Solution Naval Numerical 12

Numerical 12: A ship of 15000 tonne displacement has an Admiralty Coefficient, based on shaft power, of 420. The mechanical efficiency of the machinery is 83%, shaft losses 6%, propeller efficiency 65% and QPC 0.71. At a particular speed the thrust power is 2550kW. Calculate: (i) Indicated power; (ii) Effective power; (iii) Ship speed.

Solution: Given, Δ = 15000 tonne.
Admiralty coefficient, C = 420 η(mech) = 83%
Shaft loss= 6%,
thus η(shaft) = 94%
η = 65%
QPC = 0.71
tp = 2550 kW 
To find,
(i) Indicate power
(ii) Effective power
(iii) Ship speed.
We know that Delivered power = Thrust power/ propeller efficiency 
= 2550/0.65 = 3923 KW.
The effective power = delivered power x QPC
= 3923 x 0.71 =2785.33 kW.
The indicated power = (dp x 1.06)/0.83
= (3923 x1.06)/0.83 = 5010 kW. 
Admiralty Coefficient, C =

 

thus, ship speed (V) = 15knots.