Certification Assistance for Marine Engineers

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Propulsion Plant Simulator, Level One
British Columbia Institute of Technology
Pacific Marine Training Campus
Version 1 by: Martin Leduc, July 1998
Version 2 by: Myles OBrien, Sept 2002

Persons wanting to obtain the Canadian marine engineering license, or chief endorsement, need to complete the PPS L1 and L2 courses, in which they will have to carry out five labs. These are results for one lab for your comparison. Findings are based on observation carried out and graphs not all presented here.

 

Modec 100
Generator Load Sharing
Lab 7 - Version 1

 

Introduction

This lab was performed to study of the relationship between speed and governor droop on the load sharing of two diesel generator.

Method of investigation

Procedure . The procedures set forth in the PPS Level 1 Course booklet were followed. The main ideas are as follows: Having two machines under load and observing load sharing as load is applied. Three separate procedures were performed. Each time, diesel generator #1 was under different governor droop settings, 50%, 25% and Isochronous.

Results

Equipment Used. The simulator used for this particular experiment was the Modeq 100 . This simulator has two diesel generators, synchronizing panel, and common bus. Variable inductive, resistive, and capacitive load are served by the bus. Governor droop is set using the system's PC. Load sharing is also represented by the graphic display.

Data. The parameters for each steps were recorded by observations, on tables provided with the PPS Level 1 course booklet. They are represented here, on the following pages.

 

These test were made with single generator operation. We can observe both diesel generator responding in a similar fashion.

Droop is as expected, about 4%, as the load increase.

STEP ONE

In this step, both generator shared the load equally because both governors were set at the same four percent.

STEP TWO

In this step, we can observe DG1 taking more of the load because of the droop setting which is at two percent. The engine is responding a little quicker which is taking a larger load than DG2 with it's droop at four percent.

STEP THREE

In this step we can observe DG1 with no droop. The governor is reacting quickly to oncoming loads. This response is faster than DG2 with it's droop still at four percent. Therefore DG1 takes more of the electrical load. The same will hold true when load is being taking off; DG1 will shed load faster because of the rapid response the droop is providing.

Conclusion

For trouble free operation it may be concluded that two machine share equally, with equal response, an electrical load. In order to do that, diesel generator droop settings should be equal. Unless, of course, a person wishes a particular machine to 'lead' which would require the adjustment of the droop settings for one machine.

Modec 100
Generator Load Sharing
Lab 7 - Version 2

Objective of Lab: To investigate the relationship between speed and voltage droop on the load sharing characteristics of two diesel generators.

Part 1 

Objective of part 1: The objective is to see that the frequency will change by 4% when we apply load to a generator when its governor is set for 4% droop.

Set up: Generator 1 and Generator 2 running 60 Hz with 225 V with voltage regulators in auto and 4% droop. Each generator is put on line with out the other and ramped up with a resistive load ranging from 400 KW to 1000 KW.

Results:  

Gen. #1	Load KW	Current A	Freq. HZ	Load %	Drop in freq.	% of drop
	0 KW	0 A	60 HZ	0%	0 HZ	0 %
	400 KW	1152 A	60.00 HZ	40%	0 HZ	0 %
	500 KW	1312 A	58.70 HZ	50%	1.30 HZ	2.16 %
	600 KW	1525 A	58.40 HZ	60%	1.60 HZ	2.67 %
	700 KW	1806 A	58.05 HZ	70%	1.95 HZ	3.25 %
	800 KW	2054 A	57.90 HZ	80%	2.10 HZ	3.5 %
	900 KW	2357 A	57.70 HZ	90%	2.30 HZ	3.83 %
	1000 KW	2612 A	57.40 HZ	100%	2.60 HZ	4.33 %

 

Gen. #2	Load KW	Current A	Freq. HZ	Load %	Drop in freq.	% of drop
	0 KW	0 A	60 HZ	0%	0 HZ	0 %
	400 KW	1175 A	60.00 HZ	40%	0 HZ	0 %
	500 KW	1316 A	58.70 HZ	50%	1.30 HZ	2.16 %
	600 KW	1550 A	58.50 HZ	60%	1.50 HZ	2.50 %
	700 KW	1801 A	58.30 HZ	70%	1.70 HZ	2.83 %
	800 KW	2060 A	58.00 HZ	80%	2.00 HZ	3.33 %
	900 KW	2306 A	57.90 HZ	90%	2.10 HZ	3.50 %
	1000 KW	2623 A	57.60 HZ	100%	2.40 HZ	4.00 %

Conclusion: From the results above a 4.33% droop was achieved when Generator 1 was loaded to 100%. Generator 2 achieved a 4% droop when loaded to 100%.  

Part 2

Objective of part 2: We are going to parallel generator 1 and generator 2; both governors will have the same droop, therefore, we should see that as the load increases, both generators would share the increase.

Set up: Generator 1 and Generator 2 running 60 Hz with 225 V with voltage regulators in auto and 4% droop. Each generator is put on line with the bus running a resistive load of 800 KW shared evenly between generators (400 KW each). An inductive load will be added to the bus for the second part of part 2.

Observations of parallel Generator control:

-To redistribute the bus load between generators, the frequency (governor) control is adjusted.

-To redistribute the reactive power (KVAR or value Q) the automatic voltage regulator (excitation or magnetization) is adjusted.

-To raise the bus frequency, both governor controls are manipulated at the same time and in the same direction.

