When DG opportunity existed, utilities are investigate the maximum size and the interconnection point (location) by conducting a "System Impact Study" through which the interconnection requirements will be assigned. So, the DG; type, capacity and proposed location is provided to utility first and then utility will come with the interconnection requirement

Calculate Size of Diesel Generator

Calculate Size of Diesel Generator having following Electrical Load. Consider Future Expansion ratio is 10%. Average use of Equipment is 0.8 (1 is Full Time Use)

• 4 No’s of 1Ph, 230V, 80Watt CFL Bulbs, Diversity Factor is 0.8; Starting & Running P.F is 0.8.
• 2 No’s of 1Ph, 230V, 3000Watt Air Condition, Diversity Factor is 1, Starting & Running P.F is 0.8.
• 2 No’s of 1Ph, 230V, 500Watt Halogen Lights Diversity Factor is 0.8 Starting & Running P.F is 0.8.
• 1 No’s of 1Ph,230V, 10KW Motor with Y-D Starter, Diversity Factor is 0.8,Starting P.F is 0.7 & Running P.F is 0.8
• 1 No’s of 3Ph,430V, 130KW Motor with Soft Starter , Diversity Factor is 0.8,Starting P.F is 0.7 & Running P.F is 0.8

Calculation:

Type of LoadEquipmentStarting CurrentLinear LoadGeneral Equipment 100% of Full Load CurrentNon-Linear LoadUPS, Inverter, Computer, Ballast160% of Full Load Current

Type of StarterStarting CurrentDOL6 X Full Load CurrentStar-Delta4 X Full Load CurrentAuto Transformer3 X Full Load CurrentSoft Starter2 X Full Load CurrentVFD1.5 X Full Load Current

Load Calculation-1:

• Full Load KW of CFL Bulb=(No X Watt X Diversity Factor) /1000
• Full Load KW of CFL Bulb=(4x80x0.8)/1000=0.3KW
• Full Load KVA of CFL Bulb=KW / P.F
• Full Load KVA of CFL Bulb=0.3 / 0.8=0.4KVA————(H)
• Full Load current of CFL Bulb=(No X Watt X Diversity Factor) / (Volt x P.F)
• Full Load current of CFL Bulb=(4x80x0.8) / (230×0.8) =2 Amp—-(M)
• Type of Load=Linear
• Starting KVA of CFL Bulb=1 X (KW / Starting P.F)
• Starting KVA of CFL Bulb=0.3 / 0.8=0.4KVA———————-(1)
• Starting Current=100% of Full Load Current.
• Starting Current=1 X 2= 2 Amp.————–(A)

Load Calculation-2:

• Total Full Load KW of A.C=(No X Watt X Diversity Factor) /1000
• Total Full Load KW of A.C =(2x3000x0.8)/1000=4.8KW
• Total Full Load KVA of A.C =KW / P.F
• Total Full Load KVA of A.C =4.8 / 0.8=6KVA————(I)
• Total Full Load current of A.C =(No X Watt X Diversity Factor) / (Volt x P.F)
• Total Full Load current of A.C =(2x3000x0.8) / (230×0.8) =26 Amp——-(N)
• Type of Load=Non Linear
• Starting KVA of A.C=1.6 X (KW / Starting P.F)
• Starting KVA of A.C =1.6 X (4.8 / 0.8)=9.6KVA———————-(2)
• Starting Current=160% of Full Load Current.
• Starting Current=1.6 X 26= 42 Amp. ————–(B)

Load Calculation-3:

• Full Load KW of Halogen Bulb=(No X Watt X Diversity Factor) /1000
• Full Load KW of Halogen Bulb=(2x500x0.8)/1000=0.8KW
• Full Load KVA of Halogen Bulb=KW / P.F
• Full Load KVA of Halogen Bulb=0.8 / 0.8=1KVA————(J)
• Full Load current of Halogen Bulb=(No X Watt X Diversity Factor) / (Volt x P.F)
• Full Load current of Halogen Bulb=(2x500x0.8) / (230×0.8) =4 Amp—-(O)
• Type of Load=Non Linear
• Starting KVA of Halogen Bulb =1.6 X (KW / Starting P.F) / Starting P.F
• Starting KVA of Halogen Bulb =1.6 X (0.8 / 0.8)=1.6KVA———————-(3)
• Starting Current=160% of Full Load Current.
• Starting Current=1.6 X 4= 7 Amp .————–(C)

