Exercise set

Ship Power Systems and Propulsion Plant Exercises

Solved ship power exercises for generator reserve, load shedding, frequency nadir, diesel-electric losses, batteries, harmonics, shaft torque and cooling.

These exercises practise ship power-system and propulsion-plant calculations as naval engineering evidence. They focus on installed machinery, electrical loading, propulsion demand, shaft torque, generator reserve, contingency response, frequency nadir, harmonic RMS current, short-circuit duty, cooling, fuel, batteries, state-of-charge limits, emergency loads, and validation records.

The goal is not only to add installed kilowatts. A vessel can have enough rated power and still fail because the wrong generators are online, load shedding is late, power factor consumes MVA capacity, battery state of charge is outside the usable window, cooling duty is underestimated, or a common auxiliary system disables redundant equipment.

Assume simplified screening models unless an exercise states otherwise. Real vessel work should also check classification rules, flag requirements, switchboard topology, protection studies, short-circuit duty, harmonic distortion, control logic, cable routes, fire zones, redundancy notation, sea-trial evidence, and crew procedures.

How to Use These Exercises

For each exercise, define:

  1. the operating mode: transit, maneuvering, dynamic positioning, port, emergency, degraded, or maintenance;
  2. the power boundary: shaft power, motor power, bus electrical power, generator output, fuel power, or cooling duty;
  3. the contingency being claimed;
  4. the load priority and shedding rule;
  5. the validation evidence needed before releasing the operating mode.

The common mistake is treating installed power as available power. Available power depends on online configuration, conversion losses, protection limits, cooling, control logic, fuel, and survivability after credible failures.

Release Evidence Notes

Ship power calculations should be tied to a named vessel operating mode and electrical boundary. Record online generators, switchboard split, propulsion state, hotel load, thruster availability, emergency bus boundary, shore-power state, battery SOC/SOH, cooling configuration, fuel condition, protection settings and blackout-recovery procedure. A reserve or load result without that boundary cannot support mode release.

Generator and bus evidence should separate active power, apparent power and fault duty. Online reserve, N-1 survivability, load sharing, power factor, voltage dip, harmonic behavior, short-circuit contribution, breaker rating and protection timing can each govern a different limit. Passing kW loading does not prove that the switchboard, protection or transient response is acceptable.

Energy-support evidence should include usable windows and recovery state. Batteries, UPS, fuel reserve, emergency generator starts, cooling loops, ventilation, pumps, strainers and auxiliaries should be checked for current state, degraded condition, common-cause failure, recovery time and crew action. A support system can limit available propulsion power even when generators are electrically capable.

Validation evidence should come from real or representative events. Load-shed tests, motor starts, generator trips, shore-transfer trials, blackout recovery, sea-trial power records, cooling temperatures, alarm proof tests, event logs and crew procedures should match the operating envelope being released. Simulation-only evidence should identify what remains to be confirmed onboard.

The practical release question is whether online configuration, power boundary, protection, cooling, energy storage, fuel, automation, crew action and event evidence all support the same operating mode. If one layer conflicts, the result should trigger mode restriction, additional generator online, load-shed tuning, motor-start block, protection review, cooling maintenance or retest.

Engineering Boundary Notes

Ship power evidence must identify which boundary is being controlled: shaft power, motor input, converter output, switchboard load, generator rating, fuel input, cooling duty or emergency-bus capacity. The same number can look acceptable or unsafe depending on where losses, power factor, transients and auxiliaries are counted.

Normal operation and degraded operation need separate boundaries. A plant can satisfy steady load with all generators healthy and still fail a generator trip, bow-thruster start, transformer limit, harmonic limit, cooling impairment, emergency-generator start sequence or blackout-recovery requirement. Exercises that look like simple reserve calculations should therefore be read as configuration checks.

Evidence also has a time boundary. Frequency nadir, load-shed timing, motor-start voltage dip, battery ride-through, UPS autonomy and blackout recovery depend on seconds or minutes, not only on final steady-state power. A release package should say which transient has been tested, which one is simulated, and which one remains an onboard proof point.

Common Release Mistakes

  • comparing active power reserve with apparent power limits or breaker fault-duty limits;
  • counting a standby generator as available before start, synchronization and load acceptance are proven;
  • using battery nameplate energy instead of usable SOC window, SOH and inverter limit;
  • accepting motor-start or thruster load without checking voltage dip, protection timing and concurrent hotel load;
  • ignoring cooling-water, fuel, ventilation or lube-oil auxiliaries that cap available propulsion;
  • treating harmonic RMS current as a cable-only problem while converter, transformer and protection limits may govern;
  • releasing an operating mode from simulation without matching event logs, sea-trial data or crew procedure.

Scenario Map

ScenarioExercisesPrimary checkEngineering decision
Generator reserve and contingency response1, 2, 8, 11, 17Online reserve, N-1 load shedding, load sharing, shedding timing and frequency nadirDecide whether the plant can survive generator trips without blackout or overload.
Electrical boundaries and shore connection3, 4, 10, 14, 15, 18Conversion losses, apparent power, power factor, transformer margin, voltage dip, harmonic RMS current and short-circuit dutyCheck whether bus, generator, switchboard and shore-power limits match the real electrical boundary.
Energy, fuel and cooling support5, 6, 7, 9, 13Battery usable energy, SOC ride-through, fuel mass, cooling flow and emergency starting peakDecide whether support systems can sustain the claimed operating mode.
Mechanical propulsion train3, 16Electrical-to-shaft conversion, gearbox ratio, shaft rpm, service-factor torque and derateCheck whether mechanical torque limits match the claimed propulsion power.
Reliability and validation evidence12RPN reduction after monitoring changesCheck whether monitoring reduces risk enough while preserving high-severity release evidence.

