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24 January 2021

99. NA October 2020 Q.8

January 24, 2021 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. Volume of displacement will be 7.918 instead of 7918 ??

    ReplyDelete
  2. little tricky. Please recheck solution.

    ReplyDelete
  3. 5.05m draft k liye kaise hua ye ans can you explain

    ReplyDelete