117.NA February 2021 Q.6

 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 2550 KW. Calculate: (i) Indicated power; (ii) Effective power; (iii) Ship speed.

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Given

Δ = 15000 t
C = 420
ղm = 0.83
Shaft loss = 0.06
ղp = 0.65
QPC = 0.71
tp = 2550 KW

To find

i) Indicated power
ii) Effective power
iii) Ship speed.

Solution

We know that Thrust power = Delivered power x Propeller efficiency
So delivered power , dp = tp / ղp
                                       = 2550 / 0.65
                                       = 3923.07 KW

Also know that delivered power = Shaft power x transmission efficiency ---------------(1)

It is given that shaft loss is 6 %. So power delivered by shaft is 94% = 0.94

Transmission efficiency, ղt  = 0.94
                                             
So from (1) -------------------------------->

Shaft power, sp = dp / ղt
                          = 3923.07 / 0.94
                          = 4173.47 KW

i) Shaft power = Indicated power x mechanical efficiency

   So Indicated power, ip = sp / ղm
                                       = 4173.47 / 0.83
                                       = 5028.3 KW

ii) Effective power, ep = dp x QPC
                                   = 3923.07 x 0.71
                                   = 2785.37 KW

iii) We know that Admirality coefficient, C = (Δ2/3 x V3) / sp
                                                           420 = (150002/3 x V3) / 4173.47
                                                1752857.4 = 608.22 x V3
                                                             V3 = 2881.95
                                           Ship speed, V = 14.23 knots