FInals Essay qn Part 2 Alam

Question 1

How is hull cleaning frequency optimised?

Answer 1

Rf= SFV^n
Rf= frictional resistance(N)
V= velocity in knots
S= wetted surface area
n = 2

1. Clean Hull–> Less frictional resistance–> Less fuel consumption
2. Potential savings from hull cleaning frequency every year vs every 2 years

Question 3

What factors make 4 S engine scavenging better than 2 S engine?

Answer 3

1. Dedicated Exhaust Stroke
   –> Allows exhaust gas to be fully expelled, ensuring efficient gas exchange and cleaner combustion in 4S.
   –> In 2S, relies on intake air during scavenging , leaving residues that reduce efficiency.
2. Separation of Intake and Exhaust
   –> 4 S engine prevent mixing of fresh air with residual gas purity.
3. More time for scavenging
   –> Exhaust and intake process of 4 S engine stretched over separate strokes, more time for complete exchange of gases
4. Higher no of revolutions per cycle

Question 4

What is the format to derive Integrated Horsepower (IHP)?
Hint: power is in what units?

Answer 4

Go by PLAN to derive IHP
P→ Pressure to push piston down (N/m^2)
L + A→ Volume (Length of Stroke and Piston Area)
N→ Revolutions Per Minute(RPM or RPS) : J/S
n→ no of cylinders( how many units of power?)
PV= N/m^2 X m^3 = Nm= J X90/60= J/s= W

Question 5

What is the respective power components to derive power of engine calculation

Answer 5

Essentially, The propeller converts rotational power into useful thrust.
The Thrust Horsepower (THP) is the power needed for this thrust, influenced by speed and thrust generated, including losses from the powertrain.

1. BHP (Brake Horsepower): Power output of the engine (***only in 4 stroke engine)
2. SHP (Shaft Horsepower): BHP minus mechanical losses in reduction gear.
3. DHP (Delivered Horsepower): Power delivered to the propeller, accounting for losses in gearbox, bearings, and stern tube seals (95%-98% of BHP).
4. THP (Thrust Horsepower)= RT X V: Power from the propeller thrust, considering losses in gearbox, shafting, and propeller. equal to the product of the speed of advance and the thrust generated by the propeller
5. EHP (Effective Horsepower): Power required to move the ship at a given speed, derived as RT×V, excluding propeller action. This power is equal to the Brake Horsepower minus losses due to the gearbox, shafting and propeller, as well as interaction between the propeller and the hull.

Question 6

What are the factors affecting propeller efficiencies?

Answer 6

1. Propeller Diameter
2. RPM
3. No. of Blades
4. Blade Outline
5. Camber & Angle of Attack
6. Pitch/Diameter Ratio

Question 7

Differentiate hull cleaning net savings vs Slow steaming net savings

Answer 7

Hull Cleaning:
Net saving = Fuel Saving (Gross)- (Diver’s cost + offhire cost( stay longer at port to do cleaning)+ speeding up fuel cost( unplanned cleaning, need speed up to next port)

Slow steaming:
Net savings = Fuel saved - charter hire
If charter hire is low, use slow steaming , charter hire high→ DO NOT use slow steaming

Question 8

What is the equation governing slow steaming, not considering time taken to reach?

Answer 8

Fc directly proportional to V^3,
fuel consumption directly proportional to speed
Fc22/Fc 20= fuel savings by reducing vessel speed by 2 knots

Question 9

Net savings using slow steaming in a voyage?

Answer 9

d= Distance
D= No of Days (which is time)
V= velocity ( which is speed) ,
so derived r/s we know Speed= Distance/TIme, and so V= d/D

FcT20 is total fuel consumption for vessel running at 20 knots speed
Distance of Voyage won’t change, time taken to reach A to B can change according to the speed u travel
so FcT20 he break down by FcD 20 (Daily fuel consumed) X D20( no of days req for ship travelling at 20 knots)

Thats why there u see how come got one D come out, by taking that out he will account for the fuel lost due to the extra days at the sea, which is not accounted for in the 1.332, your daily consumption of fuel

which is then proven as become v^2 from v^3
Fc becomes directly proportional to v^2

Question 10

What are the steps to the principle of combustion of internal engine?

Answer 10

Chemical Energy (fuel)–> Mechanical (KInetic Energy) + Heat energy

1. Consists of a fixed cylinder and a moving piston
2. The expanding combustion gases push the piston, which in turn rotates the crankshaft.
3. Ultimately, this motion drives the vehicle’s wheels through a system of gears in the powertrain.
4. Typical energy content of HFO is 41 KJ/g
5. If it goes below 160 degrees C, will fall below due point, causing condensation and corrosion. Thus cannot recover anymore
6. Overall Efficiency: 49% (ME going to propeller) + 6% (Waste Heat Recovery)= 55% (energy utilised)