Load Calculation-4:

• Full Load KW of Motor=(No X Watt X Diversity Factor) /1000
• Full Load KW of Motor =(1x10000x0.8)/1000=8KW
• Full Load KVA of Motor =KW / P.F
• Full Load KVA of Motor =8 / 0.8=10KVA————(K)
• Full Load current of Motor =(No X Watt X Diversity Factor) / (Volt x P.F)
• Full Load current of Motor =(1x10000x0.8) / (230×0.8) =43 Amp—-(P)
• Type of Starter=Star-Delta
• Starting KVA of Motor =4 X (KW / Starting P.F)
• Starting KVA of Motor=4X (8 / 0.7)=45.7KVA————————(4)
• Starting Current=4 X Full Load Current
• Starting Current=4 X 11.4 = 174 Amp.————–(D)

Load Calculation-5:

• Full Load KW of Motor=(No X Watt X Diversity Factor) /1000
• Full Load KW of Motor =(1x120000x0.8)/1000=96KW
• Full Load KVA of Motor =KW / P.F
• Full Load KVA of Motor =96 / 0.8=120KVA————(L)
• Full Load current of Motor =(No X Watt X Diversity Factor) / (Volt x P.F)
• Full Load current of Motor =(1x120000x0.8) / (1.732x430x0.8) =167 Amp—-(Q)
• Type of Starter=Auto Transformer
• Starting KVA of Motor =3 X (KW / Starting P.F)
• Starting KVA of Motor=3 X (96 / 0.7)=411.4KVA—————(5)
• Starting Current=3 X Full Load Current
• Starting Current=3 X 167 = 501 Amp.————–(E)

Total Load Calculation:

• Total Starting KVA = (1) + (2) +(3) + (4) + (5)
• Total Starting KVA =0.4+9.6+1.6+45.7+411.4 =468.7 KVA
• Total Starting Current = (A) + (B) +(C) + (D) + E
• Total Starting Current =2+42+7+174+501= 725 Amp
• Total Running KVA =(H)+(I)+(J)+(K)+(L)
• Total Running KVA =4+6+1+10+120= 137KVA
• Total Running Current=(M)+(N)+(O)+(P)+(Q)
• Total Running Current=2+26+4+43+167= 242 Amp
• Size of Diesel Generator= Starting KVA X Future Expansion X Average Use of Equipments
• Size of Diesel Generator=468.7 X 1.1 X 0.8
• Size of Diesel Generator= 412 KVA

Summary:

• Total Starting KVA =468.7 KVA
• Total Starting Current =725 Amp
• Total Running KVA = 137KVA
• Total Running Current= 242 Amp
• Size of Diesel Generator= 412 KVA

Heat Generated by Generator:

• For generator set installations, the heat radiated by the generator can be estimated by
• H (kW) =P X ((1/Eff)-1)
• H (Btu/min) =P X ((1/EFF)-1) x56.9
• Where:
• H = Heat Radiated by the Generator (kW), (Btu/min)
• P = Generator Output at Maximum Engine Rating (kW)
• Eff = Generator Efficiency % / 100%
• Example: 975 kW standby generator set has a generator efficiency of 92%. The generator radiant heat for this genset can be calculated as follows.
• P = 975 kW
• Efficiency = 92% = 0.92
• H = 975 x ((1/92%) – 1)
• H= 84.78 kW
• H = 975 x ((1/92%) – 1) x 56.9
• H = 4824 Btu/min

Types of Ventilation System:

