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### exam question

Posted: **Sat Nov 12, 2016 4:26 am**

by **kiamlee**

hello, does anybody have the spare time to teach me the formula and solution for this problem, its one of the board exam questions that im having a hard time solving. here it is

A fuel tank on your vessel is 20feet high 20feet long and 20feet wide. If it is filled 100% with fuel having an API gravity of 35.7 at a temperature of 60F, how many long tons of fuel are in the tank?

A. 177.76 B. 188.25 C. 196.47 D.210.84

ANS. B. 188.25

the answer was given by a friend though without explanation. lol

### Re: exam question

Posted: **Sat Nov 12, 2016 5:30 am**

by **Revolver**

V = LxWxH (and convert from ft^3 to m^3)

SG=141.5/(API+131.5)

m(kg) = V*SG

m(lt) = m(kg)/1.016

Haven't been doing any number puzzles yet, and did this with my cell phone calculator..., but I believe that'll be about right.

### Re: exam question

Posted: **Sat Nov 12, 2016 6:08 am**

by **D Winsor**

I can understand why it would be difficult to explain on how to get the correct answer especially if you are unfamiliar with the differences between US measuring system and the British or Imperial measuring systems especially when it comes to weights & volumes

1) Calculate Volume of tank in cubic feet - 20x20x20=8000 Cubic Ft.

2) Convert volume to US Gallons @ 7.48 Gallons/Cubic Foot - 8000x7.48 = 59844.16 US Gallons

3) Find using an API Gravity Table the weight in Pounds/US Gallon. 35.7 API=7.047lbs/US Gallon (This value was calculated with the interpolation of table values of API 35=7.076 lbs/US Gallon and API 40=6.870 lbs/US Gallon)

4) Convert the calculated volume to Pounds - 59844.16 x 7.047 = 421731.37 Pounds

5) Convert weight to Tons by dividing by 2240lbs/Long Ton - 421731.37/2240 = 188.27 Tons

Making "B" the correct answer

FYI This a very good "Trick" question because when using US standard values most people would miss the Long Tons and calculate in Short or US Tons making "D" the calculated answer. As you can see I used values based on US standards of measurement of the ratio between weight and volume except for the last calculation where the Imperial standard weight in pounds per ton was used.

Hope this helps

### Re: exam question

Posted: **Sat Nov 12, 2016 10:02 am**

by **Revolver**

My calculations actually turned out 188.69... Close, but no cigar I guess.

Both to be blamed on my cell phone calculator and, probably moreso, rust on my brain equations/conversions haha

### Re: exam question

Posted: **Sat Nov 12, 2016 12:36 pm**

by **Merlyn**

Well worked out answer without falling in the hidden pit falls.

Together being of course correct to the Nth degree.

( N of course being the unknown number. )

### Re: exam question

Posted: **Sat Nov 12, 2016 4:55 pm**

by **kiamlee**

i dont know about my calculator but when i try to solve it on my own i get 189 but im good with it since its the closest number to 188.25 which is the correct answer. As long as i get an answer close to any of the choices with .75 difference. haha thank you so much guys! i have another exam question can you guys try to look at this im pretty sure we didnt take this one with our instructor.

A vessel departed from point A at 1206 with a counter reading of 616729 and arrived at point B with a counter reading of 731929 at 1148 the following day. This vessel is equipped with a 20 foot 8 inch diameter propeller, with a pitch of 20 feet. The observed distance of 404.16 miles was covered at an observed speed of 16.85 knots. What should be the apparent slip for this trip?

A. 1.04% B. 1.29% C. -6.65% D. -11.04%

neither of my friends know the answer to this one. lol

### Re: exam question

Posted: **Sat Nov 12, 2016 10:20 pm**

by **JollyJack**

STCW Examinations are all in SI units, ie, metric. Antiquated units like feet, long tons and short tons are not used, nor is API Gravity or degrees F. Dimensions are given in metres, gravity as mass/unit volume and you never fill a fuel tank 100%. The question is a red herring.

