s

The airflow for combustion + 50 %. The total airflow to the engine room shall not be less than the airflow for combustion [engine(s) and boiler(s)] plus 50 %.


VENTILATION SYSTEM



= 26.38 m³/s + 50% of the requirement for a combustion engine

= 35.17 m³/s

Q = qc + qh

= 17.59 + 12.64

= 30.22 m³/s

5.2. Ducting Design in Engine Room

So as to supply air near the turbo charge is:

Q1 = 50% of the requirement for a combustion engine

=

= 17.59 m³/s

Q2 = 2.00 m³/s

Q3 = 2.00 m³/s

Q4 = 2.00 m³/s

Q5 = 2.00 m³/s

Q6 = 0.75 m³/s

Q7 = 0.75 m³/s

Q8 = 0.75 m³/s

Q9 = 0.75 m³/s

Q10 = 0.75 m³/s

Q11 = 0.75 m³/s

Q12 = 0.75 m³/s

Q13 = 0.75 m³/s

Q14 = 0.75 m³/s

Q15 = 0.75 m³/s

Q16 = 0.75 m³/s

Q17 = 0.75 m³/s

So value of Q' =

= 17.00 m³/s

So value of Q total = 34.59 m³/s

- Air flow velocity

These two methods of calculation, it can be of greatest value. So the largest value contained in the following calculation methods:

In planning this ducting design follows the rules of ISO 8861 (Annex A). Where, for ventilation air duct design distributed 50% in the channel close to the Turbo Charge and 50% were other appropriate ducting design in Engine Room.

50% x 35,17 m³/s

And the other place air flow capacity (Q ') is 50% of the total supply air requirements, details are as follows:

Σ Q2 up to Q9

In planning the air velocity in the system using assumptions for industrial systems so that the air flow velocity is:


VENTILATION SYSTEM



- Main duct size

Qtotal = A x v

35.17 =

d = √(4 x35.17) / (π x 18)

= 1.58 m

No Description

1 35.17 18 1.95 1577.76 Main ducting

2 17.59 12 1.47 1366.38

3 2.00 12 0.17 460.78

4 2.00 12 0.17 460.78

5 2.00 12 0.17 460.78

6 2.00 12 0.17 460.78

7 0.75 8 0.09 345.58

8 0.75 8 0.09 345.58

9 0.75 8 0.09 345.58

10 0.75 8 0.09 345.58

11 0.75 8 0.09 345.58

12 0.75 8 0.09 345.58

13 0.75 8 0.09 345.58

14 0.75 8 0.09 345.58

(1/4) π (d²) x v

Q (m3/s)v

(m/s)A cal(m2)

d(mm)

Main branch ducting(ME)

Branch ducting(AE 1)




Main branch ducting(Platform 2)

Main branch ducting(Platform 2)

Main branch ducting(Platform1)





Main branch ducting(Tanktop)


VENTILATION SYSTEM



15 0.75 8 0.09 345.58

16 0.75 8 0.09 345.58

17 0.75 8 0.09 345.58

18 0.75 8 0.09 345.58

5.3. Total Pressure / Pressure Losses

- Ducting size

Qtotal = A x v

A =

d = √(4 x A) / (π x v)

- Friction losses in the main duct system

=

where :

ρ = Density of air (at 20 ˚ C)

= 1.184

μ = viscosity of air (at 20 ˚ C)

= kg/m.s

ε / d = roughness factor

=

=

ε = 0.000002

L = length of main duct (m)

d = diameter of main duct (m)

v = fluid flow velocity (m/s)

f = friction factor in pipe

Rn = (ρvd) / μ

So the value of the friction losses in the main duct system are:





Calculation of total pressure / pressure losses in this case using the method of "equal friction". This calculation is used to determine the losses caused by the duct system design. So as to know the total value of the total pressure that must be owned by the fans, in order to supply the needs of the air in the engine room. Equal friction calculation can be determined by the following steps:

(1/4) π (d²) x v

∆P ρ x (fL / d) x (v² / 2)

kg/m³

18.413 x 10-6

roughness factor according to the materials used. The material used is aluminum with a roughness factor is 0.002 x 10-3

Searched by using diagrams moody. Using the parameters of the roughness factor with the Reynolds Number on the pipe friction factor


VENTILATION SYSTEM



No. ε / d L (m) d (m) Rn f ∆P (Pa)

