25-Nav-A1 Fundamentals of Naval Architecture · May-98-Nav-A1 2016
Question 6 of 7: Permeability & New Drafts After Adding Multiple Weights
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examinations, May 2016 — 98-Nav-A1 Fundamentals of Naval Architecture, 3 hours, closed book (5 questions constitute a complete exam paper, first five as they appear in the answer book are marked, each of equal value; all 7 answered below for full study coverage).
Reference texts: Tupper, Introduction to Naval Architecture; Lewis (ed.), Principles of Naval Architecture (PNA); IMO, International Code on Intact Stability (IS Code).
Question 6: Permeability & New Drafts After Adding Multiple Weights
Check: the pairing of weights to LCGs in the paper's weight/LCG table (page 4) is an assumption. The pairing used above — 50.00t/70.00m aft, 170.00t/36.00m aft, 100.00t/5.00m aft, 130.00t/4.00m fwd, 40.00t/63.00m fwd — is the reading that keeps every printed number attached to a plausible partner (all levers well within the 70 m half-length); it is stated here as an explicit assumption rather than solved silently.
(a) Volume and surface permeability
Volume permeability $\mu$ of a compartment is the fraction of its total (moulded) volume that can actually be occupied by floodwater, i.e. $\mu=\dfrac{\text{volume of water that can enter}}{\text{total volume of the space}}$. An empty cargo hold or void has $\mu$ close to 1 (typically 0.95–0.98 once accounting for structure); a machinery space is far lower (typically 0.80–0.85) because engines, floor plates and foundations displace much of the volume; a full store or bunker has $\mu$ close to the space's own porosity.
Surface permeability is the analogous fraction applied to a horizontal area (waterplane or deck) within a flooded space — the fraction of that area that is actually free water surface rather than solid structure — and is what actually appears in the free-surface-effect and lost-buoyancy waterplane-area calculations for a damaged, partially-fitted-out compartment.
(b) New drafts after adding several weights
Given. Five weights are added at different longitudinal positions; find the resulting bodily sinkage and the change of trim, distributed about the ship's own centre of flotation.
Approach. Sum the weights for total sinkage, sum their moments about midship then re-reference to $LCF$ for the net trimming moment, and apportion the resulting change of trim between the ends as before.
Total weight and sinkage. $W=50.00+170.00+100.00+130.00+40.00=490.0\text{ tonnef}$.
$$\text{sinkage}=\frac{490.0}{24}\approx20.42\text{ cm}=0.2042\text{ m}$$
Net moment about $LCF$. Taking aft as negative, forward as positive, about midship: $\Sigma wx=50.00(-70.00)+170.00(-36.00)+100.00(-5.00)+130.00(4.00)+40.00(63.00)=-7080\text{ tonnef}\cdot\text{m}$. Re-referencing to $LCF$ ($+2.16\text{ m}$ fwd of midship):
$$M_F=\Sigma wx-W(LCF)=-7080-490.0(2.16)\approx-8138.4\text{ tonnef}\cdot\text{m (net aft of }F\text{, so trim by the stern)}$$
$$COT=\frac{8138.4}{252}\approx32.30\text{ cm}$$
Apportion the trim. $l_F=L/2-2.16=67.84\text{ m}$, $l_A=L/2+2.16=72.16\text{ m}$:
$$\Delta d_F=32.30\left(\frac{67.84}{140}\right)\approx15.65\text{ cm (decrease)}, \quad \Delta d_A=32.30\left(\frac{72.16}{140}\right)\approx16.65\text{ cm (increase)}$$
New drafts.
$$\boxed{d_F'=7.70+0.2042-0.1565\approx7.748\text{ m}, \qquad d_A'=8.25+0.2042+0.1665\approx8.621\text{ m}}$$
