The Cutterman’s Guide to Navigation Problems Part One: Fuel Consumption and Engine Problems Fuel consumption problems are useful when calculating voyage efficiency or for logistics requirements. There are two types of problems to solve. The first type of problem involves the relationship between speed, distance, and available fuel. The second type of problem involves the relationships between pitch, engine revolutionsperminute, and speed. Equations There are five equations used to deal with in these problems: Equation (1) !"# !"#$%&’()"# !"# !"#$%&’()"# = !"# !"##$ ! !"# !"##$ ! Equation (2) !"# !"#$%&’()"# !"# !"#$%&’()"# = !"# !"##$ ! × !"# !"#$%&’( !"# !"##$ ! × !"! !"#$%&’( Equation (3) = !"#$"% !"##$!!"#$%&$’ !"##$ !"#$"% !"##$ × 100 Equation (4) = 100% − Equation (5) = !"# × !" × !"#$! × !""#$#%&$’ !"#" Speed, Distance, and Fuel Problems These problems can take several forms, but all ultimately come down to the relationship between speed, distance and fuel. All problems use either equation (1) or equation (2). Problem 11 (CG632). The following question is taken directly from the USCG test bank and illustrates the use of equation (1).
Transcript
The Cutterman’s Guide to Navigation Problems
Part One: Fuel Consumption and Engine Problems Fuel consumption problems are useful when calculating voyage efficiency or for logistics requirements. There are two types of problems to solve. The first type of problem involves the relationship between speed, distance, and available fuel. The second type of problem involves the relationships between pitch, engine revolutions-‐per-‐minute, and speed.
Equations There are five equations used to deal with in these problems: Equation (1) !"# !"#$%&'()"#
Speed, Distance, and Fuel Problems These problems can take several forms, but all ultimately come down to the relationship between speed, distance and fuel. All problems use either equation (1) or equation (2). Problem 1-‐1 (CG-‐632). The following question is taken directly from the USCG test bank and illustrates the use of equation (1).
While steaming 17.5 knots, your vessel consumes 378 barrels of fuel oil per day. In order to reduce consumption to 194 barrels of fuel oil per day, what is the maximum speed the vessel can turn for? Answer: 14.01 knots. Use equation (1) to solve for the new speed required. Step 1: !"# !"#$%&'()"#
!"# !"#$%&'()"#= !"# !"##$!
!"# !"##$!
Step 2: !"# !"##$%& !"# !"#
!"# !"##$%& !"# !"#= !"# !"##$!
(!".!!"#)!
Step 3: 0.5132 = !"# !"##$!
!"!#.!
Step 4: 2750.4 = 𝑁𝑒𝑤 𝑆𝑝𝑒𝑒𝑑!
Step 5: 𝟏𝟒.𝟎𝟏 𝒌𝒕𝒔 = 𝑵𝒆𝒘 𝑺𝒑𝒆𝒆𝒅 Problem 1-‐2 (CG-‐809). The following question is taken directly from the USCG test bank and illustrates the use of equation (2). You have steamed 174 miles and consumed 18.0 tons of fuel. If you maintain the same speed, how many tons of fuel will you consume while steaming 416 miles? Answer: 43.03 tons. Use equation (2) to solve for the new consumption. Since the speed is to remain the same, you can cancel the terms from the numerator and denominator. Step 1: !"# !"#$%&'()"#
!"# !"#$%&'()"#= !"# !"##$! × !"# !"#$%!"#
!"# !"##$! × !"# !"#$%&'(
Step 2: !"# !"#$%&'()"#
!".! !"#$ !"# !"#= !""#$#%&'( !"##$! × !"# !"#$%
!""#$#%&'( !"##$! × !"# !"#$%
Step 3: Cancel the speed terms from the numerator and denominator. Step 4: !"# !"#$%&'()"#
!".! !"#$ !"# !"#= !"# !"#$%
!"# !"#$%
Step 5: !"# !"#$%&'()"#
!".! !"#$ !"# !"#= 2.3908
Step 6: 𝑵𝒆𝒘 𝑪𝒐𝒏𝒔𝒖𝒎𝒑𝒕𝒊𝒐𝒏 = 𝟒𝟑.𝟎𝟑 𝒕𝒐𝒏𝒔 𝒑𝒆𝒓 𝒅𝒂𝒚
Pitch, RPM, and Speed Problems These problems take many forms and sometimes require multiple equations to solve a single problem. They rely on equations (3), (4), and (5). Problem 1-‐3. The following question illustrates the use of equation (3). While turning for 18.0 knots, accounting for current and wind, the ship is observed to make good 17.3 knots. What is the apparent slip of the propeller? Answer: +3.88%. The reason for multiplying the fractional value by 100 is to obtain an answer in percent form, the standard form for slip calculations. Note that positive slip and negative slip values can occur, and it is crucial to watch the sign in later, more complex problems.