Validation Package Checklist

Before treating a ship-power calculation as release evidence, collect:

  1. operating mode, online generator lineup, bus split and power-management mode;
  2. active power, apparent power, power factor, harmonic and short-circuit boundaries;
  3. load list with priority class, shedding rule and restart sequence;
  4. propulsion demand, thruster demand, hotel load, emergency load and cooling duty;
  5. battery, UPS, fuel, cooling, ventilation and auxiliary-system state;
  6. protection settings, breaker ratings, interlocks and automation assumptions;
  7. test or event evidence for trips, motor starts, shore transfer and blackout recovery;
  8. release decision, operating restriction, retest requirement or maintenance action.

Exercise 1: Online Generator Reserve

A diesel-electric vessel is maneuvering with two generators online. Each generator is rated:

P_g=1.8\ \text{MW}

The active loads are:

LoadPower
Propulsion drives1.20\ \text{MW}
Hotel and navigation load0.55\ \text{MW}
Cooling and auxiliaries0.28\ \text{MW}
Bow thruster0.45\ \text{MW}

Compute total online capacity, total load, reserve power, and reserve as a percentage of online capacity.

Solution

Online capacity:

P_{cap}=2(1.8)=3.60\ \text{MW}

Total load:

P_L=1.20+0.55+0.28+0.45=2.48\ \text{MW}

Reserve:

P_R=P_{cap}-P_L=3.60-2.48=1.12\ \text{MW}

Reserve percentage:

\displaystyle R=\frac{1.12}{3.60}=0.311

So:

R=31.1\%

Engineering Comment

The two-generator operating mode has positive reserve in normal operation. That does not prove single-failure survivability. If one generator trips, available online capacity changes immediately while the standby generator is still starting and synchronizing.

Plausibility Check

The two online generators provide 3.60 MW and the load is 2.48 MW, so the normal reserve is 1.12 MW or 31.1\%. That reserve is positive, but it is smaller than one generator rating, which explains why the following N-1 shedding case is still necessary.

Exercise 2: Single-Generator Contingency Shedding

Using the same load from Exercise 1, assume one online generator trips. The vessel must keep remaining load below:

85\%

of the surviving generator rating during recovery. Compute the maximum allowed load and the required shedding amount.

Solution

Surviving generator capacity:

P_{survive}=1.8\ \text{MW}

Maximum allowed load:

P_{allowed}=0.85(1.8)=1.53\ \text{MW}

Pre-trip load:

P_L=2.48\ \text{MW}

Required shedding:

P_{shed}=P_L-P_{allowed}=2.48-1.53=0.95\ \text{MW}

Engineering Comment

This is the load-shedding target before considering governor response, motor transients, and standby-generator timing. A power-management system that sheds less than 0.95\ \text{MW} or sheds it too late cannot support the stated contingency claim.

Plausibility Check

The surviving generator can carry only 1.53 MW under the 85\% recovery rule. Since the pre-trip load is 2.48 MW, the required shed amount is 0.95 MW, about 38\% of the pre-trip load, so the contingency cannot be handled as a small trim action.

Exercise 3: Diesel-Electric Propulsion Conversion Losses

A propulsor requires mechanical output at the shaft:

P_{shaft}=1.60\ \text{MW}

The propulsion motor efficiency is:

\eta_m=0.96

The drive efficiency is:

\eta_d=0.97

The transformer and distribution efficiency is:

\eta_{dist}=0.99

Estimate the electrical power drawn from the ship service bus.

Solution

Overall electrical-to-shaft efficiency:

\eta=\eta_m\eta_d\eta_{dist}=0.96(0.97)(0.99)=0.922

Bus power:

\displaystyle P_{bus}=\frac{P_{shaft}}{\eta}=\frac{1.60}{0.922}=1.735\ \text{MW}

Engineering Comment

The generator and switchboard must supply about 1.74\ \text{MW} for a 1.60\ \text{MW} shaft demand. The losses also become heat in equipment rooms, drives, motors, and transformers. Cooling and ventilation checks should use the electrical boundary, not only the propeller or shaft boundary.

Plausibility Check

The combined efficiency is 0.96(0.97)(0.99)=0.922, so bus power should be larger than shaft power. The difference is about 1.735-1.60=0.135 MW, which is a credible heat and loss allowance for the conversion chain.

Exercise 4: Apparent Power and Power Factor

A propulsion drive draws active power:

P=1.20\ \text{MW}

at power factor:

PF=0.82

Compute apparent power. Then compute apparent power if power factor is corrected to:

PF=0.95

The generator limit for this drive allocation is:

S_{limit}=1.50\ \text{MVA}

Solution

Apparent power:

\displaystyle S=\frac{P}{PF}

At PF=0.82:

\displaystyle S=\frac{1.20}{0.82}=1.463\ \text{MVA}

At PF=0.95:

\displaystyle S=\frac{1.20}{0.95}=1.263\ \text{MVA}

Both values are below:

1.50\ \text{MVA}

but the lower power factor leaves only:

1.50-1.463=0.037\ \text{MVA}

of margin.

