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24 August 2021

137.NA August 2021 Q.6

August 24, 2021 Posted by AK No comments

 A propeller has a pitch ratio of 0.95. When turning at 120 rev/min the real slip is 30%, the wake fraction 0.28 & the ship speed 16 knots. The thrust is found to be 400 kN, the torque 270 kNm & the QPC 0.67.Calculate:

a) The propeller diameter

b) The shaft power

c) The propeller efficiency

d) The thrust deduction factor

IF YOU ARE NOTICING ANY ERROR KINDLY COMMENT BELOW


Given

p = 0.95

N = 120

s = 0.3

w = 0.28

V = 16

T = 400

Torque = 270

QPC = 0.67


To find 

a) The propeller diameter

b) The shaft power

c) The propeller efficiency

d) The thrust deduction factor


Solution

We know that wake fraction, w = (V - Va) / V

                                           0.28 = (16 - Va) / 16

                                           4.48 = 16 - Va

                Speed of advance, Va = 16 - 4.48

                                                   = 11.52 knots

We know that real slip, s = (Vt - Va) / Vt

                                  0.3 = (Vt - 11.52) / Vt

                             0.3 Vt = Vt - 11.52

                      Vt - 0.3 Vt = 11.52

                              0.7 Vt = 11.52

      Theoretical speed, Vt = 16.457 knots

We know that theoretical speed, Vt = (P x N x 60) / 1852

                                             16.457 = (P x 120 x 60) / 1852

                                         30478.36 = 7200 x P

                                           Pitch, P = 4.23 m


a) We know that pitch ratio, p = P/D

                                        0.95 = 4.23 / D

                                    0.95 D = 4.23

             Propeller diameter, D = 4.45 m


b) We know that delivered power dp = 2π n x Torque

And shaft power = dp / transmission efficiency

In question nothing mentioned about transmission efficiency. So it considered as negligible.

So shaft power, sp = 2π n x Torque {n = rev per sec

                              = 2π x (120/60) x 270     

                              = 3391.2 kW             


c) We know that thrust power, tp = Thrust x Speed of advance

                                                     = 400 x 11.52 x (1852/3600)

                                                     = 2370.56 kW 

Also thrust power, tp = dp x propeller efficiency

So propeller efficiency = tp /dp

                                      = 2370.56 / 3391.2

                                      = 69.90 %


d) We know that effective power, ep = dp x QPC

                                                           = 3391.2 x 0.67

                                                           = 2272.10 kW

Also ep = Rt x V

The total resistance, Rt = ep / V

                                      = [2272.10 / 16] x [3600 / 1852]

                                      = 276.03 kN

 Also Rt = T (1 - t)

  276.03 = 400 (1 - t)

       1- t = 0.69

Thrust reduction factor, t = 1- 0.69

                                         = 0.31

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