Relationship between Maximum demand and size of a diesel generator

T

T.M.Haja Sahib

Guest
No, if you are sizing a generator just for a building, I think demand factors are enough, if you are sizing a generator which will feed many different buildings in that case we will use diversity factor. load factor is another case for large power plant, it plays an important part on the cost of generation per unit.
Thanks for your excellent reply.

You are correct in stating, ''if you are sizing a generator just for a building, I think demand factors are enough, if you are sizing a generator which will feed many different buildings in that case we will use diversity factor.'' I want to add that if there are sub-feeders in addition to feeders in a single building, both demand factor and diversity factor come into play.
 

mivey

Senior Member
The demand meter will read both steady load and starting KW of the motor and so the reading will be a little bit more than 75 KW.
Correct. I get 75.67 kW but I think Iwire mixed up the numbers with the point he was trying to make.

Even so: Now, using the assumption that the motor will take 6X running current at start time, calculate the power the generator will have to deliver (assume something like a 3 second start time).

For a simplified calculation (although not technically exact-a-mundo because of starting p.f., inrush, etc.):
( 3s * (480kW + 1kW) + (840s - 3s) * (80kW + 1kW) + 60s * 1kW ) / 900s = 77 kW


So the demand reading of 77 kW is a long way from the starting need of at least 481 kW (roughly calculated).
 

iwire

Moderator
Staff member
Location
Massachusetts
Correct. I get 75.67 kW but I think Iwire mixed up the numbers with the point he was trying to make.

Even so: Now, using the assumption that the motor will take 6X running current at start time, calculate the power the generator will have to deliver (assume something like a 3 second start time).

For a simplified calculation (although not technically exact-a-mundo because of starting p.f., inrush, etc.):
( 3s * (480kW + 1kW) + (840s - 3s) * (80kW + 1kW) + 60s * 1kW ) / 900s = 77 kW


So the demand reading of 77 kW is a long way from the starting need of at least 481 kW (roughly calculated).
Help me out.

Why do all the links I posted say the highest demand for 15 minutes?
 
T

T.M.Haja Sahib

Guest
So the demand reading of 77 kW is a long way from the starting need of at least 481 kW (roughly calculated).
But it is an illusion. Because starting 481 kW exists but a fraction of a second and so the average power impacting the maximum demand is very much less.
 

mivey

Senior Member
Help me out.

Why do all the links I posted say the highest demand for 15 minutes?
Not sure exactly what you are asking (why the results are what they are or why a 15 minute interval) but I'll give two responses:

#1 (the results):

The demand is actually a measure of the energy used in the demand interval (the highest total energy used in a 15 minute block). We do not care about the instantaneous demand (the highest total energy used in a block of time that is a fraction of a second).

By the motor running up to 14 minutes, the demand during the interval will approximately be the motor demand since it ran for almost all of the demand interval practically 14 of the 15 minutes).

In my previous post, the energy during the 3-second start was: 3s * (480kW + 1kW)

So we have 1443 kW over a 3-second interval = (481 kW) * (3 sec) * (1 hr) / (3600 sec/hr) = 0.4008 kWh

Similarly for the remaining 14 minutes of motor run we have 80 kW + 1 kW:
(840s - 3s) * (80kW + 1kW) * 1hr / 3600sec/hr = 67797 kW-sec = 18.8325 kWh

For the remaining 1 minute of the demand interval without the motor running:
60s * 1kW = 60 kW-sec = 0.0167 kWh

For the demand interval of 15 minutes (900 seconds), we have: 0.4008 + 18.8325 + 0.0167 = 19.25 kWh. You can see the time of the 1 kW contribution (the 0.0167 kWh) is very small.

To get the demand, the meter then calculates using the time for the demand interval (15 minutes or 0.25 hours)
19.25 kWh / 0.25 h = 77 kW




#2 (why 15 minutes):

A lot of energy is sold in hourly blocks at the wholesale level. What happens a the 1-second level is not really important there.

The power grid responds to sudden changes in load by voltages levels moving up and down. Most load changes result in minor voltage changes. Corrections to relatively minor changes in voltage levels happens over longer periods and a 15 minute interval is usually fine for the distribution system (real-time SCADA data and load-management equipment is used for large systems and at the transmission levels for severe changes).

A very large customer served at the transmission level is probably going to be different as you don't get help from load diversification.

15 minute demands are usually fine for sizing most distribution equipment, especially given the diversity that occurs.For some loads a shorter interval might be more revealing when equipment heating is an issue (like for bump loads at a saw-mill). You can usually handle the increased distribution costs (bigger transformers, power quality mitigation, etc.) with a higher demand charge on a saw-mill rate and the rest can be allocated using the 15-minute demand data.