Engineering Comment

Active power is not the whole generator loading picture. Low power factor can consume MVA capacity, increase current, raise losses, and reduce fault or transient margin. Marine electrical review should include active power, reactive power, apparent power, harmonic distortion, drive behavior, and protection settings.

Plausibility Check

At PF=0.82, the drive uses 1.463 MVA and leaves only 0.037 MVA below the 1.50 MVA allocation. Correcting to PF=0.95 reduces apparent power to 1.263 MVA, increasing the margin to about 0.237 MVA.

Exercise 5: Battery Peak-Shaving Energy

A hybrid vessel uses a battery to cover a propulsion peak of:

P_{peak}=0.60\ \text{MW}

for:

t=8\ \text{min}

Battery discharge efficiency is:

\eta=0.94

Only:

60\%

of nominal battery capacity is allowed for this operating window. Estimate required nominal battery energy.

Solution

Energy delivered to the bus:

\displaystyle E_{bus}=P_{peak}t=0.60\left(\frac{8}{60}\right)=0.080\ \text{MWh}

Battery energy drawn:

\displaystyle E_{bat}=\frac{E_{bus}}{\eta}=\frac{0.080}{0.94}=0.0851\ \text{MWh}

Nominal capacity with a 60 percent usable window:

\displaystyle E_{nom}=\frac{0.0851}{0.60}=0.1418\ \text{MWh}

So:

E_{nom}=142\ \text{kWh}

Engineering Comment

This is an energy screen, not a battery-system design. The final battery review must check power limit, C-rate, state-of-charge window, thermal management, degradation, fire protection, fault isolation, charging opportunity, class rules, and whether the peak event can repeat before recharge.

Plausibility Check

The bus event needs 0.080 MWh. Accounting for 94\% discharge efficiency raises the required battery draw to 0.0851 MWh, and limiting usable capacity to 60\% gives a nominal capacity near 142 kWh, which back-checks to the required delivered energy.

Exercise 6: Fuel Mass for a Transit Profile

A vessel operates for:

t=10\ \text{h}

with average generator active output:

P=2.40\ \text{MW}

The diesel generator specific fuel consumption is:

SFC=205\ \text{g/kWh}

Estimate fuel mass, then add an operating reserve of:

8\%

Solution

Energy produced:

E=Pt=2.40(10)=24.0\ \text{MWh}=24000\ \text{kWh}

Fuel mass:

m_f=E(SFC)=24000(205)=4920000\ \text{g}

Convert:

m_f=4920\ \text{kg}

With 8 percent reserve:

m_{total}=4920(1.08)=5313.6\ \text{kg}

Engineering Comment

This estimate assumes constant generator loading and SFC. Real fuel use depends on load sharing, auxiliary load, sea state, fouling, fuel temperature and density, generator efficiency at part load, battery cycling, hotel load, and time spent maneuvering or waiting.

Plausibility Check

The transit produces 24,000 kWh, and 205 g/kWh gives 4,920,000 g or 4920 kg of fuel. Adding 8\% reserve adds about 394 kg, giving the reported 5314 kg total.

Exercise 7: Cooling Duty and Flow

A diesel generator produces mechanical/electrical output:

P_{out}=1.20\ \text{MW}

Its thermal efficiency is:

\eta_{th}=0.42

Assume:

22\%

of fuel input power must be removed by a freshwater cooling loop. The allowable coolant temperature rise is:

\Delta T=8\ \text{K}

Use:

c_p=4.18\ \text{kJ/(kg K)}

Estimate required coolant mass flow.

Solution

Fuel input power:

\displaystyle P_{fuel}=\frac{P_{out}}{\eta_{th}}=\frac{1.20}{0.42}=2.857\ \text{MW}

Cooling duty:

\dot{Q}=0.22(2.857)=0.629\ \text{MW}=629\ \text{kW}

Cooling relation:

\dot{Q}=\dot{m}c_p\Delta T

Mass flow:

\displaystyle \dot{m}=\frac{629}{4.18(8)}=18.8\ \text{kg/s}

Engineering Comment

The cooling system must be checked at real seawater temperature, fouling condition, pump curve, strainer condition, heat-exchanger margin, redundancy state, and low-speed airflow or ventilation condition. Underestimating cooling duty can derate available power even when generator electrical capacity looks sufficient.

Plausibility Check

The fuel input is 1.20/0.42=2.857 MW, and 22\% of that is 0.629 MW of cooling duty. Dividing 629 kW by 4.18\times8 kJ/(kg K) gives 18.8 kg/s, so the unit conversion is consistent.

Exercise 8: Generator Load-Sharing Mismatch

Two generators should share a total load:

P_{total}=2.60\ \text{MW}

Actual measured loads are:

P_1=1.45\ \text{MW}

and:

P_2=1.15\ \text{MW}

The load-sharing acceptance limit is:

5\%

from equal share. Check the mismatch of the more heavily loaded generator.

Solution

Ideal equal share:

\displaystyle P_{eq}=\frac{2.60}{2}=1.30\ \text{MW}

Deviation of generator 1:

\Delta P=1.45-1.30=0.15\ \text{MW}

Relative deviation:

\displaystyle \frac{0.15}{1.30}=0.115

So:

11.5\%

Since:

11.5\%>5\%

the load-sharing criterion fails.