1 0.0000013 26.671 1.58 18 1826171.4 0.016 51.88 1.952 0.0000015 12 1.37 12 1054340.5 0.012 8.98 0.753 0.0000043 10 0.46 12 355547.8 0.011 20.35 2.044 0.0000043 14 0.46 12 355547.8 0.011 28.49 2.045 0.0000043 14 0.46 12 355547.8 0.011 28.49 2.046 0.0000043 18 0.46 12 355547.8 0.011 36.63 2.047 0.0000058 11.5 0.35 8 177773.9 0.017 21.43 1.868 0.0000058 15 0.35 8 177773.9 0.017 27.96 1.869 0.0000058 14.5 0.35 8 177773.9 0.017 27.03 1.86

10 0.0000058 14.5 0.35 8 177773.9 0.017 27.03 1.8611 0.0000058 17 0.35 8 177773.9 0.017 31.68 1.8612 0.0000058 21 0.35 8 177773.9 0.017 39.14 1.8613 0.0000058 20 0.35 8 177773.9 0.017 37.28 1.8614 0.0000058 17.2 0.35 8 177773.9 0.017 32.06 1.8615 0.0000058 18 0.35 8 177773.9 0.017 33.55 1.8616 0.0000058 17.5 0.35 8 177773.9 0.017 32.62 1.8617 0.0000058 24 0.35 8 177773.9 0.017 44.73 1.8618 0.0000058 23.5 0.35 8 177773.9 0.017 43.80 1.86

Where,

Q = 35.17 m³/s 1 m³/s = 2118.88 CFM

= 35.17 x 2118.88

= 74530.1977 CFM

∆P = 51.9 Pa

= 0.21 Inch

v(m/s)

∆P / L(Pa/m)

From the calculations above by looking at the air flow parameters with total pressure, then the fan which can be selected as follows:


VENTILATION SYSTEM



Merk : HARTZELL 44 M Marine Duct Axial

Series : 44-M-606DA---STAIR4

Power : 44.19 HP: 32.92 kW

RPM : 1175 RPM

Q : 87179 CFM

Project : Design IVDoc. No : 14 - 42 11 140 - VA

Rev. No :Page :

Introduction


This document is design air ventilation system to engine room. The propose to:• Supply air cumbution to main engine at 45°C• Preventing over heating engine and all part in engine roomDuct design's system constitutes one of system which exist at ship. Availibility of air goes to room enclosedding to meet the need occupant and / or stipubting of equipment in it. (ISO 8861). This system important enough because the function of air-refreshing and convenience problem and working safety. Collations main stipubting duct system : Stipubting Can meet the need airflow at that location, Regarded by economic flank and its efficient.

This International Standard specifies design requirements and suitable calculation methods for the ventilation of the engine room in diesel-engined ships, for normal conditions in all waters. Annex A provides guidance and good practice in the design of ventilation systems for ships' engine rooms.The outside ambient air temperature shall be taken as + 35 °C. Temperature rise from air intake to air passing from the engine room up to the casing entrance shall be max. 12,5 K. The capacity of the ventilation plant should be such as to provide comfortable working conditions in the engine room, to supply the necessary combustion air to the diesel engine(s) and boiler(s), and to prevent heat-sensitive apparatus from overheating. In order to meet these requirements, the air should be distributed to all parts of the engine room, so that pockets of stagnant hot air are avoided. Special considerations should be given to areas with great heat emission and to all normal working areas, where reasonably fresh and clean outdoor air should be provided through adjustable inlet devices. When arranging the air distribution, all normal conditions at sea and in harbour for in-service machinery shall be taken into account.In engine room approximately 50 % of the ventilation air should be delivered at the level of the top of the main propulsion diesel engine(s), close to the turbo-charger inlet(s), care being taken to ensure that no sea water can be drawn into that air inlet. No air should be blown directly onto heat-emitting components or directly onto electrical or other apparatus sensitive to water.

Project : Design IVDoc. No : 14 - 42 12 140 - VA

Rev. No :Page :


The air exhaust system should be designed to maintain a slight positive pressure in the engine room. This should normally not exceed 50 Pa. Exhaust fans should be installed where the exhaust air cannot be led through the funnel or extractor openings. The purifier room containing fuel oil separators, etc. should have a separate fan-operated exhaust system discharging to the atmosphere as remote as possible from any air inlet. Fire dampers and weather-tight closing appliances should be installed in the ventilation coamings in accordance with the SOLAS chapter II-2 and the International Load Line Convention, 1966.

Doc. No. 14 - 42 12 140 - VA

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