There are some times where a 5-second or 1-minute or 5-minute interval is needed for technical reasons but it is usually not at the utility bill level.
 
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mivey

Senior Member
But it is an illusion. Because starting 481 kW exists but a fraction of a second and so the average power impacting the maximum demand is very much less.
Starting power is not an illusion for the supplying local generation. It will not be a very stiff source and can't absorb the bump as easily as the relatively stiff utility grid.
 
T

T.M.Haja Sahib

Guest
Starting power is not an illusion for the supplying local generation. It will not be a very stiff source and can't absorb the bump as easily as the relatively stiff utility grid.
Here the question is how the starting kw of pump motor is accounted for in the utility MD and to what extent a diesel gen set with the size equal to the MD will meet the starting requirement of the pump motor (whether any reduced voltage starting arrangement required etc.,)
 

mivey

Senior Member
Here the question is how the starting kw of pump motor is accounted for in the utility MD and to what extent a diesel gen set with the size equal to the MD will meet the starting requirement of the pump motor (whether any reduced voltage starting arrangement required etc.,)
With large starting loads behind the meter, I don't think the utility metered demand alone will tell you how to size the generator. It will help with the steady-state load portion of the calcs.

PS: Just look at any generator manufacturer's recommendations and the difference in sizing a generator for sites with no motor load vs sites with large motor loads.
 
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T

T.M.Haja Sahib

Guest
With large starting loads behind the meter, I don't think the utility metered demand alone will tell you how to size the generator. It will help with the steady-state load portion of the calcs.
The starting KW's and KVA's of loads are used for sizing up the engine and alternator respectively in a diesel gen set. Such values are scaled down by the demand period, when measured by a utility maximum demand meter. The problem is solved, when a relationship between the two sets can be deduced in general.
 

mivey

Senior Member
The starting KW's and KVA's of loads are used for sizing up the engine and alternator respectively in a diesel gen set. Such values are scaled down by the demand period, when measured by a utility maximum demand meter. The problem is solved, when a relationship between the two sets can be deduced in general.
So it should be clear that the metered demand is not helpful for finding the starting loads. Isn't that what the others have been saying all along?
 
T

T.M.Haja Sahib

Guest
So it should be clear that the metered demand is not helpful for finding the starting loads.
It may or it may not be, as the starting loads are also used to find out the metered demand and as measured by the MD meter.
 

mivey

Senior Member
It may or it may not be, as the starting loads are also used to find out the metered demand and as measured by the MD meter.
Not a two-way street.

So we have 6X running load for a fraction of a percent of the demand period. How do you suppose that a demand reading from a 2% accuracy meter, comprised mostly of running load data, is going to tell you much about that fraction of a percent?
 

Hameedulla-Ekhlas

Senior Member
Location
AFG
No, if you are sizing a generator just for a building, I think demand factors are enough, if you are sizing a generator which will feed many different buildings in that case we will use diversity factor. load factor is another case for large power plant, it plays an important part on the cost of generation per unit.
Ok, thanks for clearing this. You might have large project with different equipments. Here is an example to size generator which covers both motor starting current, running current. Hope this make clear some points.

"Genset Sizing Guidelines


Some conservative rules of thumb for genset sizing include:
1. Oversize genset 20?25% for reserve capacity and for motor starting.
2. Oversize gensets for unbalanced loading or low power factor running loads.
3. Use 1/2 hp per kW for motor loads.
4. For variable frequency drives, oversize the genset by at least 40%.
5. For UPS systems, oversize the genset by 40% for 6 pulse and 15% for 6 pulse with input filters or 12 pulse.
6. Always start the largest motor first when stepping loads.
For basic sizing of a generator system, the following example could be used:
Step 1: Calculate Running Amperes
■Motor loads:
❑200 hp motor . . . . . . . . . . . . . 156A
❑100 hp motor . . . . . . . . . . . . . . 78A
❑60 hp motor . . . . . . . . . . . . . . . 48A
■Lighting load . . . . . . . . . . . . . . . . 68A
■Miscellaneous loads . . . . . . . . . . 95A
■Running amperes. . . . . . . . . . . 445A
Step 2: Calculating Starting Amperes
Using 1.25 Multiplier
■Motor loads:
❑. . . . . . . . . . . . . . . . . . . . . . . . 195A
❑. . . . . . . . . . . . . . . . . . . . . . . . . 98A
❑. . . . . . . . . . . . . . . . . . . . . . . . . 60A
■Lighting load . . . . . . . . . . . . . . . . 68A
■Miscellaneous loads . . . . . . . . . . 95A
■Starting amperes. . . . . . . . . . . 516A
Step 3: Selecting kVA of Generator
■Running kVA =
(445A x 480V x 1.732)/
1000 = 370 kVA
■Starting kVA =
(516A x 480V x 1.732)/
1000 = 428 kVA
Solution
Generator must have a minimum starting capability of 428 kVA and minimum running capability of 370 kVA. Also, please see section ?Factors Governing Voltage Drop? on generator loading and reduced voltage starting techniques for motors."