Engineering Comment

Load-sharing mismatch can overload one generator while total plant load looks acceptable. Review should check governor settings, droop or isochronous control, power-management configuration, current transformers, voltage regulator behavior, communication delay, and whether the mismatch worsens during motor starts or load shedding.

Plausibility Check

Equal share would be 1.30 MW per generator. Generator 1 is 0.15 MW above that value, or 11.5\% high, which is more than twice the 5\% acceptance limit and therefore a clear load-sharing failure.

Exercise 9: Emergency Generator Starting Load

Emergency loads are:

Emergency loadRunning power
Steering gear60\ \text{kW}
Navigation and communications40\ \text{kW}
Emergency lighting35\ \text{kW}
Emergency bilge pump40\ \text{kW}
Fire pump motor90\ \text{kW}

The fire pump has starting demand:

2.5

times its running power. If the fire pump starts while the other emergency loads are already running, estimate peak emergency generator demand. The emergency generator is rated:

450\ \text{kW}

Solution

Other loads already running:

P_{other}=60+40+35+40=175\ \text{kW}

Fire pump starting demand:

P_{start}=2.5(90)=225\ \text{kW}

Peak demand:

P_{peak}=175+225=400\ \text{kW}

Margin:

M=450-400=50\ \text{kW}

Relative margin:

\displaystyle \frac{50}{450}=0.111

So:

11.1\%

Engineering Comment

The peak demand passes this simplified screen, but the margin is not large. Emergency-generator review should include motor-start voltage dip, sequencing, breaker settings, fuel supply, ventilation, ambient temperature, battery starting system, periodic test evidence, and whether emergency loads are separated from nonessential loads.

Plausibility Check

The non-fire-pump emergency loads total 175 kW and the fire-pump start demand is 225 kW. The combined 400 kW peak leaves 50 kW, or 11.1\%, below the 450 kW rating, so the pass is plausible but not generous.

Exercise 10: Shore-Power Transfer MVA Check

A vessel connects to shore power for hotel load:

P=700\ \text{kW}

at power factor:

PF=0.88

The shore transformer is rated:

S_{rated}=900\ \text{kVA}

Compute apparent power. Then check the same system with a 15 percent load growth allowance.

Solution

Apparent power:

\displaystyle S=\frac{P}{PF}=\frac{700}{0.88}=795.5\ \text{kVA}

This is below:

900\ \text{kVA}

With 15 percent growth:

P_{growth}=700(1.15)=805\ \text{kW}

Apparent power:

\displaystyle S_{growth}=\frac{805}{0.88}=914.8\ \text{kVA}

This exceeds:

900\ \text{kVA}

Engineering Comment

The present load passes, but the growth case fails. Shore-power transfer review should also check voltage, frequency, grounding, neutral arrangement, synchronizing or break-before-make logic, harmonic content, protection coordination, cable rating, interlocks, and whether onboard generators can be safely unloaded and stopped.

Plausibility Check

The present apparent power is 795.5 kVA, leaving about 104.5 kVA below the transformer rating. With 15\% load growth, apparent power rises to 914.8 kVA, exceeding the rating by about 14.8 kVA.

Exercise 11: Load-Shedding Timing

After a generator trip, the plant must shed:

0.75\ \text{MW}

within:

1.0\ \text{s}

The programmed stages are:

StageShedded loadAction time
10.35\ \text{MW}0.2\ \text{s}
20.25\ \text{MW}0.8\ \text{s}
30.20\ \text{MW}2.5\ \text{s}

Check whether the timing requirement is met.

Solution

Load shed within 1.0 s includes stages 1 and 2:

P_{shed,1s}=0.35+0.25=0.60\ \text{MW}

Required shedding:

0.75\ \text{MW}

Deficit within the time limit:

0.75-0.60=0.15\ \text{MW}

The total eventual shedding is:

0.35+0.25+0.20=0.80\ \text{MW}

but the final stage is too late for the 1.0 s requirement.

Engineering Comment

The load-shedding design has enough total shedding but not enough fast shedding. For blackout prevention, timing is part of the capacity calculation. Validation should use event records with generator trip time, breaker status, frequency trace, drive limiter state, load-shed commands, actual load drop, and standby generator start sequence.

Plausibility Check

Stages 1 and 2 shed 0.60 MW inside 1.0 s, leaving a 0.15 MW deficit against the required 0.75 MW. The eventual total is 0.80 MW, but stage 3 occurs at 2.5 s, so capacity is sufficient only after the timing requirement has already been missed.

Exercise 12: Risk Priority After a Plant Monitoring Change

A fuel-filter blockage failure mode has initial ratings:

S=8,\quad O=4,\quad D=6

where S is severity, O is occurrence, and D is detection. A differential-pressure alarm and maintenance trigger reduce occurrence to:

O=3

and detection rating to:

D=2

Compute initial and revised RPN.

Solution

Initial RPN:

RPN_0=SOD=8(4)(6)=192

Revised RPN:

RPN_1=8(3)(2)=48

Reduction:

\Delta RPN=192-48=144

Relative reduction:

\displaystyle \frac{144}{192}=0.75

So:

75\%

Engineering Comment

The monitoring change improves detection and should reduce recurrence, but severity remains high because fuel starvation can still threaten propulsion or generator availability. The release evidence should include alarm setpoint basis, proof test, crew response procedure, maintenance interval, spare filter policy, bypass control, and event-log review after service entry.