### Re: exam question

Posted: **Mon Nov 14, 2016 8:26 am**

by **Merlyn**

Well I am struggling to work out if the S. G. Is hidden in the equation somewhere.

Ref. Metric and m.m. I still think m.m. stands for Micky Mouse.

### Re: exam question

Posted: **Mon Nov 14, 2016 12:21 pm**

by **D Winsor**

Merlyn wrote:Well I am struggling to work out if the S. G. Is hidden in the equation somewhere.

Ref. Metric and m.m. I still think m.m. stands for Micky Mouse.

In the question I answered I stuck with the API Gravity quoted and calculated the specific weight in lbs/usgal through interpolation as it was easier than to stick with one standard of measurement than do a lot converting between different standards of measurement creating errors along the way

To get SG from API you use the formula 141.5/(APIG+131.5). Using this formula 141.5/(35.7+131.5)=0.8463

If you want to read where the standard numbers come from follow this Wikipedia Link

https://en.wikipedia.org/wiki/API_gravity That explains everything .

With respect to the propeller slip question I'll have to dig out my old Southern's Verbal Notes & Sketches for Marine Engineers to make sure my old calculator brain has the formula right to figure that one out. I do know it can't be a negative number unless the trip was down a long river with the current pushing the boat the full way or it was a sailboat and they ran the engine while using the sails

### Re: exam question

Posted: **Mon Nov 14, 2016 3:56 pm**

by **D Winsor**

I found my copy of Southern's Verbal Notes and Sketches so here goes

Step 1 Question Break down

Revolutions turned 731929-616729=115200

Total Time of travel 23 Hours 42 minutes or 1422 minutes

Average RPM for trip 115200/1422 = 81 RPM

Calculate vessel actual speed over ground 404.16 miles/23.7hours = 17.05 knots

Step2

Calculate Theoretical Prop speed (Pitch(Ft) x RPM x 60)/6080 (20x81x60)/6080=15.99 Knots

Calculate % Apparent Slip ((Theoretical Prop speed - Actual SOG)/Theoretical Prop Speed) X 100 ((15.99-17.05)/15.99)x100 = -6.63%

Using the observed speed of the vessel would have given the following answer ((15.99-16.85)/15.99)x100 = -5.38%

Making Answer "C" the correct answer

Making this definitely a "Down River" trip with current pushing the vessel along

While we are on the topic of Propeller Slip calculations. This calculation can also be done on a CP Propeller as well even though others may say otherwise. All you need to know is how to find the relative pitch of the blades for a given OD Box Tail-rod reading. On a Ka-Me-Wa Prop it is fairly easy as there is usually a graph in the book that came with the prop which indicates the Pitch Ratio relative to Propeller Diameter and OD Box Tail-rod reading. From the graph reading the pitch can be calculated for a given OD Box Reading and "K" the design or optimal pitch of the propeller. This can be used along with shaft tachometer reading,to calculate the theoretical speed of the prop relative to the OD Box Reading and "K". Then using the current speed over ground of the vessel say from the GPS, one can calculate the propeller slip for an instant in time relative to the pitch from an actual OD Box Reading and relative to "K".

### Re: exam question

Posted: **Wed Nov 16, 2016 8:24 pm**

by **JollyJack**

sg is simply mass per unit volume, where Kg is mass and Cubic metre is volune, diesel is SG 0.86m which means there are 860 Kg in a cubic meter. HFO is sg 0.99, which is 990kg in a cubic meter. The calorific value of fuel is about 47 Kj/Kg. NOTE THAT CAL VAL is PER KG. It therefore follows that heavy oil at a density of 990 Kg/m3 has 47mj x 990 mj of enery. Diesel oil, on the other hand, is only 860 kg/m3.

It follows therefore, you have to burn a lot more DG oil to generate the same energy output as burning heavy 0il.

If you get stuck in the ice, do NOT burn high sulphur heavy oil, increase the rack settings and burn more gas oil. Stay in compliance, avoid a $1.000.000 fine and 2 years in jail.

