A semisubmersible at a draft of 19 feet 9 inches arrives on location planning to deploy eight mooring lines. Each anchor weighs 15 long tons and each mooring line consists of 3,000 feet of 3-inch chain (89.6 lbs/ft). If no ballast corrections are made, what is the expected draft if the average TPI is 60?
A. 17 feet 9 inches
B. 18 feet 3 inches
C. 18 feet 9 inches
D. 21 feet 3 inches
A. 17 feet 9 inches
B. 18 feet 3 inches
C. 18 feet 9 inches
D. 21 feet 3 inches
When you deploy the 8 anchors and chains, you are removing weight from the rig. The draft will decrease.
ReplyDelete8 anchors at 15 T = 8 x 15 = 120 T
8 chains x 3000 ft x 89.6 lb/ft = 2,150,400 lbs / 2240 lb/T = 960 T
Anchors + Chains = 120 + 960 = 1080 T
Change in draft = 1080 T/60 T/inch = 18 inches less
Initial draft = 19'-09" - 18 inches = 18'-03", choice B is best answer!
This question really does not consider the depth of the water and how much chain is on the bottom, so the actual decrease in draft should be less. You just have to play the cards you are delt!
Do you Agree or Disagree?
Does the chain that is not on the bottom still effect the draft of the vessel? It's still hanging off the vessel so it must effect it, right?
ReplyDeleteThat's one vague and convoluted question.
ReplyDeleteMost of the chain and anchor is on the bottom with wire running up from the seafloor
So I like where you went with the weight of the chain and anchor times eight.
Based on that, I agree with the math leading to ANS: BRAVO.
Good Question, in concept, but i still say its vague.