TEKNIK PERMESINAN KAPAL II (Minggu 5)

LS 1329 ( 3 SKS) Jurusan Teknik Sistem Perkapalan

ITS Surabaya

Keseimbangan Energi Motor Diesel Motor Diesel

Dynamometer

Air Pendingin masuk

Engine

Bahan Bakar

Udara

Gas Buang

DAYA

Air Pendingin Keluar

Pelumas Panas Konveksi

Daya Motor

Indicated Power (IP) - Kerja akibat hasil pembakaran terhadap piston - Digambarkan pada grafik P V - Pendekatannya dengan menggunakan rate of

Heat release Effective Power (brake horsepower) Daya keluaran dari motor

Brake horsepower Torsi hasil pengereman / pembebanan

T : mg x d

Daya perhitungan dari torsi

P = 2 x Rps x T

Daya dari motor

P = i x MEP x L x A x z x Rps MEP : mean effective pressure

i = 1 untuk 2-stroke

i = untuk 4-stroke

L : panjang langkah

A : penampang silinder

Z : jumlah silinder

Energi Pembakaran

Fuel type Heavy Fuel Oil Marine Diesel Oil Low Heating Value* 42.927 MJ/kg 42.7 MJ/kg

energyNyxOHyxCONOyxHC yx +

++++

++ 22222 4

76.32

)76.3(4

Reaksi Pembakaran

Energi Pembakaran dari Bahan Bakar Energi dari hasil pembakaran tergantung dari laju massa bahan bakar dan low heating value

Qfuel = mf x LHV

*Low heating value berbeda untuk setiap type dan produksi bahan bakar

200

225

250

275

300

325

350

30% 40% 50% 60% 70% 80% 90%

Engine Load

SFO

C (g

/kW

h)

0% 8% 12% 23% 30%

Konsumsi Bahan Bakar Konsumsi Bahan Bakar (Fuel Consumption / FC) - Jumlah massa bahan bakar yang dibutuhkan oleh suatu

motor dalam rentang waktu operasionalnya - Unit satuan, ex. Kg/h Konsumsi Bahan Bakar Spesifik (Specific Fuel

Consumption / SFC) - Jumlah massa bahan bakar yang dibutuhkan oleh suatu

motor untuk setiap satuan daya dan waktu pada beban dan putaran tertentu

- Unit satuan, ex. g/kWh, g/BHPh - SFC = mf / Brake power

Contoh Konsumsi Bahan Bakar dari S80MC MAN B&W

Efisiensi

Eff.mekanik (m) terjadi akibat adanya gesekan pada komponen-komponen motor dan penggunaan daya untuk keperluan operasional motor sendiri

Efisiensi Eff. Volumetrik (v) Perbandingan antara massa udara yang masuk

ke dalam silinder terhadap massa udara ideal pada volume silinder

(v) = ma / (L A z a N) Eff. Efektif (e) Besarnya energi panas dari bahan bakar yang

dapat dimanfaatkan untuk menghasilkan kerja bersih

e = brake power / Qfuel

Contoh Soal Sebuah motor 4 langkah dengan 6 buah silinder

beroperasi pada putaran 3000 rpm dan indicated mean effective pressure (IMEP) 5 atm. Motor dioperasikan menggunakan octane fuel dengan equivalence ratio of 0.9. The brake torque pada kondisi ini adalah 339 Nm., sedangkan eff. volumetriknya adalah 85%. Setiap silinder berdiameter 12.7 cm bore dengan panjang langkah 15.2 cm. Kondisi udara lingkungan pada 1 atm dan 300K.

Tentukan indicated horsepower, brake horsepower, friction horsepower, eff. Mekanik, laju aliran bahan bakar, dan SFC (specific fuel consumption) dari motor tersebut.

