Solution Naval Numerical 26

Numerical 26: A box shaped vessel is 80 m long, 12 m wide and floats at a draught of 4m. A full width midships compartment 15 m long is bilged and this results in the draught increasing to 4.5 m. Calculate EACH of the following, 

(a) The permeability of the compartment; 

(b) The change in metacentric height due to bilging.

Solution:Given

B = 12m

L = 80m

d = 4m

l = 15m

$d_1$ =4.5m

to find μ and $GM_1$

(a) using the relation

$d_1-d=\frac{\mu ld}{L-\mu(l)}$ 

$4.5-4=\frac{\mu\times 15\times 4}{80-\mu(15)}$ 

μ =0.5926 

(b) In such cases it is consider KG is constant

before bilging:

KG = KB +BM -GM

KB = $\frac{d}{2}$ 

KB = $\frac{4}{2}$ = 2m

and BM = $\frac{1}{12}\times \frac{B^{2}}{d}$ 

BM = $\frac{1}{12}\times \frac{12^{2}}{4}$

BM = 3m

thus KG = 2+3-GM

GM = 5 -KG

After bilging:

KG = $KB_1$ + $BM_1$ - $GM_1$ 

$KB_1$ = $\frac{d_1}{2}$ 

$KB_1$ = $\frac{4.5}{2}$

$KB_1$ = 2.25m

$BM_1$ = $\frac{I\times \rho }{\Delta }$

$I =\frac{L\times B^3}{12}-\frac{\mu \times l\times b^3}{12}$ 

$I =\frac{80\times 12^3}{12}-\frac{0.5926 \times 15\times 12^3}{12}$

I = 10,239 $m^4$

$BM_1$ = $\frac{10239\times \rho }{L\times B\times d\times \rho }$

$BM_1$ = $\frac{10239 }{80\times 12\times 4 }$

$BM_1$ = 2.66m

Thus KG = 2.25 + 2.66 - $GM_1$

$GM_1$ = 4.91 -KG

thus, change in GM

$GM_1$ - GM = 0.09m