Tech Explanation on Diesel Engine in AC Generator Set

[IOC-AUS]

Hello All,
I was hoping to get a Technical Explanation on the Power ( kW Input ) requirements of an Industrial Diesel Engine 1500rpm in a 4 Pole AC Generator Set.

Is there any impact on kW Rating ( or another way to ask kW Efficiency ) of a Diesel Engine when the Output Supply and AC Alternator configuration are different.

for example,
● CASE # 1 - A Diesel Engine is used in a Diesel Generator Set configured to Supply 3 Phase 4 Wire Output 415V 50Hz ( typically the AC Alternator is connected in Series Star Cofiguration ).
● CASE # 2 - The same Diesel Engine is used in a Diesel Generator Set to Supply 3 Phase 4 Wire Output 690V 50Hz ( typically the AC Alternator is connected in Three Phase Zig Zag Cofiguration ).

I understand the Current Rating will be lower on the 690V Supply compared to 415V Supply.

Not sure if the AC Alternator Efficiency will be different for each Winding Confguration but I would assume the Efficiency should be in the order of 92 to 94%.

Welcome any comments RE this enquiry that may help understand the Diesel Engine kW Rating within each System.

 

[LittleInch]

Errr, some idea of the power required in each option / amps generated and the rated power for the diesel engine at 1500 rpm might be useful.

 

[IOC-AUS]

Yeah good point.
Load will be a 400 kW Electric Motor with VFD Driving a Water Pump.
690V Motor - Full Load = 450 Amps per Phase.
● Case 1 - Power Draw 347 kW ( 1018 rpm )
● Case 2 - Power Draw 370 kW ( 1025 rpm )
● Case - Power Draw 395 kW ( 1034 rpm )

 

[waross]

First, the engine develops enough power to support the load plus losses.
The engine is not always loaded to 100% of rated power.
Each winding will have a current rating, and a voltage rating.
The maximum burden in KVA that the generator will supply is Max Volts times Max Amps time Root 3 divided by 1000, or A x V x 1.73 / 1000. = KVA Max.
It is common practice to supply enough kW to support 80% of the rated KVA.
So, for example.
Winding Max Volts = 240 Volts.
Winding Max Amps = 104 Amps
6 Windings, 2 per phase.
Max KVA = 240 V x 104 A x 6 x / 1000 = 150 KVA.
For standby service, this would be driven with an engine rated at 150 KVA x 0.8 PF = 120 kW.
For prime power service it is typical to over-size the engine by 10% (132 kW) so that after 15,000 hours, the engine will still be able to output 120 kW.
Let's stick to standby sets for now.
The different voltages are realize by series or parallel connections, star or delta connections and voltage regulator adjustments.

Example #1, Terminal voltage = 415 Volts,
Connection #1, series delta, 208/415 Volts, center tapped.
Voltage regulator set to 208 Volts.
KVA = 208 V x 104 Amps x 6 / 1000 = 130 KVA Max.
120 kW / 130 KVA = 0.92 Allowable PF.

Example #2 Terminal voltage = 415 Volts
Connection #2. Parallel star, 240/415 Volts.
Voltage regulator set to 240 Volts.
KVA = 240 Volts x 104 Amps x 6 windings / 1000 = 150 KVA
120 kW /150 KVA = 0.8 Allowable PF

Example #3 Terminal voltage = 690 Volts.
Connection, series star, 400/690 Volts
Voltage regulator set to 200 Volts.
KVA = 200 Volts x 104 Amps x 6 windings / 1000 = 125 KVA

Generator ends are rated in KVA.
Rated Maximum Allowable KVA = Winding Rated Voltage X Winding Rated Current X Number of Windings / 1000.
Actual Maximum Allowable KVA = Winding Actual Voltage (Voltage Regulator Setting) X Winding Rated Current X Number of Windings / 1000.
Note #1: When using winding values to calculate KVA, the root three factor is replaced by the number of windings.
Note #2: Both kW and KVA must be considered when sizing a gen-set.
The set must be capable of delivering the required current at the rated voltage (KVA)
The prime mover must be capable of meeting the required kW demand of the load.
If the load PF is above 0.8, the set must be oversized relative to the KVA rating.