Solution Naval Numerical 23

Numerical 23: 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 3540 KW. If the real slip is 20% and the thrust deduction factor O.198, calculate the Quasi Propulsive Coefficient (QPC) and the wake fraction.

Solution: V= 15.5 knots

Pitch =5.5m

rpm =95

T =380kN

dp = 3540kW

slip(s) =20%

thrust deduction factor (t)=0.198

QPC=?

Wake fraction =?

$QPC =\frac{e_p}{d_p}$ 

Wake fraction =

Theoretical speed of the ship= $V_t=\frac{PN60}{1852}$

$V_t=\frac{5.5\times95\times60}{1852}$

=16.927knots

speed of advance = $V_a = V_t (1-s)$

$V_a = 16.927 (1-0.2)$

$V_a = 13.5416$Knots

hence wake fraction=$\frac{15.5-13.5416}{15.5}$ =0.126

Resistance; $R_t=T(1-t)$

thus $R_t=380(1-0.198)$

$R_t=304.76$

Effective power = $R_t \times V$

$e_p=304.76\times 15.5\times \frac{1852}{3600}$ 

$e_p$=2430.122kW

Thus QPC = $\frac{2430.122}{3540}$

QPC=0.6864