Plausibility Check

The initial RPN is 8(4)(6)=192 and the revised RPN is 8(3)(2)=48. The reduction is 144, or 75\%, but severity remains 8, so the monitoring change lowers risk priority without removing the high-consequence failure mode.

Exercise 13: Essential-Load Ride-Through and SOC Gate

During blackout recovery, essential loads must remain supplied for 12\ \text{min} while the standby generator starts, synchronizes, and accepts load. The essential-load bus has:

Essential loadPower
Steering control30\ \text{kW}
Navigation and communications45\ \text{kW}
Power-management and automation controls20\ \text{kW}
Emergency lighting25\ \text{kW}

The UPS/battery system has nominal energy E_{nom}=60\ \text{kWh}, current state of charge SOC_0=72\%, minimum permitted state of charge SOC_{min}=30\%, and inverter efficiency \eta=0.92.

Check whether the current SOC can support the event without crossing the minimum SOC. Then estimate the minimum starting SOC needed for the event.

Solution

Total essential load:

P_{ess}=30+45+20+25=120\ \text{kW}

Energy delivered to the essential-load bus:

\displaystyle E_{bus}=P_{ess}t=120\left(\frac{12}{60}\right)=24.0\ \text{kWh}

Battery energy drawn:

\displaystyle E_{bat}=\frac{E_{bus}}{\eta}=\frac{24.0}{0.92}=26.1\ \text{kWh}

Usable energy above the minimum SOC:

E_{usable}=E_{nom}(SOC_0-SOC_{min})
E_{usable}=60(0.72-0.30)=25.2\ \text{kWh}

Energy margin:

M_E=25.2-26.1=-0.9\ \text{kWh}

Final SOC after the event:

\displaystyle SOC_f=SOC_0-\frac{E_{bat}}{E_{nom}}
\displaystyle SOC_f=0.72-\frac{26.1}{60}=0.285=28.5\%

Minimum starting SOC:

\displaystyle SOC_{0,min}=SOC_{min}+\frac{E_{bat}}{E_{nom}}
\displaystyle SOC_{0,min}=0.30+\frac{26.1}{60}=0.735=73.5\%

Engineering Comment

The current SOC fails the ride-through gate even though the shortfall is small. The vessel should recharge the UPS/battery, reduce nonessential demand on the essential-load bus, shorten the standby-generator recovery time, or restrict the operating mode until the measured SOC and battery health support the recovery sequence.

Plausibility Check

The essential bus needs 120\ \text{kW} for one fifth of an hour, or 24\ \text{kWh} delivered. Inverter losses increase battery draw to about 26\ \text{kWh}. A 60\ \text{kWh} battery at 72\% SOC has only 25.2\ \text{kWh} above the 30\% floor, so a small SOC shortfall is expected.

Exercise 14: Switchboard Short-Circuit Duty With Paralleled Generators

A shipboard low-voltage switchboard can be operated with three diesel generators in parallel. Each online generator can contribute an initial symmetrical RMS fault current of:

I_g=4.6\ \text{kA}

Connected motors contribute:

I_m=1.3\ \text{kA}

The switchboard breaker interrupting rating is:

I_{br}=16.0\ \text{kA}

The engineering screen applies an uncertainty and model guard of:

g=8\%

Calculate prospective short-circuit current with three generators online, raw breaker margin, guarded breaker margin, and the guarded margin if the operating mode is limited to two generators online.

Solution

Prospective three-generator short-circuit current:

I_{sc,3}=3I_g+I_m
I_{sc,3}=3(4.6)+1.3=15.1\ \text{kA}

Raw breaker margin:

M_{raw}=I_{br}-I_{sc,3}=16.0-15.1=0.9\ \text{kA}

Raw utilization:

\displaystyle U_{raw}=\frac{15.1}{16.0}=94.4\%

Guarded fault current:

I_{guard,3}=I_{sc,3}(1+g)
I_{guard,3}=15.1(1.08)=16.3\ \text{kA}

Guarded margin:

M_{guard,3}=16.0-16.3=-0.3\ \text{kA}

The three-generator mode fails the guarded short-circuit duty screen.

With two generators online:

I_{sc,2}=2(4.6)+1.3=10.5\ \text{kA}

Guarded two-generator current:

I_{guard,2}=10.5(1.08)=11.3\ \text{kA}

Guarded two-generator margin:

M_{guard,2}=16.0-11.3=4.7\ \text{kA}

The two-generator mode passes this simplified short-circuit screen, but it must still be checked against reserve, load shedding, blackout recovery and propulsion demand.

Engineering Comment

More online generators can improve active-power reserve while increasing fault duty at the switchboard. That trade-off is easy to miss if the review only tracks kW and blackout recovery. A release decision should use the approved protection study, generator decrement data, motor contribution assumptions, breaker making and interrupting ratings, switchboard withstand rating, protection settings, bus-tie state, operating-mode restrictions and commissioning records.

Plausibility Check

Three generators contribute 13.8\ \text{kA} before motor contribution, so a total of 15.1\ \text{kA} is already close to a 16.0\ \text{kA} breaker rating. An 8 percent guard naturally pushes the screened current above the rating. Removing one generator lowers the guarded value to about 11.3\ \text{kA}, which restores a large margin but may reduce operational redundancy.