Source: http://www.zone4info.com/articles/350/a-proper-generator-sizing-guid-line
 
T

T.M.Haja Sahib

Guest
Not a two-way street.

So we have 6X running load for a fraction of a percent of the demand period. How do you suppose that a demand reading from a 2% accuracy meter, comprised mostly of running load data, is going to tell you much about that fraction of a percent?
Of course, the meter may or may not show any increase in reading. But I am talking about the general case: Can there be a relationship between the MD and size of a diesel gen set in general?
 
T

T.M.Haja Sahib

Guest
Ok, thanks for clearing this. You might have large project with different equipments. Here is an example to size generator which covers both motor starting current, running current. Hope this make clear some points.

"Genset Sizing Guidelines


Some conservative rules of thumb for genset sizing include:
1. Oversize genset 20–25% for reserve capacity and for motor starting.
2. Oversize gensets for unbalanced loading or low power factor running loads.
3. Use 1/2 hp per kW for motor loads.
4. For variable frequency drives, oversize the genset by at least 40%.
5. For UPS systems, oversize the genset by 40% for 6 pulse and 15% for 6 pulse with input filters or 12 pulse.
6. Always start the largest motor first when stepping loads.
For basic sizing of a generator system, the following example could be used:
Step 1: Calculate Running Amperes
■Motor loads:
❑200 hp motor . . . . . . . . . . . . . 156A
❑100 hp motor . . . . . . . . . . . . . . 78A
❑60 hp motor . . . . . . . . . . . . . . . 48A
■Lighting load . . . . . . . . . . . . . . . . 68A
■Miscellaneous loads . . . . . . . . . . 95A
■Running amperes. . . . . . . . . . . 445A
Step 2: Calculating Starting Amperes
Using 1.25 Multiplier
■Motor loads:
❑. . . . . . . . . . . . . . . . . . . . . . . . 195A
❑. . . . . . . . . . . . . . . . . . . . . . . . . 98A
❑. . . . . . . . . . . . . . . . . . . . . . . . . 60A
■Lighting load . . . . . . . . . . . . . . . . 68A
■Miscellaneous loads . . . . . . . . . . 95A
■Starting amperes. . . . . . . . . . . 516A
Step 3: Selecting kVA of Generator
■Running kVA =
(445A x 480V x 1.732)/
1000 = 370 kVA
■Starting kVA =
(516A x 480V x 1.732)/
1000 = 428 kVA
Solution
Generator must have a minimum starting capability of 428 kVA and minimum running capability of 370 kVA. Also, please see section “Factors Governing Voltage Drop” on generator loading and reduced voltage starting techniques for motors."

Source: http://www.zone4info.com/articles/350/a-proper-generator-sizing-guid-line
One point to be taken note of is this:

The diesel engine in a gen set has no overload capability. So its size should be at least equal to the starting KW corresponding to the starting 428 KVA. Unfortunately, it is not given in the above calculation.
 

mivey

Senior Member
Of course, the meter may or may not show any increase in reading. But I am talking about the general case: Can there be a relationship between the MD and size of a diesel gen set in general?
In general, the bigger the demand the bigger the generator.

Does the demand reading alone tell us the size generator we need? No. At most it can provide only part of the information we need for sizing the generator.
 
T

T.M.Haja Sahib

Guest
At most it can provide only part of the information we need for sizing the generator.
Exactly. How that part of the information is related analytically or empirically to the whole is the relationship between maximum demand and size of a diesel generator. I hope the route is now clear for you to produce result. ;)
 

mivey

Senior Member
Exactly. How that part of the information is related analytically or empirically to the whole is the relationship between maximum demand and size of a diesel generator. I hope the route is now clear for you to produce result. ;)
No clear path since, in general, there can't be a good correlation between metered demand and generator size. You have to introduce other limiting factors first.
 

mivey

Senior Member
What are these, please?
You would have to limit the applicability of the function that describes the relationship between metered demand and generator size because of the various things discussed previously in this thread.

For example:

You could limit the application to certain type load shapes that did not include any motor loads behind the meter.

You could limit the application to a certain % motor load as long as the motors were certain types and/or had certain soft start requirements.

You could limit the application to exclude intermittent loads that ran for less than a certain % of the demand interval.

You could limit the application to certain harmonic constraints.

You could limit the application to include only certain parts of the load.

You could limit the application to apply only during certain times and/or seasons.

I'm sure the list could go on but you should get the idea: there is not a single general function that will provide good correlation between metered demand and generator size because there are too many different type load scenarios and requirements.
 
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