Step 5: 𝑺𝒍𝒊𝒑 = + 𝟑.𝟖𝟖% Problem 1-‐4 (CG-‐566). The following question is taken directly from the USCG test bank and illustrates the use of equations (3) and (5) together. The propeller on a vessel has a diameter of 20.6 feet and a pitch of 23.4 feet. What would be the apparent slip if the vessel cruised 538 miles in a 24-‐hour day (observed distance) at an average RPM of 87? Answer: -‐11.59%. The propeller diameter is “distractor” information and unnecessary for the calculation. The goal is to calculate slip, so the “efficiency” portion of equation (5) is not required.
Step 5: This gives us our engine speed for the next calculation. In the
problem, the vessel cruised 538 miles in a 24-‐hour day. Therefore from a basic speed=distance/time calculation, the observed speed is 22.42kts.
Step 6: 𝑆𝑙𝑖𝑝 = !"#$"% !"##$!!"#$%&!" !"##$
!"#$"% !"##$ × 100
Step 7: 𝑆𝑙𝑖𝑝 = !".!"!"#!!!.!"!"#
!".!"!"# × 100
Step 8: 𝑆𝑙𝑖𝑝 = !!.!"!"#
!".!"!"# × 100
Step 9: 𝑆𝑙𝑖𝑝 = −0.1159 × 100
Step 10: 𝑺𝒍𝒊𝒑 = −𝟏𝟏.𝟓𝟗%
Problem 1-‐5 (CG-‐146). The following question is taken directly from the USCG test bank and illustrates the use of equations (4) and (5) together. If the speed necessary for reaching port at a designated time is 18.5 knots and the pitch of the propeller is 21.7 feet, how many revolutions per minute will the shaft have to turn assuming a 4% negative slip? Answer: 83.1 RPM.
Step 1: 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 100%− 𝑆𝑙𝑖𝑝
Step 2: 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 100%− −4%
Step 3: 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 104%
Step 4: 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 1.04
Step 5: After calculating the efficiency, the next step is to use equation (5) to calculate the required RPM of the shaft.
Step 6: 𝑆𝑝𝑒𝑒𝑑 = !"# × !" × !"#$! × !""#$#%&$'
!"#"
Step 7: 18.5𝑘𝑡𝑠 = !"# × !" × !".!!" × !.!"
!"#"
Step 8: 18.5𝑘𝑡𝑠 = !"# × !"#$.!"
!"#"
Step 9: 112480 = 𝑅𝑃𝑀 × 1354.08
Step 10: 𝑹𝑷𝑴 = 𝟖𝟑.𝟎𝟔𝟕
The purpose of the “60” in equation (5) is to turn RPM into an hourly figure. If a question gives a daily revolutions count, then the “60” is not required. Problem 1-‐6 (CG-‐127). The following question is taken directly from the USCG test bank and illustrates the concept of daily revolutions. If the pitch of the propeller is 19.7 feet and the revolutions per day are 86,178, calculate the day’s run allowing 3% negative slip. Answer 287.60 nautical miles.
Step 1: 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 100%− 𝑆𝑙𝑖𝑝
Step 2: 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 100%− (−3%)
Step 3: 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 103%
Step 4: 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 1.03
Step 5: After calculating the efficiency, the next step is to use equation (5) to calculate the required RPM of the shaft. Since the daily revolutions are given, the “60” in equation (5) is not necessary, and instead of “speed,” the answer will be given in “day’s run.”