Exercise 15: Bow-Thruster Start Voltage-Dip Gate

A bow-thruster induction motor has full-load apparent power:

S_{FL}=620\ \text{kVA}

Across-the-line starting current is estimated as:

k_{LR}=5.5

times full-load current. With two generators online, the source base is:

S_{b,2}=3.0\ \text{MVA}

and the equivalent source impedance is:

Z_{2}=0.065\ \text{pu}

With one generator online, the source base is:

S_{b,1}=2.0\ \text{MVA}

and the equivalent source impedance is:

Z_{1}=0.085\ \text{pu}

Sensitive controls require the bus voltage to remain at or above:

V_{min}=0.90\ \text{pu}

Use the simplified screen:

\Delta V_{pu}\approx I_{start,pu}Z_{source,pu}

where starting current in per unit is approximated from starting apparent power divided by the source base. Check the two-generator and one-generator operating modes.

Solution

Starting apparent power:

S_{start}=k_{LR}S_{FL}
S_{start}=5.5(620)=3410\ \text{kVA}=3.41\ \text{MVA}

Two-generator starting current on the source base:

\displaystyle I_{start,2,pu}=\frac{3.41}{3.0}=1.14\ \text{pu}

Approximate two-generator voltage dip:

\Delta V_2=1.14(0.065)=0.074\ \text{pu}

Screened bus voltage:

V_{bus,2}=1.00-0.074=0.926\ \text{pu}

The two-generator mode passes the simplified 0.90\ \text{pu} voltage gate.

One-generator starting current on the source base:

\displaystyle I_{start,1,pu}=\frac{3.41}{2.0}=1.705\ \text{pu}

Approximate one-generator voltage dip:

\Delta V_1=1.705(0.085)=0.145\ \text{pu}

Screened bus voltage:

V_{bus,1}=1.00-0.145=0.855\ \text{pu}

The one-generator mode fails the simplified voltage gate because:

0.855<0.90

Engineering Comment

This is why a propulsion-plant operating envelope must include source mode, not only motor rating. The same bow-thruster start can be acceptable with two generators online and unacceptable after a generator is stopped for fuel saving or maintenance. Release options include blocking direct thruster start in one-generator mode, starting another generator first, using a soft starter or drive limit, changing load sequence, or validating a more detailed motor-starting study with measured bus voltage.

Plausibility Check

The locked-rotor start asks for 3.41\ \text{MVA}, which is larger than either source base. With two generators the stronger source and lower impedance give a moderate 7.4\% dip, leaving 0.926\ \text{pu}. With one generator the per-unit start current is higher and the source is weaker, so the dip nearly doubles to 14.5\% and the bus falls to 0.855\ \text{pu}. The pass/fail difference is therefore consistent with the operating-mode change.

Exercise 16: Shaft Torque and Gearbox Service-Factor Gate

A propulsion diesel delivers mechanical input power to a reduction gearbox:

P_{in}=1.90\ \text{MW}

at engine speed:

n_{in}=720\ \text{rpm}

The gearbox ratio is:

i=4.0:1

so the output shaft speed is n_{out}=n_{in}/i. Gearbox efficiency is:

\eta_g=0.97

The shaftline release rule applies service factor:

SF=1.15

to operating torque and requires the guarded torque to remain below:

T_{allow}=110\ \text{kN m}

Calculate input torque, output shaft speed, output power, output shaft torque, guarded torque, torque margin, maximum output power allowed by the torque gate and required input-power derate.

Solution

Input angular speed:

\displaystyle \omega_{in}=2\pi\frac{720}{60}=75.4\ \text{rad/s}

Input torque:

\displaystyle T_{in}=\frac{P_{in}}{\omega_{in}}
\displaystyle T_{in}=\frac{1.90\times10^6}{75.4}=25.2\ \text{kN m}

Output shaft speed:

\displaystyle n_{out}=\frac{720}{4.0}=180\ \text{rpm}

Output power after gearbox losses:

P_{out}=\eta_gP_{in}=0.97(1.90)=1.843\ \text{MW}

Output angular speed:

\displaystyle \omega_{out}=2\pi\frac{180}{60}=18.85\ \text{rad/s}

Output shaft torque:

\displaystyle T_{out}=\frac{P_{out}}{\omega_{out}}
\displaystyle T_{out}=\frac{1.843\times10^6}{18.85}=97.8\ \text{kN m}

Guarded service-factor torque:

T_{guard}=SF\,T_{out}=1.15(97.8)=112.5\ \text{kN m}

Torque margin:

M_T=110-112.5=-2.5\ \text{kN m}

The operating point fails the shaftline torque gate.

Maximum operating torque before service factor:

\displaystyle T_{out,max}=\frac{110}{1.15}=95.7\ \text{kN m}

Maximum output power at 180\ \text{rpm}:

P_{out,max}=T_{out,max}\omega_{out}
P_{out,max}=95.7\times10^3(18.85)=1.803\ \text{MW}

Maximum gearbox input power:

\displaystyle P_{in,max}=\frac{1.803}{0.97}=1.859\ \text{MW}

Required input-power derate:

\Delta P_{in}=1.900-1.859=0.041\ \text{MW}=41\ \text{kW}

The propulsion mode should be derated by at least about 41\ \text{kW} at the gearbox input, or the shaftline torque limit and service-factor basis should be revalidated before release.

Engineering Comment

This mechanical gate can be missed when the electrical plant has enough kW. A lower shaft speed increases torque for the same delivered power, and gearbox losses change the boundary between engine input and shaft output. Release evidence should include gearbox ratio, measured rpm, torsional-vibration review, shaft allowable torque, clutch state, bearing condition, alignment, thermal growth, propeller loading, sea state, control limits and any class-approved service factor.

