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09 June 2022

179.NA April 2022 Q.8

June 09, 2022 Posted by AK 3 comments

 The following data are available from the hydrostatic curve of a vessel.

Draught (m)

KB (m)

KM(m)

I (m4)

4.9

2.49

10.73

65.25

5.2

2.61

10.79

68.86

 Calculate the TPC at a draught of 5.05 m.

 

 IF YOU ARE NOTICING ANY ERROR KINDLY COMMENT BELOW

 

Solution

We know that TPC = (Area of water plane / 100) x Density of water

So first we need to find volume of displacement at 5.2 m draught

We know that BM = I / Vol. of displacement 

BM = I / ▽


▽ = I / BM

     = 65.25 / 8.24             (BM = KM - KB)

     = 7918   at 5.2 m draught



volume of displacement at 4.9 m draught

We know that BM = I / Vol. of displacement 

BM = I / ▽


▽ = I / BM

     = 68.86 / 8.18             (BM = KM - KB)

     = 8418   at 4.9 m draught

Difference in vol of displacement = 8418 - 7918

                                                       = 500

Difference in draught = 5.2 - 4.9

                                   = 0.3

Area of water plane = Diff in vol of displacement / Difference in draught

                                 = 500 / 0.3

                                 = 1666.67 m2

TPC = (1666.67 / 100) x 1.025

        = 17.08 

3 comments:

  1. question-
    how
    (volume of displacement)at draught 5.2=68.86 / 8.18= 8.418 became 8418.09

    (volume of displacement)at draught 4.9= 65.25 / 8.24= 7.918 became 7918

    correction to be made:-
    volume of displacement 7918 is at at 4.9m draught

    volume of displacement 8418 is at at 5.2m draught

    ReplyDelete
  2. Please reply on above

    ReplyDelete
  3. it seems some corrections are to be made

    ReplyDelete