Additional Problems and Answers All of the following questions were taken directly from the 2013 USCG test bank and illustrate the concepts in this Part. Note – not all problems have been worked and are subject to occasional errors in the database. For more problems and answers, see the USCG database of questions (database information located in the preface). Problem CG-‐621. While steaming 13.5 knots, your vessel consumes 251 barrels of fuel oil per day. In order to reduce consumption to 129 barrels of fuel oil per day, what is the maximum speed the vessel can turn for?
a) 6.9 kts b) 9.7 kts c) 10.8 kts-‐ correct d) 12.7 kts
Problem CG-‐624. While steaming at 14.5 knots, your vessel consumes 319 barrels of fuel oil per day. In order to reduce consumption to 217 barrels of fuel oil per day, what is the maximum speed the vessel can turn for?
a) 9.8 kts b) 11.9 kts c) 12.8 kts-‐ correct d) 13.5 kts
Problem CG-‐625. While steaming 15 knots, your vessel burns 326 bbls of fuel oil per day. What will be the rate of fuel consumption if you decrease speed to 12.2 knots?
a) 175 bbls per day-‐ correct b) 215 bbls per day c) 277 bbls per day d) 300 bbls per day
Problem CG-‐812. You have steamed 201 miles and consumed 18.0 tons of fuel. If you maintain the same speed, how many tons of fuel will you consume while steaming 482 miles?
a) 25.2 tons b) 43.2 tons-‐ correct c) 52.6 tons d) 103.5 tons
Problem CG-‐817. You have steamed 300 miles and consumed 34 tons of fuel. If you maintain the same speed, how many tons of fuel will you consume while steaming 700 miles?
a) 79.3 tons-‐ correct b) 74.3 tons c) 68.4 tons d) 66.2 tons
Problem CG-‐820. You have steamed 449 miles at 19 knots and burning 476 barrels of fuel per day. You must decrease your consumption to 185 barrels per day with 362 miles left in your voyage. What must you reduce your speed (kts) to in order to burn this amount of fuel?
a) 13.2-‐ correct b) 14.3 c) 17.1 d) 18.2
Problem CG-‐826. You have steamed 525 miles at 16 knots and consumed 105 tons of fuel. If you have 308 tons of usable fuel remaining, how far can you steam at 19 knots?
a) 920 miles b) 1092 miles-‐ correct c) 1297 miles d) 2172 miles
Problem CG-‐827. You have steamed 540 miles at 22 knots and burning 618 barrels of fuel oil per day. You must decrease your consumption to 372 barrels with 299 miles left in your voyage. What must you reduce your speed (kts) to in order to burn this amount of fuel?
a) 22.9-‐ correct b) 20.0 c) 19.1 d) 17.6
Problem CG-‐838. You have steamed 916 miles at 13 knots, and consumed 166 tons of fuel. If you have to steam 1325 miles to complete the voyage, how many tons of fuel will be consumed while steaming at 14 knots?
a) 133 tons b) 181 tons c) 207 tons d) 278 tons-‐ correct
Problem CG-‐902. Your vessel arrives in port with sufficient fuel to steam 726 miles at 16 knots. If you are unable to take on bunkers, at what speed must you proceed to reach your next port, 873 miles distant?
a) 14.6 kts-‐ correct b) 15.1 kts c) 16.3 kts d) 16.8 kts
Problem CG-‐912. Your vessel consumes 274 barrels of fuel per day at a speed of 17.5 knots. What will the fuel consumption be at 13.5 knots?
a) 126 bbls-‐ correct b) 163 bbls c) 211 bbls d) 253 bbls
Problem CG-‐90. At your current speed of 20 knots, you only have enough fuel remaining to travel 218 miles. You must travel 395 miles to reach your destination. What should you reduce your speed (knots) to in order to reach your destination?
a) 17.4 b) 16.2 c) 14.9-‐ correct d) 13.7
Problem CG-‐96. At your current speed of 23 knots, you only have enough fuel remaining to steam 386 miles. You must travel 785 miles to reach your destination. What should you reduce your speed (knots) to in order to reach your destination?
a) 19.3 b) 17.7 c) 16.1-‐ correct d) 14.5
Problem CG-‐635. While steaming at 19.5 knots, your vessel burns 297 bbls of fuel per day. What will be the rate of fuel consumption if you decrease speed to 15 knots?
a) 135 bbls-‐ correct b) 176 bbls c) 229 bbls d) 243 bbls
Problem CG-‐798. You have steamed 1124 miles at 21 knots and consumed 326 tons of fuel. If you have 210 tons of usable fuel remaining, how far can you steam at 17 knots?