IMEP = 5 atm = 506.6 kPa Rps = Rpm / 60 = 3000 / 60 = 50 rps A = 1/4 (0.127)2 = 0.012661 m2 L = 0.152 m z = 6 = 0.9 T = 339 Nm v = 85% Udara ; 1 atm, 300K a = 1.177 kg/m3

Indicated horsepower (IP) IP = i x IMEP x L x A x z x Rps = (1/2)(506.6 kPa)(0.152)(0.012661)(6)(50) = 146.24 kW The brake horsepower (BP) BP = 2 x rps x T = 2 x 50 x 339 = 106446 w = 106.45 kW The friction power FP = IP BP = 146.24 106.45 = 39.79 kW

Jawaban

Eff. Mekanis (m) m = BP/IP = 106.45 / 146.24 = 0.728

Laju aliran massa udara yang masuk ke silinder (ma) ma = (v) (L A z a N/2) 4 langkah = 0.85 x 0.152 x 0.012661 x 6 x 1.177 x 50/2 = 0.289 kg/s Kesetimbangan pembakaran octane (stoichiometric) C8H18 + 12.5O2 + (12.5 x 3.76)N2 8CO2 + 9H2O + (12.5 x 3.76)N2 Fuel-air ratio F/A = x (F/A)s = 0.9[(812) + (181)] / [12.5(32 + 3.7628)] = 0.0598 kg-octane /kg-udara

Jawaban

Laju aliran massa bahan bakar mf = ma (F/A) = 0.289 0.0598 = 0.0173 kg/s = 62.17 kg/hr Specific fuel consumption SFC = mf /BP = 62.17/ 106.45 = 0.584 kg/kWh = 584 g/kWh

Jawaban

Firing order Silinder Urutan Pengapian

3 1-3-2 4 1-3-4-2

5 1-2-4-5-3

6 1-5-3-6-2-4 1-6-5-4-3-2 7 1-3-5-7-2-4-6

8 1-8-4-3-6-5-7-2 1-8-7-2-6-5-4-3

10 1-10-9-4-3-6-5-8-7-2 12 1-7-5-11-3-9-6-12-2-8-4-10 16 1-12-8-11-7-14-5-16-4-15-3-10-6-9-2-13

Latihan soal A twelve-cylinder four-stroke-cycle Diesel engine has a 4-in. bore, a

4.5-in. stroke, and a compression ratio of 20. The mechanical efficiency is 89%, the cutoff ratios is 2, and the engine speed is 1200 rpm. The air entering the cylinder is at 14.5 psia and 60 F. Assuming Air Standard cycle performance, determine the cycle temperatures, indicated power, IMEP, and engine brake horsepower.

An eight-cylinder four-stroke-cycle engine has a bore of 10 cm and a

stroke of 12 cm. At a shaft speed of 53 rps, the brake power is 300kW and the mechanical efficiency is 85%. Fuel with a heating value of 40,000 kJ/kg is supplied at a rate of 40 kg/hr. What are the engine displacement, BMEP, brake torque, and indicated specific fuel consumption in kg/kW-hr?

A single-cylinder four-stroke-cycle spark-ignition engine has a BSFC of 0.4 kg/kW-hr and a volumetric efficiency of 78% at a speed of 45 rps. The bore is 6 cm and the stroke is 8.5 cm. What is the fuel flow rate, fuel-air ratio, and brake torque if the brake power output is 6 kW with ambient conditions of 100 kPa and 22C?

Engine Performance

Maximum Rating The practical maximum output of a diesel engine may

be said to have been reached when one or more of the following factors operate:

1. The maximum percentage of fuel possible is being

burned effectively in the cylinder volume available 2. The stresses in the component parts of the engine

generally, for the mechanical and thermal conditions prevailing, have attained the highest safe level for continuous working

3. The piston speed and thus revolutions per minute cannot safely be increased

SFOC

Engine Propeller Matching

HULL

PROPELLER

ENGINE

Engine Propeller Matching

HULL

Speed, Vs

Resistance, R

2

21

st SVCR =

Ct : Resistance Coefficient

: sea water density

S : wetted surface area

Vs : ship speed

Thrust, T

(required) T = R / (1-t)

T = 0.5 Ct S Vs2 / (1-t)