• Type:1 (Preferred Design) (Routing Factor of 1)
• Outside air is brought into the engine room through a system of ducts. These ducts should be routed between engines, at floor level, and discharge air near the bottom of the engine and generator. .
• Ventilation air exhaust fans should be mounted or ducted at the highest point in the engine room. They should be directly over heat sources. This system provides the best ventilation with the least amount of air required.
• Type 2 (Skid Design) (Routing Factor of 1)
• Outside air into the engine room through a system of ducts and routes it between engines.
• Type 2, however, directs airflow under the engine and generator so the air is discharged upward at the engines
• The most economical method to achieve this design is to use a service platform. The platform is built up around the engines and serves as the top of the duct
• Ventilation air exhaust fans should be mounted or ducted at the highest point in the engine room. They should be directly over heat sources.
• This system provides the best ventilation with the least amount of air required.
• Type 3 (Alternate Design) (Routing Factor of 1.5)
• If Ventilation Type 1or Type 2 is not feasible, an alternative is Type 3; however, this routing configuration will require approximately 50% more airflow than Type 1.
• Outside air is brought into the engine room utilizing fans or large intake ducts. The inlet is placed as far away as practical from heat sources and discharged into the engine room as low as possible. The air them flows across the engine room
• Ventilation air exhaust fans should be mounted or ducted at the highest point in the engine room. Preferably, they should be directly over heat sources
• Type 4 (Less Effective Design) (Routing Factor of 2.5)
• If Ventilation Type 1, Type 2 and Type 3 are not feasible, then Type-4 method can be used; however, it provides the least efficient ventilation and requires approximately two and a half times the airflow of Ventilation Type 1
• Outside air is brought into the engine room using supply fans, and discharged toward the turbocharger air inlets on the engines.
• Ventilation exhaust fans should be mounted or duct from the corners of the engine room
• This system mixes the hottest air in the engine room with the incoming cool air, raising the temperature of all air in the engine room.
• It also interferes with the natural convection flow of hot air rising to exhaust fans.
• Engine rooms can be ventilated this way, but it requires extra large capacity ventilating fans.

Ventilation for Generator:

• When Generator set installations in Room proper ventilation is required for Generator set.
• A properly designed engine room ventilation system will maintain engine room air temperatures within 8.5 to 12.5°C (15 to 22.5°F) above the ambient air temperature.
• For example, If the engine room temperature is 24°C (75°F) without the engine running, the ventilation system should maintain the room temperature between 32.5°C (90°F) and 36.5°C (97.5°F) while the engine is in operation.
• Ensures engine room temperature does not exceed 49°C (120°F).
• Required Ventilating Air is calculated as
• V=((H / D x Cp x T)+ Combustion Air) X F
• Where:
• V = Ventilating Air (m3/min), (cfm)
• H = Heat Radiation i.e. engine, generator, aux (kW),(Btu/min)
• D = Density of Air at air temperature 38°C (100°F). The density is 1.099 kg/m3 (0.071 lb/ft3)
• CP = Specific Heat of Air (0.017 kW x min/kg x °C),(0.24 Btu/LBS/°F)
• T = Permissible temperature rise in engine room (°C), (°F)
• F = Routing factor based on the ventilation type
• Example: The engine room for generator set has a Type 1 ventilation routing configuration and a dedicated duct for combustion air. It has a heat rejection value of 659 kW (37,478 Btu/min) and a permissible rise in engine room temperature of 11°C (20°F).
• V=((659 / 1.0099X0.017X11)+ 0)X1
• V = 3206.61 m3/min
• V=((659 / 0.071 X 0.024 X 20)+ 0)X1
• V = 109970.7 cfm

Dear Umair,

When you talk about "optimal" there is no single best technique. It all depends on identification of associated parameters, how many parameters you are considering, which technique you are using, how you implement the algorithm, etc... Since, you've asked for a publication, i suggest that you should go through the following publications: (Please go through section 4 in this paper, which provides detailed survey on various DGs planning methodologies and their comparison.

Article Optimal planning of distributed generation systems in distri...

Hello Umair,

I suggest to use the combintorial optimization solution approach as it fits the mentioned problem, and it produces relatively rigorous solutions.

Hello Umair,

Sizing and siting of DGs/ESSs is a meta-heuristics problem. Thus, there is no "Best" way to do this. However, thanks to the evolution in the multi-objective optimization techniques, I recommend you to check papers on IEEE xplorer using NSGAII or AUGECON2 for the MOOP Power system/smart grid problems.

Here are relatively old two examples of the use of constrained optimization using combintorial optimization approaches for placement of specific equipment in power systems.

Article OPTIMAL NUMBER AND LOCATIONS OF SECTIONALIZING DEVICE IN MED...

Article Multi-objective Placement of TCSC for Enhancement of Steady-...

You may search the RG questions to understand the scope, limitations, advantages, and disadvantages of various optimization methods. For example,

https://www.researchgate.net/post/Why_are_so_many_people_interested_in_multiple_objectives_in_general_and_metaheuristics_in_particular

and

https://www.researchgate.net/post/Computing_bounds_for_a_minimization_problem

Anyway, I suggest that you consult mathematician(s).

Looking ahead 20 years, the interesting research question is what are the optimum networks to support DG everywhere?

As described by Raafat Megahed, in a market economy the location of the DG is determined by the owner/operators of the DG. The market is a type of optimization.