A ship of 14900 tonnes displacement have a shaft power of 4460 KW at 14.55 knots. The shaft power is reduced to 4120 KW and fuel consumption at same displacement is 541 Kg/h. Calculate the fuel coefficient of the ship.
Given
△ = 14900 t
SP2 = 4460 KW
V2 = 14.55 knots
SP1 = 4120 KW
Fuel consumption = 541 Kg/h
To find
Fuel coefficient
Solution
As we know that SP directly proportional to Speed3
SP directly proportional to V3
SP2 / SP1 = (V2 / V1)3
4460 / 4120 = (14.55 / V1)3
V1 = 14.17 knots
At 14.17 knots fuel consumption = 541 Kg/h
= 541 x 24 x 10-3
= 12.98 t / day
Fuel coefficient = (△2/3 x V13 ) / Fuel consumption tonne per day
= (149002/3 x 14.173) / 12.98
If you are noticing some error in problems kindly comment below.Thanks
△ = 14900 t
SP2 = 4460 KW
V2 = 14.55 knots
SP1 = 4120 KW
Fuel consumption = 541 Kg/h
To find
Fuel coefficient
Solution
As we know that SP directly proportional to Speed3
SP directly proportional to V3
SP2 / SP1 = (V2 / V1)3
4460 / 4120 = (14.55 / V1)3
V1 = 14.17 knots
At 14.17 knots fuel consumption = 541 Kg/h
= 541 x 24 x 10-3
= 12.98 t / day
Fuel coefficient = (△2/3 x V13 ) / Fuel consumption tonne per day
= (149002/3 x 14.173) / 12.98
Fuel coefficient = 132726
A ship travelling at 15.5 knots has a propeller of 5.5 m pitch turning at 95 rev/min. The thrust of the propeller is 380 KN and the delivered power 3510 KW. If The real slip is 20% and the thrust deduction factor 0.198, calculate the Quasi Propulsive Coefficient (QPC) and the wake fraction
ReplyDeletePlease refer 71.NA october 2019 Q.10(b)
DeletePlease provide answer for above
ReplyDeletehttps://kochimmdclass4.blogspot.com/2020/01/71na-oct-2019-q10b.html
Deleteplease edit tp - to...little confusing..
ReplyDeleteUpdated.Thanks
Delete10^3 not in negative
ReplyDelete