Engine Propeller Matching

Advance velocity, Va

T = kT n2 D4 Thrust, T

(produced)

Va = Vs (1-w)

Engine Propeller Matching

Thrust, T

(produced)

Thrust, T

(required)

T = kT n2 D4 T = 0.5 Ct S Vs2 / (1-t)

Va = Vs (1-w)

T = 0.5 Ct S Va2 / (1-t)(1-w)2

KT = (0.5 Ct S / (1-t)(1-w)2D2) (Va/ nD)2 KT = J2 Propeller load characteristic

0,00

0,05

0,10

0,15

0,20

0,25

0,30

0,35

0,40

0,45

0,50

0,55

0,60

0,65

0,70

0,00 0,05 0,10 0,15 0,20 0,25 0,30 0,35 0,40 0,45 0,50 0,55 0,60 0,65 0,70 0,75 0,80 0,85 0,90 0,95 1,00

Kt

10Kq

Efficiency

Prop load char.

J

n = Va / JD

Kt

Kq

T = kT n2 D4 Q = kq n2 D5

Propeller Curve

Vs Vs

T : Thrust

THP DHP = 2Qn SHP = DHP/s BHP = SHP/g

EHP : Effective Horse Power THP : Thrust Horse Power DHP : Delivered Horse Power SHP : Shaft Horse Power BHP : Brake Horse Power

Propeller Curve

0

2000

4000

6000

8000

10000

12000

14000

40 50 60 70 80 90 100 110 120 130

N

BHP f{n3}

Pressure Charging A naturally aspirated engine draws air of the same density as the ambient atmosphere. Since this air density determines the maximum weight of fuel that can be effectively burned per working stroke in the cylinder, it also determines the maximum power that can be developed by the engine. Increasing the density of the charge air by applying a suitable compressor between the air intake and the cylinder increases the weight of air induced per working stroke, thereby allowing a greater weight of fuel to be burned with a consequent rise in specific power output.

Pressure Charging Method The power expended in driving the compressor has an important influence on the operating efficiency of the engine. It is relatively uneconomical to drive the compressor direct from the engine by chain or gear drive because some of the additional power is thereby absorbed and there is an increase in specific fuel consumption for the extra power obtained. About 35 per cent of the total heat energy in the fuel is wasted to the exhaust gases, so by using the energy in these gases to drive the compressor an increase in power is obtained in proportion to the increase in the charge air density.

The power expended in driving the compressor has an important influence on the operating efficiency of the engine. It is relatively uneconomical to drive the compressor direct from the engine by chain or gear drive because some of the additional power is thereby absorbed and there is an increase in specific fuel consumption for the extra power obtained. About 35 per cent of the total heat energy in the fuel is wasted to the exhaust gases, so by using the energy in these gases to drive the compressor an increase in power is obtained in proportion to the increase in the charge air density.

A substantial increase in engine power output for any stated size and piston speed, or conversely a substantial reduction in engine dimensions and weight for any stated horsepower.

An appreciable reduction in the specific fuel consumption rate at all engine loads

A reduction in initial engine cost. Increased reliability and reduced maintenance

costs,

There are a number of advantages of pressure charging by means of an exhaust gas turboblower system

Four stroke timing diagram

TEKNIK PERMESINAN KAPAL II(Minggu 5)Keseimbangan Energi Motor DieselDaya MotorBrake horsepowerSlide Number 5Energi PembakaranKonsumsi Bahan BakarContoh Konsumsi Bahan Bakar dari S80MC MAN B&WEfisiensiEfisiensiContoh SoalJawabanJawabanJawabanFiring orderLatihan soalEngine PerformanceSlide Number 18Maximum RatingSlide Number 20SFOCEngine Propeller MatchingEngine Propeller MatchingEngine Propeller MatchingSlide Number 25Engine Propeller MatchingSlide Number 27Propeller CurvePropeller CurveSlide Number 30Slide Number 31Slide Number 32Slide Number 33Slide Number 34Pressure ChargingPressure Charging MethodSlide Number 37Slide Number 38Four stroke timing diagramSlide Number 40