a) 1096 miles b) 1105 miles-‐ correct c) 1218 miles d) 1304 miles
Problem CG-‐126. If the pitch of the propeller is 19.4 feet, and the revolutions per day are 96,713, calculate the day’s run allowing 6% positive slip.
a) 266.4 miles b) 290.1 miles-‐ correct c) 308.6 miles d) 327.1 miles
Problem CG-‐132. If the pitch of the propeller is 21.5 feet, and the revolutions per day are 96,666, calculate the day’s run allowing 9% negative slip.
a) 311.1 miles b) 341.8 miles c) 357.9 miles d) 372.6 miles-‐ correct
Problem CG-‐136. If the pitch of the propeller is 25.1 feet, and the revolutions per day are 91,591, calculate the day’s run allowing 7 % positive slip.
a) 351.6 miles-‐ correct b) 378.1 miles c) 390.0 miles d) 404.6 miles
Problem CG-‐139. If the speed necessary for reaching port at a designated time is 12.6 knots and the pitch of the propeller is 13.6 feet, how many revolutions per minute will the shaft have to turn, assuming no slip?
a) 81 b) 85 c) 90 d) 94-‐ correct
Problem CG-‐140. If the speed necessary for reaching port at a designated time is 15.7 knots and the pitch of the propeller is 23.4 feet, how many revolutions per minute will the shaft have to turn, assuming a 6% negative slip?
a) 64-‐ correct b) 68 c) 72 d) 76
Problem CG-‐144. If the speed necessary for reaching port at a designated time is 17.8 knots and the pitch of the propeller is 24.7 feet, how many revolutions per minute will the shaft have to turn, assuming a 7% positive slip?
a) 67 b) 71 c) 75 d) 79-‐ correct
Problem CG-‐150. If the speed necessary for reaching port at a designated time is 20.7 knots and the pitch of the propeller is 23.8 feet, how many revolutions per minute will the shaft have to turn, assuming a 3% negative slip?
a) 74 b) 79 c) 86-‐ correct d) 98
Problem CG-‐562. The propeller of a vessel has a pitch of 19.0 feet. If the vessel traveled 183.5 miles (observed distance) in 24 hours at an average of 44 RPM, what was the slip?
a) +7.4%-‐ correct b) -‐7.4% c) +11.6% d) -‐11.6%
Problem CG-‐565. The propeller on a vessel has a diameter of 20.2 feet and a pitch of 19.0 feet. What would be the apparent slip if the vessel cruised 367 miles (observed distance) in a 24-‐hour day at an average RPM of 84?
a) +2.9%-‐ correct b) -‐2.9% c) +5.2% d) -‐5.2%
Problem CG-‐561. The pitch of the propeller on your vessel is 19’ 09”. You estimate slip to be -‐3%. If you averaged 82 RPM for the day’s run, how many miles did you steam?
a) 370.8 b) 373.6 c) 393.7 d) 395.3-‐ correct
Problem CG-‐571. The propeller on a vessel has a diameter of 24.0 feet and a pitch of 21.3 feet. What would be the slip if the vessel cruised 510 miles in a 24 hour day (observed distance) at an average RPM of 86?
a) -‐12.2% b) +12.2%
c) -‐17.5%-‐ correct d) +17.5%
Problem CG-‐579. The speed of advance necessary to reach port at a designated time is 15.8 knots. The pitch of the propeller is 20.75 feet. You estimate positive 5% slip. How many RPM must you turn to make the necessary speed?
a) 73.5 b) 76.2 c) 79.9 d) 81.2-‐ correct
Problem CG-‐730. You are turning 100 RPM, with a propeller pitch of 25 feet, and an estimated slip of -‐5%. What is the speed of advance?
a) 24.7 kts b) 23.5 kts c) 25.9 kts-‐ correct d) 22.3 kts
Problem CG-‐735. You are turning 82 RPM, with a propeller pitch of 23 feet, and an estimated slip of +6%. What is the speed of advance?
a) 17.5 kts-‐ correct b) 17.9 kts c) 18.4 kts d) 19.7 kts
Problem CG-‐991. Your vessel’s propeller has a pitch of 22’06”. From 0530, 19 March, to 1930, 20 March, the average RPM was 82. The distance run by observation was 721.5 miles. What was the slip?
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