(edited when sober)

### Re: exam question

Posted: **Thu Nov 17, 2016 12:58 am**

by **kiamlee**

JollyJack wrote:STCW Examinations are all in SI units, ie, metric. Antiquated units like feet, long tons and short tons are not used, nor is API Gravity or degrees F. Dimensions are given in metres, gravity as mass/unit volume and you never fill a fuel tank 100%. The question is a red herring.

those questions were taken from the website of maritime department in our country, which they, every year suppose to release an updated sample test questions for board exams for aspiring fourth marine engineers. sadly we have no choice but to take up questions with mixed measurement units. Thinking those were just sample questions, i cant start to imagine how the actual board exam questions be like.

### Re: exam question

Posted: **Thu Nov 17, 2016 1:01 am**

by **kiamlee**

D Winsor wrote:I found my copy of Southern's Verbal Notes and Sketches so here goes

Step 1 Question Break down

Revolutions turned 731929-616729=115200

Total Time of travel 23 Hours 42 minutes or 1422 minutes

Average RPM for trip 115200/1422 = 81 RPM

Calculate vessel actual speed over ground 404.16 miles/23.7hours = 17.05 knots

Step2

Calculate Theoretical Prop speed (Pitch(Ft) x RPM x 60)/6080 (20x81x60)/6080=15.99 Knots

Calculate % Apparent Slip ((Theoretical Prop speed - Actual SOG)/Theoretical Prop Speed) X 100 ((15.99-17.05)/15.99)x100 = -6.63%

Using the observed speed of the vessel would have given the following answer ((15.99-16.85)/15.99)x100 = -5.38%

Making Answer "C" the correct answer

Making this definitely a "Down River" trip with current pushing the vessel along

While we are on the topic of Propeller Slip calculations. This calculation can also be done on a CP Propeller as well even though others may say otherwise. All you need to know is how to find the relative pitch of the blades for a given OD Box Tail-rod reading. On a Ka-Me-Wa Prop it is fairly easy as there is usually a graph in the book that came with the prop which indicates the Pitch Ratio relative to Propeller Diameter and OD Box Tail-rod reading. From the graph reading the pitch can be calculated for a given OD Box Reading and "K" the design or optimal pitch of the propeller. This can be used along with shaft tachometer reading,to calculate the theoretical speed of the prop relative to the OD Box Reading and "K". Then using the current speed over ground of the vessel say from the GPS, one can calculate the propeller slip for an instant in time relative to the pitch from an actual OD Box Reading and relative to "K".

thank you so much for this detailed answer! my head is still shaking after just reading Step 1. lol May i know where can i get a copy of that Southern's Verbal Notes and Sketches you have? that would greatly help.

### Re: exam question

Posted: **Thu Nov 17, 2016 4:55 am**

by **Merlyn**

Metric Versus Imperial Cavitation calcs. eh?

Blimey or what?

But what about surface piercing props?

### Re: exam question

Posted: **Thu Nov 17, 2016 6:02 am**

by **D Winsor**

JollyJack wrote:STCW Examinations are all in SI units, ie, metric. Antiquated units like feet, long tons and short tons are not used, nor is API Gravity or degrees F. Dimensions are given in metres, gravity as mass/unit volume and you never fill a fuel tank 100%. The question is a red herring.

Jolly Jack may be right about using these "Antiquated" units of measurement but as long as our neighbors to the south of us here in Canada insist marching to their own drummer ignoring the rest of the world. We will still have to know how to work with these units of measurement especially if you have to deal with them on a regular basis

With respect to finding a copy of Southern's Verbal Notes & Sketches for Marine Engineers, I think that book has been out of print for many years maybe someone in the UK could help you find a copy or you can find a copy in a library somewhere. It's a great reference book for learning old technologies especially steam and early diesel technology and ship construction. It would also make a great instruction manual if you wanted to rebuild/restore and or run an original early to mid 20th century ship.