Results:

	Load KW	Current A	Frequency HZ	Droop
Generator #1	400 KW	1052 A	60 HZ	4 %
Generator #2	400 KW	1033 A	60 HZ	4 %

Observation: While paralleling generator 1 and generator 2 with just a resistive load on the bus, the objective is to achieve a power triangle that is flat. In this there is no reactive power (Q) and both generators run with a power factor of 1 on the power factor meter.

Now closing the breaker to our inductive load increases load and introduces reactive power.

	Load KW	Current A	Freq. HZ	Droop	Load %	Drop in Freq.	% of drop
Gen. #1	812 KW	2380 A	59 HZ	4 %	81.2%	1 HZ	1.67 %
Gen. #2	805 KW	2363 A	59 HZ	4 %	80.5%	1 HZ	1.67 %

 

Observation: With the introduction of an inductive load setting of 5 on the bus, the power factors of the generators changed. Generator 1 power factor 0.91 lag and generator 2 power factor 0.90 lag. The power factor meter read in the upper left quadrant which indicates both generators are generating and current is lagging behind voltage as in the diagram below. The power triangles for both generators with an inductive load are as follows; 

Generator #1	Generator #2
812 KW (P)	805 KW (P)
378 KVAR (Q)	378 KVAR (Q)
217 V AC	217 V AC

Conclusion: As the load (KW) increases the load is shared by both generators. As well the wattless power (reactive load) is equally shared between generators.

Part 3

Objective of part 3: The objective is to reduce droop on governor #1 and to see if doing so results in #1 taking more or any increase in load.

Set up: Generator 1 and Generator 2 running 60 Hz with 225 V with voltage regulators in auto. Governor #1 is set to 2% droop and governor #2 is set to 4% droop. Each generator is put on line with the bus running a resistive load of 800 KW shared evenly between generators (400 KW each). An inductive load will be added to the bus for the second part of part 3.

Results:

	Load KW	Current A	Frequency HZ	Droop
Generator #1	400 KW	1033 A	60 HZ	2 %
Generator #2	400 KW	1066 A	60 HZ	4 %

Now an inductive load is added to the bus.

	Load KW	Current A	Freq. HZ	Droop	Load %	Drop in Freq.	% of drop
Gen. #1	924 KW	2655 A	59.4 HZ	2 %	92.4%	0.6 HZ	1.0 %
Gen. #2	674 KW	2055 A	59.4 HZ	4 %	67.4%	0.6 HZ	1.0 %

There is a 25% difference in load of active power.

Observations: Even though Generator #1 has more load (KW), the wattless (reactive) power is still distributed evenly as the following data shows

Generator #1		Generator #2
2% Droop		4% Droop
925 KW (P)		674 KW (P)
380 KVAR (Q)	--------------------------------------------------------	380 KVAR (Q)
Power factor 0.93		Power factor 0.87

Conclusion:  Reducing droop on Governor #1 results in Generator #1 taking on more load in an increased load condition due to smaller frequency droop allowance.

Part 4 

Objective: The objective of part 4 is to see what happens when one governor is set to 0% droop (isochronous) and the other governor is set to droop 4%. It should be observed that when the generators are operating in parallel and sharing the load, the isochronous generator takes any increase in load, as well any decrease in load will be from the isochronous generator too.

Set up: Generator 1 and Generator 2 running 60 Hz with 225 V with voltage regulators in auto. Generator #1 is et to 0% droop and Generator #2 is set to 4% droop. Each generator is put on line with the bus running a resistive load of 800 KW shared evenly between generators (400 KW each). An inductive load will be added to the bus for the second part of part 2.

Results:

	Load KW	Current A	Frequency HZ	Droop
Generator #1	400 KW	1020 A	60 HZ	0 %
Generator #2	400 KW	1054 A	60 HZ	4 %

 

Now an inductive load is added to the bus.

	Load KW	Current A	Freq. HZ	Droop	Load %	Drop in Freq.	% of drop
Gen. #1	1235 KW	3383 A	60.0 HZ	0 %	123.5%	0 HZ	0 %
Gen. #2	371 KW	1454 A	60.0 HZ	4 %	37.1%	0 HZ	0 %

Observation: Even though Generator #1 grossly has more load (KW) than generator #2, the wattless (reactive) power is still distributed evenly as the following data shows

Generator #1		Generator #2
0% Droop		4% Droop
1235 KW (P)		371 KW (P)
387 KVAR (Q)	--------------------------------------------------------	387 KVAR (Q)
Power factor 0.95		Power factor 0.71

Conclusion: With one generator droop setting 0% (isochronous), and operating in parallel, it was observed that in an increased load condition this generator took any increase in load due to the torque exerted on generator by no droop allowance. As well, any decrease in load came from this generator.

Part 5

Objective: The objective of part 5 is to investigate the effects of one isochronous generator and one droop generator in a decreased load condition. It should be observed that the droop machine will motor the isochronous machine until it trips off the bus on reverse power.

Set up: Generator 1 and Generator 2 running 60 Hz with 225 V with voltage regulators in auto. Generator #1 is to 0% droop and Generator #2 is set to 4% droop. Each generator is put on line with the bus running a resistive load of 800 KW shared evenly between generators (400 KW each) and an inductive load of 5. Then the load is decreased.

Results: After the resistive load and inductive load were decreased the following was observed;

Inductive load 10 (minimum)

Generator #1		Generator #2
Droop 0%		Droop 4%
-55 KW (P) Being driven by Gen #2		371 KW
Power Factor 0.18		Power Factor 0.74
Current leading voltage		Current lagging voltage

 

Conclusion: As the load decreased the isochronous generator shed its load before the drooped generator to the point of motoring due to frequency differences.

 

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