Plausibility Check

The gearbox reduces speed by a factor of four, so output torque should be much larger than input torque even after a small power loss. The calculated output torque of 97.8\ \text{kN m} is below the raw 110\ \text{kN m} limit, but the 1.15 service factor pushes the guarded value to 112.5\ \text{kN m}, explaining the small derate requirement.

Exercise 17: Generator-Trip Frequency Nadir and Fast Load-Shed Gate

A diesel-electric vessel is operating on a 60\ \text{Hz} bus. After one online generator trips, the surviving source and drive limiters can initially support:

P_{avail}=1.35\ \text{MW}

while the connected load is still:

P_{load}=1.82\ \text{MW}

The equivalent rotating source base for the transient screen is:

S_b=2.4\ \text{MW}

and the equivalent inertia constant is:

H=3.2\ \text{s}

A fast load-shed stage removes:

P_{shed}=0.32\ \text{MW}

after:

t_s=0.45\ \text{s}

Governor and drive-limiter recovery are assumed to begin at:

t_g=1.20\ \text{s}

The blackout-prevention screen requires frequency to remain above:

f_{min}=58.5\ \text{Hz}

before governor recovery starts. Use the simplified energy form:

\displaystyle \Delta f=\frac{f_0}{2HS_b}\int \Delta P(t)\,dt

where positive \Delta P is active-power deficit in MW. Check the frequency nadir before governor response, then find the latest acceptable fast load-shed time.

Solution

Initial active-power deficit after the trip:

\Delta P_0=P_{load}-P_{avail}
\Delta P_0=1.82-1.35=0.47\ \text{MW}

Residual deficit after the fast load-shed stage:

\Delta P_1=\Delta P_0-P_{shed}
\Delta P_1=0.47-0.32=0.15\ \text{MW}

Frequency sensitivity for the simplified screen:

\displaystyle K_f=\frac{f_0}{2HS_b}
\displaystyle K_f=\frac{60}{2(3.2)(2.4)}=3.906\ \frac{\text{Hz}}{\text{MW s}}

Power-deficit impulse before the load-shed stage:

I_0=\Delta P_0t_s=0.47(0.45)=0.2115\ \text{MW s}

Power-deficit impulse after shedding and before governor response:

I_1=\Delta P_1(t_g-t_s)=0.15(1.20-0.45)=0.1125\ \text{MW s}

Total screened deficit impulse:

I=I_0+I_1=0.2115+0.1125=0.3240\ \text{MW s}

Estimated frequency drop:

\Delta f=K_fI=3.906(0.3240)=1.27\ \text{Hz}

Estimated frequency just before governor recovery:

f_{nadir}=60-1.27=58.73\ \text{Hz}

The screen passes because:

58.73>58.5

The maximum allowed frequency drop is:

\Delta f_{max}=60-58.5=1.5\ \text{Hz}

So the maximum allowed deficit impulse is:

\displaystyle I_{max}=\frac{\Delta f_{max}}{K_f}=\frac{1.5}{3.906}=0.384\ \text{MW s}

If the fast load-shed time is unknown, the deficit impulse before governor response is:

I(t_s)=\Delta P_0t_s+\Delta P_1(t_g-t_s)
I(t_s)=0.47t_s+0.15(1.20-t_s)
I(t_s)=0.18+0.32t_s

Set this equal to the maximum allowed impulse:

0.18+0.32t_s=0.384
\displaystyle t_s=\frac{0.384-0.18}{0.32}=0.638\ \text{s}

The fast load-shed stage should therefore occur no later than about:

0.64\ \text{s}

The programmed 0.45\ \text{s} stage is inside that limit. A delayed 1.0\ \text{s} stage would give:

I(1.0)=0.18+0.32(1.0)=0.50\ \text{MW s}
\Delta f=3.906(0.50)=1.95\ \text{Hz}
f=60-1.95=58.05\ \text{Hz}

so the delayed stage would fail the 58.5\ \text{Hz} frequency-nadir gate.

Engineering Comment

This transient gate is different from simply checking total load-shed quantity. The plant can have enough staged shedding in steady arithmetic and still fail if the first block is late relative to inertia, governor delay and drive limiter response. Release evidence should include event-recorder frequency traces, generator breaker time, bus-tie state, measured load-shed command and breaker-open times, propulsion drive ride-through settings, governor response, blackout-prevention logic and any class or owner frequency limits.

Plausibility Check

The initial deficit is about half a megawatt on a 2.4\ \text{MW} rotating base, so a noticeable frequency drop in the first second is expected. Fast shedding at 0.45\ \text{s} limits the deficit impulse to 0.324\ \text{MW s} and keeps the screen above 58.5\ \text{Hz}. If the same stage is delayed to 1.0\ \text{s}, the impulse rises to 0.50\ \text{MW s} and the simplified nadir falls to 58.05\ \text{Hz}, which is consistent with a timing-driven blackout-prevention failure.

Exercise 18: Propulsion-Drive Harmonic RMS Current Gate

A propulsion-drive feeder is reviewed after a power-quality survey. The measured fundamental RMS line current is:

I_1=410\ \text{A}

The current total harmonic distortion is:

THD_I=34\%=0.34

The feeder cable and breaker continuous thermal rating for this installation is:

I_{rating}=435\ \text{A}

The release rule applies a 5 percent guard for measurement uncertainty, installation temperature, bundling and harmonic heating:

I_{release}=1.05I_{rms}

Use the simplified relation:

I_{rms}=I_1\sqrt{1+THD_I^2}

Check whether the feeder can be released from the measured current. Then repeat the check after a filter and drive-parameter change reduce the operating point to:

I_1=398\ \text{A},\quad THD_I=18\%=0.18

Solution

If harmonic current were ignored, the fundamental-current margin would appear to be:

M_{I,1}=435-410=25\ \text{A}

So the feeder might look acceptable from fundamental current alone.

Total RMS current including harmonic content:

I_{rms}=410\sqrt{1+0.34^2}
I_{rms}=410(1.056)=433.0\ \text{A}

Guarded release current:

I_{release}=1.05(433.0)=454.7\ \text{A}

Guarded current margin:

M_I=435-454.7=-19.7\ \text{A}

The feeder should not be released in the measured condition because the guarded RMS current exceeds the installed cable and breaker rating.

After the correction:

I_{rms,new}=398\sqrt{1+0.18^2}
I_{rms,new}=398(1.016)=404.4\ \text{A}

Guarded revised current:

I_{release,new}=1.05(404.4)=424.6\ \text{A}

Revised current margin:

M_{I,new}=435-424.6=10.4\ \text{A}

The corrected drive configuration passes the simplified current gate, provided the harmonic spectrum, thermal condition, cable installation, breaker setting, protection coordination and drive operating modes match the measured condition.

Engineering Comment

Shipboard drives can overload feeders thermally even when active power and fundamental current look acceptable. The relevant heating quantity is total RMS current, and the release evidence should include the harmonic spectrum, power analyzer setup, operating mode, motor load, cooling ventilation, cable grouping, ambient temperature, breaker trip curve and whether the same distortion appears during thruster transients or low-speed operation.

This calculation is not a full harmonic study. It is a release screen that prevents a drive feeder from being accepted from clean kW arithmetic while harmonic current consumes the real thermal margin.

Plausibility Check

The measured fundamental current is only 25\ \text{A} below the rating, so there is little room for harmonic current and guards. A 34\% current THD increases RMS current by about 5.6\%, and the release guard pushes the checked current to 454.7\ \text{A}, above the 435\ \text{A} rating. Reducing both fundamental current and THD brings the guarded current down to 424.6\ \text{A}, which explains why the corrected case passes with a modest margin.

Review Checklist

When reviewing a ship power or propulsion-plant calculation, ask:

  • Is the boundary explicit: shaft, motor, drive, switchboard, generator, fuel, cooling, UPS or emergency bus?
  • Are online reserve, N-1 survivability, load-shedding quantity and load-shedding timing checked as separate gates?
  • Is generator-trip frequency nadir checked against inertia, fast load-shed delay, governor response and event-recorder evidence?
  • Are power factor, apparent power, harmonics and protection limits checked instead of using active power alone?
  • Are propulsion-drive harmonic RMS currents checked against cable, breaker, transformer, ventilation and installation-temperature ratings?
  • Are short-circuit duty, breaker rating and generator paralleling restrictions checked for the selected operating mode?
  • Are motor-starting voltage dip and sensitive-control ride-through checked for the selected generator online mode?
  • Are shaft torque, gearbox ratio, rpm, service factor and mechanical derate checked at the same propulsion boundary as the electrical power calculation?
  • Are battery and UPS checks based on current SOC, SOH, usable window, efficiency, recovery time and essential-load priority?
  • Are cooling, fuel, ventilation and auxiliary systems treated as constraints on available power?
  • Are shore-power transfers, emergency generator starts and standby-generator synchronization validated with real event records?
  • Are high-severity monitoring improvements supported by alarm proof tests, crew procedures, spare policy and post-entry review?

Common Mistakes

  • Treating installed generator rating as available power without naming which machines are online and synchronized.
  • Checking normal reserve while skipping N-1 survivability, load-shedding quantity and load-shedding timing.
  • Passing total load-shed quantity while ignoring frequency nadir, inertia, governor delay and first-stage timing.
  • Using active power alone when apparent power, power factor, harmonics, current limit or voltage dip controls the bus.
  • Releasing a drive feeder from fundamental current while harmonic RMS current, thermal grouping and breaker settings consume the real margin.
  • Starting a large thruster or pump in one-generator mode because the same load passed in two-generator mode.
  • Releasing propulsion from available electrical kW while shaft torque, gearbox service factor, bearing condition or torsional limits are outside margin.
  • Claiming battery or UPS ride-through from nominal kWh while SOC, SOH, usable window, efficiency and recovery time are unresolved.
  • Treating cooling and ventilation as auxiliary details even though they can derate generators, drives and propulsion motors.
  • Accepting short-circuit duty from a single source case while additional paralleled generators raise switchboard fault current.
  • Relying on RPN reduction while severity remains high and alarm proof tests or crew response procedures are missing.
  • Releasing shore-power transfer without confirming grounding, interlocks, protection coordination, cable rating and harmonic behavior.
  • Using simulation results for blackout recovery without event logs, breaker timing, load-shed records and crew procedure evidence.

Ship power and propulsion-plant engineering is strongest when the operating mode, electrical boundary, thermal support, fuel state, control logic, contingency claim and validation evidence all describe the same vessel condition.

REF

See also